login
A362732
a(n) = [x^n] E(x)^n, where E(x) = exp( Sum_{k >= 1} A006480(k)*x^k/k ).
3
1, 6, 162, 5082, 170274, 5920506, 210808494, 7631158674, 279617726754, 10341283241130, 385275082939662, 14439312879759378, 543815325940475694, 20565700004741265900, 780470358196543271622, 29708379800729905316832, 1133811403010621704628514, 43371319655978568356324868
OFFSET
0,2
COMMENTS
The De Bruijn numbers A006480(n) = (3*n)!/n!^3 satisfy the supercongruences A006480(n*p^r) == A006480(n*p^(r-1)) (mod p^(3*r)) for all primes p >= 5 and positive integers n and r (write a(n) as binomial(3*n,2*n)*binomial(2*n,n) and apply Mestrovic, equation 39, p. 12).
We inductively define a family of sequences {a(i,n) : n >= 0}, i >= 0, by setting a(0,n) = A006480(n) and, for i >= 1, a(i,n) = [x^n] ( exp(Sum_{k >= 1} a(i-1,k)*x^k/k) )^n. The present sequence is {a(1,n)}. See A362733 for {a(2,n)}.
We conjecture that the sequences {a(i,n) : n >= 0}, i >= 1, satisfy the same supercongruences as above.
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.
MAPLE
E(n, x) := series(exp(n*add(((3*k)!/k!^3*x^k)/k, k = 1..20)), x, 21):
seq(coeftayl(E(n, x), x = 0, n), n = 0..20);
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Bala, May 06 2023
STATUS
approved