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a(n) = coefficient of sqrt(3) in the expansion of (3 + sqrt(2) + sqrt(3))^n.
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%I #6 Oct 23 2024 01:06:26

%S 0,1,6,36,216,1304,7920,48320,295680,1812672,11124864,68320000,

%T 419719680,2579051008,15849305088,97406521344,598661038080,

%U 3679444570112,22614556631040,138994100486144,854291341737984,5250689954316288,32272093691707392,198352703517884416

%N a(n) = coefficient of sqrt(3) 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 4,8,10,10 and these periods:

%C p = 2: (2, 1, 1, 2)

%C p = 3: (6, 1, 1, 3, 1, 4, 2, 6)

%C p = 5: (6, 2, 7, 3, 10, 2, 12, 3, 9, 6)

%C p = 7: (14, 4, 4, 2, 12, 1, 11, 5, 1, 18)

%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)=1, a(3)=6, a(4)=36.

%F G.f.: (x (-1 + 6 x - 8 x^2))/(-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) = 6.

%t (* Program 1 generates sequences A377113-A37716. *)

%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 s3 (* _Peter J. C. Moses_, Oct 16 2024 *)

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

%t LinearRecurrence[{12, -44, 48, 8}, {0, 1, 6, 36}, 15].

%Y Cf. A377090, A377113, A377114, A377116.

%K nonn

%O 0,3

%A _Clark Kimberling_, Oct 21 2024