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 A092565 Triangle of coefficients T(n,k) (n>=0, 0<=k<=2n), read by rows, where the n-th row polynomial equals the numerator of the n-th convergent of the continued fraction [1+x+x^2;1+x+x^2,1+x+x^2,...] for n>0, with the zeroth row defined as T(0,0)=1. 2
 1, 1, 1, 1, 2, 2, 3, 2, 1, 3, 5, 8, 7, 6, 3, 1, 5, 10, 19, 22, 22, 16, 10, 4, 1, 8, 20, 42, 58, 69, 63, 49, 30, 15, 5, 1, 13, 38, 89, 142, 191, 206, 191, 146, 95, 50, 21, 6, 1, 21, 71, 182, 327, 491, 602, 637, 573, 447, 296, 167, 77, 28, 7, 1, 34, 130, 363, 722, 1191, 1626 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,5 COMMENTS T(n,k) is the number of lattice paths from (0,0) to (n,k) using steps (1,0),(2,0),(1,1),(1,2). [Joerg Arndt, Jul 01 2011] Diagonal forms A092566, row sums form A006190. Column T(n,0) forms Fibonacci numbers A000045, T(n,1) forms A001629. LINKS Alois P. Heinz, Rows n = 0..100, flattened FORMULA n-th row polynomial R(n) = sum_{k=0..n} A037027(n, k)*x^k*(1+x)^k; R(n+1)/R(n) = [1+x+x^2;1+x+x^2, ...(n+1)times..., 1+x+x^2] for n>=0; R(0)=1. EXAMPLE Ratio of row polynomials R(3)/R(2) = (3+5*x+8*x^2+7*x^3+6*x^4+3*x^5+x^6)/(2+2*x+3*x^2+2*x^3+x^4) = [1+x+x^2;1+x+x^2,1+x+x^2]. Rows begin: 1; 1, 1, 1; 2, 2, 3, 2, 1; 3, 5, 8, 7, 6, 3, 1; 5, 10, 19, 22, 22, 16, 10, 4, 1; 8, 20, 42, 58, 69, 63, 49, 30, 15, 5, 1; 13, 38, 89, 142, 191, 206, 191, 146, 95, 50, 21, 6, 1; 21, 71, 182, 327, 491, 602, 637, 573, 447, 296, 167, 77, 28, 7, 1; 34, 130, 363, 722, 1191, 1626, 1921, 1958, 1752, 1366, 931, 546, 273, 112, 36, 8, 1; ... MAPLE T:= proc(x, y) option remember; `if`(y<0 or y>2*x, 0, `if`(x=0, 1,       add(T(x-l[1], y-l[2]), l=[[1, 0], [2, 0], [1, 1], [1, 2]])))     end: seq(seq(T(n, k), k=0..2*n), n=0..10); # Alois P. Heinz, Apr 16 2013 MATHEMATICA A037027[n_, k_] := Sum[Binomial[k+j, k]*Binomial[j, n-j-k], {j, 0, n-k}]; A037027[n_, 0] = Fibonacci[n + 1]; row[n_] := CoefficientList[ Sum[A037027[n, k] x^k (1+x)^k, {k, 0, n}], x]; Flatten[Table[row[n], {n, 0, 8}]][[1 ;; 70]] (* From Jean-François Alcover, Jul 18 2011 *) PROG (PARI) T(n, k)=if(2*n

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