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 A100232 Triangle, read by rows, of the coefficients of [x^k] in G100231(x)^n such that the row sums are 5^n-1 for n>0, where G100231(x) is the g.f. of A100231. 3
 1, 1, 3, 1, 6, 17, 1, 9, 39, 75, 1, 12, 70, 220, 321, 1, 15, 110, 470, 1165, 1363, 1, 18, 159, 852, 2895, 5922, 5777, 1, 21, 217, 1393, 5943, 16807, 29267, 24475, 1, 24, 284, 2120, 10822, 38536, 93468, 141688, 103681, 1, 27, 360, 3060, 18126, 77274, 236748 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS The main diagonal forms A100233. Secondary diagonal is: T(n+1,n) = (n+1)*A033887(n) = (n+1)*Fibonacci(3*n+1). More generally, if g.f. F(x) satisfies: m^n-b^n = Sum_{k=0..n} [x^k]F(x)^n, then F(x) also satisfies: (m+z)^n - (b+z)^n + z^n = Sum_{k=0..n} [x^k](F(x)+z*x)^n for all z and F(x)=(1+(m-1)*x+sqrt(1+2*(m-2*b-1)*x+(m^2-2*m+4*b+1)*x^2))/2; the triangle formed from powers of F(x) will have the g.f.: G(x,y)=(1-2*x*y+m*x^2*y^2)/((1-x*y)*(1-(m-1)*x*y-x^2*y^2-x*(1-x*y))). LINKS Table of n, a(n) for n=0..51. FORMULA G.f.: A(x, y)=(1-2*x*y+5*x^2*y^2)/((1-x*y)*(1-4*x*y-x^2*y^2-x*(1-x*y))). EXAMPLE Rows begin: , [1,3], [1,6,17], [1,9,39,75], [1,12,70,220,321], [1,15,110,470,1165,1363], [1,18,159,852,2895,5922,5777], [1,21,217,1393,5943,16807,29267,24475], [1,24,284,2120,10822,38536,93468,141688,103681],... where row sums form 5^n-1 for n>0: 5^1-1 = 1+3 = 4 5^2-1 = 1+6+17 = 24 5^3-1 = 1+9+39+75 = 124 5^4-1 = 1+12+70+220+321 = 624 5^5-1 = 1+15+110+470+1165+1363 = 3124. The main diagonal forms A100233 = [1,3,17,75,321,1363,5777,...], where Sum_{n>=1} A100233(n)/n*x^n = log((1-x)/(1-4*x-x^2)). PROG (PARI) T(n, k, m=5)=if(n

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Last modified November 28 22:38 EST 2023. Contains 367422 sequences. (Running on oeis4.)