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a(0) = 1 and a(n) = Sum_{k = 0..2*n} n/(n + 2*k)*binomial(n + 2*k,k) for n >= 1.
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%I #19 Oct 23 2022 11:55:32

%S 1,4,64,1429,35072,898129,23571781,628750217,16965558016,461752375705,

%T 12652302369439,348552604899778,9644571491252069,267852878928912034,

%U 7462156684641697991,208446714456132946429,5836259481820028112640,163741162073796817779389,4602160147618819467316159

%N a(0) = 1 and a(n) = Sum_{k = 0..2*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, A005809(n) = binomial(3*n,n) = Sum_{k = 0..2*n} n/(n + k)*binomial(n + k,k).

%C The binomial coefficients A005809(n) are known to satisfy the supercongruences A005809(n*p^r) == A005809(n*p^(r-1)) (mod p^(3*r)) for primes p >= 5 and positive integers n and r (see Meštrović, equation 39). 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 = 2. See A333093 (case m = 1) and A352276 case (m = 3).

%H Paolo Xausa, <a href="/A352275/b352275.txt">Table of n, a(n) for n = 0..674</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) ~ 5^(5*n + 3/2) / (19 * sqrt(Pi*n) * 2^(2*n + 1) * 3^(3*n + 1/2)). - _Vaclav Kotesovec_, Mar 15 2022

%e Examples of supercongruences:

%e a(3*5) - a(3) = 208446714456132946429 - 1429 = (2^3)*3*(5^4)*13*41* 26072134391011 == 0 (mod 5^4)

%e a(17) - a(1) = 163741162073796817779389 - 4 = 5*(17^3)*1506943* 4423278397003 == 0 (mod 17^3)

%p seq(add(n/(n + 2*k)*binomial(n + 2*k,k), k = 0..2*n), n = 1..25);

%t nterms=25;Join[{1},Table[Sum[n/(n+2k)Binomial[n+2k,k],{k,0,2n}],{n,nterms-1}]] (* _Paolo Xausa_, Apr 11 2022 *)

%o (PARI) a(n) = if (n==0, 1, sum(k=0, 2*n, binomial(n + 2*k,k)*n/(n+2*k))); \\ _Michel Marcus_, Mar 17 2022

%Y Cf. A005809, A333093, A352276.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Mar 10 2022