A023554 Convolution of natural numbers >= 3 and (Fib(2), Fib(3), Fib(4), ...).

3, 10, 22, 43, 78, 136, 231, 386, 638, 1047, 1710, 2784, 4523, 7338, 11894, 19267, 31198, 50504, 81743, 132290, 214078, 346415, 560542, 907008, 1467603, 2374666, 3842326, 6217051, 10059438, 16276552, 26336055, 42612674, 68948798, 111561543, 180510414

Offset: 1

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

a(n) is the sum of row n in the triangle T(n,k) defined by: T(n,1) = T(n,n) = 2*n+1 for n>=1 and T(n,k) = 3*T(n-1,k-1) - 2*T(n-1,k) + T(n-2,k-1) for n>2, 2<=k<=n-1. - Lechoslaw Ratajczak, Nov 07 2020

Floretion Algebra Multiplication Program, FAMP code: (a(n)) = 4jesleftforcycseq[ - .25'i + .5'k - .25i' - .5j' + .5k' - .75'ii' + .75'jj' - .25'kk' + .25'jk' - .5'ki' + .25'kj' + .25e ], apart from initial terms. 4jesrightforcycseq = A022308; 2jesforcycseq(n+2) = n+2; identity: jesleft + jesright = jes; vesforcycseq was set to the constant sequence = (-1,-1,-1,-1,-1...). (Dement)

Links

• Colin Barker, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-2,-1,1).

Crossrefs

Cf. A000045, A213584.

Programs

GAP

F:=Fibonacci; List([1..40], n-> F(n+5)+2*F(n+3)-(2*n+9)); # G. C. Greubel, Jul 08 2019

Magma

F:=Fibonacci; [F(n+5)+2*F(n+3)-(2*n+9): n in [1..40]]; // G. C. Greubel, Jul 08 2019

Mathematica

Table[Fibonacci[n+5] + 2*Fibonacci[n+3] -2*n-9, {n, 40}] (* G. C. Greubel, Jul 08 2019 *)

PARI

Vec(x*(1+x)*(3-2*x) / ((1-x)^2*(1-x-x^2)) + O(x^60)) \\ Colin Barker, Feb 20 2017

PARI

vector(40, n, f=fibonacci; f(n+5)+2*f(n+3)-(2*n+9)) \\ G. C. Greubel, Jul 08 2019

SageMath

f=fibonacci; [f(n+5)+2*f(n+3)-(2*n+9) for n in (1..40)] # G. C. Greubel, Jul 08 2019

Formula

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

2*(n+5) = A022308(n+4) - a(n+1) (conjectured). Note offset of A022308 is 0. - Creighton Dement, Feb 02 2005

From Colin Barker, Feb 20 2017: (Start)

a(n) = -7 + (2^(-1-n)*((1-t)^n*(-19+9*t) + (1+t)^n*(19+9*t)))/t - 2*(1+n) where t=sqrt(5).

a(n) = 3*a(n-1) - 2*a(n-2) - a(n-3) + a(n-4) for n>4. (End)

a(n) = Fibonacci(n+5) + 2*Fibonacci(n+3) - (2*n + 9). - G. C. Greubel, Jul 08 2019

a(n) = a(n-1) + a(n-2) + 2*n + 3 for n>2. - Lechoslaw Ratajczak, Nov 07 2020