A188571 a(n) = coefficient of sqrt(2) in the expansion of (1 + sqrt(2) + sqrt(3))^n.
0, 1, 2, 14, 48, 224, 880, 3760, 15360, 64192, 265088, 1101440, 4561920, 18925568, 78447616, 325313536, 1348730880, 5592420352, 23187169280, 96141172736, 398624489472, 1652807303168, 6852965761024, 28414229807104, 117812861337600, 488483370827776
Offset: 0
Examples
a(3) = 14 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
- Sela Fried, On the coefficients of (r + sqrt(p) + sqrt(q))^n
- Index entries for linear recurrences with constant coefficients, signature (4,4,-16,8).
Programs
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Mathematica
a[n_] := Sum[Sum[2^(Floor[(n - 1)/2] - k - j) 3^j Multinomial[2 Floor[(n - 1)/2] + 1 - 2 j - 2 k, 2 j, 2 k + 1 - n + 2 Floor[n/2]], {j, 0, Floor[(n - 1)/2] - k + 1}], {k, 0, Floor[(n - 1)/2]}]; Table[a[n], {n, 0, 25}] a[n_] := Coefficient[ Expand[(1 + Sqrt[2] + Sqrt[3])^n], Sqrt[2]] /. Sqrt[3] -> 0; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jan 08 2013 *)
Formula
Conjectures from Colin Barker, Jan 08 2013: (Start)
a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3) + 8*a(n-4).
G.f.: -x*(2*x^2-2*x+1) / (8*x^4-16*x^3+4*x^2+4*x-1). (End)
The conjectures by Barker are true. See link. - Sela Fried, Jan 01 2025
Extensions
Edited by Clark Kimberling, Oct 20 2024
Comments