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A377114 a(n) = coefficient of sqrt(2) in the expansion of (3 + sqrt(2) + sqrt(3))^n.

%I A377114 #13 Aug 09 2025 04:00:12
%S A377114 0,1,6,38,240,1504,9360,57968,357888,2205376,13574784,83503232,
%T A377114 513469440,3156723712,19404782592,119276106752,733133340672,
%U A377114 4506134745088,27696241336320,170229576458240,1046279833190400,6430725296226304,39524980495024128
%N A377114 a(n) = coefficient of sqrt(2) in the expansion of (3 + sqrt(2) + sqrt(3))^n.
%C A377114 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 1,8,10,7 and these periods:
%C A377114  p = 2: (2)
%C A377114  p = 3: (4, 2, 6, 6, 1, 1, 3, 1)
%C A377114  p = 5: (12, 3, 9, 6, 6, 2, 7, 3, 10, 2)
%C A377114  p = 7: (9, 15, 3, 18, 3, 15, 9)
%C A377114 See A377109 for a guide to related sequences.
%H A377114 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (12,-44,48,8).
%F A377114 a(n) = 12*a(n-1) - 44*a(n-2) + 48*a(n-3) + 8*a(n-4), with a(0)=0, a(1)=1, a(3)=6, a(4)=38.
%F A377114 G.f.: x*(-1 + 6*x - 10*x^2)/(-1 + 12*x - 44*x^2 + 48*x^3 + 8*x^4).
%e A377114 (3 + sqrt(2) + sqrt(3))^3 = 14 + 6*sqrt(2) + 6*sqrt(3) + 2*sqrt(6), so a(3) = 6.
%t A377114 (* Program 1 generates sequences A377113-A377116. *)
%t A377114 tbl = Table[Expand[(3 + Sqrt[2] + Sqrt[3])^n], {n, 0, 24}];
%t A377114 u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &,
%t A377114    Map[({#1, #1 /. _^_ -> 1} &), Map[(Apply[List, #1] &), tbl]]];
%t A377114 {s1,s2,s3,s4}=Transpose[(PadRight[#1,4]&)/@Last/@u][[1;;4]];
%t A377114 s2  (* _Peter J. C. Moses_, Oct 16 2024 *)
%t A377114 (* Program 2 generates this sequence. *)
%t A377114 LinearRecurrence[{12, -44, 48, 8}, {0, 1, 6, 38}, 25]
%Y A377114 Cf. A377090, A377113, A377115, A377116.
%K A377114 nonn,easy
%O A377114 0,3
%A A377114 _Clark Kimberling_, Oct 21 2024