cp's OEIS Frontend

Empirical, for all rows and columns: a(n)=3*a(n-1)-6*a(n-3)+4*a(n-4).

From Andrew Howroyd, Mar 09 2024: (Start)

The above empirical formula is correct.

T(n,k) = -14 + 9*(2^(n-1) + 2^(k-1)) + 3*(2^(floor((n-1)/2)) + 2^(floor(n/2)) + 2^(floor((k-1)/2)) + 2^(floor(k/2))). (End)

Empirical, for all rows and columns: a(n)=3*a(n-1)-6*a(n-3)+4*a(n-4).

From Andrew Howroyd, Mar 09 2024: (Start)

The above empirical formula is correct.

T(n,k) = -7 + 4*(2^(n-1) + 2^(k-1)) + 2*(2^(floor((n-1)/2)) + 2^(floor(n/2)) + 2^(floor((k-1)/2)) + 2^(floor(k/2))). (End)

Empirical: a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4)

Based on the conjectured recursion formula, we may prove (by a tedious induction) that a(n) = (2^ceiling(n/2) + 1) * (2^floor(n/2) + 1) = A051032(n) * A051032(n-1) for n >= 1. - Philippe Gibone, Jul 06 2016, corrected by Robert Israel, May 21 2019

Empirical: G.f.: -x*(4-6*x-9*x^2+12*x^3) / ( (x-1)*(2*x-1)*(2*x^2-1) ). - R. J. Mathar, Jul 15 2016

Empirical formulas verified (see link): Robert Israel, May 21 2019.

2*a(n) = 2+2^n+A029744(n+3). - R. J. Mathar, Jul 19 2024

From Andrew Howroyd, Mar 09 2024: (Start)

a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4).

G.f.: 2*x*(2 - 2*x - 5*x^2 + 4*x^3)/((1 - x)*(1 - 2*x)*(1 - 2*x^2)). (End)

E.g.f.: 3*cosh(sqrt(2)*x) + cosh(2*x) - 2*cosh(x) - 2 - 2*sinh(x) + sinh(2*x) + 2*sqrt(2)*sinh(sqrt(2)*x). - Stefano Spezia, Oct 16 2024

Empirical: a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4).

Conjectures from Colin Barker, Apr 07 2018: (Start)

G.f.: x*(6 - 10*x - 13*x^2 + 20*x^3) / ((1 - x)*(1 - 2*x)*(1 - 2*x^2)).

a(n) = (3*2^(n/2) + 2^n + 6) / 2 for n even.

a(n) = 2^(n-1) + 2^((n+1)/2) + 3 for n odd.

(End)

The above empirical formula is correct. See note from Andrew Howroyd in A183986.

Empirical: a(n)=3*a(n-1)-6*a(n-3)+4*a(n-4).

Conjectures from Colin Barker, Apr 07 2018: (Start)

G.f.: x*(9 - 16*x - 19*x^2 + 32*x^3) / ((1 - x)*(1 - 2*x)*(1 - 2*x^2)).

a(n) = (3*2^(n/2) + 2^n + 12) / 2 for n even.

a(n) = 2^((n-5)/2+3) + 2^(n-1) + 6 for n odd.

(End)

The above empirical formula is correct. See note from Andrew Howroyd in A183986.

Empirical: a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4).

Conjectures from Colin Barker, Apr 08 2018: (Start)

G.f.: x*(15 - 28*x - 31*x^2 + 56*x^3) / ((1 - x)*(1 - 2*x)*(1 - 2*x^2)).

a(n) = 3*2^(n/2-1) + 2^(n-1) + 12 for n even.

a(n) = 2^(n-1) + 2^((n+1)/2) + 12 for n odd.

(End)

The above empirical formula is correct. See note from Andrew Howroyd in A183986.

Empirical: a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4).

Conjectures from Colin Barker, Apr 08 2018: (Start)

G.f.: x*(25 - 48*x - 51*x^2 + 96*x^3) / ((1 - x)*(1 - 2*x)*(1 - 2*x^2)).

a(n) = 3*2^(n/2-1) + 2^(n-1) + 22 for n even.

a(n) = 2^(n-1) + 2^((n+1)/2) + 22 for n odd.

(End)

The above empirical formula is correct. See note from Andrew Howroyd in A183986.

Conjectures from Colin Barker, Apr 09 2018: (Start)

G.f.: x*(45 - 88*x - 91*x^2 + 176*x^3) / ((1 - x)*(1 - 2*x)*(1 - 2*x^2)).

a(n) = 3*2^(n/2-1) + 2^(n-1) + 42 for n even.

a(n) = 2^(n-1) + 2^((n+1)/2) + 42 for n odd.

a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4) for n>4.

(End)

The above empirical formula is correct. See note from Andrew Howroyd in A183986.

Empirical: a(n) = 3*a(n-1) - 6*a(n-3) + 4*a(n-4).

Conjectures from Colin Barker, Apr 09 2018: (Start)

G.f.: x*(81 - 160*x - 163*x^2 + 320*x^3) / ((1 - x)*(1 - 2*x)*(1 - 2*x^2)).

a(n) = 3*2^(n/2-1) + 2^(n-1) + 78 for n even.

a(n) = 2^(n-1) + 2^((n+1)/2) + 78 for n odd.

(End)

The above empirical formula is correct. See note from Andrew Howroyd in A183986.