Displaying 1-10 of 17 results found.
Table A142978 (figurate numbers for n-dimensional cross polytope) extended by a top row.
+20
0
1, 1, 0, 1, 2, 1, 1, 4, 3, 0, 1, 6, 9, 4, 1, 1, 8, 19, 16, 5, 0, 1, 10, 33, 44, 25, 6, 1, 1, 12, 51, 96, 85, 36, 7, 0, 1, 14, 73, 180, 225, 146, 49, 8, 1, 1, 16, 99, 304, 501, 456, 231, 64, 9, 0, 1, 18, 129, 476, 985, 1182, 833, 344, 81, 10, 1
COMMENTS
Looking at table A142978, it seems natural to extend it by a first row, which turns out to be 1,0,1,0,1,0,.... Indeed, these are the values obtained when the polynomial which defines each of the columns, is "extrapolated" to n=0.
EXAMPLE
The table may be written as:
1,_ 0,__ v---- A058331(n) = 2*n^2 + 1. 1,_ 2,__ 1,__ v---- 4* A006527(n) = 4n(n^2 + 2)/3. 1,_ 4,__ 3,__ 0,__ v---- 2n^2(n^2 + 5)/3 + 1.
1,_ 6,__ 9,__ 4,__ 1,___ v---- 2n(2n^4 + 20n^2 + 23)/15.
1,_ 8,_ 19,_ 16,__ 5,___ 0,
1, 10,_ 33,_ 44,_ 25,___ 6,__ 1,
1, 12,_ 51,_ 96,_ 85,__ 36,__ 7,__ 0,
1, 14,_ 73, 180, 225,_ 146,_ 49,__ 8,_ 1,
1, 16,_ 99, 304, 501,_ 456, 231,_ 64,_ 9, 0,
1, 18, 129, 476, 985, 1182, 833, 344, 81, 10, 1,...
Sequence A142978 is the table obtained by deleting the uppermost row 1,0,1,0,1,... (One could also add a 0th column, with all zeros.)
Number of ways n appears as a cross-polytope number ( A142978).
+20
0
1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 3, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 2, 2, 2, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 2, 1, 2
COMMENTS
Every entry in the first column (of A142978) is 1, so this sequence starts at a(2). a(n) is always positive, as the first row lists the positive integers.
a(n) >= 3 infinitely often. This happens, in particular, at every even square > 4. (The second row contains the squares, and the second column the positive even numbers.)
For n <= 10000, the only instance of a(n) > 3 is a(1156) = 4. This occurs because 1156 is even, square, and octahedral (third row of A142978).
PROG
(Sage) def a(n) : return len([K for K in [2..n] if n == next( A142978(N, K) for N in (1..) if A142978(N, K) >= n)])
Expansion of 1/((1 - x)*(1 - 2*x - x^2)).
+10
68
1, 3, 8, 20, 49, 119, 288, 696, 1681, 4059, 9800, 23660, 57121, 137903, 332928, 803760, 1940449, 4684659, 11309768, 27304196, 65918161, 159140519, 384199200, 927538920, 2239277041, 5406093003, 13051463048, 31509019100, 76069501249
COMMENTS
Partial sums of Pell numbers A000129. W(n){1,3;2,-1,1} = Sum_{i=1..n} W(i){1,2;2,-1,0}, where W(n){a,b; p,q,r} implies x(n) = p*x(n-1) - q*x(n-2) + r; x(0)=a, x(1)=b.
Number of 2 X (n+1) binary arrays with path of adjacent 1's from upper left to lower right corner. - R. H. Hardin, Mar 16 2002 Number of (s(0), s(1), ..., s(n+2)) such that 0 < s(i) < 4 and |s(i) - s(i-1)| <= 1 for i = 1,2,...,n+2, s(0) = 1, s(n+2) = 3. - Herbert Kociemba, Jun 16 2004 Also, the lexicographically earliest sequence of positive integers such that for n > 3, {sqrt(2)*a(n)} is located strictly between {sqrt(2)*a(n-1)} and {sqrt(2)*a(n-2)} where {} denotes the fractional part. - Ivan Neretin, May 02 2017 a(n+1) is the number of weak orderings on {1,...,n} that are weakly single-peaked w.r.t. the total ordering 1 < ... < n. - J. Devillet, Oct 06 2017
REFERENCES
Allombert, Bill, Nicolas Brisebarre, and Alain Lasjaunias. "On a two-valued sequence and related continued fractions in power series fields." The Ramanujan Journal 45.3 (2018): 859-871. See Theorem 3, d_{4n+3}.
FORMULA
a(n) = 2*a(n-1) + a(n-2) + 1 with n > 1, a(0)=1, a(1)=3.
a(n) = ((2 + (3*sqrt(2))/2)*(1 + sqrt(2))^n - (2 - (3*sqrt(2))/2)*(1 - sqrt(2))^n )/(2*sqrt(2)) - 1/2.
a(0)=1, a(n+1) = ceiling(x*a(n)) for n > 0, where x = 1+sqrt(2). - Paul D. Hanna, Apr 22 2003 a(n) = 3*a(n-1) - a(n-2) - a(n-3). With two leading zeros, e.g.f. is exp(x)(cosh(sqrt(2)x)-1)/2. a(n) = Sum_{k=0..floor((n+2)/2)} binomial(n+2, 2k+2)2^k. - Paul Barry, Aug 16 2003 E.g.f.: exp(x)(cosh(x/sqrt(2)) + sqrt(2)sinh(x/sqrt(2)))^2. - N. J. A. Sloane, Sep 13 2003 a(n) = (((1-sqrt(2))^(n+2) + (1+sqrt(2))^(n+2) - 2) / 4). - Altug Alkan, Mar 16 2016 a(n) = Sum_{k=0..n} binomial(n+1,k+1)*2^floor(k/2). - Tony Foster III, Oct 12 2017
MAPLE
a:=n->sum(fibonacci(i, 2), i=0..n): seq(a(n), n=1..29); # Zerinvary Lajos, Mar 20 2008
MATHEMATICA
CoefficientList[Series[1/(1-3x+x^2+x^3), {x, 0, 30}], x] (* or *) LinearRecurrence[{3, -1, -1}, {1, 3, 8}, 30] (* Harvey P. Dale, Jun 13 2011 *)
PROG
(PARI) a(n)=local(w=quadgen(8)); -1/2+(3/4+1/2*w)*(1+w)^n+(3/4-1/2*w)*(1-w)^n
(PARI) vector(100, n, n--; floor((1+sqrt(2))^(n+2)/4)) \\ Altug Alkan, Oct 07 2015 (PARI) Vec(1/((1-x)*(1-2*x-x^2)) + O(x^40)) \\ Michel Marcus, May 06 2017
EXTENSIONS
Corrected and extended by Larry Reeves (larryr(AT)acm.org), Jun 11 2002
a(n) = (1/3)*(n^2 + 2*n + 3)*(n+1)^2.
+10
30
1, 8, 33, 96, 225, 456, 833, 1408, 2241, 3400, 4961, 7008, 9633, 12936, 17025, 22016, 28033, 35208, 43681, 53600, 65121, 78408, 93633, 110976, 130625, 152776, 177633, 205408, 236321, 270600, 308481
COMMENTS
a(n) is the number of 4 X 4 pandiagonal magic squares with sum 2n. - Sharon Sela (sharonsela(AT)hotmail.com), May 10 2002
Figurate numbers based on the 4-dimensional regular convex polytope called the 16-cell, hexadecachoron, 4-cross polytope or 4-hyperoctahedron with Schlaefli symbol {3,3,4}. a(n)=(n^2*(n^2+2))/3 if the offset were 1. - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004, R. J. Mathar, Jul 18 2009 If X is an n-set and Y_i (i=1,2,3) mutually disjoint 2-subsets of X then a(n-6) is equal to the number of 7-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Aug 26 2007 Equals binomial transform of [1, 7, 18, 20, 8, 0, 0, 0, ...], where (1, 7, 18, 20, 8) = row 4 of the Chebyshev triangle A081277. Also = row 4 of the array in A142978. - Gary W. Adamson, Jul 19 2008
REFERENCES
T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
LINKS
Eric Weisstein's World of Mathematics, 16-Cell
FORMULA
Or, a(n-1) = n^2*(n^2+2)/3. - Corrected by R. J. Mathar, Jul 18 2009 G.f.: (1+x)^3/(1-x)^5.
Recurrence: a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)
a(n-1) = C(n+3,4) + 3 C(n+2,4) + 3 C(n+1,4) + C(n,4).
Sum_{n>=0} 1/((1/3*(n^2 + 2*n + 3))*(n+1)^2) = (1/4)*Pi^2 - 3*sqrt(2)*Pi*coth(Pi*sqrt(2))*(1/8) + 3/8 = 1.1758589... - Stephen Crowley, Jul 14 2009 a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), with n > 4, a(0)=1, a(1)=8, a(2)=33, a(3)=96, a(4)=225. - Yosu Yurramendi, Sep 03 2013 E.g.f.: (3 + 21*x + 27*x^2 + 10*x^3 + x^4)*exp(x)/3. - G. C. Greubel, Feb 10 2019 Sum_{n >= 0} (-1)^n/(a(n)*a(n+1)) = 17/3 - 8*log(2) = 1/(8 + 2/(8 + 6/(8 + ... + n*(n-1)/(8 + ...)))). See A142983. - Peter Bala, Mar 06 2024
MAPLE
al:=proc(s, n) binomial(n+s-1, s); end; be:=proc(d, n) local r; add( (-1)^r*binomial(d-1, r)*2^(d-1-r)*al(d-r, n), r=0..d-1); end; [seq(be(4, n), n=0..100)];
MATHEMATICA
LinearRecurrence[{5, -10, 10, -5, 1}, {1, 8, 33, 96, 225}, 31] (* Jean-François Alcover, Jan 17 2018 *)
PROG
(R)
a <- c(1, 8, 33, 96, 225)
for(n in (length(a)+1):30) a[n] <- 5*a[n-1]-10*a[n-2]+10*a[n-3]-5*a[n-4]+a[n-5]
(Sage) [((n+1)^2+2)*(n+1)^2/3 for n in range(40)] # G. C. Greubel, Feb 10 2019 (GAP) List([0..40], n -> (n+1)^2*((n+1)^2 +2)/3); # G. C. Greubel, Feb 10 2019
CROSSREFS
Cf. A000332, A000583, A005900, A069038, A069039, A070212, A081277, A092181, A092182, A092183, A099175, A099193, A099195, A099196, A099197, A142978.
a(n) = Sum_{k=0..n-1} binomial(n-1,k)*binomial(n+k,k). Also a(n) = T(n,n), array T as in A049600.
+10
21
0, 1, 4, 19, 96, 501, 2668, 14407, 78592, 432073, 2390004, 13286043, 74160672, 415382397, 2333445468, 13141557519, 74174404608, 419472490257, 2376287945572, 13482186743203, 76598310928096, 435730007006341, 2481447593848524, 14146164790774359
COMMENTS
Also main diagonal of array: m(i,1)=1, m(1,j)=j, m(i,j)=m(i,j-1)+m(i-1,j-1)+m(i-1,j): 1 2 3 4 ... / 1 4 9 16 ... / 1 6 19 44 ... / 1 8 33 96 ... /. - Benoit Cloitre, Aug 05 2002 This array is now listed as A142978, where some conjectural congruences for the present sequence are given. - Peter Bala, Nov 13 2008 Define a finite triangle T(r,c) with T(r,0) = binomial(n,r) for 0 <= r <= n and the other terms recursively with T(r,c) = T(r-1,c-1) + 2*T(r,c-1). The sum of the last terms in the rows is Sum_{r=0..n} T(r,r) = a(n+1). Example: For n=4 the triangle has the rows 1; 4 9; 6 16 41; 4 14 44 129; 1 6 26 96 321 having sum of last terms 1 + 9 + 41 + 129 + 321 = 501 = a(5). - J. M. Bergot, Feb 15 2013 a(n-1) for n > 1 is the number of assembly trees with the connected gluing rule for cycle graphs with n vertices. - Nick Mayers, Aug 16 2018
FORMULA
D-finite with recurrence n*(2*n-3)*a(n) - (12*n^2-24*n+8)*a(n-1) + (2*n-1)*(n-2)*a(n-2) = 0. - Vladeta Jovovic, Aug 29 2004 a(n+1) = Sum_{k=0..n} binomial(n, k)*binomial(n+1, k+1)*2^k. - Paul Barry, Sep 20 2004 a(n) = Sum_{k=0..n} T(n, k), array T as in A008288. If shifted one place left, the third binomial transform of A098660. - Paul Barry, Sep 20 2004 E.g.f. for sequence shifted left: Sum_{n>=0} a(n+1)*x^n/n! = exp(3*x)*(BesselI(0, 2*sqrt(2)*x)+BesselI(1, 2*sqrt(2)*x)/sqrt(2)). - Paul Barry, Sep 20 2004 a(n) = Sum_{k=0..n-1} C(n,k)*C(n-1,k)*2^(n-k-1); a(n+1) = 2^n*Hypergeometric2F1(-n,-n-1;1;1/2). - Paul Barry, Feb 08 2011 Recurrence (an alternative): n*a(n) = (6-n)*a(n-6) + 2*(5*n-27)*a(n-5) + (84-15*n)*a(n-4) + 52*(3-n)*a(n-3) + 3*(2-5*n)*a(n-2) + 2*(5*n-3)*a(n-1), n >= 7. - Fung Lam, Feb 05 2014 a(n) = Hyper2F1([-n, n], [1], -1)/2 for n > 0. - Peter Luschny, Aug 02 2014 n^2*a(n) = Sum_{k=0..n-1} (2*k^2+2*k+1)*binomial(n-1,k)*binomial(n+k,k). By the Zeilberger algorithm, both sides of the equality satisfy the same recurrence. - Zhi-Wei Sun, Aug 30 2014 a(n) = [x^n] (1/2) * ((1+x)/(1-x))^n for n > 0. - Seiichi Manyama, Jun 07 2018
MAPLE
a := proc(n) local k; add(binomial(n-1, k)*binomial(n+k, k), k=0..n-1); end;
MATHEMATICA
Table[SeriesCoefficient[x*((1+x)-Sqrt[1-6*x+x^2])/(4*x*Sqrt[1-6*x+x^2]), {x, 0, n}], {n, 0, 20}] (* Vaclav Kotesovec, Oct 08 2012 *) a[n_] := Hypergeometric2F1[1-n, n+1, 1, -1]; a[0] = 0; Table[a[n], {n, 0, 23}] (* Jean-François Alcover, Feb 26 2013 *) a[n_] := Sum[ Binomial[n - 1, k] Binomial[n + k, k], {k, 0, n - 1}]; Array[a, 25] (* Robert G. Wilson v, Aug 08 2018 *)
PROG
(Maxima) makelist(if n=0 then 0 else sum(binomial(n-1, k)*binomial(n+k, k), k, 0, n-1), n, 0, 22); /* Bruno Berselli, May 19 2011 */ (Magma) [n eq 0 select 0 else &+[Binomial(n-1, k)*Binomial(n+k, k): k in [0..n-1]]: n in [0..22]]; // Bruno Berselli, May 19 2011 (Haskell)
(Python)
from sympy import binomial
def a(n):
return sum(binomial(n - 1, k) * binomial(n + k, k) for k in range(n))
(Python)
from math import comb
def A047781(n): return sum(comb(n, k)**2*k<<k-1 for k in range(1, n+1))//n if n else 0 # Chai Wah Wu, Mar 22 2023
Square array of unsigned coefficients of Chebyshev polynomials of the first kind.
+10
20
1, 1, 1, 1, 3, 2, 1, 5, 8, 4, 1, 7, 18, 20, 8, 1, 9, 32, 56, 48, 16, 1, 11, 50, 120, 160, 112, 32, 1, 13, 72, 220, 400, 432, 256, 64, 1, 15, 98, 364, 840, 1232, 1120, 576, 128, 1, 17, 128, 560, 1568, 2912, 3584, 2816, 1280, 256, 1, 19, 162, 816, 2688, 6048, 9408, 9984, 6912
COMMENTS
Formatted as a triangular array, this is [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 1, 0, 0, 0, 0, 0, 0, 0, 0, ...] (see construction in A084938 ). - Philippe Deléham, Aug 09 2005 Binomial transform of n-th row of the triangle (followed by zeros) = n-th row of the A142978 array and n-th column of triangle A104698. - Gary W. Adamson, Jul 17 2008 When formatted as a triangle, A038763=fusion of polynomial sequences (x+1)^n and (x+1)^n; see A193722 for the definition of fusion of two polynomial sequences or triangular arrays. Row n of A038763, as a triangle, consists of coefficients of the product (x+1)*(x+2)^n. - Clark Kimberling, Aug 04 2011
FORMULA
T(n, k) = (n+2k)*binomial(n+k-1, k-1)*2^(n-1)/k, k > 0.
T(n, 0) defined by g.f. (1-x)/(1-2x). Other rows are defined by (1-x)/(1-2x)^n.
T(n, 0) = 0 if n < 0, T(0, k) = 0 if k < 0, T(0, 0) = T(1, 0) = 1, T(n, k) = T(n, k-1) + 2*T(n-1, k); for example, 160 = 48 + 2*56 for n = 4 and k = 2. - Philippe Deléham, Aug 12 2005 G.f. of the triangular interpretation: (-1+x*y)/(-1+2*x*y+x). - R. J. Mathar, Aug 11 2015
EXAMPLE
Rows begin
1, 1, 2, 4, 8, ...
1, 3, 8, 20, 48, ...
1, 5, 18, 56, 160, ...
1, 7, 32, 120, 400, ...
1, 9, 50, 220, 840, ...
...
As a triangle:
1;
1, 1;
1, 3, 2;
1, 5, 8, 4;
1, 7, 18, 20, 8;
MATHEMATICA
(* Program generates triangle A081277 as the self-fusion of Pascal's triangle *) z = 8; a = 1; b = 1; c = 1; d = 1;
p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n
t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1
g[n_] := CoefficientList[w[n, x], {x}]
TableForm[Table[Reverse[g[n]], {n, -1, z}]]
Flatten[Table[Reverse[g[n]], {n, -1, z}]] (* A081277 *) TableForm[Table[g[n], {n, -1, z}]]
Flatten[Table[g[n], {n, -1, z}]] (* abs val of A118800 *) Factor[w[6, x]]
a(1) = 1, a(2) = 2, a(n+2) = 2*a(n+1) + (n + 1)*(n + 2)*a(n).
+10
18
1, 2, 10, 44, 288, 1896, 15888, 137952, 1419840, 15255360, 186693120, 2387093760, 33898314240, 502247692800, 8123141376000, 136785729024000, 2483065912320000, 46822564905984000, 942853671825408000, 19678282007924736000, 435355106182520832000
COMMENTS
This is the case m = 1 of the general recurrence a(1) = 1, a(2) = 2*m, a(n+2) = 2*m*a(n+1) + (n + 1)*(n + 2)*a(n) (we suppress the dependence of a(n) on m), which arises when accelerating the convergence of the series 1/2 + 1/2*Sum_{k > 1} (-1)^(k+1)/(k*(k + 1)) = log(2). For other cases see A142984 (m = 2), A142985 (m = 3), A142986 (m = 4) and A142987 (m = 5). The solution to the general recurrence may be expressed as a sum: a(n) = n!*p_m(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p_m(k)*p_m(k+1)), where p_m(x) = Sum_{k = 1..m} 2^(k-1)*C(m-1,k-1)*C(x,k) is the polynomial that gives the regular polytope numbers for the m-dimensional cross polytope as defined by [Kim] (see A142978). The first few values are p_1(x) = x, p_2(x) = x^2, p_3(x) = (2*x^3 + x)/3 and p_4(x) = (x^4 + 2*x^2)/3. The polynomial p_m(x) is the unique polynomial solution of the difference equation x*(f(x+1) - f(x-1)) = 2*m*f(x), normalized so that f(1) = 1.
The o.g.f. for the p_m(x) is 1/2*((1 + t)/(1 - t))^x = 1/2 + x*t + x^2*t^2 + (2*x^3 + x)/3*t^3 + .... Thus p_m(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_m(x;b,c) at b = 0, c = -1, also known as a Mittag-Leffler polynomial.
The general recurrence in the first paragraph above has a second solution b(n) = n!*p_m(n+1) with b(1) = 2*m, b(2) = m^2 + 2. Hence the behavior of a(n) for large n is given by Limit_{n-> oo} a(n)/b(n) = Sum_{k >= 1} (-1)^(k+1)/(p_m(k)*p_m(k+1)) = 1/((2*m) + 1*2/((2*m) + 2*3/((2*m) + 3*4/((2*m) + ... + n*(n + 1)/((2*m) + ...))))) = 1 + (-1)^(m+1) * (2*m)*(log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(m+1)/m)), where the final equality follows by a result of Ramanujan (see [Berndt, Chapter 12, Entry 32(i)]).
REFERENCES
Bruce C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag.
FORMULA
a(n) = n!*p(n+1)*Sum_{k = 1..n} (-1)^(k+1)/(p(k)*p(k+1)), where p(n) = n.
Recurrence: a(1) = 1, a(2) = 2, a(n+2) = 2*a(n+1) + (n + 1)*(n + 2)*a(n).
The sequence b(n) := n!*p(n+1) satisfies the same recurrence with b(1) = 2 and b(2) = 6.
Hence we obtain the finite continued fraction expansion a(n)/b(n) = 1/(2 + 1*2/(2 + 2*3/(2 + 3*4/(2 + ... + (n - 1)*n/2)))), for n >= 2.
The behavior of a(n) for large n is given by Limit_{n -> oo} a(n)/b(n) = 1/(2 + 1*2/(2 + 2*3/(2 + 3*4/(2 + ... + n*(n+1)/(2 + ...))))) = Sum_{k >= 1} (-1)^(k+1)/(k*(k + 1)) = 2*log(2) - 1.
MAPLE
a := n -> (n+1)!*sum ((-1)^(k+1)/(k*(k+1)), k = 1..n): seq(a(n), n = 1..20);
MATHEMATICA
Rest[CoefficientList[Series[(-x+2*Log[x+1])/(x-1)^2, {x, 0, 20}], x]*Range[0, 20]!] (* Vaclav Kotesovec, Oct 21 2012 *)
PROG
(Haskell)
a142983 n = a142983_list !! (n-1)
a142983_list = 1 : 2 : zipWith (+)
(map (* 2) $ tail a142983_list)
(zipWith (*) (drop 2 a002378_list) a142983_list)
Expansion of g.f. x*(1+x)^4/(1-x)^6.
+10
16
0, 1, 10, 51, 180, 501, 1182, 2471, 4712, 8361, 14002, 22363, 34332, 50973, 73542, 103503, 142544, 192593, 255834, 334723, 432004, 550725, 694254, 866295, 1070904, 1312505, 1595906, 1926315, 2309356, 2751085, 3258006, 3837087, 4495776, 5242017, 6084266
COMMENTS
Hyun Kwang Kim asserts that every nonnegative integer can be represented by the sum of no more than 14 of these numbers. - Jonathan Vos Post, Nov 16 2004 If Y_i (i=1,2,3,4) are 2-blocks of a (n+4)-set X then a(n-4) is the number of 9-subsets of X intersecting each Y_i (i=1,2,3,4). - Milan Janjic, Oct 28 2007 Starting with 1 = binomial transform of [1, 9, 32, 56, 48, 16, 0, 0, 0, ...], where (1, 9, 32, 56, 48, 16) = row 5 of the Chebyshev triangle A081277. Also = row 5 of the array in A142978. - Gary W. Adamson, Jul 19 2008
REFERENCES
H. S. M. Coxeter, Regular Polytopes, New York: Dover Publications, 1973.
FORMULA
Recurrence: a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
a(n) = C(n+4,5) + 4*C(n+3,5) + 6*C(n+2,5) + 4*C(n+1,5) + C(n,5).
Sum_{n>=1} 1/((1/15)*n*(2*n^4 + 10*n^2 + 3)*n!) = hypergeom([1, 1, 1+i*sqrt(10-2*sqrt(19))*(1/2), 1-i*sqrt(10-2*sqrt(19))*(1/2), 1+i*sqrt(10+2*sqrt(19))*(1/2), 1-i*sqrt(10+2*sqrt(19))*(1/2)], [2, 2, 2+i*sqrt(10-2*sqrt(19))*(1/2), 2-i*sqrt(10-2*sqrt(19))*(1/2), 2+i*sqrt(10+2*sqrt(19))*(1/2), 2-i*sqrt(10+2*sqrt(19))*(1/2)], 1) = 1.05351734968093116819345664995829700099916... - Stephen Crowley, Jul 14 2009 Euler transform of length 2 sequence [10, -4]. - Michael Somos, Jun 19 2018 Sum_{k >= 1} (-1)^k/(a(k)*a(k+1)) = 10*log(2) - 41/6 = 1/(10 + 2/(10 + 6/(10 + ... + n*(n-1)/(10 + ...)))). See A142983. Cf. A005900 and A014820. - Peter Bala, Mar 08 2024 E.g.f.: exp(x)*x*(15 + 60*x + 60*x^2 + 20*x^3 + 2*x^4)/15. - Stefano Spezia, Mar 10 2024
MAPLE
al:=proc(s, n) binomial(n+s-1, s); end; be:=proc(d, n) local r; add( (-1)^r*binomial(d-1, r)*2^(d-1-r)*al(d-r, n), r=0..d-1); end; [seq(be(5, n), n=0..100)];
MATHEMATICA
CoefficientList[Series[x (1 + x)^4/(1 - x)^6, {x, 0, 32}], x] (* Michael De Vlieger, Sep 02 2016 *) LinearRecurrence[{6, -15, 20, -15, 6, -1}, {0, 1, 10, 51, 180, 501}, 40] (* Harvey P. Dale, Jun 19 2021 *)
PROG
(PARI) concat(0, Vec(x*(1+x)^4/(1-x)^6 + O(x^99))) \\ Altug Alkan, Sep 02 2016 (PARI) {a(n) = n * (2*n^4 + 10*n^2 + 3) / 15}; /* Michael Somos, Jun 17 2018 */
Expansion of g.f. x*(1+x)^5/(1-x)^7.
+10
14
0, 1, 12, 73, 304, 985, 2668, 6321, 13504, 26577, 48940, 85305, 142000, 227305, 351820, 528865, 774912, 1110049, 1558476, 2149033, 2915760, 3898489, 5143468, 6704017, 8641216, 11024625, 13933036, 17455257, 21690928, 26751369, 32760460, 39855553, 48188416, 57926209
COMMENTS
Figurate numbers based on the 6-dimensional regular convex polytope called the 6-dimensional cross-polytope, or 6-dimensional hyperoctahedron, which is represented by the Schlaefli symbol {3, 3, 3, 3, 4}. It is the dual of the 6-dimensional hypercube. Kim asserts that every nonnegative integer can be represented by the sum of no more than 19 of these 6-crosspolytope numbers. - Jonathan Vos Post, Nov 16 2004 Starting with 1 = binomial transform of [1, 11, 50, 120, 160, 112, 32, 0, 0, 0, ...] where (1, 11, 50, 120, 160, 112, 32) = row 6 of the Chebyshev triangle A081277. Also = row 6 of the array in A142978. - Gary W. Adamson, Jul 19 2008
REFERENCES
H. S. M. Coxeter, Regular Polytopes, New York: Dover, 1973.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 240.
Jonathan Vos Post, "4-Dimensional Jonathan numbers: polytope numbers and Centered polytope numbers of Higher Than 3 Dimensions", Draft 1.5 of 9 a.m., 12 March 2004, circulated by e-mail.
FORMULA
Recurrence: a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7).
Sum_{n >= 1} 1/a(n) = -5*(Sum(_alpha*(77*_alpha^2+655)*Psi(1-_alpha), _alpha = RootOf(2*_Z^4+20*_Z^2+23)))*(1/3174)+15*Pi^2*(1/46)=1.10203455013915915542552577192042916250524...
Sum_{n>=1} 1/(a(n)*n!) = hypergeom([1, 1, 1, 1-a, 1+b, 1-b, 1+a], [2, 2, 2, 2+b, 2-b, 2+a, 2-a], 1) = 1.04409584723862654376639417281585634150689... where a = (i/2)*sqrt(20+6*sqrt(6)), b = (i/2)*sqrt(20-6*sqrt(6)), and i = sqrt(-1). (End)
E.g.f.: exp(x)*x*(45 + 225*x + 300*x^2 + 150*x^3 + 30*x^4 + 2*x^5)/45. - Stefano Spezia, Mar 10 2024
MAPLE
al:=proc(s, n) binomial(n+s-1, s); end; be:=proc(d, n) local r; add( (-1)^r*binomial(d-1, r)*2^(d-1-r)*al(d-r, n), r=0..d-1); end; [seq(be(6, n), n=0..100)];
MATHEMATICA
CoefficientList[Series[x (1+x)^5/(1-x)^7, {x, 0, 40}], x] (* or *) LinearRecurrence[ {7, -21, 35, -35, 21, -7, 1}, {0, 1, 12, 73, 304, 985, 2668}, 40] (* Harvey P. Dale, Aug 05 2018 *)
PROG
(PARI) x='x+O('x^100); concat(0, Vec(x*(1+x)^5/(1-x)^7)) \\ Altug Alkan, Dec 14 2015
a(n) = n*(4*n^6 + 70*n^4 + 196*n^2 + 45)/315.
+10
14
0, 1, 14, 99, 476, 1765, 5418, 14407, 34232, 74313, 149830, 284075, 511380, 880685, 1459810, 2340495, 3644272, 5529233, 8197758, 11905267, 16970060, 23784309, 32826266, 44673751, 60018984, 79684825, 104642486, 136030779, 175176964, 223619261, 283131090, 355747103
COMMENTS
Kim asserts that every nonnegative integer can be represented by the sum of no more than 21 of these numbers.
Starting with 1 = binomial transform of [1, 13, 72, 220, 400, 432, 256, 0, 0, 0, ...], where (1, 13, 72, 220, 400, 432, 256) = row 7 of the Chebyshev triangle A081277. Also = row 7 of the array in A142978. - Gary W. Adamson, Jul 19 2008
FORMULA
a(n) = n*(4*n^6 + 70*n^4 + 196*n^2 + 45)/315.
MATHEMATICA
Table[SeriesCoefficient[x (1 + x)^6/(1 - x)^8, {x, 0, n}], {n, 0, 31}] (* Michael De Vlieger, Dec 14 2015 *)
PROG
(PARI) concat(0, Vec(x*(1+x)^6/(1-x)^8 + O(x^40))) \\ Michel Marcus, Dec 14 2015
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