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 A251573 E.g.f.: exp(3*x*G(x)^2) / G(x)^2 where G(x) = 1 + x*G(x)^3 is the g.f. of A001764. 12
 1, 1, 3, 21, 261, 4833, 120303, 3778029, 143531433, 6404711553, 328447585179, 19037277446949, 1230842669484717, 87829738967634849, 6856701559496841159, 581343578623728854397, 53196439113856500195537, 5225543459274294130169601, 548468830470032135590262067, 61258398893626609968686844597 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 LINKS Paul D. Hanna, Table of n, a(n) for n = 0..200 FORMULA Let G(x) = 1 + x*G(x)^3 be the g.f. of A001764, then the e.g.f. A(x) of this sequence satisfies: (1) A'(x)/A(x) = G(x)^2. (2) A'(x) = exp(3*x*G(x)^2). (3) A(x) = exp( Integral G(x)^2 dx ). (4) A(x) = exp( Sum_{n>=1} A006013(n-1)*x^n/n ), where A006013(n-1) = binomial(3*n-2,n)/(2*n-1). (5) A(x) = F(x/A(x)) where F(x) is the e.g.f. of A251583. (6) A(x) = Sum_{n>=0} A251583(n)*(x/A(x))^n/n! where A251583(n) = 3^(n-1) * (n+1)^(n-3) * (n+3). (7) [x^n/n!] A(x)^(n+1) = 3^(n-1) * (n+1)^(n-2) * (n+3). a(n) = Sum_{k=0..n} 3^k * n!/k! * binomial(3*n-k-3, n-k) * (k-1)/(n-1) for n>1. Recurrence (for n>3): 2*(n-3)*(2*n-3)*a(n) = 3*(3*n-8)*(3*n^2 - 13*n + 15)*a(n-1) - 27*(n-2)*a(n-2). - Vaclav Kotesovec, Dec 07 2014 a(n) ~ 3^(3*n-7/2) * n^(n-2) / (2^(2*n-5/2) * exp(n-1)). - Vaclav Kotesovec, Dec 07 2014 EXAMPLE E.g.f.: A(x) = 1 + x + 3*x^2/2! + 21*x^3/3! + 261*x^4/4! + 4833*x^5/5! +... such that A(x) = exp(3*x*G(x)^2) / G(x)^2 where G(x) = 1 + x*G(x)^3 is the g.f. of A001764: G(x) = 1 + x + 3*x^2 + 12*x^3 + 55*x^4 + 273*x^5 + 1428*x^6 +... The e.g.f. satisfies: A(x) = 1 + x/A(x) + 5*x^2/(2!*A(x)^2) + 54*x^3/(3!*A(x)^3) + 945*x^4/(4!*A(x)^4) + 23328*x^5/(5!*A(x)^5) + 750141*x^6/(6!*A(x)^6) + 29859840*x^7/(7!*A(x)^7) +...+ 3^(n-1)*(n+1)^(n-3)*(n+3) * x^n/(n!*A(x)^n) +... Note that A'(x) = exp(3*x*G(x)^2) = 1 + 3*x + 21*x^2/2! + 261*x^3/3! + 4833*x^4/4! +... LOGARITHMIC DERIVATIVE. The logarithm of the e.g.f. begins: log(A(x)) = x + 2*x^2/2 + 7*x^3/3 + 30*x^4/4 + 143*x^5/5 +... and so A'(x)/A(x) = G(x)^2. TABLE OF POWERS OF E.G.F. Form a table of coefficients of x^k/k! in A(x)^n as follows. n=1: [1, 1,  3,   21,   261,   4833,  120303,   3778029, ...]; n=2: [1, 2,  8,   60,   744,  13536,  330912,  10232928, ...]; n=3: [1, 3, 15,  123,  1557,  28179,  680427,  20771235, ...]; n=4: [1, 4, 24,  216,  2832,  51552, 1237248,  37404288, ...]; n=5: [1, 5, 35,  345,  4725,  87285, 2094975,  62949825, ...]; n=6: [1, 6, 48,  516,  7416, 139968, 3378528, 101278944, ...]; n=7: [1, 7, 63,  735, 11109, 215271, 5250987, 157613463, ...]; n=8: [1, 8, 80, 1008, 16032, 320064, 7921152, 238878720, ...]; ... in which the main diagonal begins (see A251583): [1, 2, 15, 216, 4725, 139968, 5250987, 238878720, ...] and is given by the formula: [x^n/n!] A(x)^(n+1) = 3^(n-1) * (n+1)^(n-2) * (n+3) for n>=0. MATHEMATICA Flatten[{1, 1, Table[Sum[3^k * n!/k! * Binomial[3*n-k-3, n-k] * (k-1)/(n-1), {k, 0, n}], {n, 2, 20}]}] (* Vaclav Kotesovec, Dec 07 2014 *) Flatten[{1, 1, RecurrenceTable[{27*(n-2)*a[n-2]-3*(3*n-8)*(15-13*n+3*n^2)*a[n-1]+2*(n-3)*(2*n-3)*a[n]==0, a[2]==3, a[3]==21}, a, {n, 20}]}] (* Vaclav Kotesovec, Dec 07 2014 *) PROG (PARI) {a(n) = local(G=1); for(i=1, n, G=1+x*G^3 +x*O(x^n)); n!*polcoeff(exp(3*x*G^2)/G^2, n)} for(n=0, 20, print1(a(n), ", ")) (PARI) {a(n) = if(n==0|n==1, 1, sum(k=0, n, 3^k * n!/k! * binomial(3*n-k-3, n-k) * (k-1)/(n-1) ))} for(n=0, 20, print1(a(n), ", ")) CROSSREFS Cf. A251583, A251663, A001764, A006013. Cf. Variants: A243953, A251574, A251575, A251576, A251577, A251578, A251579, A251580. Sequence in context: A317059 A262939 A232470 * A265002 A012131 A322224 Adjacent sequences:  A251570 A251571 A251572 * A251574 A251575 A251576 KEYWORD nonn AUTHOR Paul D. Hanna, Dec 06 2014 STATUS approved

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Last modified November 29 21:32 EST 2021. Contains 349416 sequences. (Running on oeis4.)