OFFSET
0,3
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 7 primes, with respective period lengths 1,8,18,17,3,11,50 and these periods:
p = 2: (4)
p = 3: (4,2,6,6,1,1,3,1)
p = 5: (12,2,10,7,3,2,3,9,12,1,8,2,1,12,12,5,7,12)
p = 7: (5, 16, 26, 22, 13, 2, 1, 24, 16, 57, 2, 30, 9, 19, 6, 28, 12)
p = 11: (29, 70, 61)
p = 13: (21, 3, 24, 15, 9, 24, 9, 15, 24, 3, 21)
p = 17: (19, 11, 30, 4, 21, 2, 3, 22, 8, 7, 23, 24, 6, 30, 30, 11, 10, 9, 26, 4, 30, 18, 12, 30, 30, 15, 15, 30, 30, 8, 4, 11, 7, 30, 30, 9, 7, 14, 30, 5, 9, 16, 6, 24, 30, 30, 3, 27, 30, 30)
See A377109 for a guide to related sequences.
LINKS
Index entries for linear recurrences with constant coefficients, signature (8,-14,-8,23).
FORMULA
a(n) = 8*a(n-1) - 14*a(n-2) - 8*a(n-3) + 23*a(n-4), with a(0)=0, a(1)=1, a(3)=4, a(4)=23.
G.f.: (x (-1 + 4 x - 5 x^2))/(-1 + 8 x - 14 x^2 - 8 x^3 + 23 x^4).
a(n) = ((2 + sqrt(2) + sqrt(3))^n - (2 - sqrt(2) - sqrt(3))^n + (2 + sqrt(2) - sqrt(3))^n - (2 - sqrt(2) + sqrt(3))^n) / 2^(5/2). - Vaclav Kotesovec, Oct 21 2024
EXAMPLE
(2 + sqrt(2) + sqrt(3))^3 = 9 + 4*sqrt(2) + 4*sqrt(3) + 2*sqrt(6), so a(3) = 4.
MATHEMATICA
tbl = Table[Expand[(2 + Sqrt[2] + Sqrt[3])^n], {n, 0, 24}];
u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
Map[({#1, #1 /. _^_ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
{s1, s2, s3, s4}=Transpose[(PadRight[#1, 4]&)/@Last/@u][[1;; 4]];
s2 (* Peter J. C. Moses, Oct 16 2024 *)
(* Program 2 generates this sequence. *)
LinearRecurrence[{8, -14, -8, 23}, {0, 1, 4, 23}, 15].
CROSSREFS
KEYWORD
nonn,new
AUTHOR
Clark Kimberling, Oct 20 2024
STATUS
approved