OFFSET
1,2
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..500
Index entries for linear recurrences with constant coefficients, signature (4,-4,-2,4,0,-1).
FORMULA
a(n) = 4*a(n-1) - 4*a(n-2) - 2*a(n-3) + 4*a(n-4) - a(n-6).
G.f.: (1 + x^2)/(1 - 2*x + x^3)^2.
a(n) = n*F(n+4) - 2*(F(n+5) - n - 5), F = A000045. - Ehren Metcalfe, Jul 05 2019
MATHEMATICA
b[n_]:= n; c[n_]:= Fibonacci[n];
t[n_, k_]:= Sum[b[k-i] c[n+i], {i, 0, k-1}]
TableForm[Table[t[n, k], {n, 1, 10}, {k, 1, 10}]]
Flatten[Table[t[n-k+1, k], {n, 12}, {k, n, 1, -1}]] (* A213576 *)
r[n_] := Table[t[n, k], {k, 40}] (* columns of antidiagonal triangle *)
d = Table[t[n, n], {n, 1, 40}] (* A213577 *)
s[n_] := Sum[t[i, n+1-i], {i, 1, n}]
s1 = Table[s[n], {n, 1, 50}] (* A213578 *)
(* alternate program *)
LinearRecurrence[{4, -4, -2, 4, 0, -1}, {1, 4, 13, 34, 80, 174}, 40] (* Harvey P. Dale, Jul 04 2019 *)
PROG
(Magma) [n*Fibonacci(n+4)-2*(Fibonacci(n+5)-n-5): n in [1..40]]; // Vincenzo Librandi, Jul 05 2019
(PARI) vector(40, n, n*fibonacci(n+4)-2*(fibonacci(n+5)-n-5)) \\ G. C. Greubel, Jul 05 2019
(Sage) [n*Fibonacci(n+4)-2*(Fibonacci(n+5)-n-5) for n in (1..40)] # G. C. Greubel, Jul 05 2019
(GAP) List([1..40], n-> n*Fibonacci(n+4)-2*(Fibonacci(n+5)-n-5)) # G. C. Greubel, Jul 05 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 18 2012
STATUS
approved