A377314 - OEIS

%I #15 Oct 31 2024 21:55:34

%S 1,1,5,19,73,281,1081,4159,16001,61561,236845,911219,3505753,13487761,

%T 51891761,199644319,768096001,2955112721,11369270485,43741245619,

%U 168286661033,647452990441,2490960200041,9583526232479,36870912288001,141854275761481

%N 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.

%C See A377109 for a guide to related sequences.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,3,1).

%F 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]

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

%e ((1 + 2^(1/3) + 2^(2/3)))^3 = 19 + 15 2^(1/3) + 12 2^(2/3), so a(3) = 19.

%t (* Program 1 generates sequences A377314-A377315 and A108368. *)

%t tbl = Table[Expand[(1 + 2^(1/3) + 2^(2/3))^n], {n, 0, 24}];

%t Take[tbl, 6]

%t u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,

%t    Map[({#1, #1 /. _^_ -> 1} &), Map[(Apply[List, #1] &), tbl]]];

%t {s1, s2, s3} = Transpose[(PadRight[#1, 3] &) /@ Last /@ u][[1 ;; 3]];

%t s1  (* _Peter J. C. Moses_, Oct 16 2024 *)

%t (* Program 2 generates (a(n)) for n>=1. *)

%t LinearRecurrence[{3,3,1}, {1, 1, 5}, 15]

%Y Cf. A377109, A377117, A377315, A108368 (coefficients of 2^(2/3)).

%K nonn

%O 0,3

%A _Clark Kimberling_, Oct 26 2024