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%I #10 Oct 25 2024 09:34:57
%S 0,0,2,18,128,840,5328,33264,206080,1271808,7833472,48200064,
%T 296423424,1822459392,11203152896,68863546368,423273267200,
%U 2601614180352,15990421856256,98282063536128,604069867552768,3712780777586688,22819757583302656,140256346936639488
%N a(n) = coefficient of sqrt(6) in the expansion of (3 + 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 4 primes, with respective period lengths 4,5,9,10 and these periods:
%C p = 2: (2, 1, 1, 2)
%C p = 3: (1, 4, 3, 8, 8)
%C p = 5: (1, 6, 4, 1, 5, 7, 12, 12, 12)
%C p = 7: (1, 11, 5, 1, 18, 14, 4, 4, 2, 12)
%C See A377109 for a guide to related sequences.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (12,-44,48,8).
%F a(n) = 12*a(n-1) - 44*a(n-2) + 48*a(n-3) + 8*a(n-4), with a(0)=0, a(1)=0, a(3)=2, a(4)=18.
%F G.f.: (2 x^2 (-1 + 3 x))/(-1 + 12 x - 44 x^2 + 48 x^3 + 8 x^4).
%e (3 + sqrt(2) + sqrt(3))^3 = 14 + 6*sqrt(2) + 6*sqrt(3) + 2*sqrt(6), so a(3) = 2.
%t (* Program 1 generates sequences A377113-A377116. *)
%t tbl = Table[Expand[(3 + 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 s4 (* _Peter J. C. Moses_, Oct 16 2024 *)
%t (* Program 2 generates this sequence. *)
%t LinearRecurrence[{12, -44, 48, 8}, {0, 0, 2, 18}, 15].
%Y Cf. A377109, A377113, A377114, A377115, A377117.
%K nonn,easy
%O 0,3
%A _Clark Kimberling_, Oct 23 2024