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 A114525 Triangle of coefficients of the Lucas (w-)polynomials. 12
 2, 0, 1, 2, 0, 1, 0, 3, 0, 1, 2, 0, 4, 0, 1, 0, 5, 0, 5, 0, 1, 2, 0, 9, 0, 6, 0, 1, 0, 7, 0, 14, 0, 7, 0, 1, 2, 0, 16, 0, 20, 0, 8, 0, 1, 0, 9, 0, 30, 0, 27, 0, 9, 0, 1, 2, 0, 25, 0, 50, 0, 35, 0, 10, 0, 1, 0, 11, 0, 55, 0, 77, 0, 44, 0, 11, 0, 1, 2, 0, 36, 0, 105, 0, 112, 0, 54, 0, 12, 0, 1, 0 (list; table; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS Unsigned version of A108045. The row reversed triangle is A162514. - Paolo Bonzini, Jun 23 2016 LINKS G. C. Greubel, Rows n = 0..100 of triangle, flattened B. Johnson, Fibonacci numbers and matrices, 2009. Jong Hyun Kim, Hadamard products and tilings, JIS 12 (2009) 09.7.4. M. Pétréolle, Characterization of Cyclically Fully commutative elements in finite and affine Coxeter Groups, arXiv preprint arXiv:1403.1130 [math.GR], 2014. Eric Weisstein's World of Mathematics, Fibonacci Polynomial Eric Weisstein's World of Mathematics, Lucas Polynomial FORMULA From Peter Bala, Mar 18 2015 The Lucas polynomials L(n,x) satisfy the recurrence L(n+1,x) = x*L(n,x) + L(n-1,x) with L(0,x) = 2 and L(1,x) = x. O.g.f.: Sum_{n >= 0} L(n,x)*t^n = (2 - x*t)/(1 - t^2 - x*t) = 2 + x*t +(x^2 + 2)*t^2 + (3*x + x^3)*t^3 + .... L(n,x) = trace( [ x, 1; 1, 0 ]^n ). exp( Sum_{n >= 1} L(n,x)*t^n/n ) = Sum_{n >= 0} F(n+1,x)*t^n, where F(n,x) denotes the n-th Fibonacci polynomial. (see Appendix A3 in Johnson). exp( Sum_{n >= 1} L(n,x)*L(2*n,x)*t^n/n ) = 1/( F(1,x)*F(2*x)*F(3,x) ) * Sum_{n >= 0} F(n+1,x)*F(n+2,x)*F(n+3,x)*t^n. exp( Sum_{n >= 1} L(3*n,x)/L(n,x)*t^n/n ) = Sum_{n >= 0} L(2*n + 1,x)*t^n. L(n,1) = Lucas(n) = A000032(n); L(n,4) = Lucas(3*n) = A014448(n); L(n,11) = Lucas(5*n) = A001946(n); L(n,29) = Lucas(7*n) = A087281(n); L(n,76) = Lucas(9*n) = A087287(n); L(n,199) = Lucas(11*n) = A089772(n). The general result is L(n,Lucas(2*k + 1)) = Lucas((2*k + 1)*n). (End) From Jeremy Dover, Jun 10 2016: (Start) Read as a triangle T(n,k), n >= 0, n >= k >= 0, T(n,k) = (Binomial((n+k)/2,k) + Binomial((n+k-2)/2,k))*(1+(-1)^(n-k))/2. T(n,k) = A046854(n-1,k-1) + A046854(n-1,k) + A046854(n-2,k) for even n+k with n+k > 0, assuming A046854(n,k) = 0 for n < 0, k < 0, k > n. T(n,k) is the number of binary strings of length n with exactly k pairs of consecutive 0's and no pair of consecutive 1's, where the first and last bits are considered consecutive. (End) EXAMPLE 2, x, 2 + x^2, 3*x + x^3, 2 + 4*x^2 + x^4, 5*x + 5*x^3 + x^5, ... give   2;   0, 1;   2, 0, 1;   0, 3, 0, 1;   2, 0, 4, 0, 1;   0, 5, 0, 5, 0, 1;   2, 0, 9, 0, 6, 0, 1; MAPLE Lucas := proc(n, x)     option remember;     if  n=0 then         2;     elif n =1 then         x ;     else         x*procname(n-1, x)+procname(n-2, x) ;     end if; end proc: A114525 := proc(n, k)     coeftayl(Lucas(n, x), x=0, k) ; end proc: seq(seq(A114525(n, k), k=0..n), n=0..12) ; # R. J. Mathar, Aug 16 2019 MATHEMATICA row[n_] := CoefficientList[LucasL[n, x], x]; Table[row[n], {n, 0, 12}] // Flatten (* Jean-François Alcover, Aug 11 2018 *) CROSSREFS Cf. A108045 (signed version). Cf. A034807, A162514. Cf. Sequences L(n,x): A000032(x = 1), A002203 (x = 2), A006497 (x = 3), A014448 (x = 4), A087130 (x = 5), A085447 (x = 6), A086902 (x = 7), A086594 (x = 8), A087798 (x = 9), A086927 (x = 10), A001946 (x = 11), A086928 (x = 12), A088316 (x = 13), A090300 (x = 14), A090301 (x = 15), A090305 (x = 16), A090306 (x = 17), A090307 (x = 18), A090308 (x = 19), A090309 (x = 20), A090310 (x = 21), A090313 (x = 22), A090314 (x = 23), A090316 (x = 24), A087281 (x = 29), A087287 (x = 76), A089772 (x = 199). Sequence in context: A178524 A321731 A212357 * A127672 A294168 A299196 Adjacent sequences:  A114522 A114523 A114524 * A114526 A114527 A114528 KEYWORD nonn,tabl,easy AUTHOR Eric W. Weisstein, Dec 06 2005 STATUS approved

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Last modified October 18 20:28 EDT 2019. Contains 328197 sequences. (Running on oeis4.)