Displaying 1-10 of 76 results found.
1, 5, 63, 429, 12155, 88179, 1300075, 9694845, 583401555, 4418157975, 67282234305, 514589420475, 15801325804719, 121683714103007, 1879204156221315, 14544636039226909, 1804857108504066435
FORMULA
Numerators of binomial(2*n-3/2, -1/2).
Because A334907(n)/n! = a(n+1)/ A060818(n) for n >= 0, the o.g.f. of a(n+1)/ A060818(n), for n >= 0, is (sqrt(1 + sqrt(8*s)) - sqrt(1 - sqrt(8*s)))/sqrt(8*s * (1 - 8*s)), which is the e.g.f. of A334907 (see the link above for a proof). - Petros Hadjicostas, May 16 2020
MAPLE
seq(numer(binomial(2*n-3/2, -1/2)), n=1..20);
MATHEMATICA
Numerator[Binomial[2Range[20]-3/2, -(1/2)]] (* Harvey P. Dale, Feb 27 2012 *)
1, -1, -5, -3, -93, 95, -793, -211, -5853, 27003, -215955, 57459, -3518265, 3602027, 16811055, -4362627, -1846943453, 293601363, -14911085359, 4487888279, 144251733709, 245294787521, -1936010885087, 228009405371, -11179552565305, 63485965327535, -48562641580527
COMMENTS
Consider a lower triangular matrix T(n,k) defined by T(n,1)= A001790/ A046161, k>1: T(n,k) = (Sum_{i=1..k-1} T(n-i,k-1)) - (Sum_{i=1..k-1} T(n-i,k)). The first column in the matrix inverse of T(n,k) will have the fraction A180403/ A046161 in its first column.
0, 1, 1, 9, 5, 175, 189, 1617, 429, 57915, 60775, 508079, 264537, 8788507, 9100525, 75218625, 9694845, 5109183315, 5250613995, 43106892675, 22090789875, 723694276305, 740104577355, 6049284520695, 1543768261425, 201547523019375
COMMENTS
We write the fractions a(n)/b(n) and higher order differences as a matrix:
0, 1, 1, 9/8, 5/4,...
-1, 1/8, 0, -1/128, -1/128,... = (Lorbeta(n+1) + A161200(n+1)/ A046161(n+1)) / 2 9/8, -1/8, -1/128, 0, 1/1024,...
-5/4, 15/128, 1/128, 1/1024, 0,...
Here, Lorbeta(0)=1 and Lorbeta(n) = - A098597(n-1)/ A046161(n) for n>0 is the inverse of the Lorentz factor. The first line with numerators a(n) and denominators b(n) is 0, 1, 1, 9/8, 5/4, 175/128, 189/128, 1617/1024, 429/256, 57915/32768, 60775/32768,... It is an autosequence: Its inverse binomial transform is the signed sequence.
a(n+1)/(2*n-1)= 1, 3, 1, 25, 21, 147, 33, 3861, 3575, 26741,... .
a(n+1)/ A146535(n) = 9, 5, 35, 27, 539, 39, 4455,... . A001790(n)/ A046161(n) yields the coefficients of the Lorentz factor (or Lorentz gamma factor). With b for beta and g for gamma:
g = (1-b^2)^-1 = 1 + (b^2)/2 + 3*(b^4)/8 + 5*(b^6)/16 + ... .
b = (1-g^-2)^-1 = 1 - (g^-2)/2 - (g^-4)/8 - (g^-6)/16 - ... .
Are the denominators of the first subdiagonal 1, 1/8, -1/128, 1/1024,... A061549(n) ? a(n+1)/( A000108(n)*b(n)) = 1, 1, 9/16, 1/4, 25/256, 9/256, 49/4096, 1/256, 81/65536, 25/65536, 121/1048576,... = A191871(n+1)/ A084623(n+1)^2 ?
EXAMPLE
From the first formula: a(1)=1*1, a(2)=1*1, a(3)=3*3, a(4)=1*5, a(5)=5*35, a(6)=3*63.
MATHEMATICA
max = 25; A001803 = CoefficientList[Series[(1 - x)^(-3/2), {x, 0, max}], x] // Numerator; A001790 = CoefficientList[Series[1/Sqrt[(1 - x)], {x, 0, max}], x] // Numerator; A046161 = Table[Binomial[2n, n]/4^n, {n, 0, max}] // Denominator; a[n_] := ( A001803[[n]] + A001790[[n]])/(2* A046161[[n]]) // Numerator; a[0] = 0; Table[a[n], {n, 0, max}] (* or (from 1st formula) : *) Table[ n*Numerator[4^(1-n)*Binomial[2n-2, n-1]]/2^IntegerExponent[n, 2], {n, 0, max}]
(* or (from 2nd formula) : *) Table[ Numerator[ CatalanNumber[n-1]/2^(2n-1)]*Numerator[n^2/2^n], {n, 0, max}] (* Jean-François Alcover, Jan 31 2013 *)
PROG
(Magma) /* By definition: */ m:=25; R<x>:=PowerSeriesRing(Rationals(), m); p:=Coefficients(R!(1/(1-x)^(1/2))); q:=Coefficients(R!((1-x)^(-3/2))); A001790:=[Numerator(p[i]): i in [1..m]]; A001803:=[Numerator(q[i]): i in [1..m]]; A046161:=[Denominator(Binomial(2*n, n)/4^n): n in [0..m-1]]; [0] cat [Numerator(( A001803[n]+ A001790[n])/(2* A046161[n])): n in [1..m]]; // Bruno Berselli, Mar 11 2013
Triangle read by rows: T(n,k) is the coefficient of x^k in the polynomial P[n](x) = b(n)Q[n](x), where b(n) = numerator of binomial(2n,n)/2^n = A001790(n) and Q[n](x) = F(-n,1; 1/2-n; x) (hypergeometric function); 0 <= k <= n.
+20
0
1, 1, 2, 3, 4, 8, 5, 6, 8, 16, 35, 40, 48, 64, 128, 63, 70, 80, 96, 128, 256, 231, 252, 280, 320, 384, 512, 1024, 429, 462, 504, 560, 640, 768, 1024, 2048, 6435, 6864, 7392, 8064, 8960, 10240, 12288, 16384, 32768, 12155, 12870, 13728, 14784, 16128, 17920, 20480, 24576, 32768, 65536
COMMENTS
The polynomials Q[n](x) arise in a contact problem in elasticity theory.
REFERENCES
E. G. Deich (E. Deutsch), On an axially symmetric contact problem for a non-plane stamp with a circular cross-section (in Russian), Prikl. Mat. Mekh., 26, No. 5, 1962, 931-934.
FORMULA
Q[n](x) = (2n+1)*(Integral_{t=0..sqrt(1-x)} (x+t^2)^n dt)/sqrt(1-x).
Q[n](x) = 1 + 2*n*x*Q[n-1](x)/(2n-1).
EXAMPLE
Triangle begins:
1,
1, 2,
3, 4, 8,
5, 6, 8, 16,
35, 40, 48, 64, 128,
63, 70, 80, 96, 128, 256,
...
MAPLE
p:=proc(n) options operator, arrow: numer(simplify(hypergeom([ -n, 1], [1/2-n], x))) end proc: for n from 0 to 9 do P[n]:=p(n) end do: for n from 0 to 9 do seq(coeff(P[n], x, k), k=0..n) end do;
MATHEMATICA
b[n_] := Numerator[Binomial[2n, n]/2^n];
Q[n_][x_] := HypergeometricPFQ[{-n, 1}, {1/2 - n}, x];
T[n_, k_] := Coefficient[b[n]*Q[n][x], x, k];
1's-counting sequence: number of 1's in binary expansion of n (or the binary weight of n).
(Formerly M0105 N0041)
+10
1987
0, 1, 1, 2, 1, 2, 2, 3, 1, 2, 2, 3, 2, 3, 3, 4, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 1, 2, 2, 3, 2, 3, 3, 4, 2, 3, 3, 4, 3, 4, 4, 5, 2, 3, 3, 4, 3, 4, 4, 5, 3, 4, 4, 5, 4, 5, 5, 6, 2, 3, 3, 4, 3, 4, 4, 5, 3
COMMENTS
The binary weight of n is also called Hamming weight of n. [The term "Hamming weight" was named after the American mathematician Richard Wesley Hamming (1915-1998). - Amiram Eldar, Jun 16 2021] a(n) is also the largest integer such that 2^a(n) divides binomial(2n, n) = A000984(n). - Benoit Cloitre, Mar 27 2002 To construct the sequence, start with 0 and use the rule: If k >= 0 and a(0), a(1), ..., a(2^k-1) are the first 2^k terms, then the next 2^k terms are a(0) + 1, a(1) + 1, ..., a(2^k-1) + 1. - Benoit Cloitre, Jan 30 2003 An example of a fractal sequence. That is, if you omit every other number in the sequence, you get the original sequence. And of course this can be repeated. So if you form the sequence a(0 * 2^n), a(1 * 2^n), a(2 * 2^n), a(3 * 2^n), ... (for any integer n > 0), you get the original sequence. - Christopher.Hills(AT)sepura.co.uk, May 14 2003
The n-th row of Pascal's triangle has 2^k distinct odd binomial coefficients where k = a(n) - 1. - Lekraj Beedassy, May 15 2003 Fixed point of the morphism 0 -> 01, 1 -> 12, 2 -> 23, 3 -> 34, 4 -> 45, etc., starting from a(0) = 0. - Robert G. Wilson v, Jan 24 2006 a(n) is the number of solutions of the Diophantine equation 2^m*k + 2^(m-1) + i = n, where m >= 1, k >= 0, 0 <= i < 2^(m-1); a(5) = 2 because only (m, k, i) = (1, 2, 0) [2^1*2 + 2^0 + 0 = 5] and (m, k, i) = (3, 0, 1) [2^3*0 + 2^2 + 1 = 5] are solutions. - Hieronymus Fischer, Jan 31 2006 Sequence is given by T^(infinity)(0) where T is the operator transforming any word w = w(1)w(2)...w(m) into T(w) = w(1)(w(1)+1)w(2)(w(2)+1)...w(m)(w(m)+1). I.e., T(0) = 01, T(01) = 0112, T(0112) = 01121223. - Benoit Cloitre, Mar 04 2009 For n >= 2, the minimal k for which a(k(2^n-1)) is not multiple of n is 2^n + 3. - Vladimir Shevelev, Jun 05 2009 Triangle inequality: a(k+m) <= a(k) + a(m). Equality holds if and only if C(k+m, m) is odd. - Vladimir Shevelev, Jul 19 2009 The number of occurrences of value k in the first 2^n terms of the sequence is equal to binomial(n, k), and also equal to the sum of the first n - k + 1 terms of column k in the array A071919. Example with k = 2, n = 7: there are 21 = binomial(7,2) = 1 + 2 + 3 + 4 + 5 + 6 2's in a(0) to a(2^7-1). - Brent Spillner (spillner(AT)acm.org), Sep 01 2010, simplified by R. J. Mathar, Jan 13 2017 Let m be the number of parts in the listing of the compositions of n as lists of parts in lexicographic order, a(k) = n - length(composition(k)) for all k < 2^n and all n (see example); A007895 gives the equivalent for compositions into odd parts. - Joerg Arndt, Nov 09 2012 Just tally up row k (binary weight equal k) from 0 to 2^n - 1 to get the binomial coefficient C(n,k). (See A007318.) 0 1 3 7 15
0: O | . | . . | . . . . | . . . . . . . . |
1: | O | O . | O . . . | O . . . . . . . |
2: | | O | O O . | O O . O . . . |
3: | | | O | O O O . |
4: | | | | O |
Due to its fractal nature, the sequence is quite interesting to listen to.
(End)
The binary weight of n is a particular case of the digit sum (base b) of n. - Daniel Forgues, Mar 13 2015 The mean of the first n terms is 1 less than the mean of [a(n+1),...,a(2n)], which is also the mean of [a(n+2),...,a(2n+1)]. - Christian Perfect, Apr 02 2015 a(n) is also the largest part of the integer partition having viabin number n. The viabin number of an integer partition is defined in the following way. Consider the southeast border of the Ferrers board of the integer partition and consider the binary number obtained by replacing each east step with 1 and each north step, except the last one, with 0. The corresponding decimal form is, by definition, the viabin number of the given integer partition. "Viabin" is coined from "via binary". For example, consider the integer partition [2, 2, 2, 1]. The southeast border of its Ferrers board yields 10100, leading to the viabin number 20. - Emeric Deutsch, Jul 20 2017 a(n) is also known as the population count of the binary representation of n. - Chai Wah Wu, May 19 2020
REFERENCES
Jean-Paul Allouche and Jeffrey Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 119.
Donald E. Knuth, The Art of Computer Programming, vol. 4A, Combinatorial Algorithms, Section 7.1.3, Problem 41, p. 589. - N. J. A. Sloane, Aug 03 2012 Manfred R. Schroeder, Fractals, Chaos, Power Laws. W.H. Freeman, 1991, p. 383.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
P. J. Grabner, P. Kirschenhofer, H. Prodinger and R. F. Tichy, On the moments of the sum-of-digits function, Applications of Fibonacci numbers, Vol. 5 (St. Andrews, 1992), Kluwer Acad. Publ., Dordrecht, 1993, 263-271. J.-L. Mauclaire and Leo Murata, On q-additive functions, I, Proc. Japan Acad. Ser. A Math. Sci., Vol. 59, No. 6 (1983), pp. 274-276. Sam Northshield, Stern's Diatomic Sequence 0,1,1,2,1,3,2,3,1,4,..., Amer. Math. Month., Vol. 117, No. 7 (2010), pp. 581-598. Nanci Smith, Problem B-82, Fib. Quart., Vol. 4, No. 4 (1966), pp. 374-375.
FORMULA
a(0) = 0, a(2*n) = a(n), a(2*n+1) = a(n) + 1.
a(0) = 0, a(2^i) = 1; otherwise if n = 2^i + j with 0 < j < 2^i, a(n) = a(j) + 1.
G.f.: Product_{k >= 0} (1 + y*x^(2^k)) = Sum_{n >= 0} y^a(n)*x^n. - N. J. A. Sloane, Jun 04 2009 G.f.: (1/(1-x)) * Sum_{k>=0} x^(2^k)/(1+x^(2^k)). - Ralf Stephan, Apr 19 2003 a(0) = 0, a(n) = a(n - 2^floor(log_2(n))) + 1. Examples: a(6) = a(6 - 2^2) + 1 = a(2) + 1 = a(2 - 2^1) + 1 + 1 = a(0) + 2 = 2; a(101) = a(101 - 2^6) + 1 = a(37) + 1 = a(37 - 2^5) + 2 = a(5 - 2^2) + 3 = a(1 - 2^0) + 4 = a(0) + 4 = 4; a(6275) = a(6275 - 2^12) + 1 = a(2179 - 2^11) + 2 = a(131 - 2^7) + 3 = a(3 - 2^1) + 4 = a(1 - 2^0) + 5 = 5; a(4129) = a(4129 - 2^12) + 1 = a(33 - 2^5) + 2 = a(1 - 2^0) + 3 = 3. - Hieronymus Fischer, Jan 22 2006 A fixed point of the mapping 0 -> 01, 1 -> 12, 2 -> 23, 3 -> 34, 4 -> 45, ... With f(i) = floor(n/2^i), a(n) is the number of odd numbers in the sequence f(0), f(1), f(2), f(3), f(4), f(5), ... - Philippe Deléham, Jan 04 2004 When read mod 2 gives the Morse-Thue sequence A010060. Let floor_pow4(n) denote n rounded down to the next power of four, floor_pow4(n) = 4 ^ floor(log4 n). Then a(0) = 0, a(1) = 1, a(2) = 1, a(3) = 2, a(n) = a(floor(n / floor_pow4(n))) + a(n % floor_pow4(n)). - Stephen K. Touset (stephen(AT)touset.org), Apr 04 2007
a(n) = n - Sum_{k=2..n} Sum_{j|n, j >= 2} (floor(log_2(j)) - floor(log_2(j-1))). - Hieronymus Fischer, Jun 18 2007 For odd m >= 1, a((4^m-1)/3) = a((2^m+1)/3) + (m-1)/2 (mod 2). - Vladimir Shevelev, Sep 03 2010 a(n) - a(n-1) = { 1 - a(n-1) if and only if A007814(n) = a(n-1), 1 if and only if A007814(n) = 0, -1 for all other A007814(n) }. - Brent Spillner (spillner(AT)acm.org), Sep 01 2010 a(n) = Sum_{j = 1..m+1} (floor(n/2^j + 1/2) - floor(n/2^j)), where m = floor(log_2(n)).
General formulas for the number of digits >= d in the base p representation of n, where 1 <= d < p: a(n) = Sum_{j = 1..m+1} (floor(n/p^j + (p-d)/p) - floor(n/p^j)), where m=floor(log_p(n)); g.f.: g(x) = (1/(1-x))*Sum_{j>=0} (x^(d*p^j) - x^(p*p^j))/(1-x^(p*p^j)). (End)
a(n) = log_2(C(2*n,n) - (C(2*n,n) AND C(2*n,n)-1)). - Gary Detlefs, Jul 10 2014 Sum_{n >= 1} a(n)/2n(2n+1) = (gamma + log(4/Pi))/2 = A344716, where gamma is Euler's constant A001620; see Sondow 2005, 2010 and Allouche, Shallit, Sondow 2007. - Jonathan Sondow, Mar 21 2015 For any integer base b >= 2, the sum of digits s_b(n) of expansion base b of n is the solution of this recurrence relation: s_b(n) = 0 if n = 0 and s_b(n) = s_b(floor(n/b)) + (n mod b). Thus, a(n) satisfies: a(n) = 0 if n = 0 and a(n) = a(floor(n/2)) + (n mod 2). This easily yields a(n) = Sum_{i = 0..floor(log_2(n))} (floor(n/2^i) mod 2). From that one can compute a(n) = n - Sum_{i = 1..floor(log_2(n))} floor(n/2^i). - Marek A. Suchenek, Mar 31 2016 Sum_{k>=1} a(k)/2^k = 2 * Sum_{k >= 0} 1/(2^(2^k)+1) = 2 * A051158. - Amiram Eldar, May 15 2020 a(m*(2^n-1)) >= n. Equality holds when 2^n-1 >= A000265(m), but also in some other cases, e.g., a(11*(2^2-1)) = 2 and a(19*(2^3-1)) = 3. - Pontus von Brömssen, Dec 13 2020 G.f.: A(x) satisfies A(x) = (1+x)*A(x^2) + x/(1-x^2). - Akshat Kumar, Nov 04 2023
EXAMPLE
Using the formula a(n) = a(floor(n / floor_pow4(n))) + a(n mod floor_pow4(n)):
a(4) = a(1) + a(0) = 1,
a(8) = a(2) + a(0) = 1,
a(13) = a(3) + a(1) = 2 + 1 = 3,
a(23) = a(1) + a(7) = 1 + a(1) + a(3) = 1 + 1 + 2 = 4.
Gary W. Adamson points out (Jun 03 2009) that this can be written as a triangle:
0,
1,
1,2,
1,2,2,3,
1,2,2,3,2,3,3,4,
1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,
1,2,2,3,2,3,3,4,2,3,3,4,3,4,4,5,2,3,3,4,3,4,4,5,3,4,4,5,4,5,5,6,
1,2,2,3,2,3,...
where the rows converge to A063787. Connection to the compositions of n as lists of parts (see comment):
[ #]: a(n) composition
[ 0]: [0] 1 1 1 1 1
[ 1]: [1] 1 1 1 2
[ 2]: [1] 1 1 2 1
[ 3]: [2] 1 1 3
[ 4]: [1] 1 2 1 1
[ 5]: [2] 1 2 2
[ 6]: [2] 1 3 1
[ 7]: [3] 1 4
[ 8]: [1] 2 1 1 1
[ 9]: [2] 2 1 2
[10]: [2] 2 2 1
[11]: [3] 2 3
[12]: [2] 3 1 1
[13]: [3] 3 2
[14]: [3] 4 1
[15]: [4] 5
(End)
MAPLE
A000120 := proc(n) local w, m, i; w := 0; m := n; while m > 0 do i := m mod 2; w := w+i; m := (m-i)/2; od; w; end: wt := A000120;
with(Bits): p:=n->ilog2(n-And(n, n-1)): seq(p(binomial(2*n, n)), n=0..200) # Gary Detlefs, Jan 27 2019
MATHEMATICA
Table[DigitCount[n, 2, 1], {n, 0, 105}]
Table[Plus @@ IntegerDigits[n, 2], {n, 0, 104}]
Log[2, Nest[Join[#, 2#] &, {1}, 14]] (* gives 2^14 term, Carlos Alves, Mar 30 2014 *)
PROG
(PARI) {a(n) = if( n<0, 0, 2*n - valuation((2*n)!, 2))};
(PARI) {a(n) = if( n<0, 0, subst(Pol(binary(n)), x , 1))};
(PARI) {a(n) = if( n<1, 0, a(n\2) + n%2)}; /* Michael Somos, Mar 06 2004 */ (PARI) a(n)=norml2(binary(n)) \\ better use { A000120=hammingweight}. - M. F. Hasler, Oct 09 2012, edited Feb 27 2020 (Common Lisp) (defun floor-to-power (n pow) (declare (fixnum pow)) (expt pow (floor (log n pow)))) (defun enabled-bits (n) (if (< n 4) (n-th n (list 0 1 1 2)) (+ (enabled-bits (floor (/ n (floor-to-power n 4)))) (enabled-bits (mod n (floor-to-power n 4)))))) ; Stephen K. Touset (stephen(AT)touset.org), Apr 04 2007
(Haskell)
import Data.Bits (Bits, popCount)
a000120 :: (Integral t, Bits t) => t -> Int
a000120 = popCount
a000120_list = 0 : c [1] where c (x:xs) = x : c (xs ++ [x, x+1])
(Haskell)
a000120 = concat r
where r = [0] : (map.map) (+1) (scanl1 (++) r)
(SageMath)
if n <= 1: return Integer(n)
(Python)
import numpy as np
A000120 = np.array([0], dtype="uint8")
(Python) # Also see links.
(Scala) (0 to 127).map(Integer.bitCount(_)) // Alonso del Arte, Mar 05 2019 (Magma) [Multiplicity(Intseq(n, 2), 1): n in [0..104]]; // Marius A. Burtea, Jan 22 2020
CROSSREFS
Sum of digits of n written in bases 2-16: this sequence, A053735, A053737, A053824, A053827, A053828, A053829, A053830, A007953, A053831, A053832, A053833, A053834, A053835, A053836. This is Guy Steele's sequence GS(3, 4) (see A135416). Cf. A230952 (boustrophedon transform). Cf. A070939 (length of binary representation of n).
KEYWORD
nonn, easy, core, nice, hear, look, base, changed
Decimal expansion of square root of 2.
(Formerly M3195 N1291)
+10
317
1, 4, 1, 4, 2, 1, 3, 5, 6, 2, 3, 7, 3, 0, 9, 5, 0, 4, 8, 8, 0, 1, 6, 8, 8, 7, 2, 4, 2, 0, 9, 6, 9, 8, 0, 7, 8, 5, 6, 9, 6, 7, 1, 8, 7, 5, 3, 7, 6, 9, 4, 8, 0, 7, 3, 1, 7, 6, 6, 7, 9, 7, 3, 7, 9, 9, 0, 7, 3, 2, 4, 7, 8, 4, 6, 2, 1, 0, 7, 0, 3, 8, 8, 5, 0, 3, 8, 7, 5, 3, 4, 3, 2, 7, 6, 4, 1, 5, 7
COMMENTS
Sometimes called Pythagoras's constant.
The discovery of irrational numbers is attributed to Hippasus of Metapontum, who may have proved that sqrt(2) is not a rational number; thus sqrt(2) is often regarded as the earliest known irrational number. - Clark Kimberling, Oct 12 2017 In the first million digits,
0 occurs 99814 times;
1 occurs 99925 times;
2 occurs 100436 times;
3 occurs 100190 times;
4 occurs 100024 times;
5 occurs 100155 times;
6 occurs 99886 times;
7 occurs 100008 times;
8 occurs 100441 times;
9 occurs 100121 times. (End)
Diameter of a sphere whose surface area equals 2*Pi. More generally, the square root of x is also the diameter of a sphere whose surface area equals x*Pi. - Omar E. Pol, Nov 10 2018 Sqrt(2) = 1 + area of region bounded by y = sin x, y = cos x, and x = 0. - Clark Kimberling, Jul 03 2020 Also aspect ratio of the ISO 216 standard for paper sizes. - Stefano Spezia, Feb 24 2021 The standard deviation of a roll of a 5-sided die. - Mohammed Yaseen, Feb 23 2023 The length of a unit square diagonal.
The infinite tetration (power tower) sqrt(2)^(sqrt(2)^(sqrt(2)^(...))) equals 2 from the identity (x^(1/x))^((x^(1/x))^((x^(1/x))^(...))) = x where 1/e <= x <= e. (End)
REFERENCES
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, Section 1.1.
David Flannery, The Square Root of 2, Copernicus Books Springer-Praxis Pub. 2006.
Martin Gardner, Gardner's Workout, Chapter 2 "The Square Root of 2=1.414213562373095..." pp. 9-19 A. K. Peters MA 2002.
Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §3.4 Irrational Numbers and §4.4 Powers and Roots, pp. 84, 145.
B. Rittaud, Le fabuleux destin de sqrt(2), Le Pommier, Paris 2006.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 60, page 605.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, pp. 34-35.
FORMULA
Sqrt(2) = 14 * Sum_{n >= 0} ( A001790(n)/2^ A005187(floor(n/2)) * 10^(-2n-1)) where A001790(n) are numerators in expansion of 1/sqrt(1-x) and the denominators in expansion of 1/sqrt(1-x) are 2^ A005187(n). 14 = 2*7, see A010503 (expansion of 1/sqrt(2)). - Gerald McGarvey, Jan 01 2005 Limit_{n -> +oo} (1/n)*(Sum_{k = 1..n} frac(sqrt(1+zeta(k+1)))) = 1/(1+sqrt(2)). - Yalcin Aktar, Jul 14 2005 sqrt(2) = Product_{l=0..k-1} 2*cos((2*l+1)*Pi/(4*k)) = (Product_{l=0..k-1} R(2*l+1,rho(4*k)) - 1), identical for k >= 1, with the row polynomials R(n, x) from A127672 and rho(4*k) := 2*cos(Pi/(4*k)) is the length ratio (smallest diagonal)/side in a regular (4*k)-gon. From the product formula given in a Oct 21 2013 formula contribution to A056594, with n -> 2*k, using cos(Pi-alpha) = - cos(alpha) to obtain 2 for the square of the present product. - Wolfdieter Lang, Oct 22 2013 Equals Product_{k>=0} (1 + (-1)^k/(2*k + 1)).
Equals Sum_{k>=0} binomial(2*k,k)/8^k. (End)
Equals (sqrt(2) + (sqrt(2) + (sqrt(2) + ...)^(1/3))^(1/3))^(1/3). - Michal Paulovic, Mar 22 2023 Equals 1 + Sum_{k>=1} (-1)^(k-1)/(2^(2*k)*(2*k - 1))*binomial(2*k,k) [Newton]. - Stefano Spezia, Oct 15 2024 Equals Sum_{k>=0} 2*k*binomial(2*k,k)/8^k.
Equals Product_{k>=2} k/sqrt(k^2 + 1).
Equals Product_{k>=0} (6*k + 3)/((6*k + 3) - (-1)^k).
Equals Product_{k>=1} (2*k + 1)/((2*k + 1) + (-1)^k).
Equals Product_{k>=0} ((4*k + 3)*(4*k + 1 + (-1)^k))/((4*k + 1)*(4*k + 3 + (-1)^k)). (End)
Equals hypergeom([1/2, 1/2], [1/2], 1/2). - Stefano Spezia, Jan 05 2025
EXAMPLE
1.41421356237309504880168872420969807856967187537694807317667...
PROG
(PARI) default(realprecision, 20080); x=sqrt(2); for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b002193.txt", n, " ", d)); \\ Harry J. Smith, Apr 21 2009 (PARI) r=0; x=2; /* Digit-by-digit method */
for(digits=1, 100, {d=0; while((20*r+d)*d <= x, d++);
d--; /* while loop overshoots correct digit */
(PARI) \\ Works in v2.15.0; n = 100 decimal places
(Maxima) fpprec: 100$ ev(bfloat(sqrt(2))); /* Martin Ettl, Oct 17 2012 */ a002193 n = a002193_list !! (n-1)
a002193_list = w 2 0 where
w x r = dig : w (100 * (x - (20 * r + dig) * dig)) (10 * r + dig)
where dig = head (dropWhile (\d -> (20 * r + d) * d < x) [0..]) - 1
a(n) = a(floor(n/2)) + n; also denominators in expansion of 1/sqrt(1-x) are 2^a(n); also 2n - number of 1's in binary expansion of 2n.
(Formerly M2330)
+10
238
0, 1, 3, 4, 7, 8, 10, 11, 15, 16, 18, 19, 22, 23, 25, 26, 31, 32, 34, 35, 38, 39, 41, 42, 46, 47, 49, 50, 53, 54, 56, 57, 63, 64, 66, 67, 70, 71, 73, 74, 78, 79, 81, 82, 85, 86, 88, 89, 94, 95, 97, 98, 101, 102, 104, 105, 109, 110, 112, 113, 116, 117, 119, 120, 127, 128
COMMENTS
Also exponent of the largest power of 2 dividing (2n)! ( A010050) and (2n)!! ( A000165). Write n in binary: 1ab..yz, then a(n) = 1ab..yz + ... + 1ab + 1a + 1. - Ralf Stephan, Aug 27 2003 Wikipedia's article on the "Whitney Immersion theorem" mentions that the a(n)-dimensional sphere arises in the Immersion Conjecture proved by Ralph Cohen in 1985. - Jonathan Vos Post, Jan 25 2010 For n > 0, denominators for consecutive pairs of integral numerator polynomials L(n+1,x) for the Legendre polynomials with o.g.f. 1 / sqrt(1-tx+x^2). - Tom Copeland, Feb 04 2016 a(n) is the total number of pointers in the first n elements of a perfect skip list. - Alois P. Heinz, Dec 14 2017 a(n) is the position of the n-th a (indexing from 0) in the fixed point of the morphism a -> aab, b -> b. - Jeffrey Shallit, Dec 24 2020
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Laurent Alonso, Edward M. Reingold, and René Schott, Determining the majority, Inform. Process. Lett. 47 (1993), no. 5, 253-255.
FORMULA
For m>0, let q = floor(log_2(m)); a(2m+1) = 2^q + 3m + Sum_{k>=1} floor((m-2^q)/2^k); a(2m) = a(2m+1) - 1. - Len SmileyG.f.: A(x) = Sum_{k>=0} x^(2^k)/((1-x)*(1-x^(2^k))). - Ralf Stephan, Dec 24 2002 a(n) = Sum_{k=1..n} A001511(k), sum of binary Hamming distances between consecutive integers up to n. - Gary W. Adamson, Jun 15 2003 Conjecture: a(n) = 2n + O(log(n)). - Benoit Cloitre, Oct 07 2003 [true as a(n) = 2*n - hamming_weight(2*n). Joerg Arndt, Jun 10 2019] Recurrence: a(n) = n + a(floor(n/2)); a(2n) = 2n + a(n); a(n*2^m) = 2*n*(2^m-1) + a(n).
a(2^m) = 2^(m+1) - 1, m >= 0.
Asymptotic behavior: a(n) = 2n + O(log(n)), a(n+1) - a(n) = O(log(n)); this follows from the inequalities below.
a(n) <= 2n-1; equality holds for powers of 2.
a(n) >= 2n-1-floor(log_2(n)); equality holds for n = 2^m-1, m > 0.
lim inf (2n - a(n)) = 1, for n-->oo.
lim sup (2n - log_2(n) - a(n)) = 0, for n-->oo.
lim sup (a(n+1) - a(n) - log_2(n)) = 1, for n-->oo. (End)
PURRS demo results: Bounds for a(n) = n + a(n/2) with initial conditions a(1) = 1: a(n) >= -2 + 2*n - log(n)*log(2)^(-1), a(n) <= 1 + 2*n for each n >= 1. - Alexander R. Povolotsky, Apr 06 2008 If n = 2^a_1 + 2^a_2 + ... + 2^a_k, then a(n) = n-k. This can be used to prove that 2^n maximally divides (2n!)/n!. - Jon Perry, Jul 16 2009 a(n) = log_2(denominator(binomial(-1/2,n))). - Peter Luschny, Nov 25 2011 G.f.: (1/(1 - x))*Sum_{k>=0} (2^(k+1) - 1)*x^(2^k)/(1 + x^(2^k)). - Ilya Gutkovskiy, Jul 23 2017
EXAMPLE
G.f. = x + 3*x^2 + 4*x^3 + 7*x^4 + 8*x^5 + 10*x^6 + 11*x^7 + 15*x^8 + ...
MAPLE
A005187 := n -> 2*n - add(i, i=convert(n, base, 2)):
MATHEMATICA
a[0] = 0; a[n_] := a[n] = a[Floor[n/2]] + n; Table[ a[n], {n, 0, 50}] (* or *)
Table[2n-DigitCount[2n, 2, 1], {n, 0, 70}] (* Harvey P. Dale, May 24 2014 *)
PROG
(PARI) {a(n) = if( n<0, 0, valuation((2*n)!, 2))}; /* Michael Somos, Oct 24 2002 */ (PARI) {a(n) = if( n<0, 0, sum(k=1, n, (2*n)\2^k))}; /* Michael Somos, Oct 24 2002 */ (PARI) {a(n) = if( n<0, 0, 2*n - subst( Pol( binary( n ) ), x, 1) ) }; /* Michael Somos, Aug 28 2007 */ (Haskell)
a005187 n = a005187_list !! n
a005187_list = 0 : zipWith (+) [1..] (map (a005187 . (`div` 2)) [1..])
(Sage)
@CachedFunction
(Magma) [n + Valuation(Factorial(n), 2): n in [0..70]]; // Vincenzo Librandi, Jun 11 2019 (Python)
a(n) = denominator of binomial(2n,n)/4^n.
+10
99
1, 2, 8, 16, 128, 256, 1024, 2048, 32768, 65536, 262144, 524288, 4194304, 8388608, 33554432, 67108864, 2147483648, 4294967296, 17179869184, 34359738368, 274877906944, 549755813888, 2199023255552, 4398046511104, 70368744177664, 140737488355328, 562949953421312
COMMENTS
Also denominator of e(0,n) (see Maple line). - N. J. A. Sloane, Feb 16 2002 Denominator of coefficient of x^n in (1+x)^(k/2) or (1-x)^(k/2) for any odd integer k. - Michael Somos, Sep 15 2004 Numerator of binomial(2n,n)/4^n = A001790(n). Denominators in expansion of sqrt(c(x)), c(x) the g.f. of A000108. - Paul Barry, Jul 12 2005 Denominator of 2^m*Gamma(m+3/4)/(Gamma(3/4)*Gamma(m+1)). - Stephen Crowley, Mar 19 2007 Denominator in expansion of Jacobi_P(n,1/2,1/2,x). - Paul Barry, Feb 13 2008 This sequence equals the denominators of the coefficients of the series expansions of (1-x)^((-1-2*n)/2) for all integer values of n; see A161198 for detailed information. - Johannes W. Meijer, Jun 08 2009 Numerators of binomial transform of 1, -1/3, 1/5, -1/7, 1/9, ... (Madhava-Gregory-Leibniz series for Pi/4): 1, 2/3, 8/15, 16/35, 128/315, 256/693, .... First differences are -1/3, -2/15, -8/105, -16/315, -128/3465, -256/9009, ... which contain the same numerators, negated. The second differences are 1/5, 2/35, 8/315, 16/1155, 128/15015, ... again with the same numerators. Second column: 2/3, -2/15, 2/35, -2/63, 2/99; see A000466(n+1) = A005563(2n+1). Third column: 8*(1/15, -1/105, 1/315, -1/693, ...), see A061550. See A173294 and A173296. - Paul Curtz, Feb 16 2010 0, 1, 5/3, 11/5, 93/35, 193/63, 793/231, ... = (0 followed by A120778(n))/ A001790(n) is the binomial transform of 0, 1, -1/3, 1/5, -1/7, 1/9, ... . See A173755 and formula below. - Paul Curtz, Mar 13 2013 Numerator of power series of arcsin(x)/sqrt(1-x^2), centered at x=0. - John Molokach, Aug 02 2013 Denominators of coefficients in the Taylor series expansion of Sum_{n>=0} exp((-1)^n * Euler(2*n)*x^n/(2*n)), see A280442 for the numerators. - Johannes W. Meijer, Jan 05 2017 Denominators of Pochhammer(n+1, -1/2)/sqrt(Pi). - Adam Hugill, Sep 11 2022 a(n) is the denominator of the mean value of cos(x)^(2*n) from x = 0 to 2*Pi. - Stefano Spezia, May 16 2023 4^n/binomial(2n,n) is the expected value of the number of socks that are randomly drawn out of a drawer of n different pairs of socks, when one sock is drawn out at a time until a matching pair is found (King and King, 2005). - Amiram Eldar, Jul 02 2023 a(n) is the denominator of (1/Pi) * Integral_{x=-oo..+oo} sech(x)^(2*n+1) dx. The corresponding numerator is A001790(n). - Mohammed Yaseen, Jul 29 2023 a(n) is the numerator of Integral_{x=0..Pi/2} sin(x)^(2*n+1) dx. The corresponding denominator is A001803(n). - Mohammed Yaseen, Sep 22 2023
REFERENCES
W. Feller, An Introduction to Probability Theory and Its Applications, Vol. 1, 2nd ed. New York: Wiley, 1968; Chap. III, Eq. 4.1.
B. D. Hughes, Random Walks and Random Environments, Oxford 1995, vol. 1, p. 513, Eq. (7.282).
Eli Maor, e: The Story of a Number. Princeton, New Jersey: Princeton University Press (1994), p. 72.
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, equations 6:14:5 - 6:14:9 at pages 50-51.
LINKS
Jeremy and Patricia King, Problem 89.G, Problem Corner, The Mathematical Gazette, Vol. 90, No. 515 (2005), p. 314; Solution, ibid., Vol. 90, No. 517 (2006), pp. 163-164.
FORMULA
a(n) = 2^(2*n - 1 - A048881(n-1)), if n > 0. a(n) = denominator(binomial(-1/2,n)). - Peter Luschny, Nov 21 2012
EXAMPLE
sqrt(1+x) = 1 + (1/2)*x - (1/8)*x^2 + (1/16)*x^3 - (5/128)*x^4 + (7/256)*x^5 - (21/1024)*x^6 + (33/2048)*x^7 + ...
binomial(2n,n)/4^n => 1, 1/2, 3/8, 5/16, 35/128, 63/256, 231/1024, 429/2048, 6435/32768, ...
The sequence e(0,n) begins 1, 3/2, 21/8, 77/16, 1155/128, 4389/256, 33649/1024, 129789/2048, 4023459/32768, ...
MAPLE
e := proc(l, m) local k; add(2^(k-2*m)*binomial(2*m-2*k, m-k)* binomial(m+k, m) *binomial(k, l), k=l..m); end: seq(denom(e(0, n)), n = 0..24);
Z[0]:=0: for k to 30 do Z[k]:=simplify(1/(2-z*Z[k-1])) od: g:=sum((Z[j]-Z[j-1]), j=1..30): gser:=series(g, z=0, 27): seq(denom(coeff(gser, z, n)), n=-1..23); # Zerinvary Lajos, May 21 2008 A046161 := n -> 4^n/2^add(i, i=convert(n, base, 2)):
MATHEMATICA
a[n_, m_] := Binomial[n - m/2 + 1, n - m + 1] - Binomial[n - m/2, n - m + 1]; s[n_] := Sum[ a[n, k], {k, 0, n}]; Table [Denominator[s[n]], {n, 0, 26}] (* Michele Dondi (bik.mido(AT)tiscalinet.it), Jul 11 2002 *)
Denominator[Table[Binomial[2n, n]/4^n, {n, 0, 30}]] (* Harvey P. Dale, Oct 29 2012 *) Table[Denominator@LegendreP[2n, 0], {n, 0, 24}] (* Andres Cicuttin, Jan 22 2018 *)
PROG
(PARI) a(n)=if(n<0, 0, denominator(binomial(2*n, n)/4^n)) /* Michael Somos, Sep 15 2004 */ (Sage)
(Maxima)
a(n) := denom(binomial(-1/2, n));
(Magma) [Denominator(Binomial(2*n, n)/4^n): n in [0..30]]; // Vincenzo Librandi, Jul 18 2015 (Python)
CROSSREFS
Cf. A161198 triangle related to the series expansions of (1-x)^((-1-2*n)/2) for all values of n. Cf. A000108, A000120, A000466, A001511, A005563, A048881, A060818, A061550, A120778, A173294, A173296, A173755, A280442.
KEYWORD
nonn,easy,nice,frac,changed
a(n) = 2^( A000120(n+1) - 1), n >= 0.
+10
52
1, 1, 2, 1, 2, 2, 4, 1, 2, 2, 4, 2, 4, 4, 8, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 2, 4, 4
COMMENTS
a(n), n >= 1: Numerators of Maclaurin series for 1 - ((sin x)/x)^2, A117972(n), n >= 2: Denominators of Maclaurin series for 1 - ((sin x)/x)^2, the correlation function in Montgomery's pair correlation conjecture. - Daniel Forgues, Oct 16 2011 Also the number of coarsenings of the (n+1)-th composition in standard order. The k-th composition in standard order (graded reverse-lexicographic, A066099) is obtained by taking the set of positions of 1's in the reversed binary expansion of k, prepending 0, taking first differences, and reversing again. This gives a bijective correspondence between nonnegative integers and integer compositions. See link for sequences related to standard compositions. For example, the a(10) = 4 coarsenings of (2,1,1) are: (2,1,1), (2,2), (3,1), (4). Also the number of times n+1 appears in A357134. For example, 11 appears at positions 11, 20, 33, and 1024, so a(10) = 4. (End)
LINKS
Neil J. Calkin, Eunice Y. S. Chan, Robert M. Corless, David J. Jeffrey, and Piers W. Lawrence, A Fractal Eigenvector, arXiv:2104.01116 [math.DS], 2021.
FORMULA
a(n) = 2^k if 2^k divides A000108(n) but 2^(k+1) does not divide A000108(n). It appears that a(n) = Sum_{k=0..n} binomial(2*(n+1), k) mod 2. - Christopher Lenard (c.lenard(AT)bendigo.latrobe.edu.au), Aug 20 2001
a(0) = 1; a(2*n) = 2*a(2*n-1); a(2*n+1) = a(n).
a(n) = (1/2) * A001316(n+1). - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 26 2004 It appears that a(n) = Sum_{k=0..2n} floor(binomial(2n+2, k+1)/2)(-1)^k = 2^n - Sum_{k=0..n+1} floor(binomial(n+1, k)/2). - Paul Barry, Dec 24 2004 a(n) = numerator(b(n)), where sin(x)^2/x = Sum_{n>0} b(n)*(-1)^n x^(2*n-1). - Vladimir Kruchinin, Feb 06 2013
EXAMPLE
If written as a triangle:
1;
1,2;
1,2,2,4;
1,2,2,4,2,4,4,8;
1,2,2,4,2,4,4,8,2,4,4,8,4,8,8,16;
1,2,2,4,2,4,4,8,2,4,4,8,4,8,8,16,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32;
...,
the first half-rows converge to Gould's sequence A001316. (End)
MAPLE
a := n -> 2^(add(i, i=convert(n+1, base, 2))-1): seq(a(n), n=0..97); # Peter Luschny, May 01 2009
MATHEMATICA
NestList[Flatten[#1 /. a_Integer -> {a, 2 a}] &, {1}, 4] // Flatten (* Robert G. Wilson v, Aug 01 2012 *) Denominator[Table[BernoulliB[2*n] / (Zeta[2*n]/Pi^(2*n)), {n, 1, 100}]] (* Terry D. Grant, May 29 2017 *) Table[Denominator[((2 n)!/2^(2 n + 1)) (-1)^n], {n, 1, 100}]/4 (* Terry D. Grant, May 29 2017 *) 2^IntegerExponent[CatalanNumber[Range[0, 100]], 2] (* Harvey P. Dale, Apr 30 2018 *)
PROG
(PARI) a(n)=if(n<1, 1, if(n%2, a(n/2-1/2), 2*a(n-1)))
(PARI) a(n) = 1 << (hammingweight(n+1)-1); \\ Kevin Ryde, Feb 19 2022 (Haskell)
a048896 n = a048896_list !! n
a048896_list = f [1] where f (x:xs) = x : f (xs ++ [x, 2*x])
(Haskell)
import Data.List (transpose)
a048896 = a000079 . a000120
a048896_list = 1 : concat (transpose
[zipWith (-) (map (* 2) a048896_list) a048896_list,
map (* 2) a048896_list])
(Magma) [Numerator(2^n / Factorial(n+1)): n in [0..100]]; // Vincenzo Librandi, Apr 12 2014
CROSSREFS
This is Guy Steele's sequence GS(3, 5) (see A135416). Equals first right hand column of triangle A160468. Standard compositions are listed by A066099. The opposite version (counting refinements) is A080100. The version for Heinz numbers of partitions is A317141.
Numerators in expansion of (1 - x)^(-3/2).
(Formerly M2986 N1207)
+10
50
1, 3, 15, 35, 315, 693, 3003, 6435, 109395, 230945, 969969, 2028117, 16900975, 35102025, 145422675, 300540195, 9917826435, 20419054425, 83945001525, 172308161025, 1412926920405, 2893136075115, 11835556670925
COMMENTS
a(n) is the denominator of the integral from 0 to Pi of (sin(x))^(2*n+1). - James R. Buddenhagen, Aug 17 2008 a(n) is the denominator of (2n)!!/(2*n + 1)!! = 2^(2*n)*n!*n!/(2*n + 1)! (see Andersson). - N. J. A. Sloane, Jun 27 2011 a(n) = (2*n + 1)* A001790(n). A046161(n)/a(n) = 1, 2/3, 8/15, 16/35, 128/315, 256/693, ... is binomial transform of Madhava-Gregory-Leibniz series for Pi/4 (i.e., 1 - 1/3 + 1/5 - 1/7 + ... ). See A173384 and A173396. - Paul Curtz, Feb 21 2010 a(n) is the denominator of Integral_{x=-oo..oo} sech(x)^(2*n+2) dx. The corresponding numerator is A101926(n). - Mohammed Yaseen, Jul 25 2023
REFERENCES
M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 798.
G. Prévost, Tables de Fonctions Sphériques. Gauthier-Villars, Paris, 1933, pp. 156-157.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 6, equation 6:14:9 at page 51.
LINKS
Milton Abramowitz and Irene A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972 [alternative scanned copy].
FORMULA
a(n) is the numerator of (2*n + 1)*binomial(2*n,n)/(4^n).
(1 - x)^(-3/2) = Sum_{n>=0} ((2*n + 1)*binomial(2*n,n)/(4^n)*x^n)
(End)
Truncations of rational expressions like those given by the numerator or denominator operators are artifacts in integer formulas and have many disadvantages. A pure integer formula follows. Let n$ denote the swinging factorial and sigma(n) = number of '1's in the base-2 representation of floor(n/2). Then a(n) = (2*n+1)$ / sigma(2*n+1) = A056040(2*n+1) / A060632(2*n+2). Simply said: This sequence gives the odd part of the swinging factorial at odd indices. - Peter Luschny, Aug 01 2009 a(n) = denominator(Pi*binomial(n, -1/2)). - Peter Luschny, Dec 06 2024
MAPLE
swing := proc(n) option remember; if n = 0 then 1 elif irem(n, 2) = 1 then swing(n-1)*n else 4*swing(n-1)/n fi end:
sigma := n -> 2^(add(i, i= convert(iquo(n, 2), base, 2))):
a := n -> swing(2*n+1)/sigma(2*n+1); # Peter Luschny, Aug 01 2009 A001803 := proc(n) (2*n+1)*binomial(2*n, n)/4^n ; numer(%) ; end proc: # R. J. Mathar, Jul 06 2011
a := n -> denom(Pi*binomial(n, -1/2)): seq(a(n), n = 0..22); # Peter Luschny, Dec 06 2024
MATHEMATICA
Numerator/@CoefficientList[Series[(1-x)^(-3/2), {x, 0, 25}], x] (* Harvey P. Dale, Feb 19 2011 *) Table[Denominator[Beta[1, n + 1, 1/2]], {n, 0, 22}] (* Gerry Martens, Nov 13 2016 *)
PROG
(PARI) a(n) = numerator((2*n+1)*binomial(2*n, n)/(4^n)); \\ Altug Alkan, Sep 06 2018 (Julia)
CROSSREFS
The denominator is given in A046161. Cf. A002596 (numerators in expansion of (1-x)^(1/2)). Cf. A161198 (triangle related to the series expansions of (1-x)^((-1-2*n)/2)). (End)
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