OFFSET
0,4
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 2,2,4,5,2,5 and these periods:
p = 2: (2,1)
p = 3: (7, 1)
p = 5: (21, 5, 13, 1)
p = 7: (15, 17, 20, 43, 1)
p = 11: (9, 1)
p = 13: (102, 5, 17, 15, 1)
See A377109 for a guide to related sequences.
LINKS
FORMULA
a(n) = 6*a(n-2) + 6*a(n-3) for n>=1, with a(0)=0, a(1)=1, a(3)=1.
G.f.: (x (1 + x))/(1 - 6 x^2 - 6 x^3).
EXAMPLE
((2^(1/3) + 2^(2/3)))^3 = 4 + 2*2^(1/3) + 2^(2/3), so a(3) = 1.
MATHEMATICA
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]];
s3 (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates (a(n)) for n>=1. *)
LinearRecurrence[{0, 6, 6}, {0, 1, 1}, 15].
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 26 2024
STATUS
approved