Displaying 1-10 of 41 results found.
Numerators of partial sums of the Madhava series for Pi/(2*sqrt(3)) = A093766.
+20
1
1, 8, 41, 856, 23147, 254512, 3309041, 29780368, 168757087, 28857376792, 259716622073, 5973480691064, 89602217802389, 7257779456082784, 210475605899597261, 6524743766713282016, 19574231315333822573, 6524743770186190936, 2172739675639135323463, 19554657080276529569192
COMMENTS
This Madhava series results from the arctan series for tan(Pi/6) = sqrt(3)/3 = A020760.
REFERENCES
L. B. W. Jolley, Summation of Series, Dover (1961), eq. (273), pp. 16 and 17.
Ian Stewart, Grössen der Mathematik, Rowohlt Tachenbuch Verlag, Nr. 63394, 2020, p. 74. [English Original: Significant Figures. Lives and Works of Traiblazing Mathematicians, Profile Books, London, 2017]
FORMULA
a(n) = numerator(Sum_{j=0..n} (-1)^j/((2*j+1)*3^j)), for n >= 0.
EXAMPLE
The partial sums begin: 1/1, 8/9, 41/45, 856/945, 23147/25515, 254512/280665, 3309041/3648645, 29780368/32837805, 168757087/186080895, ...
For n = 100 the partial sum is 0.9068996821171089252970391288210778661420331240463726... compared to 0.9068996821171089252970391288210778661420331240463702...(the first 53 digits coincide).
MATHEMATICA
Numerator @ Accumulate @ Table[(-1)^j/((2*j + 1)*3^j), {j, 0, 20}] (* Amiram Eldar, Apr 08 2022 *)
PROG
(PARI) a(n) = numerator(sum(j=0, n, (-1)^j/((2*j+1)*3^j))); \\ Michel Marcus, Apr 08 2022
Denominators of partial sums of the Madhava series for Pi/(2*sqrt(3)) = A093766.
+20
1
1, 9, 45, 945, 25515, 280665, 3648645, 32837805, 186080895, 31819833045, 286378497405, 6586705440315, 98800581604725, 8002847109982725, 232082566189499025, 7194559551874469775, 21583678655623409325, 7194559551874469775, 2395788330774198435075, 21562094976967785915675
COMMENTS
For a comment and references see A352397.
FORMULA
a(n) = denominator(Sum_{j=0..n} (-1)^j/((2*j+1)*3^j)), for n >= 0.
MATHEMATICA
Denominator @ Accumulate @ Table[(-1)^j/((2*j + 1)*3^j), {j, 0, 20}] (* Amiram Eldar, Apr 08 2022 *)
PROG
(PARI) a(n) = denominator(sum(j=0, n, (-1)^j/((2*j+1)*3^j))); \\ Michel Marcus, Apr 08 2022
Octagonal numbers: n*(3*n-2). Also called star numbers.
(Formerly M4493 N1901)
+10
262
0, 1, 8, 21, 40, 65, 96, 133, 176, 225, 280, 341, 408, 481, 560, 645, 736, 833, 936, 1045, 1160, 1281, 1408, 1541, 1680, 1825, 1976, 2133, 2296, 2465, 2640, 2821, 3008, 3201, 3400, 3605, 3816, 4033, 4256, 4485, 4720, 4961, 5208, 5461
COMMENTS
Write 1,2,3,4,... in a hexagonal spiral around 0; then a(n) is the sequence found by reading the line from 0 in the direction 0,1,....
The spiral begins:
.
85--84--83--82--81--80
/ \
86 56--55--54--53--52 79
/ / \ \
87 57 33--32--31--30 51 78
/ / / \ \ \
88 58 34 16--15--14 29 50 77
/ / / / \ \ \ \
89 59 35 17 5---4 13 28 49 76
/ / / / / \ \ \ \ \
90 60 36 18 6 0 3 12 27 48 75
/ / / / / / / / / / /
91 61 37 19 7 1---2 11 26 47 74
\ \ \ \ \ . / / / /
92 62 38 20 8---9--10 25 46 73
\ \ \ \ . / / /
93 63 39 21--22--23--24 45 72
\ \ \ . / /
94 64 40--41--42--43--44 71
\ \ . /
95 65--66--67--68--69--70
\ .
96
.
Also the number of distinct three-cell blocks that may be removed out of A000217(n+1) square cells arranged in a stepping triangular array of side (n+1). A 5-layer triangular array of square cells, for instance, has vertices outlined thus: x x
x x x
x x x x
x x x x x
x x x x x x
x x x x x x (End)
Starting from n=1, the sequence corresponds to the Wiener index of K_{n,n} (the complete bipartite graph wherein each independent set has n vertices). - Kailasam Viswanathan Iyer, Mar 11 2009
a(n) = A001399(6n-5), number of partitions of 6*n - 5 into parts < 4. For example a(2)=8 and partitions of 6*2 - 5 = 7 into parts < 4 are: [1,1,1,1,1,1,1], [1,1,1,1,1,2],[1,1,1,1,3], [1,1,1,2,2], [1,1,2,3], [1,2,2,2], [1,3,3], [2,2,3]. - Adi Dani, Jun 07 2011 Also, sequence found by reading the line from 0 in the direction 0, 8, ..., and the parallel line from 1 in the direction 1, 21, ..., in the square spiral whose vertices are the generalized octagonal numbers A001082. - Omar E. Pol, Sep 10 2011 Generate a Pythagorean triple using Euclid's formula with (n, n-1) to give A,B,C. a(n) = B + (A + C)/2. - J. M. Bergot, Jul 13 2013 The number of active (ON, black) cells in n-th stage of growth of two-dimensional cellular automaton defined by "Rule 773", based on the 5-celled von Neumann neighborhood. - Robert Price, May 23 2016 For n >= 1, the continued fraction expansion of sqrt(27*a(n)) is [9n-4; {1, 2n-2, 3, 2n-2, 1, 18n-8}]. For n=1, this collapses to [5; {5, 10}]. - Magus K. Chu, Oct 10 2022 a(n)*a(n+1) + 1 = (3n^2 + n - 1)^2. In general, a(n)*a(n+k) + k^2 = (3n^2 + (3k-2)n - k)^2. - Charlie Marion, May 23 2023
REFERENCES
Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 1.
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).
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 123.
FORMULA
a(n) = n*(3*n-2).
a(n) = (3n-2)*(3n-1)*(3n)/((3n-1) + (3n-2) + (3n)), i.e., (the product of three consecutive numbers)/(their sum). a(1) = 1*2*3/(1+2+3), a(2) = 4*5*6/(4+5+6), etc. - Amarnath Murthy, Aug 29 2002 E.g.f.: exp(x)*(x+3*x^2). - Paul Barry, Jul 23 2003 a(n) = Sum_{k=1..n} (5*n - 4*k). - Paul Barry, Sep 06 2005 a(n) = C(n+1,2) + 5*C(n,2).
Starting (1, 8, 21, 40, 65, ...) = binomial transform of [1, 7, 6, 0, 0, 0, ...]. - Gary W. Adamson, Apr 30 2008 a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0)=0, a(1)=1, a(2)=8. - Jaume Oliver Lafont, Dec 02 2008 a(n) = 2*a(n-1) - a(n-2) + 6. - Ant King, Sep 01 2011 a(6*a(n) + 16*n + 1) = a(6*a(n) + 16*n) + a(6*n + 1). - Vladimir Shevelev, Jan 24 2014 Sum_{n>=1} 1/a(n) = (sqrt(3)*Pi + 9*log(3))/12 = 1.2774090575596367311949534921... . - Vaclav Kotesovec, Apr 27 2016 Inverse binomial transform of A084857. Sum_{n>=1} (-1)^(n+1)/a(n) = Pi/(2*sqrt(3)) = A093766. (End) Product_{n>=2} (1 - 1/a(n)) = 3/4. - Amiram Eldar, Jan 21 2021 P(4k+4,n) = ((k+1)*n - k)^2 - (k*n - k)^2 where P(m,n) is the n-th m-gonal number (a generalization of the Apr 10 2013 formula, a(n) = (2*n-1)^2 - (n-1)^2). - Charlie Marion, Oct 07 2021 a(n) = A000384(n) + 2* A000217(n-1). See Twin Rectangular Rays illustration.
MATHEMATICA
LinearRecurrence[{3, -3, 1}, {1, 8, 21}, {0, 20}] (* Eric W. Weisstein, Sep 07 2017 *)
PROG
(PARI) vector(50, n, n--; n*(3*n-2)) \\ G. C. Greubel, Nov 15 2018 (GAP) List([0..50], n -> n*(3*n-2)); # G. C. Greubel, Nov 15 2018 (Haskell)
(Sage) [n*(3*n-2) for n in range(50)] # G. C. Greubel, Nov 15 2018 (Python) # Intended to compute the initial segment of the sequence, not isolated terms.
def aList():
x, y = 1, 1
yield 0
while True:
yield x
x, y = x + y + 6, y + 6
Numbers congruent to 1 or 5 mod 6.
+10
230
1, 5, 7, 11, 13, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 121, 125, 127, 131, 133, 137, 139, 143, 145, 149, 151, 155, 157, 161, 163, 167, 169, 173, 175
COMMENTS
Or, numbers relatively prime to 2 and 3, or coprime to 6, or having only prime factors >= 5; also known as 5-rough numbers. (Edited by M. F. Hasler, Nov 01 2014: merged with comments from Zak Seidov, Apr 26 2007 and Michael B. Porter, Oct 09 2009) Apart from initial term(s), dimension of the space of weight 2n cuspidal newforms for Gamma_0( 38 ).
Numbers k such that k mod 2 = 1 and (k+1) mod 3 <> 1. - Klaus Brockhaus, Jun 15 2004 Also numbers n such that the sum of the squares of the first n integers is divisible by n, or A000330(n) = n*(n+1)*(2*n+1)/6 is divisible by n. - Alexander Adamchuk, Jan 04 2007 Numbers n such that the sum of squares of n consecutive integers is divisible by n, because A000330(m+n) - A000330(m) = n*(n+1)*(2*n+1)/6 + n*(m^2+n*m+m) is divisible by n independent of m. - Kaupo Palo, Dec 10 2016 Terms of this sequence (starting from the second term) are equal to the result of the expression sqrt(4!*(k+1) + 1) - but only when this expression yields integral values (that is when the parameter k takes values, which are terms of A144065). - Alexander R. Povolotsky, Sep 09 2008 Numbers n such that ChebyshevT(x, x/2) is not an integer (is integer/2). - Artur Jasinski, Feb 13 2010 If 12*k + 1 is a perfect square (k = 0, 2, 4, 10, 14, 24, 30, 44, ... = A152749) then the square root of 12*k + 1 = a(n). - Gary Detlefs, Feb 22 2010 Cf. property described by Gary Detlefs in A113801 and in Comment: more generally, these numbers are of the form (2*h*n+(h-4)*(-1)^n-h)/4 (with h, n natural numbers), therefore ((2*h*n+(h-4)*(-1)^n-h)/4)^2-1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 6). Also a(n)^2 - 1 == 0 (mod 12). - Bruno Berselli, Nov 05 2010 - Nov 17 2010 Numbers n such that ( Sum_{k = 1..n} k^14 ) mod n = 0. (Conjectured) - Gary Detlefs, Dec 27 2011 The above conjecture is true. Apply Ireland and Rosen, Proposition 15.2.2. with m = 14 to obtain the congruence 6*( Sum_{k = 1..n} k^14 )/n = 7 (mod n), true for all n >= 1. Suppose n is coprime to 6, then 6 is a unit in Z/nZ, and it follows from the congruence that ( Sum_{k = 1..n} k^14 )/n is an integer. On the other hand, if either 2 divides n or 3 divides n then the congruence shows that ( Sum_{k = 1..n} k^14 )/n cannot be integral. (End)
a(n) are values of k such that Sum_{m = 1..k-1} m*(k-m)/k is an integer. Sums for those k are given by A062717. Also see Detlefs formula below based on A062717. - Richard R. Forberg, Feb 16 2015 a(n) are exactly those positive integers m such that the sequence b(n) = n*(n + m)*(n + 2*m)/6 is integral, and also such that the sequence c(n) = n*(n + m)*(n + 2*m)*(n + 3*m)/24 is integral. Cf. A007775. - Peter Bala, Nov 13 2015 Along with 2, these are the numbers k such that the k-th Fibonacci number is coprime to every Lucas number. - Clark Kimberling, Jun 21 2016 This sequence is the Engel expansion of 1F2(1; 5/6, 7/6; 1/36) + 1F2(1; 7/6, 11/6; 1/36)/5. - Benedict W. J. Irwin, Dec 16 2016 The sequence a(n), n >= 4 is generated by the successor of the pair of polygonal numbers {P_s(4) + 1, P_(2*s - 1)(3) + 1}, s >= 3. - Ralf Steiner, May 25 2018 The asymptotic density of this sequence is 1/3. - Amiram Eldar, Oct 18 2020 Also, the only vertices in the odd Collatz tree A088975 that are branch values to other odd nodes t == 1 (mod 2) of A005408. - Heinz Ebert, Apr 14 2021 For any two terms j and k, the product j*k is also a term (the same property as p^n and smooth numbers).
From a(2) to a(phi( A033845(n))), or a(( A033845(n))/3), the terms are the totatives of the A033845(n) itself. (End) If k is in the sequence, then k*2^m + 3 is also in the sequence, for all m > 0. - Jules Beauchamp, Aug 29 2024
REFERENCES
K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, Springer-Verlag, 1980.
FORMULA
a(n) = 3*n - 1 - (n mod 2). - Zak Seidov, Jan 18 2006 a(1) = 1 then alternatively add 4 and 2. a(1) = 1, a(n) = a(n-1) + 3 + (-1)^n. - Zak Seidov, Mar 25 2006 1 + 1/5^2 + 1/7^2 + 1/11^2 + ... = Pi^2/9 [Jolley]. - Gary W. Adamson, Dec 20 2006 For n >= 3 a(n) = a(n-2) + 6. - Zak Seidov, Apr 18 2007 Expand (x+x^5)/(1-x^6) = x + x^5 + x^7 + x^11 + x^13 + ...
O.g.f.: x*(1+4*x+x^2)/((1+x)*(1-x)^2). (End)
a(n) = 6*floor(n/2) + (-1)^(n+1). - Gary Detlefs, Dec 29 2011 a(n) = 2*n + 1 + 2*floor((n-2)/2) = 2*n - 1 + 2*floor(n/2), leading to the o.g.f. given by R. J. Mathar above. - Wolfdieter Lang, Jan 20 2012 1 - 1/5^3 + 1/7^3 - 1/11^3 + - ... = Pi^3*sqrt(3)/54 (L. Euler). - Philippe Deléham, Mar 09 2013 a(n) = 2*floor(3*n/2) - 1.
a(k*n) = k*a(n) + (4*k + (-1)^k - 3)/2 for k>0 and odd n, a(k*n) = k*a(n) + k - 1 for even n. Some special cases:
k=2: a(2*n) = 2*a(n) + 3 for odd n, a(2*n) = 2*a(n) + 1 for even n;
k=3: a(3*n) = 3*a(n) + 4 for odd n, a(3*n) = 3*a(n) + 2 for even n;
k=4: a(4*n) = 4*a(n) + 7 for odd n, a(4*n) = 4*a(n) + 3 for even n;
k=5: a(5*n) = 5*a(n) + 8 for odd n, a(5*n) = 5*a(n) + 4 for even n, etc. (End)
a(2*m) = 6*m - 1, m >= 1; a(2*m + 1) = 6*m + 1, m >= 0. - Ralf Steiner, May 17 2018 Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(3) ( A002194). Product_{n>=2} (1 + (-1)^n/a(n)) = Pi/3 ( A019670). (End)
EXAMPLE
G.f. = x + 5*x^2 + 7*x^3 + 11*x^4 + 13*x^5 + 17*x^6 + 19*x^7 + 23*x^8 + ...
MAPLE
seq(seq(6*i+j, j=[1, 5]), i=0..100); # Robert Israel, Sep 08 2014
MATHEMATICA
Select[Range[200], MemberQ[{1, 5}, Mod[#, 6]] &] (* Harvey P. Dale, Aug 27 2013 *) Table[2*Floor[3*n/2] - 1, {n, 1000}] (* Mikk Heidemaa, Feb 11 2016 *)
PROG
(PARI) \\ given an element from the sequence, find the next term in the sequence.
(Sage) [i for i in range(150) if gcd(6, i) == 1] # Zerinvary Lajos, Apr 21 2009 (Haskell)
a007310 n = a007310_list !! (n-1)
a007310_list = 1 : 5 : map (+ 6) a007310_list
(GAP) Filtered([1..150], n->n mod 6=1 or n mod 6=5); # Muniru A Asiru, Dec 19 2018 (Python)
CROSSREFS
Cf. A000330, A001580, A002194, A019670, A032528 (partial sums), A038509 (subsequence of composites), A047209, A047336, A047522, A056020, A084967, A090771, A091998, A144065, A175885- A175887. Row 3 of A260717 (without the initial 1).
AUTHOR
C. Christofferson (Magpie56(AT)aol.com)
Product t2(q^d); d | 3, where t2 = theta2(q) / (2 * q^(1/4)).
+10
39
1, 1, 0, 2, 1, 0, 2, 0, 0, 2, 2, 0, 1, 1, 0, 2, 0, 0, 2, 2, 0, 2, 0, 0, 3, 0, 0, 0, 2, 0, 2, 2, 0, 2, 0, 0, 2, 1, 0, 2, 1, 0, 0, 0, 0, 4, 2, 0, 2, 0, 0, 2, 0, 0, 2, 2, 0, 0, 2, 0, 1, 0, 0, 2, 2, 0, 4, 0, 0, 2, 0, 0, 0, 3, 0, 2, 0, 0, 2, 0, 0, 2, 0, 0, 3, 2, 0
COMMENTS
Number of solutions of 8*n + 4 = x^2 + 3*y^2 in positive odd integers. - Michael Somos, Sep 18 2004 Half the number of integer solutions of 4*n + 2 = x^2 + y^2 + z^2 where 0 = x + y + z and x and y are odd. - Michael Somos, Jul 03 2011 Given g.f. A(x), then q^(1/2) * 2 * A(q) is denoted phi_1(z) where q = exp(Pi i z) in Conway and Sloane.
Half of theta series of planar hexagonal lattice (A2) with respect to an edge.
Bisection of A002324. Number of ways of writing n as a sum of a triangular plus three times a triangular number [Hirschhorn]. - R. J. Mathar, Mar 23 2011
REFERENCES
Burce C. Berndt, Ramanujan's Notebooks Part III, Springer-Verlag, 1991, see p. 223 Entry 3(i).
J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer-Verlag, 1999, p. 103. See Eq. (13).
Nathan J. Fine, Basic Hypergeometric Series and Applications, Amer. Math. Soc., 1988; p. 78, Eq. (32.27).
FORMULA
Expansion of q^(-1/2) * (eta(q^2) * eta(q^6))^2 / (eta(q) * eta(q^3)) in powers of q. - Michael Somos, Apr 18 2004 Expansion of q^(-1) * (a(q) - a(q^4)) / 6 in powers of q^2 where a() is a cubic AGM theta function. - Michael Somos, Oct 24 2006 Expansion of psi(x) * psi(x^3) in powers of x where psi() is a Ramanujan theta function. - Michael Somos, Jul 03 2011 Euler transform of period 6 sequence [ 1, -1, 2, -1, 1, -2, ...]. - Michael Somos, Apr 18 2004 Given g.f. A(x), then B(x) = (x * A(x^2))^2 satisfies 0 = f(B(x), B(x^2), B(x^4)) where f(u, v, w) = v^3 + 4*u*v*w + 16*v*w^2 - 8*w*v^2 - w*u^2.
a(n) = b(2*n + 1) where b() is multiplicative with b(2^e) = 0^e, b(3^e) = 1, b(p^e) = (1 + (-1)^e) / 2 if p==5 (mod 6) otherwise b(p^e) = e+1. (Clarification: the g.f. A(x) is not the primary function of interest, but rather B(x) = x * A(x^2), which is an eta-quotient and is the generating function of a multiplicative sequence.)
G.f.: (Sum_{j>0} x^((j^2 - j) / 2)) * (Sum_{k>0} x^(3(k^2 - k) / 2)) = Product_{k>0} (1 + x^k) * (1 - x^(2*k)) * (1 + x^(3*k)) * (1 - x^(6*k)).
G.f.: Sum_{k>=0} a(k) * x^(2*k + 1) = Sum_{k>0} x^k * (1 - x^k) * (1 - x^(4*k)) * (1 - x^(5*k)) / (1 - x^(12*k)). (End)
G.f.: s(4)^2*s(12)^2/(s(2)*s(6)), where s(k) := subs(q=q^k, eta(q)), where eta(q) is Dedekind's function, cf. A010815. [Fine] G.f.: Sum_{k>=0} a(k) * x^(2*k + 1) = Sum_{k>0} x^k / (1 + x^k + x^(2*k)) - x^(4*k) / (1 + x^(4*k) + x^(8*k)). - Michael Somos, Nov 04 2005 Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Pi/(2*sqrt(3)) = 0.906899... ( A093766). - Amiram Eldar, Nov 23 2023
EXAMPLE
G.f. = 1 + x + 2*x^3 + x^4 + 2*x^6 + 2*x^9 + 2*x^10 + x^12 + x^13 + 2*x^15 + ...
G.f. = q + q^3 + 2*q^7 + q^9 + 2*q^13 + 2*q^19 + 2*q^21 + q^25 + q^27 + 2*q^31 + ...
a(6) = 2 since 8*6 + 4 = 52 = 5^2 + 3*3^2 = 7^2 + 3*1^2.
MATHEMATICA
a[ n_] := If[ n < 0, 0, DivisorSum[ 2 n + 1, Mod[(3 - #)/2, 3, -1] &]]; (* Michael Somos, Jul 03 2011 *) QP = QPochhammer; s = (QP[q^2]*QP[q^6])^2/(QP[q]*QP[q^3]) + O[q]^100; CoefficientList[s, q] (* Jean-François Alcover, Nov 27 2015, adapted from PARI *) a[ n_] := If[ n < 1, Boole[n == 0], Times @@ (Which[# < 2, 0^#2, Mod[#, 6] == 5, 1 - Mod[#2, 2], True, #2 + 1] & @@@ FactorInteger@(2 n + 1))]; (* Michael Somos, Mar 06 2016 *) %t A033762 a[ n_] := SeriesCoefficient[ (1/4) x^(-1/2) EllipticTheta[ 2, 0, x^(1/2)] EllipticTheta[ 2, 0, x^(3/2)], {x, 0, n}]; (* Michael Somos, Mar 06 2016 *)
PROG
(PARI) {a(n) = my(A); if( n<0, 0, A = x * O(x^n); polcoeff( (eta(x^2 + A) * eta(x^6 + A))^2 / (eta(x + A) * eta(x^3 + A)), n))}; /* Michael Somos, Sep 18 2004 */ (PARI) {a(n) = if( n<0, 0, n = 2*n + 1; sumdiv( n, d, kronecker( -12, d) * (n / d % 2)))}; /* Michael Somos, Nov 04 2005 */ (PARI) {a(n) = if( n<0, 0, n = 8*n + 4; sum( j=1, sqrtint( n\3), (j%2) * issquare(n - 3*j^2)))} /* Michael Somos, Nov 04 2005 */ (PARI) {a(n) = if( n<0, 0, sumdiv(2*n + 1, d, kronecker(-3, d)))}; /* Michael Somos, Mar 06 2016 */ (Magma) A := Basis( ModularForms( Gamma1(12), 1), 202); A[2] + A[4]; /* Michael Somos, Jul 25 2014 */
CROSSREFS
Cf. A002324, A004016, A005881, A035178, A091393, A093766, A093829, A096936, A112298, A113447, A113661, A113974, A115979, A122860, A123331, A123484, A136748, A137608.
Decimal expansion of Pi/sqrt(3) = sqrt(2*zeta(2)).
+10
28
1, 8, 1, 3, 7, 9, 9, 3, 6, 4, 2, 3, 4, 2, 1, 7, 8, 5, 0, 5, 9, 4, 0, 7, 8, 2, 5, 7, 6, 4, 2, 1, 5, 5, 7, 3, 2, 2, 8, 4, 0, 6, 6, 2, 4, 8, 0, 9, 2, 7, 4, 0, 5, 7, 5, 5, 6, 9, 8, 8, 4, 9, 3, 5, 3, 8, 8, 1, 2, 3, 1, 8, 1, 1, 2, 6, 3, 5, 3, 8, 8, 3, 6, 8, 4, 1, 2, 4, 9, 8, 8, 2, 1, 2, 0, 6, 0, 1, 6, 8, 8, 5, 6, 2, 2
COMMENTS
Volume of a cube with edge length 1 rotated about a space diagonal. See MathWorld Cube page. - Francis Wolinski, Mar 10 2019 Volume of a cone with unit radius and 60-degree opening angle, and so height sqrt(3). Equivalently, the volume of the cone formed by rotating a 30-60-90 degree triangle with unit short leg about the long leg. - Christoph B. Kassir, Sep 17 2022
LINKS
Eric Weisstein's World of Mathematics, Cube
FORMULA
Equals (3/2)*Integral_{x=0..oo} 1/(1+x+x^2) dx. - Bruno Berselli, Jul 23 2013 Equals Sum_{n >= 0} (1/(6*n+1) - 4/(6*n+2) - 5/(6*n+3) - 1/(6*n+4) + 4/(6*n+5) + 5/(6*n+6)). - Mats Granvik, Sep 23 2013 Equals (1/2) * Sum_{n >= 0} (14*n + 11)*(-1/3)^n/((4*n + 1)*(4*n + 3)*binomial(4*n,2*n)). For more series representations of this type see the Bala link. - Peter Bala, Feb 04 2015 Equals 3*Sum_{n >= 1} 1/( (3*n - 1)*(3*n - 2) ).
Equals 2 - 6*Sum_{n >= 1} 1/( (3*n - 1)*(3*n + 1)*(3*n + 2) ).
Equals 5!*Sum_{n >= 1} 1/( (3*n - 1)*(3*n - 2)*(3*n + 2)*(3*n + 4) ).
Equals 3*( 1 - 2*Sum_{n >= 1} 1/(9*n^2 - 1) ).
Equals 1 + Sum_{n >=1 } (-1)^(n+1)*(6*n + 1)/(n*(n + 1)*(3*n + 1)*(3*n - 2)).
Equals (27/2)*Sum_{n >= 1} (2*n + 1)/( (3*n - 1)*(3*n + 1)*(3*n + 2)*(3*n + 4) ).
Equals 3*Integral_{x = 0..1} 1/(1 + x + x^2) dx.
Equals 3*Integral_{x = 0..1} (1 + x)/(1 - x + x^2) dx.
Equals 3*Integral_{x = 0..oo} cosh(x)/cosh(3*x) dx. (End)
Equals Integral_{x = 0..oo} log(1+x^3)/x^3 dx. - Amiram Eldar, Aug 20 2020 For any integer k, Pi/sqrt(3) = Sum_{n >= 0} (1/(n + k + 1/3) - 1/(n - k + 2/3)) = (1/3)*Sum_{n >= 0} (1/(n - k + 1/6) - 1/(n + k + 5/6)).
Equals (3/2)*Sum_{n >= 0} 1/((2*n + 1)*binomial(2*n, n)). (End)
EXAMPLE
Pi/sqrt(3) = 1.8137993642342178505940782576421557322840662480927405755...
MATHEMATICA
RealDigits[Pi/Sqrt[3], 10, 120][[1]] (* Harvey P. Dale, Mar 04 2012 *)
PROG
(PARI) default(realprecision, 20080); x=Pi*sqrt(3)/3; for (n=1, 20000, d=floor(x); x=(x-d)*10; write("b093602.txt", n, " ", d)); \\ Harry J. Smith, Jun 19 2009 (Magma) SetDefaultRealField(RealField(100)); R:= RealField(); Pi(R)/Sqrt(3); // G. C. Greubel, Mar 10 2019 (Sage) numerical_approx(pi/sqrt(3), digits=100) # G. C. Greubel, Mar 10 2019
Number of representations of n as a sum of three times a square and two times a triangular number.
+10
27
1, 0, 1, 2, 0, 2, 1, 0, 0, 2, 0, 0, 3, 0, 2, 2, 0, 0, 2, 0, 1, 0, 0, 2, 2, 0, 0, 2, 0, 2, 1, 0, 2, 4, 0, 0, 0, 0, 0, 2, 0, 0, 3, 0, 0, 2, 0, 2, 2, 0, 2, 0, 0, 0, 4, 0, 1, 2, 0, 2, 2, 0, 0, 0, 0, 0, 0, 0, 4, 2, 0, 0, 1, 0, 0, 4, 0, 2, 2, 0, 0, 2, 0, 2, 2, 0, 0, 2, 0, 0, 3, 0, 0, 2, 0, 2, 0, 0, 0, 2, 0, 0, 2, 0, 2
COMMENTS
Ramanujan theta functions: f(q) := Prod_{k>=1} (1-(-q)^k) (see A121373), phi(q) := theta_3(q) := Sum_{k=-oo..oo} q^(k^2) ( A000122), psi(q) := Sum_{k=0..oo} q^(k*(k+1)/2) ( A010054), chi(q) := Prod_{k>=0} (1+q^(2k+1)) ( A000700). Number of representations of 2n as a sum of three times a triangular number and a triangular number.
FORMULA
a(n) = A002324(4n+1) = A033762(2n) = d_{1, 3}(4n+1) - d_{2, 3}(4n+1) where d_{a, m}(n) equals the number of divisors of n which are congruent to a mod m. Expansion of (psi(q)psi(q^3) + psi(-q)psi(-q^3))/2 in powers of q^2 where psi() is a Ramanujan theta function.
G.f.: (Sum_{k} x^k^2)^3*(Sum_{k>0} x^((k^2-k)/2))^2 = Product_{k>0} (1-x^(4k))(1-x^(6k))(1+x^(2k))(1+x^(3k))^2/(1+x^(6k))^2.
Euler transform of period 12 sequence [0, 1, 2, -1, 0, -2, 0, -1, 2, 1, 0, -2, ...]. (End)
G.f. is a period 1 Fourier series which satisfies f(-1 / (48 t)) = 3^(1/2) (t/i) g(t) where q = exp(2 Pi i t) and g() is g.f. for A164272. a(3*n + 1) = 0. (End)
Asymptotic mean: Limit_{m->oo} (1/m) * Sum_{k=1..m} a(k) = Pi/(2*sqrt(3)) = 0.906899... ( A093766). - Amiram Eldar, Nov 24 2023
EXAMPLE
a(12) = 3 since we can write 12 = 3(2)^2 + 0 = 3(-2)^2 + 0 = 0 + 2*6.
2*12 = 24 = 3*1+21 = 3*3+15 = 3*6+6 so a(12) = 3.
G.f. = 1 + x^2 + 2*x^3 + 2*x^5 + x^6 + 2*x^9 + 3*x^12 + 2*x^14 + 2*x^15 + ... - Michael Somos, Aug 11 2009 G.f. = q + q^9 + 2*q^13 + 2*q^21 + q^25 + 2*q^37 + 3*q^49 + 2*q^57 + 2*q^61 + ... - Michael Somos, Aug 11 2009
MATHEMATICA
a[n_] := DivisorSum[4n+1, Switch[Mod[#, 3], 1, 1, 2, -1, 0, 0]&]; Table[ a[n], {n, 0, 104}] (* Jean-François Alcover, Dec 04 2015, adapted from PARI *)
PROG
(PARI) {a(n) = if(n<0, 0, n=4*n+1; sumdiv(n, d, (d%3==1) - (d%3==2)))};
(PARI) {a(n) = my(A); if(n<0, 0, A=x*O(x^n); polcoeff( eta(x^6+A)^5 / eta(x^2+A)*(eta(x^4+A) / eta(x^3+A) / eta(x^12+A))^2, n))}; /* Michael Somos, Feb 14 2006 */
Decimal expansion of (1/16)*Pi^2.
+10
26
6, 1, 6, 8, 5, 0, 2, 7, 5, 0, 6, 8, 0, 8, 4, 9, 1, 3, 6, 7, 7, 1, 5, 5, 6, 8, 7, 4, 9, 2, 2, 5, 9, 4, 4, 5, 9, 5, 7, 1, 0, 6, 2, 1, 2, 9, 5, 2, 5, 4, 9, 4, 1, 4, 1, 5, 0, 8, 3, 4, 3, 3, 6, 0, 1, 3, 7, 5, 2, 8, 0, 1, 4, 0, 1, 2, 0, 0, 3, 2, 7, 6, 8, 7, 6, 1, 0, 8, 3, 7, 7, 3, 2, 4, 0, 9, 5, 1, 4, 4, 8, 9, 0, 0
COMMENTS
Conjectured to be density of densest packing of equal spheres in four dimensions (achieved for example by the D_4 lattice).
Also decimal expansion of Sum_{k>=0} (-1)^k*d(2*k+1)/(2*k+1), where d(n) is the number of divisors of n A000005(n). Ramanujan's question 770 in the Journal of the Indian Mathematical Society (VIII, 120) asked "If d(n) denotes the number of divisors of n, show that d(1) - d(3)/3 + d(5)/5 - d(7)/7 + d(9)/9 - ... is a convergent series ...".
A summation of the first 2*10^9 terms performed by Hans Havermann yields 0.6168503077..., which is close to (Pi/4)^2=0.616850275... (End)
Modulo questions about rearrangement of conditionally convergent series, which I expect a more careful treatment would handle, Sum_{k>=0} (-1)^k*d(2*k+1)/(2*k+1) should indeed be Pi^2/16.
Sum_{k>=0} (-1)^k d(2k+1)/(2k+1)
= Sum_{k>=0} Sum_{2i+1 | 2k+1} (-1)^k/(2k+1)
(letting 2k+1=(2i+1)(2j+1): note that k == i+j (mod 2))
= Sum_{i>=0} Sum_{j>=0} (-1)^(i+j)/((2i+1)(2j+1))
= (Sum_{i>=0} (-1)^i/(2i+1))^2 = (Pi/4)^2. (End)
Volume bounded by the surface (x+y+z)^2-2(x^2+y^2+z^2)=4xyz, the ellipson (see Wildberger, p. 287). - Patrick D McLean, Dec 03 2020
REFERENCES
S. D. Chowla, Solution and Remarks on Question 770, J. Indian Math. Soc. 17 (1927-28), 166-171.
J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer, 3rd. ed., 1998. See p. xix.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.7, p. 507.
S. Ramanujan, Coll. Papers, Chelsea, 1962, Question 770, page 333.
G. N. Watson, Solution to Question 770, J. Indian Math. Soc. 18 (1929-30), 294-298.
FORMULA
Equals Sum_{k>=1} zeta(2*k)*k/4^k. - Amiram Eldar, May 29 2021
EXAMPLE
0.6168502750680849136771556874922594459571...
MATHEMATICA
Integrate[Boole[(x+y+z)^2-2(x^2+y^2+z^2)>4x y z], {x, 0, 1}, {y, 0, 1}, {z, 0, 1}] (* Patrick D McLean, Dec 03 2020 *)
PROG
(Magma) pi:=Pi(RealField(110)); Reverse(Intseq(Floor((1/16)*10^100*pi^2))); // Vincenzo Librandi, Feb 20 2017
CROSSREFS
Related constants: A020769, A020789, A093766, A093825, A222066, A222067, A222069, A222070, A222071, A222072, A260646.
Decimal expansion of 1/sqrt(12) = 1/(2*sqrt(3)).
+10
25
2, 8, 8, 6, 7, 5, 1, 3, 4, 5, 9, 4, 8, 1, 2, 8, 8, 2, 2, 5, 4, 5, 7, 4, 3, 9, 0, 2, 5, 0, 9, 7, 8, 7, 2, 7, 8, 2, 3, 8, 0, 0, 8, 7, 5, 6, 3, 5, 0, 6, 3, 4, 3, 8, 0, 0, 9, 3, 0, 1, 1, 6, 3, 2, 4, 1, 9, 8, 8, 8, 3, 6, 1, 5, 1, 4, 6, 6, 6, 7, 2, 8, 4, 6, 8, 5, 7, 6, 9, 7, 7, 9, 2, 8, 7, 4, 7, 6, 2
COMMENTS
Center density of densest packing of equal circles in two dimensions (achieved for example by the A2 lattice).
Let a equal the length of one side of an equilateral triangle and let b equal the radius of the circle inscribed in that triangle. This sequence gives the decimal expansion of b/a. - Christopher M. Tomaszewski (cmt1288(AT)comcast.net), Feb 20 2004
The constant (3+sqrt 3)/6, which is 0.5 larger than this, plays a role in Borsuk's conjecture. - Arkadiusz Wesolowski, Mar 17 2014
REFERENCES
J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer, 3rd. ed., 1998. See p. xix.
EXAMPLE
0.28867513459481288225457439025097872782380087563506343800930116324198883615...
CROSSREFS
Related constants: A020789, A093766, A093825, A222066, A222067, A222068, A222069, A222070, A222071, A222072, A260646.
Decimal expansion of Pi/(3*sqrt(2)).
+10
21
7, 4, 0, 4, 8, 0, 4, 8, 9, 6, 9, 3, 0, 6, 1, 0, 4, 1, 1, 6, 9, 3, 1, 3, 4, 9, 8, 3, 4, 3, 4, 4, 8, 9, 4, 9, 7, 6, 9, 1, 0, 3, 6, 1, 4, 8, 9, 5, 9, 4, 8, 3, 7, 0, 5, 1, 4, 2, 3, 2, 6, 0, 1, 1, 5, 9, 4, 0, 5, 7, 9, 8, 8, 4, 9, 9, 1, 2, 3, 1, 8, 4, 2, 9, 2, 2, 1, 1, 5, 5, 7, 9, 4, 1, 2, 7, 5, 3, 9, 5, 6, 0
COMMENTS
Density of densest packing of equal spheres in three dimensions (achieved for example by the fcc lattice).
Atomic packing factor (APF) of the face-centered-cubic (fcc) and the hexagonal-close-packed (hcp) crystal lattices filled with spheres of the same diameter. - Stanislav Sykora, Sep 29 2014
REFERENCES
J. H. Conway and N. J. A. Sloane, "Sphere Packings, Lattices and Groups", Springer, 3rd. ed., 1998. See p. 15, line n = 3.
Steven R. Finch, Mathematical Constants, Encyclopedia of Mathematics and its Applications, vol. 94, Cambridge University Press, 2003, Section 8.7, p. 506.
Clifford A. Pickover, The Math Book: From Pythagoras to the 57th Dimension, 250 Milestones in the History of Mathematics (2009), at p. 126.
David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987, p. 29.
FORMULA
Equals Integral_{x >= 0} (4*x^2 + 1)/((2*x^2 + 1)*(8*x^2 + 1)) dx. - Peter Bala, Feb 12 2025
EXAMPLE
0.74048048969306104116931349834344894976910361489594837...
MATHEMATICA
RealDigits[Pi/(3 Sqrt[2]), 10, 120][[1]] (* Harvey P. Dale, Feb 03 2012 *)
PROG
(PARI) default(realprecision, 20080); x=10*Pi*sqrt(2)/6; for (n=0, 20000, d=floor(x); x=(x-d)*10; write("b093825.txt", n, " ", d)); \\ Harry J. Smith, Jun 18 2009
CROSSREFS
Cf. APF's of other crystal lattices: A019673 (simple cubic), A247446 (diamond cubic). Related constants: A020769, A020789, A093766, A222066, A222067, A222068, A222069, A222070, A222071, A222072, A260646.
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