A362732 a(n) = [x^n] E(x)^n, where E(x) = exp( Sum_{k >= 1} A006480(k)*x^k/k ).

1, 6, 162, 5082, 170274, 5920506, 210808494, 7631158674, 279617726754, 10341283241130, 385275082939662, 14439312879759378, 543815325940475694, 20565700004741265900, 780470358196543271622, 29708379800729905316832, 1133811403010621704628514, 43371319655978568356324868

Offset: 0

Author

Peter Bala, May 06 2023

Keywords

• nonn
• easy

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.

For i >= 1, a(i, n) = 1/i *[x^n] E(x)^(i*n), where E(x) = exp(Sum_{k >= 1} (3*k)!/k!^3 * x^k/k) is the g.f. of A229451. - Peter Bala, Oct 30 2024

Links

• Vaclav Kotesovec, Table of n, a(n) for n = 0..620
• Romeo Meštrović, Wolstenholme's theorem: Its Generalizations and Extensions in the last hundred and fifty years (1862-2011), arXiv:1111.3057 [math.NT], 2011.

Crossrefs

Cf. A006480, A229451, A362722 - A362733.

Programs

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);

Mathematica

nmax = 20; Table[SeriesCoefficient[E^(n*Sum[(3*k)!/k!^3*x^k/k, {k, 1, n}]), {x, 0, n}], {n, 0, nmax}] (* Vaclav Kotesovec, Nov 26 2024 *)

Python

from sympy import symbols, factorial, exp, series
x = symbols('x')
nmax = 10
result = []
for n in range(0, nmax + 1):
    summation = sum(factorial(3 * k) / (factorial(k)**3 * k) * x**k for k in range(1, n + 1))
    exp_series = exp(n * summation)
    coefficient = exp_series.series(x, 0, n + 1).coeff(x, n)
    result.append(coefficient)
print(result) # Robert C. Lyons, Jan 27 2025

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.

a(n) ~ c * d^n / sqrt(n), where d = 39.5963012687026... and c = 0.107979249748...