0,1

These polynomials arise in the following setup. Suppose G and H are power series satisfying G+H=G*H=1/x. Then G^n+H^n = (1/x^n)*L_n(-x).

Apart from signs, triangle of coefficients when 2cos(nt) is expanded in terms of x=2cos(t). For example, 2cos(2t)=x^2-2, 2cos(3t)=x^3-3x and 2cos(4t)=x^4-4x^2+2. - Anthony Robin (anthony_robin(AT)hotmail.com), Jun 02 2004

Triangle of coefficients of expansion of Z_{nk} in terms of Z_k.

Row n has 1+floor(n/2) terms. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Dec 25 2004

T(n,k)=number of k-matchings of the cycle C_n (n>1). Example: T(6,2)=9 because the 2-matchings of the hexagon with edges a,b,c,d,e,f are ac, ad, ae, bd, be, bf, ce, cf and df. - Emeric Deutsch (deutsch(AT)duke.poly.edu), Dec 25 2004

An example for the first comment: G=c(x), H=1/(x*c(x)) with c(x) the o.g.f. Catalan numbers A000108: (x*c(x))^n + (1/c(x))^n = L(n,-x)= sum(T(n,k)*(-x)^k,k=0..floor(n/2)).

This triangle also supplies the absolute values of the coefficients in the multiplication formulae for the Lucas numbers A000032.

A. Brousseau, Fibonacci and Related Number Theoretic Tables. Fibonacci Association, San Jose, CA, 1972, p. 148.

C. D. Godsil, Algebraic Combinatorics, Chapman and Hall, New York, 1993.

T. J. Osler, Cardan polynomials and the reduction of radicals, Math. Mag., 74 (No. 1, 2001), 26-32.

Thomas Koshy, Fibonacci and Lucas Numbers with Applications. New York, etc.: John Wiley & Sons, 2001. (Chapter 13, "Pascal-like Triangles," is devoted to the present triangle.)

T. D. Noe, Rows n=0..100 of triangle, flattened

Moussa Benoumhani, A Sequence of Binomial Coefficients Related to Lucas and Fibonacci Numbers, J. Integer Seqs., Vol. 6, 2003.

Eric Weisstein's World of Mathematics, Lucas Polynomial

Lucas polynomial coefficients: 1, -n, [n(n-3)]/2!, - [n(n-4)(n-5)]/3!, [n(n-5)(n-6)(n-7)]/4!, - [n(n-6)(n-7)(n-8)(n-9)]/5!... - Herb Conn, HCR 83, Box 93, Custer, SD 57730 and Gary W. Adamson (qntmpkt(AT)yahoo.com), May 28 2003

G.f.: (2-x)/(1-x-x^2*y). - Vladeta Jovovic (vladeta(AT)eunet.rs), May 31 2003

T(n, k) = T(n-1, k)+T(n-2, k-1), n>1. T(n, 0) = 1, n>0. T(n, k) = binomial(n-k, k)+binomial(n-k-1, k-1) = n*binomial(n-k-1, k-1)/k, 0< = 2*k< = n except T(0, 0) = 2.

T(n,k)=(n*(n-1-k)!)/(k!*(n-2*k)!),n>0,k>=0. - Alexander Elkins (alexander_elkins(AT)hotmail.com), Jun 09 2007

I have seen two versions of these polynomials: One version begins L_0 = 2, L_1 = 1, L_2 = 1+2*x, L_3 = 1+3*x, L_4 = 1+4*x+2*x^2, L_5 = 1+5*x+5*x^2, L_6 = 1+6*x+9*x^2+2*x^3, L_7 = 1+7*x+14*x^2+7*x^3, L_8 = 1+8*x+20*x^2+16*x^3+2*x^4, L_9 = 1+9*x+27*x^2+30*x^3+9*x^4, ...

The other version (probably the more official one) begins L_0(x) = 2, L_1(x) = x, L_2(x) = 2+x^2, L_3(x) = 3*x+x^3, L_4(x) = 2+4*x^2+x^4, tc

L5 = x^5 - 5x^3 + 5x = 1, -5, 5 = 1, -n, [n(n-3)]/2.

Comment from John Blythe Dobson, Oct 11 2007: Triangle begins:

2;

1;

1, 2;

1, 3;

1, 4, 2;

1, 5, 5;

1, 6, 9, 2;

1, 7, 14, 7;

1, 8, 20, 16, 2;

1, 9, 27, 30, 9;

1, 10, 35, 50, 25, 2;

1, 11, 44, 77, 55, 11;

1, 12, 54, 112, 105, 36, 2;

1, 13, 65, 156, 182, 91, 13;

1, 14, 77, 210, 294, 196, 49, 2;

1, 15, 90, 275, 450, 378, 140, 15;

T:=proc(n, k) if n=0 and k=0 then 2 elif k>floor(n/2) then 0 else n*binomial(n-k, k)/(n-k) fi end: for n from 0 to 15 do seq(T(n, k), k=0..floor(n/2)) od; # yields sequence in triangular form (Deutsch)

(PARI) T(n, k)=if(k<0|2*k>n, 0, binomial(n-k, k)+binomial(n-k-1, k-1)+(n==0&k==0))

Row sums = A000032 (Lucas numbers). T(2n, n-1)=A000290(n), T(2n+1, n-1)=A000330(n), T(2n, n-2)=A002415(n). T(n, k)=A029635(n-k, k), if n>0. See also A061896.

Cf. A114525

Sequence in context: A055893 A050221 A113279 this_sequence A135062 A088428 A025838

Adjacent sequences: A034804 A034805 A034806 this_sequence A034808 A034809 A034810

tabf,easy,nonn

N. J. A. Sloane (njas(AT)research.att.com).

Improved description, more terms, etc., from Michael Somos