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A361888 a(n) = S(5,n)/S(1,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

%I A361888 #20 Mar 24 2025 10:19:02
%S A361888 1,1,1,11,46,415,3265,30955,299500,3173626,33576266,386672861,
%T A361888 4340714886,52846226091,620906440961,7857161332715,95704821415240,
%U A361888 1246162831674580,15624127945644100,207990691516965886,2669841775757784796,36176886727828945286,473508685502539872586
%N A361888 a(n) = S(5,n)/S(1,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.
%C A361888 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) conjectured that S(3,n) was always divisible by S(1,n). See A183069 for {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). The present sequence is {S(5,n)/S(1,n)}.
%H A361888 Seiichi Manyama, <a href="/A361888/b361888.txt">Table of n, a(n) for n = 0..840</a>
%H A361888 H. W. Gould, <a href="http://www.jstor.org/stable/2976965">Problem E2384</a>, Amer. Math. Monthly, 81 (1974), 170-171.
%F A361888 a(n) = 1/binomial(n,floor(n/2)) * Sum_{k = 0..floor(n/2)} ( (n - 2*k + 1)/(n - k + 1) * binomial(n,k) )^5.
%F A361888 a(n) ~ 2^(4*n + 9) / (125 * Pi^2 * n^4). - _Vaclav Kotesovec_, Mar 24 2025
%p A361888 seq(add( ( binomial(n,k) - binomial(n,k-1) )^5/binomial(n,floor(n/2)), k = 0..floor(n/2)), n = 0..20);
%t A361888 Table[Sum[(Binomial[n, k]-Binomial[n, k-1])^5/Binomial[n, Floor[n/2]], {k, 0, Floor[n/2]}], {n, 0, 20}] (* _Vaclav Kotesovec_, Mar 24 2025 *)
%o A361888 (PARI) s(r, n) = sum(k=0, n\2, (binomial(n, k)-binomial(n, k-1))^r);
%o A361888 a(n) = s(5, n)/s(1, n); \\ _Seiichi Manyama_, Mar 24 2025
%Y A361888 Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A183069 ( S(3,2*n-1)/ S(1,2*n-1) ), A361887 ( S(5,n) ), A361889 ( S(5,2*n-1)/S(1,2*n-1) ), A361890 ( S(7,n) ), A361891 ( S(7,n)/S(1,n) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).
%K A361888 nonn,easy
%O A361888 0,4
%A A361888 _Peter Bala_, Mar 29 2023