A114525 - OEIS

%I #95 Feb 17 2023 22:33:16

%S 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,

%T 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,

%U 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

%N Triangle of coefficients of the Lucas (w-)polynomials.

%C Unsigned version of A108045.

%C The row reversed triangle is A162514. - _Paolo Bonzini_, Jun 23 2016

%H G. C. Greubel, <a href="/A114525/b114525.txt">Rows n = 0..100 of triangle, flattened</a>

%H B. Johnson, <a href="https://web.archive.org/web/20201130211206/http://maths.dur.ac.uk/~dma0rcj/PED/fib.pdf">Fibonacci numbers and matrices</a>, 2009.

%H Jong Hyun Kim, <a href="https://cs.uwaterloo.ca/journals/JIS/VOL12/Kim/kim18.html">Hadamard products and tilings</a>, JIS 12 (2009) 09.7.4.

%H M. Pétréolle, <a href="http://arxiv.org/abs/1403.1130">Characterization of Cyclically Fully commutative elements in finite and affine Coxeter Groups</a>, arXiv preprint arXiv:1403.1130 [math.GR], 2014.

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FibonacciPolynomial.html">Fibonacci Polynomial</a>

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/LucasPolynomial.html">Lucas Polynomial</a>

%F From _Peter Bala_, Mar 18 2015

%F 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.

%F 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 + ....

%F L(n,x) = trace( [ x, 1; 1, 0 ]^n ).

%F 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).

%F 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.

%F exp( Sum_{n >= 1} L(3*n,x)/L(n,x)*t^n/n ) = Sum_{n >= 0} L(2*n + 1,x)*t^n.

%F 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)

%F From _Jeremy Dover_, Jun 10 2016: (Start)

%F 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.

%F 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.

%F 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)

%F From _Peter Bala_, Sep 03 2022: (Start)

%F L(n,x) = 2*(i)^n*T(n,-i*x/2), where i = sqrt(-1) and T(n,x) is the n-th Chebyshev polynomial of the first kind.

%F d/dx(L(n,x)) = n*F(n,x), where F(n,x) denotes the n-th Fibonacci polynomial.

%F Let P_n(x,y) = (L(n,x) - L(n,y))/(x - y). Then {P_n(x,y): n >= 1} is a fourth-order linear divisibility sequence of polynomials in the ring Z[x,y]: if m divides n in Z then P_m(x,y) divides P_n(x,y) in Z[x,y].

%F P_n(1,1) = A045925(n); P_n(1,4) = A273622; P_n(2,2) = A093967(n).

%F L(2*n,x)^2 - L(2*n-1,x)*L(2*n+1,x) = x^2 + 4 for n >= 1.

%F Sum_{n >= 1} L(2*n,x)/( L(2*n-1,x) * L(2*n+1,x) ) = 1/x^2 and

%F Sum_{n >= 1} (-1)^(n+1)/( L(2*n,x) + x^2/L(2*n,x) ) = 1/(x^2 + 4), both valid for all nonzero real x. (End)

%F From _Peter Bala_, Nov 18 2022: (Start)

%F L(n,x) = Sum_{k = 0..floor(n/2)} (n/(n-k))*binomial(n-k,k)*x^(n-2*k) for n >= 1.

%F For odd m, L(n, L(m,x)) = L(n*m, x).

%F For integral x, the sequence {u(n)} := {L(n,x)} satisfies the Gauss congruences: u(m*p^r) == u(m*p^(r-1)) (mod p^r) for all positive integers m and r and all primes p.

%F Let p be an odd prime and let 0 <= k <= p - 1. Let alpha_k = the p-adic limit_{n -> oo} L(p^n,k). Then alpha_k is a well-defined p-adic-integer and the polynomial L(p,x) - x of degree p factorizes as L(p,x) - x = Product_{k = 0..p-1} (x - alpha_k). For example, L(5,x) - x = x^5 + 5*x^3 + 4*x = x*(x - A269591)*(x - A210850)*(x - A210851)*(x - A269592) in the ring of 5-adic integers. (End)

%F The formula for L(n,x) given in the first line of the preceding section, with L(0, x) = 2, is rewritten L(n, x) = Sum_{k = 0..floor(n/2)} A034807(n, k)*x^(n - 2*k). See the formula by Alexander Elkins in A034807. - _Wolfdieter Lang_, Feb 10 2023

%e 2, x, 2 + x^2, 3*x + x^3, 2 + 4*x^2 + x^4, 5*x + 5*x^3 + x^5, ... give triangle

%e n\k   0  1  2  3  4  5  6  7  8  9 10 ...

%e 0:    2

%e 1:    0  1

%e 2:    2  0  1

%e 3:    0  3  0  1

%e 4:    2  0  4  0  1

%e 5:    0  5  0  5  0  1

%e 6:    2  0  9  0  6  0  1

%e 7:    0  7  0 14  0  7  0  1

%e 8:    2  0 16  0 20  0  8  0  1

%e 9:    0  9  0 30  0 27  0  9  0  1

%e 10:   2  0 25  0 50  0 35  0 10  0  1

%e n\k   0  1  2  3  4  5  6  7  8  9 10 ...

%e .... reformatted by _Wolfdieter Lang_, Feb 10 2023

%p Lucas := proc(n,x)

%p     option remember;

%p     if  n=0 then

%p         2;

%p     elif n =1 then

%p         x ;

%p     else

%p         x*procname(n-1,x)+procname(n-2,x) ;

%p     end if;

%p end proc:

%p A114525 := proc(n,k)

%p     coeftayl(Lucas(n,x),x=0,k) ;

%p end proc:

%p seq(seq(A114525(n,k),k=0..n),n=0..12) ; # _R. J. Mathar_, Aug 16 2019

%t row[n_] := CoefficientList[LucasL[n, x], x];

%t Table[row[n], {n, 0, 12}] // Flatten (* _Jean-François Alcover_, Aug 11 2018 *)

%Y Cf. A108045 (signed version).

%Y Cf. A034807, A162514.

%Y 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).

%K nonn,tabl,easy

%O 0,1

%A _Eric W. Weisstein_, Dec 06 2005