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A352276 a(0) = 1 and a(n) = Sum_{k = 0..3*n} n/(n + 2*k)*binomial(n + 2*k,k) for n >= 1. 2

%I #15 Oct 23 2022 11:55:14

%S 1,9,625,58785,5986993,633580634,68611922731,7545931449401,

%T 839183314181297,94112350842056469,10623982584664109750,

%U 1205644823097085684641,137414820511792364274091,15718880489100999321142976,1803621273322664188151352631,207499462144488863314062180035

%N a(0) = 1 and a(n) = Sum_{k = 0..3*n} n/(n + 2*k)*binomial(n + 2*k,k) for n >= 1.

%C The following identity can be easily verified using Maple's SumTools:-Summation procedure: for n >= 1, A005810(n) = binomial(4*n,n) = Sum_{k = 0..3*n} n/(n + k)*binomial(n + k,k).

%C The binomial coefficients A005810(n) are known to satisfy the supercongruences A005810(n*p^r) == A005810(n*p^(r-1)) (mod p^(3*r)) for primes p >= 5 and positive integers n and r (see Meštrović, equation 39).

%C Calculation suggests that the present sequence satisfies the same congruences.

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

%C More generally, for m a positive integer, define a sequence u_m by setting u_m(n) = Sum_{k = 0..m*n} n/(n + 2*k)*binomial(n + 2*k,k) for n >= 1.

%C Then we conjecture that each sequence u_m satisfies the above supercongruences. This is the case m = 3. See A333093 (case m = 1) and A352275 (m = 2).

%H Paolo Xausa, <a href="/A352276/b352276.txt">Table of n, a(n) for n = 0..475</a>

%H R. Meštrović, <a href="https://arxiv.org/abs/1111.3057">Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862--2012)</a>, arXiv:1111.3057 [math.NT], 2011.

%F a(n) ~ 7^(7*n + 3/2) / (37 * sqrt(Pi*n) * 2^(8*n + 3/2) * 3^(3*n + 1/2)). - _Vaclav Kotesovec_, Mar 15 2022

%e Examples of supercongruences:

%e a(11) - a(1) = 1205644823097085684641 - 9 = (2^3)*3*(11^3)*43*2887*5059* 60096637 == 0 (mod 11^3)

%e a(3*5) - a(3) = 207499462144488863314062180035 - 58785 = 2*(5^4)*1801* 4959701*18583938263214197 == 0 (mod 5^4)

%t nterms=25;Join[{1},Table[Sum[n/(n+2k)Binomial[n+2k,k],{k,0,3n}],{n,nterms-1}]] (* _Paolo Xausa_, Apr 10 2022 *)

%Y Cf. A005810, A333093, A352275.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Mar 10 2022

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