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%I #26 Aug 31 2021 01:09:41
%S 1,1,1,1,2,1,0,3,3,1,-2,2,6,4,1,-4,-3,7,10,5,1,-3,-12,0,16,15,6,1,5,
%T -18,-21,11,30,21,7,1,20,-4,-50,-24,35,50,28,8,1,29,48,-51,-98,-9,78,
%U 77,36,9,1,1,124,45,-164,-150,42,147,112,45,10,1,-94,128,282,-67,-365,-177,154,250,156,55,11,1
%N Triangle read by rows T(n, k) such that Fib(n, x+1) = Sum_{k=1..n} T(n, k)*Fib(k, x) where Fib(n, x) is the n-th Fibonacci polynomial.
%H Michel Marcus, <a href="/A346038/b346038.txt">Rows n=1..100 of triangle, flattened</a>
%H Math StackExchange, <a href="https://math.stackexchange.com/questions/2731420/variable-transformation-x-to-x1-in-fibonacci-polynomial-f-nx">Variable transformation x->x+1 in Fibonacci-polynomial Fn(x)</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/FibonacciPolynomial.html">Fibonacci Polynomial</a>.
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 2, 1;
%e 0, 3, 3, 1;
%e -2, 2, 6, 4, 1;
%e -4, -3, 7, 10, 5, 1;
%e ...
%e The first 3 Fibonacci polynomials are 1, x, x^2 + 1. So F3(n, x+1) = x^2 + 2*x + 2 = 1*1 + 2*x + 1*(x^2+1) = 1*F(1,x) + 2*F(2, x) + 1*F(3,x), so the 3rd row is [1, 2, 1].
%o (PARI) rowV(n) = my(v= if (n==0, [0], n--; vector(n+1, k, k--; if (k%2==0, binomial(n-k/2, k/2))))); Pol(v); \\ A162515
%o rowT(n, vfp, vfp1) = {my(vp1 = vfp1[n], vc = vector(n), i=n); forstep (k = poldegree(vp1), 0, -1, vc[i] = polcoef(vp1, k)/polcoef(vfp[k+1], k); vp1 -= vfp[k+1]*vc[i]; i--;); vc;}
%o tabl(nn) = {my(vfp = vector(nn, k, rowV(k))); my(vfp1 = vector(nn, k, subst(vfp[k], x, x+1))); for(n=1, nn, print((rowT(n, vfp, vfp1))););}
%Y Cf. A000012, A000027, A000217, A005581: diagonals.
%Y Cf. A162515 and A168561 (Fibonacci polynomials coefficients).
%K sign,tabl
%O 1,5
%A _Michel Marcus_, Jul 02 2021
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