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A361889 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.

%I A361889 #25 Mar 25 2025 15:14:07
%S A361889 1,11,415,30955,3173626,386672861,52846226091,7857161332715,
%T A361889 1246162831674580,207990691516965886,36176886727828945286,
%U A361889 6510211391453319830461,1205449991704260042021490,228686327051301858363357905,44299708036441260810228742915,8738765548899621077157770551275
%N A361889 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 A361889 Odd bisection of A361888.
%C A361889 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 A361889 Seiichi Manyama, <a href="/A361889/b361889.txt">Table of n, a(n) for n = 1..420</a>
%H A361889 H. W. Gould, <a href="http://www.jstor.org/stable/2976965">Problem E2384</a>, Amer. Math. Monthly, 81 (1974), 170-171.
%F A361889 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.
%F A361889 a(n) ~ 2^(8*n + 1) / (125 * Pi^2 * n^4). - _Vaclav Kotesovec_, Mar 24 2025
%e A361889 Examples of supercongruences:
%e A361889 a(13) - a(1) = 1205449991704260042021490 - 1 = 3*(13^3)*182893338143568508879 == 0 (mod 13^3).
%e A361889 a(2*5) - a(2) = 207990691516965886 - 11 = (5^3)*7*237703647447961 == 0 (mod 5^3)
%p A361889 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);
%t A361889 Table[Sum[(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}] (* _Vaclav Kotesovec_, Mar 24 2025 *)
%o A361889 (Python)
%o A361889 from math import comb
%o A361889 def A361889(n): return sum((comb((n<<1)-1,j)*(m:=n-j<<1)//(m+j))**5 for j in range(n))//comb((n<<1)-1,n-1) # _Chai Wah Wu_, Mar 25 2025
%Y A361889 Cf. A003161 ( S(3,n) ), A003162 ( S(3,n)/S(1,n) ), A382394 ( S(3,2*n-1) ), 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 A361889 nonn,easy
%O A361889 1,2
%A A361889 _Peter Bala_, Mar 29 2023