A375441 Expansion of g.f. A(x) satisfying 0 = Sum_{k=0..n} (-1)^k * binomial(3*n-2*k, k) * ([x^k] A(x)^n) for n >= 1.
1, 1, 3, 27, 520, 17461, 924123, 72565316, 8092491188, 1237738519836, 252223179026317, 66826143258640668, 22547253821858556366, 9516722710620123685768, 4947980149610424524104600, 3126554347854062023615490499, 2372761989077914432028426533800, 2140079932579758851404630992687571
Offset: 0
Keywords
Examples
G.f.: A(x) = 1 + x + 3*x^2 + 27*x^3 + 520*x^4 + 17461*x^5 + 924123*x^6 + 72565316*x^7 + 8092491188*x^8 + ... RELATED TABLES. The table of coefficients of x^k in A(x)^n begins: n=1: [1, 1, 3, 27, 520, 17461, 924123, ...]; n=2: [1, 2, 7, 60, 1103, 36124, 1887017, ...]; n=3: [1, 3, 12, 100, 1758, 56097, 2890755, ...]; n=4: [1, 4, 18, 148, 2495, 77500, 3937572, ...]; n=5: [1, 5, 25, 205, 3325, 100466, 5029880, ...]; n=6: [1, 6, 33, 272, 4260, 125142, 6170284, ...]; ... from which we may illustrate the defining property given by 0 = Sum_{k=0..n} (-1)^k * binomial(3*n-2*k, k) * ([x^k] A(x)^n). Using the coefficients in the table above, we see that n=1: 0 = 1*1 - 1*1; n=2: 0 = 1*1 - 4*2 + 1*7; n=3: 0 = 1*1 - 7*3 + 10*12 - 1*100; n=4: 0 = 1*1 - 10*4 + 28*18 - 20*148 + 1*2495; n=5: 0 = 1*1 - 13*5 + 55*25 - 84*205 + 35*3325 - 1*100466; n=6: 0 = 1*1 - 16*6 + 91*33 - 220*272 + 210*4260 - 56*125142 + 1*6170284; ... The triangle A193636(n,k) = binomial(3*n-2*k, k) begins: n=0: 1; n=1: 1, 1; n=2: 1, 4, 1; n=3: 1, 7, 10, 1; n=4: 1, 10, 28, 20, 1; n=5: 1, 13, 55, 84, 35, 1; n=6: 1, 16, 91, 220, 210, 56, 1; ...
Links
- Paul D. Hanna, Table of n, a(n) for n = 0..200
Programs
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PARI
{a(n) = my(A=[1],m); for(i=1, n, A=concat(A, 0); m=#A-1; A[#A] = sum(k=0, m, (-1)^(m-k+1) * binomial(3*m-2*k, k) * polcoef(Ser(A)^m, k) )/m ); A[n+1]} for(n=0, 20, print1(a(n), ", "))
Formula
a(n) ~ c * d^n * n!^3 * n^alpha, where d = 0.1579852929267375678916376580224..., alpha = 2.6601429516008505168108..., c = 0.86048778713891683578001... - Vaclav Kotesovec, Sep 12 2024
Comments