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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

A128966 Triangle read by rows of coefficients of polynomials P[n](x) defined by P[0]=0, P[1]=x+1; for n >= 2, P[n]=(x+1)*P[n-1]+x*P[n-2].

Original entry on oeis.org

0, 1, 1, 1, 2, 1, 1, 4, 4, 1, 1, 6, 10, 6, 1, 1, 8, 20, 20, 8, 1, 1, 10, 34, 50, 34, 10, 1, 1, 12, 52, 104, 104, 52, 12, 1, 1, 14, 74, 190, 258, 190, 74, 14, 1, 1, 16, 100, 316, 552, 552, 316, 100, 16, 1, 1, 18, 130, 490, 1058, 1362, 1058, 490, 130, 18, 1, 1, 20, 164
Offset: 0

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Author

N. J. A. Sloane, May 10 2007

Keywords

Comments

A variant of A008288 (they satisfy the same recurrence).

Examples

			Triangle begins:
0
1, 1
1, 2, 1
1, 4, 4, 1
1, 6, 10, 6, 1
1, 8, 20, 20, 8, 1
1, 10, 34, 50, 34, 10, 1
1, 12, 52, 104, 104, 52, 12, 1
1, 14, 74, 190, 258, 190, 74, 14, 1
1, 16, 100, 316, 552, 552, 316, 100, 16, 1
		

Crossrefs

Cf. A163271 (row sums), A110170 (central terms).
Cf. A102413.

Programs

  • Haskell
    a128966 n k = a128966_tabl !! n !! k
    a128966_row n = a128966_tabl !! n
    a128966_tabl = map fst $ iterate
       (\(us, vs) -> (vs, zipWith (+) ([0] ++ us ++ [0]) $
                          zipWith (+) ([0] ++ vs) (vs ++ [0]))) ([0], [1, 1])
    -- Reinhard Zumkeller, Jul 20 2013
  • Maple
    P[0]:=0;
    P[1]:=x+1;
    for n from 2 to 14 do
    P[n]:=expand((x+1)*P[n-1]+x*P[n-2]);
    lprint(P[n]);
    lprint(seriestolist(series(P[n],x,200)));
    od:
  • Mathematica
    t[n_, k_] := 2^(1-n)*Binomial[n, k]*Sum[Binomial[n, 2*m+1]*HypergeometricPFQ[{-k, -m, k-n}, {1/2-n/2, -n/2}, -1], {m, 0, (n-1)/2}]; Table[t[n, k], {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jan 09 2014, after Max Alekseyev *)
  • PARI
    { T(n,k) = sum(m=0,(n-1)\2, binomial(n,2*m+1) * sum(j=0,m, binomial(m,j) * binomial(n-2*j,k-j) * 2^(2*j+1-n) ) ) } \\ Max Alekseyev, Mar 10 2008
    

Formula

P[n](x) = (x+1) * ( ((x+1+sqrt(x^2+6x+1))/2)^n - ((x+1-sqrt(x^2+6x+1))/2)^n ) / sqrt(x^2+6x+1) - Max Alekseyev, Mar 10 2008
P[n](x) = (x+1) * (sqrt(x)*I)^(n-1) * U[n-1](-I*(x+1)/sqrt(x)/2), where U[n](t) is Chebyshev polynomial of the 2nd kind. - Max Alekseyev, Mar 10 2008