A346038 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.

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, -18, -21, 11, 30, 21, 7, 1, 20, -4, -50, -24, 35, 50, 28, 8, 1, 29, 48, -51, -98, -9, 78, 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

Offset

1,5

Links

Michel Marcus, Rows n=1..100 of triangle, flattened

Math StackExchange, Variable transformation x->x+1 in Fibonacci-polynomial Fn(x)

Eric Weisstein's World of Mathematics, Fibonacci Polynomial.

Example

Triangle begins:
   1;
   1,  1;
   1,  2, 1;
   0,  3, 3,  1;
  -2,  2, 6,  4, 1;
  -4, -3, 7, 10, 5, 1;
  ...
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].

Prog

(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
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; }
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))); ); }

Crossrefs

Cf. A000012, A000027, A000217, A005581: diagonals.

Cf. A162515 and A168561 (Fibonacci polynomials coefficients).

Sequence in context: A079123 A121548 A180179 * A215062 A215063 A316781

Adjacent sequences: A346035 A346036 A346037 * A346039 A346040 A346041

Keyword

sign,tabl

Author

Michel Marcus, Jul 02 2021

Status

approved