OFFSET
1,2
COMMENTS
Conjecture: a(n) is divisible by prime(n) for n > 1.
Conjecture: all terms are odd.
LINKS
Paul D. Hanna, Table of n, a(n) for n = 1..500
EXAMPLE
L.g.f.: L(x) = x + 3*x^2/2 + 25*x^3/3 + 287*x^4/4 + 7601*x^5/5 + 116337*x^6/6 + 3740017*x^7/7 + 80014415*x^8/8 + 2949026857*x^9/9 + 133787267943*x^10/10 + ...
where exponentiation yields F(x), the g.f. of A393866:
exp(L(x)) = F(x) = 1 + x + 2*x^2 + 10*x^3 + 82*x^4 + 1610*x^5 + 21103*x^6 + 557487*x^7 + 10593462*x^8 + ... + A393866(n)*x^n + ...
which satisfies [x^n] F(x)^prime(n) * (1 - prime(n)*x) = 0 for n >= 1.
Dividing a(n) by prime(n) for n > 1 yields
a(n)/prime(n) = [1, 5, 41, 691, 8949, 220001, 4211285, 128218559, 4613354067, 135486553647, 6540086267969, 271221354269665, ...].
PROG
(PARI) \\ Set N to desired number of terms
N=30; F=[1]; for(n=1, N, F=concat(F, 0); V=Vec(Ser(F)^prime(n)); F[#F] = V[#F-1] - V[#F]/prime(n) ); A = Vec(x*Ser(F)'/Ser(F))
\\ Test: a(n) == 0 (mod prime(n)) for n > 1 - returns '1' if true
for(n=2, N, print1( denominator( A[n]/prime(n) ), ", "))
CROSSREFS
KEYWORD
nonn
AUTHOR
Paul D. Hanna, Mar 29 2026
STATUS
approved
