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A377118 a(n) = coefficient of 2^(1/3) in the expansion of (2^(1/3) + 2^(2/3))^n.

%I A377118 #11 Aug 09 2025 05:00:04
%S A377118 0,1,2,6,18,48,144,396,1152,3240,9288,26352,75168,213840,609120,
%T A377118 1734048,4937760,14059008,40030848,113980608,324539136,924068736,
%U A377118 2631118464,7491647232,21331123200,60736594176,172936622592,492406304256,1402039300608,3992057561088
%N A377118 a(n) = coefficient of 2^(1/3) in the expansion of (2^(1/3) + 2^(2/3))^n.
%C A377118 Conjecture: every prime divides a(n) for infinitely many n, and if K(p) = (k(1), k(2),...) is the maximal subsequence of indices n such that p divides a(n), then the difference sequence of K(p) is eventually periodic; indeed, K(p) is purely periodic for the first 6 primes, with respective period lengths 1,5,4,17,1,13 and these periods:
%C A377118 p = 2: (1)
%C A377118 p = 3: (9, 4, 1, 2, 8)
%C A377118 p = 5: (7, 6, 5, 22)
%C A377118 p = 7: (60, 23, 5, 9, 16, 8, 42, 7, 19, 1, 2, 10, 31, 4, 11, 6, 34)
%C A377118 p = 11: (30)
%C A377118 p = 13: (119, 13, 9, 25, 15, 51, 45, 1, 2, 17, 41, 28, 54)
%C A377118 See A377109 for a guide to related sequences.
%H A377118 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (0,6,6).
%F A377118 a(n) = 6*a(n-2) + 6*a(n-3) for n>=1, with a(0)=0, a(1)=1, a(3)=2.
%F A377118 G.f.: x*(1 + 2*x)/(1 - 6*x^2 - 6*x^3).
%e A377118 ((2^(1/3) + 2^(2/3)))^3 = 4 + 2*2^(1/3) + 2^(2/3), so a(3) = 2.
%t A377118 (* Program 1 generates sequences A377117-A377119. *)
%t A377118 tbl = Table[Expand[(2^(1/3) + 2^(2/3))^n], {n, 0, 30}];
%t A377118 u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
%t A377118    Map[({#1, #1 /. _^_ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
%t A377118 {s1, s2, s3} = Transpose[(PadRight[#1, 3] &) /@ Last /@ u][[1 ;; 3]];
%t A377118 s2   (* _Peter J. C. Moses_, Oct 16 2024 *)
%t A377118 (* Program 2 generates (a(n)) for n>=1. *)
%t A377118 LinearRecurrence[{0,6,6}, {0,1,2}, 30]
%Y A377118 Cf. A377109, A377117, A377119.
%K A377118 nonn,easy
%O A377118 0,3
%A A377118 _Clark Kimberling_, Oct 26 2024