A015219 -id:A015219 - cp's OEIS frontend

A203778 a(n) = -24A015219(n-2)a(n-1), with a(1) = 2.

2, -48, 40320, -159667200, 1743565824000, -40548366802944000, 1723467782592331776000, -120987438337981690675200000, 13052124847901464790040576000000, -2050227771108362089219573678080000000, 449688758403823707201064412255354880000000

Offset: 1

John M. Campbell, Jan 05 2012

Sums of coefficients from (4n)th moments of binomial(m,k)*binomial(3*m,k): see Maple code below.

			The evaluation of sum(binomial(n,k)*binomial(3*n,k)*k^8,k=0..n) involves the polynomial 729*n^13+729*n^12-12879*n^11+9801*n^10+50247*n^9-84825*n^8-105*n^7+74167*n^6-36968*n^5-2296*n^4+1472*n^3-120*n^2, the sum of the coefficients of which is a(2)=-48.

Eric W. Weisstein, MathWorld: Binomial Sums
Index to divisibility sequences

Cf. A015219, A202948, A202946.

with(PolynomialTools);polyn:=q->expand(simplify((1/(GAMMA(n-((2*floor((q+1)/4)-1))/(2))))*(1/sqrt(3))*GAMMA(n+1/3)*GAMMA(n+2/3)*(1/3)*(1/(27^(-n)))*GAMMA(n)*1/64^n*sum(binomial(n,k)*binomial(3*n,k)*k^q,k=0..n)*(1/(GAMMA(2*n-((2*floor(q/2)-1)/(2)))))*(2^((floor((1/2)*q+1/2)-1)+q))));coefl:=h->CoefficientList(expand(polyn(h)),n);coe:=(d,b)->coefl(d)[b];seq(sum(coe((4*g),a),a=1..(2*(4*g)-floor(((4*g)+3)/4))),g=1..6);seq(simplify(-(1/32)*GAMMA(2*n-3/2)*GAMMA(n-1/2)*(-1)^n*64^n/Pi),n=1..6);

a(n) = -(1/32)*Gamma(2*n-3/2)*Gamma(n-1/2)*(-1)^n*64^n/Pi.

A000447 a(n) = 1^2 + 3^2 + 5^2 + 7^2 + ... + (2n-1)^2 = n(4*n^2 - 1)/3.

Original entry on oeis.org

0, 1, 10, 35, 84, 165, 286, 455, 680, 969, 1330, 1771, 2300, 2925, 3654, 4495, 5456, 6545, 7770, 9139, 10660, 12341, 14190, 16215, 18424, 20825, 23426, 26235, 29260, 32509, 35990, 39711, 43680, 47905, 52394, 57155, 62196, 67525, 73150, 79079, 85320, 91881, 98770, 105995, 113564, 121485

Offset: 0

N. J. A. Sloane

4 times the variance of the area under an n-step random walk: e.g., with three steps, the area can be 9/2, 7/2, 3/2, 1/2, -1/2, -3/2, -7/2, or -9/2 each with probability 1/8, giving a variance of 35/4 or a(3)/4. - Henry Bottomley, Jul 14 2003
Number of standard tableaux of shape (2n-1,1,1,1) (n>=1). - Emeric Deutsch, May 30 2004
Also a(n) = (1/6)*(8*n^3-2*n), n>0: structured octagonal diamond numbers (vertex structure 9). Cf. A059722 = alternate vertex; A000447 = structured diamonds; and structured tetragonal anti-diamond numbers (vertex structure 9). Cf. A096000 = alternate vertex; A100188 = structured anti-diamonds. Cf. A100145 for more on structured numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
The n-th tetrahedral (or pyramidal) number is n(n+1)(n+2)/6. This sequence contains the tetrahedral numbers of A000292 obtained for n= 1,3,5,7,... (see A015219). - Valentin Bakoev, Mar 03 2009
Using three consecutive numbers u, v, w, (u+v+w)^3-(u^3+v^3+w^3) equals 18 times the numbers in this sequence. - J. M. Bergot, Aug 24 2011
This sequence is related to A070893 by A070893(2*n-1) = n*a(n)-sum(i=0..n-1, a(i)). - Bruno Berselli, Aug 26 2011
Number of integer solutions to 1-n <= x <= y <= z <= n-1. - Michael Somos, Dec 27 2011
Partial sums of A016754. - Reinhard Zumkeller, Apr 02 2012
Also the number of cubes in the n-th Haüy square pyramid. - Eric W. Weisstein, Sep 27 2017

			G.f. = x + 10*x^2 + 35*x^3 + 84*x^4 + 165*x^5 + 286*x^6 + 455*x^7 + 680*x^8 + ...
a(2) = 10 since (-1, -1, -1), (-1, -1, 0), (-1, -1, 1), (-1, 0, 0), (-1, 0, 1), (-1, 1, 1), (0, 0, 0), (0, 0, 1), (0, 1, 1), (1, 1, 1) are the 10 solutions (x, y, z) of -1 <= x <= y <= z <= 1.
a(0) = 0, which corresponds to the empty sum.

G. Chrystal, Textbook of Algebra, Vol. 1, A. & C. Black, 1886, Chap. XX, Sect. 10, Example 2.
F. E. Croxton and D. J. Cowden, Applied General Statistics. 2nd ed., Prentice-Hall, Englewood Cliffs, NJ, 1955, p. 742.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 140.
C. V. Durell, Advanced Algebra, Volume 1, G. Bell & Son, 1932, Exercise IIIe, No. 4.
L. B. W. Jolley, Summation of Series. 2nd ed., Dover, NY, 1961, p. 7.
J. Riordan, Combinatorial Identities, Wiley, 1968, p. 217.
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).

T. D. Noe, Table of n, a(n) for n = 0..1000
J. L. Bailey, A table to facilitate the fitting of certain logistic curves, Annals Math. Stat., Vol. 2 (1931), pp. 355-359. [Annotated scanned copy]
Valentin Bakoev, Algorithmic approach to counting of certain types m-ary partitions, Discrete Mathematics, Vol. 275 (2004), pp. 17-41. - _Valentin Bakoev_, Mar 03 2009
F. E. Croxton and D. J. Cowden, Applied General Statistics, 2nd Ed., Prentice-Hall, Englewood Cliffs, NJ, 1955 [Annotated scans of just pages 742-743]
Milan Janjic, Two Enumerative Functions.
T. P. Martin, Shells of atoms, Phys. Reports, Vol. 273 (1996), pp. 199-241, eq. (11).
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
Eric Weisstein's World of Mathematics, Haüy Construction.
Eric Weisstein's World of Mathematics, Square Pyramid.
Index entries for two-way infinite sequences.
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

(1/12)*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Column 1 in triangles A008956 and A008958.
Cf. A035328, A069072, A190152.
A000447 is related to partitions of 2^n into powers of 2, as it is shown in the formula, example and cross-references of A002577. - Valentin Bakoev, Mar 03 2009
Cf. A002412, A016061.

a000447 n = a000447_list !! n
a000447_list = scanl1 (+) a016754_list
-- Reinhard Zumkeller, Apr 02 2012

[n*(4*n^2-1)/3: n in [0..50]]; // Vincenzo Librandi, Jan 12 2016

A000447:=z*(1+6*z+z**2)/(z-1)**4; # Simon Plouffe, 1992 dissertation.
A000447:=n->n*(4*n^2 - 1)/3; seq(A000447(n), n=0..50); # Wesley Ivan Hurt, Mar 30 2014

Table[n (4 n^2 - 1)/3, {n, 0, 80}] (* Vladimir Joseph Stephan Orlovsky, Apr 18 2011 *)
LinearRecurrence[{4, -6, 4, -1}, {0, 1, 10, 35}, 80] (* Harvey P. Dale, May 25 2012 *)
Join[{0}, Accumulate[Range[1, 81, 2]^2]] (* Harvey P. Dale, Jul 18 2013 *)
CoefficientList[Series[x (1 + 6 x + x^2)/(-1 + x)^4, {x, 0, 20}], x] (* Eric W. Weisstein, Sep 27 2017 *)

A000447(n):=n*(4*n^2 - 1)/3$ makelist(A000447(n),n,0,20); /* Martin Ettl, Jan 07 2013 */

PARI
```
{a(n) = n * (4*n^2 - 1) / 3};
```

concat(0, Vec(x*(1+6*x+x^2)/(1-x)^4 + O(x^100))) \\ Altug Alkan, Jan 11 2016

def A000447(n): return n*((n**2<<2)-1)//3 # Chai Wah Wu, Feb 12 2023

a(n) = binomial(2*n+1, 3) = A000292(2*n-1).
G.f.: x*(1+6*x+x^2)/(1-x)^4.
a(n) = -a(-n) for all n in Z.
a(n) = A000330(2*n)-4*A000330(n) = A000466(n)*n/3 = A000578(n)+A007290(n-2) = A000583(n)-2*A024196(n-1) = A035328(n)/3. - Henry Bottomley, Jul 14 2003
a(n+1) = (2*n+1)*(2*n+2)(2*n+3)/6. - Valentin Bakoev, Mar 03 2009
a(0)=0, a(1)=1, a(2)=10, a(3)=35, a(n)=4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4). - Harvey P. Dale, May 25 2012
a(n) = v(n,n-1), where v(n,k) is the central factorial numbers of the first kind with odd indices. - Mircea Merca, Jan 25 2014
a(n) = A005917(n+1) - A100157(n+1), where A005917 are the rhombic dodecahedral numbers and A100157 are the structured rhombic dodecahedral numbers (vertex structure 9). - Peter M. Chema, Jan 09 2016
For any nonnegative integers m and n, 8*(n^3)*a(m) + 2*m*a(n) = a(2*m*n). - Ivan N. Ianakiev, Mar 04 2017
E.g.f.: exp(x)*x*(1 + 4*x  + (4/3)*x^2). - Wolfdieter Lang, Mar 11 2017
a(n) = A002412(n) + A016061(n-1), for n>0. - Bruce J. Nicholson, Nov 12 2017
From Amiram Eldar, Jan 04 2022: (Start)
Sum_{n>=1} 1/a(n) = 6*log(2) - 3.
Sum_{n>=1} (-1)^(n+1)/a(n) = 3 - 3*log(2). (End)

Chrystal and Durell references from R. K. Guy, Apr 02 2004

A060543 Triangle, read by antidiagonals, where T(n,k) = C(n+nk+k, nk+k).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 10, 5, 1, 1, 35, 28, 7, 1, 1, 126, 165, 55, 9, 1, 1, 462, 1001, 455, 91, 11, 1, 1, 1716, 6188, 3876, 969, 136, 13, 1, 1, 6435, 38760, 33649, 10626, 1771, 190, 15, 1, 1, 24310, 245157, 296010, 118755, 23751, 2925, 253, 17, 1, 1, 92378, 1562275

Offset: 0

Henry Bottomley, Apr 02 2001

Main diagonal is A108288. Antidiagonal sums is A108289. Inverse binomial transforms of each row give triangle A108290. G.f. of row n multiplied by (1-x)^(n+1) equals g.f. of row n of triangle A108267 (rows sums of A108267 equal (n+1)^n).

			row 1: (2*n+1)/1!
row 2: (3*n+1)*(3*n+2)/2!
row 3: (4*n+1)*(4*n+2)*(4*n+3)/3!
row 4: (5*n+1)*(5*n+2)*(5*n+3)*(5*n+4)/4!
row 5: (6*n+1)*(6*n+2)*(6*n+3)*(6*n+4)*(6*n+5)/5!.
Table begins:
1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,...
1,3,5,7,9,11,13,15,17,19,21,23,25,27,...
1,10,28,55,91,136,190,253,325,406,496,...
1,35,165,455,969,1771,2925,4495,6545,...
1,126,1001,3876,10626,23751,46376,82251,...
1,462,6188,33649,118755,324632,749398,...
1,1716,38760,296010,1344904,4496388,...

Harry J. Smith, Table of n, a(n) for n=0..252

Cf. A108290, A108267, A108288, A108289, A060544 (row 2), A015219 (row 3).

Rows include A000012, A001700, A025174. Columns include A000012, A005408, A060544. Main diagonal is A060545.

PARI
```
T(n,k)=binomial(n+n*k+k,n*k+k)
```

{ i=0; write("b060543.txt", "0 1"); for (m=0, 20, for (k=0, m + 1, n=m - k + 1; write("b060543.txt", i++, " ", binomial(n + n*k + k, n*k + k))); ) } \\ Harry J. Smith, Jul 06 2009

a(n) = A060539(n, k)/n = A007318(nk, k)/n = A060540(n, k)/A060540(n-1, k).

Entry revised by Paul D. Hanna, May 31 2005

Edited by N. J. A. Sloane at the suggestion of Andrew S. Plewe, Jun 17 2007

A001505 a(n) = (4n+1)(4n+2)(4n+3).

Original entry on oeis.org

6, 210, 990, 2730, 5814, 10626, 17550, 26970, 39270, 54834, 74046, 97290, 124950, 157410, 195054, 238266, 287430, 342930, 405150, 474474, 551286, 635970, 728910, 830490, 941094, 1061106, 1190910, 1330890, 1481430, 1642914, 1815726, 2000250, 2196870, 2405970

Offset: 0

N. J. A. Sloane

T. D. Noe, Table of n, a(n) for n = 0..1000
L. B. W. Jolley, Summation of Series, Dover, 1961
Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1).

Cf. A015219.

[(4*n+1)*(4*n+2)*(4*n+3): n in [0..100]]; // Vincenzo Librandi, Apr 04 2011

Table[(4n+1)(4n+2)(4n+3),{n,0,49}] (* Vladimir Joseph Stephan Orlovsky, Jan 22 2012 *)

a(n)=(4*n+1)*(4*n+2)*(4*n+3) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 6 * A015219(n).
Sum_{n>=0} 1/a(n) = log(2)/4 = 0.17328679513998... [Jolley eq. 253. Typo fixed by Jaume Oliver Lafont, Jan 09 2009]
G.f.: 6*(1+x)*(x^2+30*x+1) / (x-1)^4. - R. J. Mathar, Apr 02 2011
Sum_{n>=0} (-1)^n/a(n) = (sqrt(2)-1)*Pi/8. - Amiram Eldar, Sep 17 2022
E.g.f.: 2*exp(x)*(3 + 102*x + 144*x^2 + 32*x^3). - Stefano Spezia, Aug 24 2025

A060541 a(n) = binomial(4*n, 4).

Original entry on oeis.org

1, 70, 495, 1820, 4845, 10626, 20475, 35960, 58905, 91390, 135751, 194580, 270725, 367290, 487635, 635376, 814385, 1028790, 1282975, 1581580, 1929501, 2331890, 2794155, 3321960, 3921225, 4598126, 5359095, 6210820, 7160245, 8214570, 9381251, 10668000, 12082785

Offset: 1

Henry Bottomley, Apr 02 2001

Harry J. Smith, Table of n, a(n) for n=1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Cf. A000027, A000332, A000384, A006566, A015219, A060539.

[Binomial(4 n, 4): n in [1..40]]; // Vincenzo Librandi, Jan 20 2015

Table[Binomial[4n, 4], {n, 100}] (* Wesley Ivan Hurt, Sep 27 2013 *)
LinearRecurrence[{5,-10,10,-5,1},{1,70,495,1820,4845},40] (* Harvey P. Dale, Jan 13 2015 *)

a(n) = n*(2*n - 1)*(4*n - 1)*(4*n - 3)/3; \\ Harry J. Smith, Jul 06 2009

a(n) = n*(2n-1)*(4n-1)*(4n-3)/3.
a(n) = n * A015219(n-1) = A000332(4n) = A060539(n, 4).
G.f.: x*(1+65*x+155*x^2+35*x^3) / (1-x)^5. - R. J. Mathar, Oct 03 2011
From Amiram Eldar, Mar 08 2022: (Start)
Sum_{n>=1} 1/a(n) = 6*log(2) - Pi.
Sum_{n>=1} (-1)^(n+1)/a(n) = 2*sqrt(2)*log(sqrt(2)-1) - log(2) + (2*sqrt(2) - 3/2)*Pi. (End)

Offset changed from 0 to 1 by Harry J. Smith, Jul 06 2009

A014795 Squares of odd tetrahedral numbers.

Original entry on oeis.org

1, 1225, 27225, 207025, 938961, 3136441, 8555625, 20205025, 42837025, 83521321, 152300281, 262926225, 433680625, 688275225, 1056835081, 1576963521, 2294889025, 3266694025, 4559625625, 6253488241, 8442118161, 11234940025, 14758605225, 19158712225, 24601608801

Offset: 0

Mohammad K. Azarian

Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000292, A015219, A014796.

Table[((4*n+1)*(4*n+2)*(4*n+3)/6)^2, {n, 0, 40}] (* Amiram Eldar, Mar 07 2022 *)

From Amiram Eldar, Mar 07 2022: (Start)
a(n) = A015219(n)^2 = ((4*n+1)*(4*n+2)*(4*n+3)/6)^2.
Sum_{n>=0} 1/a(n) = 9*Pi*(Pi-3)/4. (End)

More terms from Erich Friedman

More terms from Amiram Eldar, Mar 07 2022

A015220 Even tetrahedral numbers.

Original entry on oeis.org

0, 4, 10, 20, 56, 84, 120, 220, 286, 364, 560, 680, 816, 1140, 1330, 1540, 2024, 2300, 2600, 3276, 3654, 4060, 4960, 5456, 5984, 7140, 7770, 8436, 9880, 10660, 11480, 13244, 14190, 15180, 17296, 18424, 19600, 22100, 23426, 24804, 27720, 29260, 30856, 34220, 35990

Offset: 0

Mohammad K. Azarian

Index entries for linear recurrences with constant coefficients, signature (1,0,3,-3,0,-3,3,0,1,-1).

Cf. A000292, A004772, A015219.

LinearRecurrence[{1,0,3,-3,0,-3,3,0,1,-1},{0,4,10,20,56,84,120,220,286,364},41] (* Ant King, Oct 19 2012 *)
Select[Table[(Times@@(n+{0,1,2}))/6,{n,0,60}],EvenQ] (* Harvey P. Dale, Jan 22 2013 *)

From Ant King, Oct 19 2012: (Start)
a(n) = a(n-1) + 3*a(n-3) - 3*a(n-4) - 3*a(n-6) + 3*a(n-7) + a(n-9) - a(n-10).
a(n) = 64 + 3*a(n-3) - 3*a(n-6) + a(n-9).
G.f.: 2*x*(2+3*x+5*x^2+12*x^3+5*x^4+3*x^5+2*x^6) / ((1-x)^4*(1+x+x^2)^3).
Sum_{n>=1} 1/a(n) = 3/2*(1-log(2)). (End)
From Amiram Eldar, Mar 07 2022: (Start)
a(n) = A000292(A004772(n+1)).
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*log(2) - 15/2 + 9*sqrt(2)*log(sqrt(2)+1)/2. (End)

More terms from Erich Friedman

a(0) prepended by Amiram Eldar, Mar 07 2022

A178034 a(n) = binomial(n*Omega(n),Omega(n)) / n.

Original entry on oeis.org

1, 1, 1, 7, 1, 11, 1, 253, 17, 19, 1, 595, 1, 27, 29, 39711, 1, 1378, 1, 1711, 41, 43, 1, 138415, 49, 51, 3160, 3403, 1, 3916, 1, 25637001, 65, 67, 69, 477191, 1, 75, 77, 657359, 1, 7750, 1, 8515, 8911, 91, 1, 132563501, 97, 11026, 101, 11935, 1, 1633355

Offset: 1

Michel Lagneau, May 17 2010

nonn

Omega(.) = A001222(.) is the number of prime divisors of n (counted with multiplicity).
binomial(nk,k)= n*binomial(nk-1,k-1) ensures that all entries are integers.
Subcases for this sequence:
If n is prime, Omega(n) = 1, and a(n) = binomial (n,1) / n = 1.
If n and n+1 are products of two primes (A070552), then Omega(n) = Omega(n+1) = 2, and binomial(n*Omega(n), Omega(n)) / n = binomial(2*n, 2) / n = 2*n-1 and binomial(2*(n+1), 2) / (n+1) = 2*n+1, and we obtain two consecutive numbers of the form (x, x+2), for example (17,19), (27,29), (41,43),... at n =9, 14...
Chaining this property: If n, n+1, and n+2 are semiprimes (A056809) , we find three consecutive numbers of the form (x, x+2,x+4), for example (65, 67, 69), (169, 171, 173), at n=33, 85.
At places where Omega(n)=3, we find the subsequence A060544, for example a(8) = A060544(8).
At places where Omega(n)=4, we find the subsequence A015219.

			a(8) = binomial(8*Omega(8),Omega(8))/8 = binomial(8*3,3)/8 = 2024/8 = 253.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A001358, A038456

A178034 := proc(n)
        local o ;
        o := numtheory[bigomega](n) ;
        binomial(n*o,o)/n ;
end proc: # R. J. Mathar, Jul 08 2012

bon[n_]:=Module[{o=PrimeOmega[n]},Binomial[n*o,o]/n]; Array[bon,60] (* Harvey P. Dale, Jul 22 2014 *)

a(n)=my(b=bigomega(n));binomial(n*b,b)/n \\ Charles R Greathouse IV, Oct 25 2012

A204820 a(n) = -4a(n-1)A001505(n-2), with a(1)=8.

Original entry on oeis.org

8, -192, 161280, -638668800, 6974263296000, -162193467211776000, 6893871130369327104000, -483949753351926762700800000, 52208499391605859160162304000000, -8200911084433448356878294712320000000

Offset: 1

John M. Campbell, Jan 19 2012

Sums of coefficients from (4n+1)th moments of binomial(m,k) * binomial(3*m,k); see Maple code below.

			The evaluation of sum(binomial(n, k)*binomial(3*n, k)*k^9, k=0..n) involves the polynomial 2187*n^11+6561*n^10-45927*n^9-28431*n^8+322947*n^7-257985*n^6-473445*n^5+726003*n^4-110482*n^3-189924*n^2+52624*n-4320, the sum of the coefficients of which is -192 = a(2).

Eric W. Weisstein, MathWorld: Binomial Sums
Index to divisibility sequences

Cf. A203778, A015219, A202948, A202946.

with(PolynomialTools); polyn:=q->expand(simplify((1/(GAMMA(n-((2*floor((q+1)/4)-1))/(2))))*(1/sqrt(3))*GAMMA(n+1/3)*GAMMA(n+2/3)*(1/3)*(1/(27^(-n)))*GAMMA(n)*1/64^n*sum(binomial(n, k)*binomial(3*n, k)*k^q, k=0..n)*(1/(GAMMA(2*n-((2*floor(q/2)-1)/(2)))))*(2^((floor((1/2)*q+1/2)-1)+q)))); coefl:=h->CoefficientList(expand(polyn(h)), n); coe:=(d, b)->coefl(d)[b];seq(sum(coe((4*d+1),b),b=1..(4*d+1)+floor(((4*d+1)+1)/4)+floor((4*d+1)/2)),d=1..6);seq(-(1/8)*GAMMA(2*n-3/2)*GAMMA(n-1/2)*(-1)^n*64^n/Pi,n=1..6);

a(n)=-(1/8)*GAMMA(2*n-3/2)*GAMMA(n-1/2)*(-1)^n*64^n/Pi

A205795 Sums of coefficients of polynomials from 5n-th moments of X ~ Hypergeometric(4m, 5m, m).

Original entry on oeis.org

24, 2880, 43545600, 5230697472000, 2432902008176640000, 3102242008666197196800000, 8841761993739701954543616000000, 49205466506600690141269768273920000000, 485663859076129603777149565235783270400000000, 7911522544013240381082219675638737768808448000000000

Offset: 1

John M. Campbell, Feb 09 2012

nonn

See Maple code below for formula for such polynomials.

			The evaluation of sum(binomial(n, k)*binomial(4*n, k)*k^5, k = 0 .. n) involves the polynomial  256*n^5-640*n^3+400*n^2+108*n-100, the sum of the coefficients of which is 24 = a(1).

Eric W. Weisstein, MathWorld: Binomial Sums
Wikipedia, Hypergeometric Distribution
Index to divisibility sequences

Cf. A204820, A203778, A015219, A202948, A202946.

with(PolynomialTools);polyn:=w->simplify(Pi^2*sum(binomial(n,k)*binomial(4*n,k)*k^w,k=0..n)*5^w/3125^n*csc((1/5)*Pi)*csc((2/5)*Pi)*GAMMA(4*n)/GAMMA(n-(floor((w+1)/5)*5-2)/5)/GAMMA(n-(floor(w/5)*5-1)/5)/GAMMA(n-(floor((w+2)/5)*5-3)/5)/GAMMA(n-(floor((w+3)/5)*5-4)/5));coefl:=d->CoefficientList(expand(polyn(d)),n);seq(sum(coefl(5*h)[m],m=1..nops(coefl(5*h))),h=1..5);seq(simplify(12*5^(5*n-5)*GAMMA(n-4/5)*GAMMA(n-3/5)*GAMMA(n-2/5)*GAMMA(n-1/5)*(cos((2/5)*Pi)+cos((1/5)*Pi))/Pi^2),n=1..5);

a(n) = 120*A151989(n-2)*a(n-1), with a(1)=24.

a(n) = 12*5^(5*n-5)*GAMMA(n-4/5)*GAMMA(n-3/5)*GAMMA(n-2/5)*GAMMA(n-1/5)*(cos((2/5)*Pi)+cos((1/5)*Pi))/Pi^2.

cp's OEIS Frontend

A203778 a(n) = -24*A015219(n-2)*a(n-1), with a(1) = 2.

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A000447 a(n) = 1^2 + 3^2 + 5^2 + 7^2 + ... + (2*n-1)^2 = n*(4*n^2 - 1)/3.

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A060543 Triangle, read by antidiagonals, where T(n,k) = C(n+n*k+k, n*k+k).

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A001505 a(n) = (4n+1)(4n+2)(4n+3).

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A060541 a(n) = binomial(4*n, 4).

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A014795 Squares of odd tetrahedral numbers.

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A015220 Even tetrahedral numbers.

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A203778 a(n) = -24A015219(n-2)a(n-1), with a(1) = 2.

A000447 a(n) = 1^2 + 3^2 + 5^2 + 7^2 + ... + (2n-1)^2 = n(4*n^2 - 1)/3.

A060543 Triangle, read by antidiagonals, where T(n,k) = C(n+nk+k, nk+k).

A204820 a(n) = -4a(n-1)A001505(n-2), with a(1)=8.