A377109 - OEIS

%I #21 Dec 27 2024 09:26:31

%S 1,2,9,38,185,922,4689,23998,123217,633458,3258489,16765718,86273225,

%T 443967370,2284733313,11757749038,60508271137,311391065570,

%U 1602499602537,8246883961094,42440638964825,218410733951098,1123999345270833,5784397706237854

%N a(n) = coefficient of the term that is independent of sqrt(2), sqrt(3), and sqrt(6) in the expansion of (2 + sqrt(2) + sqrt(3))^n.

%C 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:

%C p = 2: (4)

%C p = 3: (8, 1, 4, 3, 8)

%C p = 5: (10, 20, 9, 8, 32, 21, 20)

%C p = 7: (2, 30, 9, 19, 6, 28, 12, 5, 16, 26, 22, 13, 2, 1, 24, 16, 57)

%C p = 11: (61, 29, 70)

%C p = 13: (9, 15, 24, 3, 21, 21, 3, 24, 15, 9, 24)

%C 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)

%C Guide to related sequences:

%C (1 + sqrt (2) + sqrt (3))^n, coefficients of absolute terms: A188570

%C (1 + sqrt (2) + sqrt (3))^n, coefficients of sqrt (2): A188571

%C (1 + sqrt (2) + sqrt (3))^n, coefficients of sqrt (3): A188572

%C (1 + sqrt (2) + sqrt (3))^n, coefficients of sqrt (6): A188573

%C (2 + sqrt (2) + sqrt (3))^n, coefficients of independent terms: this sequence

%C (2 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(2): A377110

%C (2 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(3): A377111

%C (2 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(6): A377112

%C (3 + sqrt (2) + sqrt (3))^n, coefficients of independent terms: A377113

%C (3 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(2): A377114

%C (3 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(3): A377115

%C (3 + sqrt (2) + sqrt (3))^n, coefficients of sqrt(6): A377116

%C (2^(1/3) + 2^(2/3))^n, coefficients of independent terms: A377117

%C (2^(1/3) + 2^(2/3))^n, coefficients of 2^(1/3): A377118

%C (2^(1/3) + 2^(2/3))^n, coefficients of 2^(2/3): A377119

%C (1 + 2^(1/3) + 2^(2/3))^n, coefficients of independent terms: A377314

%C (1 + 2^(1/3) + 2^(2/3))^n, coefficients of 2^(1/3): A377315

%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (8,-14,-8,23).

%F 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.

%F G.f.: (-1 + 6 x - 7 x^2 - 2 x^3)/(-1 + 8 x - 14 x^2 - 8 x^3 + 23 x^4).

%e (2 + sqrt(2) + sqrt(3))^3 = 9 + 4*sqrt(2) + 4*sqrt(3) + 2*sqrt(6), so a(3) = 9.

%t (*Program 1 generates sequences A377109-A377112.*)

%t tbl = Table[Expand[(2 + Sqrt[2] + Sqrt[3])^n], {n, 0, 24}];

%t u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,

%t    Map[({#1, #1 /. _^_ -> 1} &), Map[(Apply[List, #1] &), tbl]]];

%t {s1, s2, s3, s4} = Transpose[(PadRight[#1, 4] &) /@ Last /@ u][[1 ;; 4]];

%t s1  (* _Peter J.C.Moses_, Oct 16 2024 *)

%t (*Program 2 generates this sequence.*)

%t LinearRecurrence[{8, -14, -8, 23}, {1, 2, 9, 38}, 15]

%t (*Program 3 confirms the periodicity properties described in Comments.*)

%t tbl = Table[Expand[(2 + Sqrt[2] + Sqrt[3])^n], {n, 0, 1000}];

%t u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,

%t   Map[({#1, #1 /. _^_ -> 1} &), Map[(Apply[List, #1] &), tbl]]];

%t v = {s1, s2, s3, s4} = Transpose[(PadRight[#1, 4] &) /@ Last /@ u][[1 ;; 4]];

%t Position[Partition[list, Length[#], 1], Flatten[{___, #, ___}]] &[seqtofind];

%t period[seq_] := (If[Last[#1] == {} || Length[#1] == Length[seq] - 1,

%t   0, Length[#1]] &)[NestWhileList[Rest, Rest[seq], #1 != Take[seq, Length[#1]] &, 1]];

%t periodicityReport[seq_] := ({Take[seq, Length[seq] - Length[#1]], period[#1],

%t       Take[#1, period[#1]]} &)[Take[seq, -Length[

%t       NestWhile[Rest[#1] &, seq, period[#1] == 0 &, 1, Length[seq]]]]];

%t seq = s1; Take[seq, 10]

%t f[n_] := Flatten[Position[Mod[s1, Prime[n]], 0]];

%t d[n_] := Differences[f[n]];

%t Table[Take[f[n], 10], {n, 2, 4}]

%t Table[Take[d[n], 10], {n, 2, 4}]

%t Column[Table[{n, Prime[n], periodicityReport[d[Prime[n]]]}, {n, 1, 8}]]

%t (* _Peter J. C. Moses_,Aug 07 2014,Oct 16 2024*)

%Y Cf. A188570, A377110, A377111, A377112.

%K nonn,easy

%O 0,2

%A _Clark Kimberling_, Oct 20 2024