OFFSET
0,2
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 1,5,7,17,3,11,35 and these periods:
p = 2: (4)
p = 3: (8, 1, 4, 3, 8)
p = 5: (10, 20, 9, 8, 32, 21, 20)
p = 7: (2, 30, 9, 19, 6, 28, 12, 5, 16, 26, 22, 13, 2, 1, 24, 16, 57)
p = 11: (61, 29, 70)
p = 13: (9, 15, 24, 3, 21, 21, 3, 24, 15, 9, 24)
p = 17: (30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 30, 1, 29, 29, 1, 30, 30, 30, 30, 2, 28, 30, 30, 28, 2, 17, 13, 30, 30, 30, 13, 17)
Guide to related sequences:
(1 + sqrt (2) + sqrt (3))^n, coefficients of absolute terms: A188570
(1 + sqrt (2) + sqrt (3))^n, coefficients of sqrt (2): A188571
(1 + sqrt (2) + sqrt (3))^n, coefficients of sqrt (3): A188572
(1 + sqrt (2) + sqrt (3))^n, coefficients of sqrt (6): A188573
(2 + sqrt (2) + sqrt (3))^n, coefficients of independent terms: this sequence
(2 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(2): A377110
(2 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(3): A377111
(2 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(6): A377112
(3 + sqrt (2) + sqrt (3))^n, coefficients of independent terms: A377113
(3 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(2): A377114
(3 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(3): A377115
(3 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(6): A377116
(2^(1/3) + 2^(2/3))^n, coefficients of independent terms: A377117
(2^(1/3) + 2^(2/3))^n, coefficients of 2^(1/3): A377118
(2^(1/3) + 2^(2/3))^n, coefficients of 2^(2/3): A377119
(1 + 2^(1/3) + 2^(2/3))^n, coefficients of independent terms: A377314
(1 + 2^(1/3) + 2^(2/3))^n, coefficients of 2^(1/3): A377315
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)=1, a(1)=2, a(3)=9, a(4)=38.
G.f.: (-1 + 6 x - 7 x^2 - 2 x^3)/(-1 + 8 x - 14 x^2 - 8 x^3 + 23 x^4).
EXAMPLE
(2 + sqrt(2) + sqrt(3))^3 = 9 + 4*sqrt(2) + 4*sqrt(3) + 2*sqrt(6), so a(3) = 9.
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]];
s1 (* _Peter J.C.Moses_, Oct 16 2024 *)
(*Program 2 generates this sequence.*)
LinearRecurrence[{8, -14, -8, 23}, {1, 2, 9, 38}, 15]
(*Program 3 confirms the periodicity properties described in Comments.*)
tbl = Table[Expand[(2 + Sqrt[2] + Sqrt[3])^n], {n, 0, 1000}];
u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
Map[({#1, #1 /. _^_ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
v = {s1, s2, s3, s4} = Transpose[(PadRight[#1, 4] &) /@ Last /@ u][[1 ;; 4]];
Position[Partition[list, Length[#], 1], Flatten[{___, #, ___}]] &[seqtofind];
period[seq_] := (If[Last[#1] == {} || Length[#1] == Length[seq] - 1,
0, Length[#1]] &)[NestWhileList[Rest, Rest[seq], #1 != Take[seq, Length[#1]] &, 1]];
periodicityReport[seq_] := ({Take[seq, Length[seq] - Length[#1]], period[#1],
Take[#1, period[#1]]} &)[Take[seq, -Length[
NestWhile[Rest[#1] &, seq, period[#1] == 0 &, 1, Length[seq]]]]];
seq = s1; Take[seq, 10]
f[n_] := Flatten[Position[Mod[s1, Prime[n]], 0]];
d[n_] := Differences[f[n]];
Table[Take[f[n], 10], {n, 2, 4}]
Table[Take[d[n], 10], {n, 2, 4}]
Column[Table[{n, Prime[n], periodicityReport[d[Prime[n]]]}, {n, 1, 8}]]
(* Peter J. C. Moses, Aug 07 2014, Oct 16 2024*)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Oct 20 2024
STATUS
approved