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A229452 G.f.: exp( Sum_{n>=1} (3*n)!/(3!*n!^3) * x^n/n ). 2
1, 1, 8, 101, 1569, 27445, 518407, 10333243, 214320244, 4583132411, 100411556533, 2243625355010, 50955869372055, 1173262656151429, 27332509319090516, 643208905017756216, 15270427859720369204, 365356267775348553277, 8801688936499808334602 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Self-convolution 6th power yields A229451.

From Peter Bala, Feb 16 2020: (Start)

The sequence b(n) = [x^n] A(x)^n for n >= 1 begins [1, 17, 352, 7969, 189876, 4676768, 117905565, 3024222753, 78607893934, 2064924478892, 54710782664836, ...]. We conjecture that b(n) satisfies the supercongruences b(n*p^k) == b(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and all positive integers n and k.

More generally, for a positive integer m, set A_m(x) = exp( Sum_{n >= 1} (m*n)!/(m!*n!^m) * x^n/n ) and define a sequence b_m(n) := [x^n] A_m(x)^n for n >=1. Then we conjecture that b_m(n) is an integer sequence satisfying the same supercongruences. (End)

LINKS

Table of n, a(n) for n=0..18.

EXAMPLE

G.f.: A(x) = 1 + x + 8*x^2 + 101*x^3 + 1569*x^4 + 27445*x^5 +...

where

log(A(x)) = x + 15*x^2/2 + 280*x^3/3 + 5775*x^4/4 + 126126*x^5/5 + 2858856*x^6/6 +...+ A060542(n)*x^n/n +...

MATHEMATICA

CoefficientList[Series[Exp[Sum[(3*k)!/(3!*k!^3)*x^k/k, {k, 1, 20}]], {x, 0, 20}], x] (* Vaclav Kotesovec, Mar 05 2020 *)

PROG

(PARI) {a(n)=polcoeff(exp(sum(k=1, n, (3*k)!/(3!*k!^3)*x^k/k) +x*O(x^n)), n)}

for(n=0, 25, print1(a(n), ", "))

CROSSREFS

Cf. A229451, A060542, A006480 (De Bruijn's S(3,n)).

Sequence in context: A317598 A238947 A291536 * A199816 A302870 A317862

Adjacent sequences:  A229449 A229450 A229451 * A229453 A229454 A229455

KEYWORD

nonn

AUTHOR

Paul D. Hanna, Sep 23 2013

STATUS

approved

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Last modified August 4 16:18 EDT 2021. Contains 346447 sequences. (Running on oeis4.)