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A357959 a(n) = 5*A005259(n-1) + 2*A005258(n).

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%I A357959 #9 Nov 06 2022 07:50:25
%S A357959 11,63,659,9727,187511,4304943,109312739,2941124607,82033399631,
%T A357959 2345394917563,68306797052879,2018580243252847,60368874298729631,
%U A357959 1823588997226603663,55558079041172790659,1705174802761490321407,52672634815976274443711,1636296942340074307669443
%N A357959 a(n) = 5*A005259(n-1) + 2*A005258(n).
%C A357959 Conjectures:
%C A357959 1) a(p) == a(1) (mod p^5) for all primes p >= 5 (checked up to p = 271).
%C A357959 2) a(p^r) == a(p^(r-1)) ( mod p^(3*r+3) ) for r >= 2, and for all primes p >= 5.
%C A357959 These are stronger supercongruences than those satisfied separately by the two types of Apéry numbers A005258 and A005259. Cf. A357959.
%C A357959 There is also a product version of these conjectures:
%C A357959 3) the sequence {u(n): n >= 1} defined by u(n) = A005259(n-1)^5 * A005258(n)^6 also satisfies the congruences in 1) and 2) above. See A357960.
%F A357959 a(n) = 5*Sum_{k = 0..n-1} binomial(n-1,k)^2*binomial(n+k-1,k)^2 + 2*Sum_{k = 0..n} binomial(n,k)^2*binomial(n+k,k).
%F A357959 a(n*p^r) == a(n*p^(r-1)) ( mod p^(3*r) ) for positive integers n and r and for all primes p >= 5.
%e A357959 Examples of supercongruences:
%e A357959 a(13) - a(1) = 60368874298729631 - 11 = (2^2)*3*5*(13^5)*131*20685869 == 0 (mod 13^5).
%e A357959 a(5^2) - a(5) = 51292638914356604042099497031437511 - 187511 = (2^4)*3*(5^10)* 37*72974432287*40526706713533 == 0 (mod 5^10).
%p A357959 seq( add( 5*binomial(n-1,k)^2*binomial(n+k-1,k)^2 + 2*binomial(n,k)^2* binomial(n+k,k), k = 0..n ), n = 1..20);
%Y A357959 Cf. A005258, A005259, A212334, A352655, A357567, A357956, A356957, A357958, A357960.
%K A357959 nonn,easy
%O A357959 1,1
%A A357959 _Peter Bala_, Oct 25 2022