A361891 - OEIS

%I #10 Mar 30 2023 13:02:46

%S 1,1,1,43,386,9451,246961,6031627,212559508,6571985126,243940325734,

%T 9140730357409,352312505157354,14801600281919487,600054439936968241,

%U 26927918031565051915,1149140935414286560040,53804800109969394477580,2401141625752684697505820

%N a(n) = S(7,n)/S(1,n), where S(r,n) = Sum_{k = 0..floor(n/2)} ( binomial(n,k) - binomial(n,k-1) )^r.

%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) 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(7,n)/S(1,n)}.

%H H. W. Gould, <a href="http://www.jstor.org/stable/2976965">Problem E2384</a>, Amer. Math. Monthly, 81 (1974), 170-171.

%F a(n) = 1/binomial(n,floor(n/2)) * Sum_{k = 0..floor(n/2)} ( (n - 2*k + 1)/(n - k + 1) * binomial(n,k) )^7.

%p seq(add( ( binomial(n,k) - binomial(n,k-1) )^7/binomial(n,floor(n/2)), k = 0..floor(n/2)), n = 0..20);

%Y 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) ), A361888( S(5,n)/S(1,n) ), A361889 ( S(5,2*n-1)/S(1,2*n-1) ), A361890 ( S(7,n) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).

%K nonn,easy

%O 0,4

%A _Peter Bala_, Mar 30 2023