A188570 a(n) = coefficient of the term that is independent of sqrt(2) and sqrt(3) in the expansion of (1 + sqrt(2) + sqrt(3))^n.

1, 1, 6, 16, 80, 296, 1296, 5216, 21952, 90304, 375936, 1555456, 6456320, 26754560, 110963712, 460015616, 1907494912, 7908659200, 32792076288, 135963148288, 563742310400, 2337417887744, 9691567030272, 40183767891968, 166612591968256, 690819710058496

Offset: 0

Author

Mateusz Szymański, Dec 28 2012

Keywords

• nonn

Comments

From Clark Kimberling, Oct 23 2024: (Start)

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 6 primes, with respective period lengths 1,5,5,3,3,5 and these periods:

p = 2: (4)

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

p = 5: (9, 10, 1, 20, 20)

p = 7: (9, 9, 18)

p = 11: (10, 11, 43)

p = 13: (7, 21, 21, 7, 28)

See A377109 for a guide to related sequences. (End)

Examples

			a(3) = 16 because (1+sqrt(2)+sqrt(3))^3 = 16 + 14*sqrt(2) + 12*sqrt(3) + 6*sqrt(6).
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Index entries for linear recurrences with constant coefficients, signature (4,4,-16,8).

Crossrefs

Cf. A188571, A188572, A188573, A377109.

Programs

Mathematica

a[n_] := Sum[Sum[2^(Floor[n/2] - k - j) 3^j Multinomial[2 k + n - 2 Floor[n/2], 2 j, 2 Floor[n/2] - 2 k - 2 j], {j, 0, Floor[n/2] - k}], {k, 0, Floor[n/2]}]; Table[a[n], {n, 0, 25}]
a[n_] := Expand[(1 + Sqrt[2] + Sqrt[3])^n] /. Sqrt[] -> 0; Table[a[n], {n, 0, 25}] (* _Jean-François Alcover, Jan 08 2013 *)
LinearRecurrence[{4,4,-16,8},{1,1,6,16},30] (* Harvey P. Dale, Jan 25 2019 *)

Formula

Recurrence: a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3) + 8*a(n-4). - Vaclav Kotesovec, Aug 13 2013

a(n) ~ (1+sqrt(3)+sqrt(2))^n/4. - Vaclav Kotesovec, Aug 13 2013

Extensions

Edited by Clark Kimberling, Oct 20 2024