A362733 a(n) = [x^n] F(x)^n, where F(x) = exp( Sum_{k >= 1} A362732(k)*x^k/k ).
1, 6, 234, 10428, 492522, 24033006, 1197423396, 60530725380, 3092592004074, 159295600885794, 8258018380659234, 430335300869496072, 22521831447746893092, 1182951246247357578348, 62325193477833011143260, 3292376206935392483917428, 174323297281680647978503146, 9248680725006429075147528150
Offset: 0
Programs
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Maple
E(n,x) := series(exp(n*add((3*k)!/k!^3*x^k/k, k = 1..20)), x, 21): A362732(n) := coeftayl(E(n,x), x = 0, n): F(n,x) := series(exp(n*add(A362732(k)*x^k/k, k = 1..20)), x, 21): seq(coeftayl(F(n,x), x = 0, n), n = 0..20); # alternative program G(n,x) := series(exp(n*add((3*k)!/k!^3*x^(2*k)/k, k = 1..40)), x, 41): seq((1/2)*coeftayl(G(2*n,x), x = 0, 2*n), n = 1..20); # Peter Bala, Oct 27 2024
Formula
a(n*p^r) == a(n*p^(r-1)) (mod p^r) (Gauss congruence) holds for all primes p and positive integers n and r.
Conjecture: the supercongruence a(n*p^r) == a(n*p^(r-1)) (mod p^(3*r)) holds for all primes p >= 3 and positive integers n and r.
For n >= 1, a(n) = (1/2) * [x^n] G(x)^(2*n), where G(x) = exp(Sum_{k >= 1} A006480(k)*x^k/k) is the g.f. of A229451. - Peter Bala, Oct 27 2024
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