

A102426


Triangle read by rows giving coefficients of polynomials defined by F(0)=0, F(1)=1, F(n+1) = F(n) + x*F(n1).


10



0, 1, 1, 1, 1, 2, 1, 1, 3, 1, 3, 4, 1, 1, 6, 5, 1, 4, 10, 6, 1, 1, 10, 15, 7, 1, 5, 20, 21, 8, 1, 1, 15, 35, 28, 9, 1, 6, 35, 56, 36, 10, 1, 1, 21, 70, 84, 45, 11, 1, 7, 56, 126, 120, 55, 12, 1, 1, 28, 126, 210, 165, 66, 13, 1, 8, 84, 252, 330, 220, 78, 14, 1, 1, 36, 210, 462, 495, 286, 91
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OFFSET

0,6


COMMENTS

Essentially the same as A098925: a(0)=0 followed by A098925. [R. J. Mathar, Aug 30 2008]
F(n) + 2x * F(n1) gives Lucas polynomials (cf. A034807).  Maxim Krikun (krikun(AT)iecn.unancy.fr), Jun 24 2007
After the initial 0, these are the nonzero coefficients of the Fibonacci polynomials; see the Mathematica section.  Clark Kimberling, Oct 10 2013


REFERENCES

Dominique Foata and GuoNiu Han, Multivariable tangent and secant qderivative polynomials, Manuscript, Mar 21 2012


LINKS

Table of n, a(n) for n=0..79.
G. Hetyei, Hurwitzian continued fractions containing a repeated constant and an arithmetic progression, arXiv preprint arXiv:1211.2494, 2012.  From N. J. A. Sloane, Jan 02 2013


FORMULA

Alternatively, as n is even or odd: T(n2, k) + T(n1, k1) = T(n, k), T(n2, k) + T(n1, k) = T(n, k)
T(n, k)=binomial(floor(n/2)+k, floor((n1)/2k) )  Paul Barry, Jun 22 2005


EXAMPLE

The first few polynomials are:
0
1
1
x + 1
2x + 1
x^2 + 3x + 1
3x^2 + 4x + 1


MATHEMATICA

Table[Fibonacci[n, x], {n, 0, 10}] (* Clark Kimberling, Oct 10 2013 *)


CROSSREFS

Upward diagonals sums are A062200. Downward rows are A102427. Row sums are A000045. Row terms reversed = A011973. Also A102427, A102428, A102429.
All of A011973, A092865, A098925, A102426, A169803 describe essentially the same triangle in different ways.
Sequence in context: A035667 A092865 A098925 * A052920 A089141 A228267
Adjacent sequences: A102423 A102424 A102425 * A102427 A102428 A102429


KEYWORD

easy,nonn,tabf


AUTHOR

Russell Walsmith, Jan 08 2005


STATUS

approved



