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%I #7 Aug 19 2024 15:11:35
%S 1,1,8,100,1556,27260,515510,10284094,213433728,4566363088,
%T 100082133066,2236952393302,50817223209451,1170319824912699,
%U 27268900054818390,641812268110993694,15239341125950643462,364655982858022960206,8785745372509009963892,212976842702489760621536
%N G.f.: exp( Sum_{n>=1} A132303(n)/3 * x^n/n ), where A132303(n) = sum of the cubes of the trinomial coefficients in row n of triangle A027907.
%C Self-convolution cube yields A168593.
%e G.f.: A(x) = 1 + x + 8*x^2 + 100*x^3 + 1556*x^4 + 27260*x^5 +...
%e where
%e log(A(x)) = 1*x + 15*x^2/2 + 277*x^3/3 + 5727*x^4/4 + 125301*x^5/5 + 2843643*x^6/6 + 66214485*x^7/7 + 1571497119*x^8/8 +...+ A132303(n)/3*x^n/n +...
%o (PARI) {A027907(n,k) = polcoeff((1+x+x^2)^n, k)}
%o {A132303(n) = sum(k=0, 2*n, A027907(n,k)^3)}
%o {a(n) = if(n==0, 1, polcoeff(exp(sum(m=1, n, A132303(m)/3 * x^m/m) +x*O(x^n)), n))}
%o for(n=0,30,print1(a(n),", "))
%Y Cf. A168593, A132303, A027907.
%K nonn
%O 0,3
%A _Paul D. Hanna_, Feb 28 2015