OFFSET
1,1
COMMENTS
See A277402 for illustrations.
LINKS
Lars Blomberg, Table of n, a(n) for n = 1..500
FORMULA
Empirically for 12 < n < 500: a(n) = a(n-2) + a(n-10) - a(n-12) + 120.
Conjectures from Colin Barker, Jun 08 2020: (Start)
G.f.: x*(3 + 4*x + 11*x^2 + 24*x^4 + 24*x^6 + 24*x^8 - 24*x^9 + 45*x^10 + 20*x^11 - 11*x^12) / ((1 - x)^3*(1 + x)^2*(1 - x + x^2 - x^3 + x^4)*(1 + x + x^2 + x^3 + x^4)).
a(n) = a(n-1) + a(n-2) - a(n-3) + a(n-10) - a(n-11) - a(n-12) + a(n-13) for n>13.
(End)
Colin Barker's recurrence conjecture holds for 13 < n <= 500. Lars Blomberg, Jun 12 2020
Empirical: a(2*k - 1) = 3*(4*k^2 - 6*k + 3), for k >= 1. - Ivan N. Ianakiev, Jul 15 2020
CROSSREFS
KEYWORD
nonn
AUTHOR
Lars Blomberg, Jun 08 2020
STATUS
approved