OFFSET
1,1
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).
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
MATHEMATICA
Table[Fibonacci[n+5] + 2*Fibonacci[n+3] -2*n-9, {n, 40}] (* G. C. Greubel, Jul 08 2019 *)
PROG
(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
(Magma) F:=Fibonacci; [F(n+5)+2*F(n+3)-(2*n+9): n in [1..40]]; // 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
(GAP) F:=Fibonacci; List([1..40], n-> F(n+5)+2*F(n+3)-(2*n+9)); # G. C. Greubel, Jul 08 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved