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 A100235 Triangle, read by rows, of the coefficients of [x^k] in G100234(x)^n such that the row sums are 6^n-1 for n>0, where G100234(x) is the g.f. of A100234. 3
 1, 1, 4, 1, 8, 26, 1, 12, 63, 139, 1, 16, 116, 436, 726, 1, 20, 185, 965, 2830, 3774, 1, 24, 270, 1790, 7335, 17634, 19601, 1, 28, 371, 2975, 15505, 52444, 106827, 101784, 1, 32, 488, 4584, 28860, 124424, 358748, 633952, 528526, 1, 36, 621, 6681, 49176, 256194 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,3 COMMENTS The main diagonal forms A100236. Secondary diagonal is: T(n+1,n) = (n+1)*A100237(n). 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..50. FORMULA G.f.: A(x, y)=(1-2*x*y+6*x^2*y^2)/((1-x*y)*(1-5*x*y-x^2*y^2-x*(1-x*y))). EXAMPLE Rows begin: , [1,4], [1,8,26], [1,12,63,139], [1,16,116,436,726], [1,20,185,965,2830,3774], [1,24,270,1790,7335,17634,19601], [1,28,371,2975,15505,52444,106827,101784], [1,32,488,4584,28860,124424,358748,633952,528526],... where row sums form 6^n-1 for n>0: 6^1-1 = 1+4 = 5 6^2-1 = 1+8+26 = 35 6^3-1 = 1+12+63+139 = 215 6^4-1 = 1+16+116+436+726 = 1295 6^5-1 = 1+20+185+965+2830+3774 = 7775. The main diagonal forms A100236 = [1,4,26,139,726,3774,...], where Sum_{n>=1} A100236(n)/n*x^n = log((1-x)/(1-5*x-x^2)). MATHEMATICA row[n_] := CoefficientList[ Series[ (1 + 5*x + Sqrt[1 + 6*x + 29*x^2])^n/2^n, {x, 0, n}], x]; Flatten[ Table[ row[n], {n, 0, 9}]](* Jean-François Alcover, May 11 2012, after PARI *) PROG (PARI) T(n, k, m=6)=if(n

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Last modified September 27 10:09 EDT 2023. Contains 365688 sequences. (Running on oeis4.)