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A034807
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Triangle T(n,k) of coefficients of Lucas (or Cardan) polynomials.
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27
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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, 1, 16, 104
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,1
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COMMENTS
| 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.
Contribution by L. Edson Jeffery, March 19, 2011. (Start):
This sequence is related to rhombus substitution tilings. A signed version of it (see A132460), formed as a triangle with interlaced zeros extending each row to n terms, begins as
{2}
{1,0}
{1,0,-2}
{1,0,-3,0}
{1,0,-4,0,2}
(1,0,-5,0,5,0}
....
For the n X n tridiagonal unit-primitive matrix G_(n,1) (n>=2) (see the L. E. Jeffery link below), defined by
G_(n,1)=
(0 1 0 ... 0)
(1 0 1 0 ... 0)
(0 1 0 1 0 ... 0)
...
(0 ... 0 1 0 1)
(0 ... 0 2 0),
Row n (i.e. {T(n,k)}, k=0..n) of the signed table gives the coefficients of its characteristic function: c_n(x)=Sum{k=0..n, T(n,k)*x^(n-k)}=0. For example, let n=3. Then
G_(3,1)=
(0 1 0)
(1 0 1)
(0 2 0),
and row 3 of the table is {1,0,-3,0}. Hence c_3(x)=x^3-3*x=0. G_(n,1) has n distinct eigenvalues (the solutions of c_n(x)=0), given by w_j=2*cos((2*j-1)*Pi/(2*n)), j=1,2,...,n. (End)
For n>0, T(n,k) is the number of k-subsets of {1,2,...,n} which contain neither consecutive integers nor both 1 and n. Equivalently, T(n,k) is the number of k-subsets without neighbors of a set of n points on a circle. - José H. Nieto S. Jan 17 2012
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REFERENCES
| 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.
EMRAH KILIC AND ELIF TAN KILIC, SOME SUBSEQUENCES OF THE GENERALIZED FIBONACCI AND LUCAS SEQUENCES, Preprint, 2011.
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.)
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LINKS
| 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.
L. E. Jeffery, Unit-primitive matrices
Eric Weisstein's World of Mathematics, Lucas Polynomial
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FORMULA
| 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
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EXAMPLE
| 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;
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MAPLE
| 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)
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PROG
| (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))
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CROSSREFS
| 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 * A182961 A135062 A088428
Adjacent sequences: A034804 A034805 A034806 * A034808 A034809 A034810
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KEYWORD
| tabf,easy,nonn,changed
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AUTHOR
| N. J. A. Sloane (njas(AT)research.att.com).
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EXTENSIONS
| Improved description, more terms, etc., from Michael Somos
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