cp's OEIS Frontend

G.f.: (1-x)/(1-x-x^2)^3. (from Turban reference eq.(2.15)).

a(n)= ((5*n^2+37*n+50)*F(n+1)+4*(n+1)*F(n))/50 with F(n)=A000045(n) (Fibonacci numbers) (from Turban reference eq. (2.17)).

From Peter Bala, Oct 25 2007 (Start):

Since F(-n) = (-1)^(n+1)*F(n), we can use the previous formula to extend the sequence to negative values of n; we find a(-n) = (-1)^n* A129707(n-3).

Recurrence relations: a(n+4) = 2*a(n+3) + a(n+2) - 2*a(n+1) - a(n) + F(n+3), with a(0) = 1, a(1) = 2, a(2) = 6 and a(3) = 13;

a(n+2) = a(n+1) + a(n) + A010049(n+3), with a(0) = 1 and a(1) = 2.

a(n-3) = Sum_{k = 2..floor((n+1)/2)} C(k,2)*C(n-k,k-1) = (1/2)*G''(n,1), where the polynomial G(n,x) := Sum_{k = 1..floor((n+1)/2)} C(n-k,k-1)*x^k = x^((n+1)/2) * F(n, 1/sqrt(x)) and where F(n,x) denotes the n-th Fibonacci polynomial. Since G(n,1) yields the Fibonacci numbers A000045 and G'(n,1) yields the second-order Fibonacci numbers A010049, a(n) may be considered as the sequence of third-order Fibonacci numbers.

For n >= 4, the polynomials Sum_{k = 0..n} C(n,k) * G''(n-k,1)*(-x)^k appear to satisfy a Riemann hypothesis; their zeros appear to lie on the vertical line Re x = 1/2 in the complex plane. Compare with the remarks in A094440 and A010049. (End)

a(n) = A076791(n+3, 2). - Michael Somos, Sep 24 2024

E.g.f.: exp(x/2)*(5*(25 + 23*x + 5*x^2)*cosh(sqrt(5)*x/2) + sqrt(5)*(29 + 65*x + 10*x^2)*sinh(sqrt(5)*x/2))/125. - Stefano Spezia, Sep 26 2024

p(n,x) = u*p(n-1,x) + v*p(n-2,x) for n >= 3, where p(1,x) = 1, p(2,x) = 1 + 3*x, u = p(2,x), and v = 1 - x^2.

p(n,x) = k*(b^n - c^n), where k = -1/sqrt(5 + 6*x + 5*x^2), b = (1/2)*(3*x + 1 - 1/k), c = (1/2)*(3*x + 1 + 1/k).

p(n,x) = u*p(n-1,x) + v*p(n-2,x) for n >=3, where p(1,x) = 1, p(2,x) = 1 + 3 x, u = p(2,x), and v = 2 - x^2.

p(n,x) = k*(b^n - c^n), where k = -1/sqrt(9 + 6 x + 5 x^2), b = (1/2) (3 x + 1 - 1/k), c = (1/2) (3 x + 1 + 1/k).

G.f. as array: ((y^2 + y - 1)*x - y + 1)/(1 + (y^2 + y - 1)*x^2 + (-2*y - 1)*x). - Robert Israel, Oct 31 2019

p(n,x) = u*p(n-1,x) + v*p(n-2,x) for n >= 3, where p(1,x) = 1, p(2,x) = 1 + 3*x, u = p(2,x), and v = 3 - x^2.

p(n,x) = k*(b^n - c^n), where k = -1/sqrt(13 + 6*x + 5*x^2), b = (1/2)*(3*x + 1 - 1/k), c = (1/2)*(3*x + 1 + 1/k).

Let A007318 (Pascal's triangle) = P. then A131048 = (1/3) * (P^2 - 1/P). Delete right border of zeros.

From Peter Bala, Oct 24 2007: (Start)

O.g.f.: 1/(1 - (2*x + 1)*t + (x^2 + x - 2)*t^2) = 1 + (1 + 2*x)*t + (3 + 3*x + 3*x^2)*t^2 + ....

T(n,n-k) = (1/3)*C(n,k)*(2^k - (-1)^k) = C(n,k)*A001045(k).

The row polynomials R(n,x) := Sum_{k = 0..n} T(n,n-k)*x^(n-k) = (1/3)*((x + 2)^n - (x - 1)^n) and have the divisibility property R(n,x) divides R(m,x) in the polynomial ring Z[x] if n divides m.

The polynomials R(n,-x), n >= 2, satisfy a Riemann hypothesis: their zeros lie on the vertical line Re x = 1/2 in the complex plane. Compare with A094440. (End)

p(n,x) = u*p(n-1,x) + v*p(n-2,x) for n >= 3, where p(1,x) = 1, p(2,x) = 1 + 2*x, u = p(2,x), and v = 1 - 3*x - x^2.

p(n,x) = k*(b^n - c^n), where k = -1/sqrt(5 - 8*x), b = (1/2)*(2*x + 1 - 1/k), c = (1/2)*(2*x + 1 + 1/k).

G.f.: (2 - (2x + 1)*t)/(1 - (2x + 1)*t + (x^2 + x - 1)*t^2) = 2 + (1 + 2*x)*t + (3 + 2*x + 2*x^2)*t^2 + (4 + 9*x + 3*x^2 + 2*x^3)*t^3 + ... .

The row polynomials L(n,x) = Sum_{k = 0..n} C(n,k)*Lucas(n-k)*x^k satisfy L(n,x)*F(n,x) = F(2*n,x), where F(n,x) = Sum_{k = 0..n} C(n,k)*Fibonacci(n-k)*x^k.

Other identities and formulas include:

L(n+1,x)^2 - L(n,x)*L(n+2,x) = -5*(x^2 + x - 1)^n;

L(n+1,x) - (x^2 + x - 1)*L(n-1,x) = 5*F(n,x) for n >= 1;

L(2*n,x) - 2*(x^2 + x - 1)^n = 5*F(n,x)^2;

L(n,2*x) = Sum_{ k = 0..n} C(n,k)*L(n-k,x)*x^k;

L(n,3*x) = Sum_{ k = 0..n} C(n,k)*L(n-k,2*x)*x^k etc.

Sum_{k = 0..n} C(n,k)*L(k,x)*F(n-k,x) = 2^n*F(n,x).

Row sums: L(n,1) = Lucas(2*n); Alternating row sums: L(n,-1) = (-1)^n*Lucas(n); L(n,1/phi) = (-1)^n*L(n,-phi) = sqrt(5)^n for n >= 1, where phi = (1 + sqrt(5))/2.

The polynomials L(n,-x) satisfy a Riemann hypothesis: the zeros of L(n,-x) lie on the vertical line Re x = 1/2 in the complex plane.

From Peter Bala, Jun 29 2016: (Start)

L(n,x) = (x + phi)^n + (x - 1/phi)^n, where phi = (1 + sqrt(5))/2. The zeros of L(n,x) are given by -1/2 - i*sqrt(5)/2*cot( (2*k - 1)*Pi/(2*n) ) for k = 1..n.

d/dx(L(n,x)) = n*L(n-1,x).

L(-n,x) = L(n,x)/(x^2 + x - 1)^n.

L(n,x - 1) = (-1)^n*L(n,-x).

L(n,x)^2 - L(2*n,x) = 2*(x^2 + x - 1)^n.

L(n,x)^3 - L(3*n,x) = 3*(x^2 + x - 1)^n*L(n,x).

L(n,x)^4 - L(4*n,x) = 4*(x^2 + x - 1)^n*L(n,x)^2 - 2*(x^2 + x - 1)^(2*n).

If n divides m and m/n is odd then L(n,x) divides L(m,x) in the polynomial ring Z[x].

L(n,x) = F(n+1,x) - (x^2 + x - 1)*F(n-1,x) = 2*F(n+1,x) - (2*x + 1)*F(n,x), where F(n,x) = Sum_{k = 0..n} binomial(n,k)*Fibonacci(n-k)*x^k denotes the n-th row polynomial of A094440 (taken with an offset of 0).

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

exp( Sum_{n >= 1} L(n,x)*L(2*n,x)*z^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)*z^n. (End)

From Peter Bala, Dec 31 2023: (Start)

For n >= 1, the n-th row polynomial L(n,x) is the numerator of 1/(n-1)! * (d/dx)^(n-1) (2*x + 1)/(1 - x - x^2).

Recurrence: for n >= 1, n*L(n+1,x) = n*(2*x + 1)*L(n,x) + (1 - x - x^2)* d/dx(L(n,x)) with L(1,x) = 2*x + 1. (End)

p(n,x) = u*p(n-1,x) + v*p(n-2,x) for n >= 3, where p(1,x) = 1, p(2,x) = 1 + 3*x, u = p(2,x), and v = 1 - 3*x - 2*x^2.

p(n,x) = k*(b^n - c^n), where k = -1/sqrt(5 - 6*x + x^2), b = (1/2)*(3*x + 1 - 1/k), c = (1/2)*(2*x + 1 + 1/k).

p(n,x) = u*p(n-1,x) + v*p(n-2,x) for n >= 3, where p(1,x) = 1, p(2,x) = 1 + 2*x, u = p(2,x), and v = 1 + x^2.

p(n,x) = k*(b^n - c^n), where k = -1/sqrt(5 + 4*x + 8*x^2), b = (1/2)*(2*x + 1 - 1/k), c = (1/2)*(2*x + 1 + 1/k).