login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A377314
a(n) = coefficient of the term that is independent of 2^(1/3) and 2^(2/3) in the expansion of (1 + 2^(1/3) + 2^(2/3))^n.
3
1, 1, 5, 19, 73, 281, 1081, 4159, 16001, 61561, 236845, 911219, 3505753, 13487761, 51891761, 199644319, 768096001, 2955112721, 11369270485, 43741245619, 168286661033, 647452990441, 2490960200041, 9583526232479, 36870912288001, 141854275761481
OFFSET
0,3
COMMENTS
See A377109 for a guide to related sequences.
FORMULA
a(n) = 3*a(n-1) + 3*a(n-2) + a(n-3), with a(0)=1, a(1)=1, a(3)=5. [Corrected by Jianing Song, Oct 31 2024]
G.f.: (-1 + 2 x + x^2)/(-1 + 3 x + 3 x^2 + x^3).
EXAMPLE
((1 + 2^(1/3) + 2^(2/3)))^3 = 19 + 15 2^(1/3) + 12 2^(2/3), so a(3) = 19.
MATHEMATICA
(* Program 1 generates sequences A377314-A377315 and A108368. *)
tbl = Table[Expand[(1 + 2^(1/3) + 2^(2/3))^n], {n, 0, 24}];
Take[tbl, 6]
u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
Map[({#1, #1 /. _^_ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
{s1, s2, s3} = Transpose[(PadRight[#1, 3] &) /@ Last /@ u][[1 ;; 3]];
s1 (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates (a(n)) for n>=1. *)
LinearRecurrence[{3, 3, 1}, {1, 1, 5}, 15]
CROSSREFS
Cf. A377109, A377117, A377315, A108368 (coefficients of 2^(2/3)).
Sequence in context: A034548 A255455 A255444 * A287805 A129166 A149763
KEYWORD
nonn
AUTHOR
Clark Kimberling, Oct 26 2024
STATUS
approved