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A393622
O.g.f. A(x) satisfies: [x^n] exp( n*(n+1) * x*A(x)^2 ) / A(x)^(n+1) = 0 for n > 0.
9
1, 1, 6, 70, 1224, 28597, 836316, 29352144, 1200797216, 56066001324, 2940845713742, 171199095229462, 10954722005150672, 764494066937495046, 57809413765164115896, 4710689021921976956708, 411690053536098606833376, 38428328136985584340246892, 3817043873159181783485193592
OFFSET
0,3
LINKS
FORMULA
a(n) ~ c * 2^(2*n) * n^(n - 3/2) / (exp(n) * w^n * (2-w)^n), where w = -LambertW(-2*exp(-2)) = -A226775 and c = 2.806979111236... - Vaclav Kotesovec, Feb 25 2026
EXAMPLE
O.g.f.: A(x) = 1 + x + 6*x^2 + 70*x^3 + 1224*x^4 + 28597*x^5 + 836316*x^6 + 29352144*x^7 + 1200797216*x^8 + ...
ILLUSTRATION OF DEFINITION.
The table of coefficients of x^k/k! in exp(n*(n+1)*x*A(x)^2) / A(x)^(n+1) for n >= 0 begins
n = 0: [1, -1, -10, -354, -25560, -3072480, ...];
n = 1: [1, 0, -14, -616, -46920, -5776768, ...];
n = 2: [1, 3, 0, -744, -68760, -8753184, ...];
n = 3: [1, 8, 68, 0, -79608, -12198528, ...];
n = 4: [1, 15, 250, 4130, 0, -13919320, ...];
n = 5: [1, 24, 630, 17736, 474480, 0, ...];
in which the main diagonal is all zeros after the initial term.
The above table is derived from the following series
n = 1: exp(1*2*x*A(x)^2)/A(x)^2 = 1 + 0*x - 14*x^2/2! - 616*x^3/3! - 46920*x^4/4! - 5776768*x^5/5! + ...
n = 2: exp(2*3*x*A(x)^2)/A(x)^3 = 1 + 3*x + 0*x^2/2! - 744*x^3/3! - 68760*x^4/4! - 8753184*x^5/5! + ...
n = 3: exp(3*4*x*A(x)^2)/A(x)^4 = 1 + 8*x + 68*x^2/2! + 0*x^3/3! - 79608*x^4/4! - 12198528*x^5/5! + ...
n = 4: exp(4*5*x*A(x)^2)/A(x)^5 = 1 + 15*x + 250*x^2/2! + 4130*x^3/3! + 0*x^4/4! - 13919320*x^5/5! + ...
n = 5: exp(5*6*x*A(x)^2)/A(x)^6 = 1 + 24*x + 630*x^2/2! + 17736*x^3/3! + 474480*x^4/4! + 0*x^5/5! + ...
...
illustrating that [x^n] exp(n*(n+1)*x*A(x)^2) / A(x)^(n+1) = 0 for n > 0.
PROG
(PARI) {a(n) = my(A=[1], m); for(i=1, n, A=concat(A, 0); m=#A; A[m] = Vec( exp((m-1)*m*x*Ser(A)^2 +x*O(x^m)) / Ser(A)^m )[m]/m ); A[n+1]}
for(n=0, 25, print1( a(n), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Feb 23 2026
STATUS
approved