A003162 A binomial coefficient summation.

1, 1, 1, 3, 6, 19, 49, 163, 472, 1626, 5034, 17769, 57474, 206487, 688881, 2508195, 8563020, 31504240, 109492960, 406214878, 1432030036, 5349255726, 19077934506, 71672186953, 258095737156, 974311431094, 3537275250214, 13408623649893

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

From Peter Bala, Mar 26 2023: (Start)

For r a positive integer define S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r. Gould (1974) proposed the problem of showing that S(3,n) was always divisible by S(1,n). The present sequence is {S(3,n)/S(1,n)}. In fact, calculation suggests that if r is odd then S(r,n) is always divisible by S(1,n). For other cases see A361888 ({S(5,n)/S(1,n)}) and A361891 ({S(7,n)/ S(1,n)}).

Conjecture: Let b(n) = a(2*n-1). Then the supercongruence b(n*p^k) == b(n*p^(k-1)) (mod p^(3*k)) holds for positive integers n and k and all primes p >= 5. See A183069. (End)

References

• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• Seiichi Manyama, Table of n, a(n) for n = 0..1000
• H. W. Gould, Problem E2384, Amer. Math. Monthly, 81 (1974), 170-171.

Crossrefs

Cf. A003161, A183069, A361887, A361888, A361889, A361890, A361891, A361892.

Programs

Maple

H := hypergeom([1/2,1/2],[1],16*x^2);
ogf := (Int(6*H*(4*x^2+5)/(4-x^2)^(3/2),x)+H*(16*x^2-1)/(4-x^2)^(1/2))*((2-x)/(2+x))^(1/2)/(4*x)+1/(8*x);
series(ogf,x=0,20);  # Mark van Hoeij, May 06 2013

Mathematica

Table[Sum[(Binomial[n, k]-Binomial[n, k-1])^3/Binomial[n, Floor[n/2]],{k,0,Floor[n/2]}],{n,0,20}] (* Vaclav Kotesovec, Mar 06 2014 *)

PARI

a(n)=if(n<0, 0, sum(k=0,n\2, (binomial(n,k)-binomial(n,k-1))^3)/binomial(n,n\2)) /* Michael Somos, Jun 02 2005 */

Formula

G.f.: hypergeometric expression with an antiderivative, see Maple program. - Mark van Hoeij, May 06 2013

Recurrence: 4*n*(n+1)^2*(196*n^3 - 819*n^2 + 530*n + 528)*a(n) = 2*n*(1372*n^4 - 3633*n^3 - 7455*n^2 + 21934*n - 8448)*a(n-1) + (12740*n^6 - 90867*n^5 + 195310*n^4 - 13277*n^3 - 452690*n^2 + 528384*n - 174960)*a(n-2) + 8*(n-2)*(686*n^4 - 3010*n^3 + 1176*n^2 + 6543*n - 4725)*a(n-3) - 16*(n-3)^2*(n-2)*(196*n^3 - 231*n^2 - 520*n + 435)*a(n-4). - Vaclav Kotesovec, Mar 06 2014

a(n) ~ 4^(n+2)/(9*Pi*n^2). - Vaclav Kotesovec, Mar 06 2014