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%I M2597 #33 Apr 06 2023 08:33:59
%S 1,1,1,3,6,19,49,163,472,1626,5034,17769,57474,206487,688881,2508195,
%T 8563020,31504240,109492960,406214878,1432030036,5349255726,
%U 19077934506,71672186953,258095737156,974311431094,3537275250214,13408623649893
%N A binomial coefficient summation.
%C From _Peter Bala_, Mar 26 2023: (Start)
%C 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)}).
%C 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)
%D N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
%H H. W. Gould, <a href="http://www.jstor.org/stable/2976965">Problem E2384</a>, Amer. Math. Monthly, 81 (1974), 170-171.
%F G.f.: hypergeometric expression with an antiderivative, see Maple program. - _Mark van Hoeij_, May 06 2013
%F 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
%F a(n) ~ 4^(n+2)/(9*Pi*n^2). - _Vaclav Kotesovec_, Mar 06 2014
%p H := hypergeom([1/2,1/2],[1],16*x^2);
%p 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);
%p series(ogf,x=0,20); # _Mark van Hoeij_, May 06 2013
%t 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 *)
%o (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 */
%Y Cf. A003161, A183069, A361888, A361891
%K nonn,easy
%O 0,4
%A _N. J. A. Sloane_
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