OFFSET
0,4
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.
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
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 *)
PROG
(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 */
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved