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A357673 a(n) = 4*Sum_{k = 0..2*n} binomial(n+k-1,k) + 3*Sum_{k = 0..2*n} binomial(n+k-1,k)^2.

%I A357673 #19 Oct 29 2022 15:40:24
%S A357673 7,21,225,5124,162657,5812521,219004812,8516056500,338508840801,
%T A357673 13679415485805,559978704877725,23162632151271480,966309241173439500,
%U A357673 40602415885424806824,1716435895297948558812,72941388509291664563124,3113826813351114598588257,133458673478315967012049245
%N A357673 a(n) = 4*Sum_{k = 0..2*n} binomial(n+k-1,k) + 3*Sum_{k = 0..2*n} binomial(n+k-1,k)^2.
%C A357673 Conjectures:
%C A357673 1) a(p) == a(1) (mod p^5) for all primes p >= 3 (checked up to p = 499).
%C A357673 2) For r >= 2, and all primes p >= 3, a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ).
%C A357673 3) More generally, let m be a positive integer and set u(n) = 2*m*Sum_{k = 0..m*n} binomial(n+k-1,k) + (m + 1)*Sum_{k = 0..m*n} binomial(n+k-1,k)^2. Then the sequence {u(n)} satisfies the supercongruence u(p) == u(1) (mod p^5) for all primes p >= 7. This is the case m = 2. See A357673 for the case m = 1.
%C A357673 4) For r >= 2, and all primes p >= 5, u(p^r) == u(p^(r-1)) ( mod p^(3*r+3) ).
%F A357673 a(n) = 4*A005809(n) + 3*Sum_{k = 0..2*n} binomial(n+k-1,k)^2.
%e A357673 Examples of supercongruences:
%e A357673 a(17) - a(1) = 133458673478315967012049245 - 21 = (2^3)*3*7*(17^5)*61*109*4441*86491*219071 == 0 (mod 17^5).
%e A357673 a(25) - a(5) = 1681058690656849873108154414589433546896 - 5812521 = 3*(5^9)*17*124471*39410141*65963867*52155532801 == 0 (mod 5^9).
%p A357673 seq(add( 4*binomial(n+k-1,k) + 3*binomial(n+k-1,k)^2, k = 0..2*n ), n = 0..20);
%t A357673 Table[4 Sum[Binomial[n+k-1,k],{k,0,2n}]+3*Sum[Binomial[n+k-1,k]^2,{k,0,2n}],{n,0,20}] (* _Harvey P. Dale_, Oct 29 2022 *)
%o A357673 (PARI) a(n) = 4*sum(k = 0, 2*n, binomial(n+k-1,k)) + 3*sum(k = 0, 2*n, binomial(n+k-1,k)^2); \\ _Michel Marcus_, Oct 24 2022
%Y A357673 Cf. A005809, A357509, A357565, A357566, A357671, A357672, A357674.
%K A357673 nonn,easy
%O A357673 0,1
%A A357673 _Peter Bala_, Oct 11 2022