A377315 a(n) = coefficient of 2^(1/3) in the expansion of (1 + 2^(1/3) + 2^(2/3))^n.
0, 1, 4, 15, 58, 223, 858, 3301, 12700, 48861, 187984, 723235, 2782518, 10705243, 41186518, 158457801, 609638200, 2345474521, 9023795964, 34717449655, 133569211378, 513883779063, 1977076420978, 7606449811501, 29264462476500, 112589813284981, 433169277095944
Offset: 0
Examples
(1 + 2^(1/3) + 2^(2/3))^3 = 19 + 15 2^(1/3) + 12 2^(2/3), so a(3) = 15.
Links
- Index entries for linear recurrences with constant coefficients, signature (3,3,1).
Programs
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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]]; s2 (* Peter J. C. Moses, Oct 16 2024 *) (* Program 2 generates (a(n)) for n>=1. *) LinearRecurrence[{3,3,1}, {0, 1, 4}, 15]
Formula
a(n) = 3*a(n-1) + 3*a(n-2) + a(n-3), with a(0)=0, a(1)=1, a(3)=4.
G.f.: -((x (1 + x))/(-1 + 3 x + 3 x^2 + x^3)).
Comments