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

0, 1, 2, 6, 18, 48, 144, 396, 1152, 3240, 9288, 26352, 75168, 213840, 609120, 1734048, 4937760, 14059008, 40030848, 113980608, 324539136, 924068736, 2631118464, 7491647232, 21331123200, 60736594176, 172936622592, 492406304256, 1402039300608, 3992057561088

Offset: 0

Author

Clark Kimberling, Oct 26 2024

Keywords

• nonn
• easy

Comments

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:

p = 2: (1)

p = 3: (9, 4, 1, 2, 8)

p = 5: (7, 6, 5, 22)

p = 7: (60, 23, 5, 9, 16, 8, 42, 7, 19, 1, 2, 10, 31, 4, 11, 6, 34)

p = 11: (30)

p = 13: (119, 13, 9, 25, 15, 51, 45, 1, 2, 17, 41, 28, 54)

See A377109 for a guide to related sequences.

Examples

			((2^(1/3) + 2^(2/3)))^3 = 4 + 2*2^(1/3) + 2^(2/3), so a(3) = 2.
		

Links

• Index entries for linear recurrences with constant coefficients, signature (0,6,6).

Crossrefs

Cf. A377109, A377117, A377119.

Programs

Mathematica

(* Program 1 generates sequences A377117-A377119. *)
tbl = Table[Expand[(2^(1/3) + 2^(2/3))^n], {n, 0, 30}];
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[{0,6,6}, {0,1,2}, 30]

Formula

a(n) = 6*a(n-2) + 6*a(n-3) for n>=1, with a(0)=0, a(1)=1, a(3)=2.

G.f.: x*(1 + 2*x)/(1 - 6*x^2 - 6*x^3).