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A136689
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Triangular sequence of q-Fibonacci polynomials for s=3: F(x,n) = x*F(x,n-1) + s*F(x,n-2).
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2
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1, 0, 1, 3, 0, 1, 0, 6, 0, 1, 9, 0, 9, 0, 1, 0, 27, 0, 12, 0, 1, 27, 0, 54, 0, 15, 0, 1, 0, 108, 0, 90, 0, 18, 0, 1, 81, 0, 270, 0, 135, 0, 21, 0, 1, 0, 405, 0, 540, 0, 189, 0, 24, 0, 1, 243, 0, 1215, 0, 945, 0, 252, 0, 27, 0, 1, 0, 1458, 0, 2835, 0, 1512, 0
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OFFSET
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1,4
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COMMENTS
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Row sums: 1, 1, 4, 7, 19, 40, 97, 217, 508, 1159, 2683, ... = A006130(n-1).
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LINKS
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FORMULA
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F(x,n) = x*F(x,n-1) + s*F(x,n-2), where F(x,0)=0, F(x,1)=1 and s=3.
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EXAMPLE
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Triangle begins:
1;
0, 1;
3, 0, 1;
0, 6, 0, 1;
9, 0, 9, 0, 1;
0, 27, 0, 12, 0, 1;
27, 0, 54, 0, 15, 0, 1;
0, 108, 0, 90, 0, 18, 0, 1;
81, 0, 270, 0, 135, 0, 21, 0, 1;
0, 405, 0, 540, 0, 189, 0, 24, 0, 1;
243, 0, 1215, 0, 945, 0, 252, 0, 27, 0, 1;
...
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MAPLE
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A136689 := proc(n) option remember: if(n<=1)then return n: else return x*procname(n-1)+3*procname(n-2): fi: end:
seq(seq(coeff(A136689(n), x, m), m=0..n-1), n=1..10); # Nathaniel Johnston, Apr 27 2011
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MATHEMATICA
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s=2; F[x_, n_]:= F[x, n]= If[n<2, n, x*F[x, n-1] + s*F[x, n-2]]; Table[
CoefficientList[F[x, n], x], {n, 10}]//Flatten
F[n_, x_, s_, q_]:= Sum[QBinomial[n-j-1, j, q]*q^Binomial[j+1, 2]*x^(n-2*j-1) *s^j, {j, 0, Floor[(n-1)/2]}]; Table[CoefficientList[F[n, x, 3, 1], x], {n, 1, 10}]//Flatten (* G. C. Greubel, Dec 16 2019 *)
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PROG
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(Sage)
def f(n, x, s, q): return sum( q_binomial(n-j-1, j, q)*q^binomial(j+1, 2)*x^(n-2*j-1)*s^j for j in (0..floor((n-1)/2)))
P.<x> = PowerSeriesRing(ZZ, prec)
return P( f(n, x, 3, 1) ).list()
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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