A361889 - OEIS

%I #13 Mar 30 2023 13:03:07

%S 1,11,415,30955,3173626,386672861,52846226091,7857161332715,

%T 1246162831674580,207990691516965886,36176886727828945286,

%U 6510211391453319830461,1205449991704260042021490,228686327051301858363357905,44299708036441260810228742915,8738765548899621077157770551275

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

%C Odd bisection of A361888.

%C Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for positive integers n and r and all primes p >= 5.

%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(2*n-1,n-1) * Sum_{k = 0..n-1} ( (2*n - 2*k)/(2*n  - k) * binomial(2*n-1,k) )^5 for n >= 1.

%e Examples of supercongruences:

%e a(13) - a(1) = 1205449991704260042021490 - 1 = 3*(13^3)*182893338143568508879 == 0 (mod 13^3).

%e a(2*5) - a(2) = 207990691516965886 - 11 = (5^3)*7*237703647447961 == 0 (mod 5^3)

%p seq(add( ( binomial(2*n-1,k) - binomial(2*n-1,k-1) )^5/binomial(2*n-1,n-1), k = 0..n-1), n = 1..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) ), A361890 ( S(7,n) ), A361891 ( S(7,n)/S(1,n) ), A361892 ( S(7,2*n-1)/S(1,2*n-1) ).

%K nonn,easy

%O 1,2

%A _Peter Bala_, Mar 29 2023