OFFSET
0,3
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 4,8,14,3,3,5 and these periods:
p = 2: (4)
p = 3: (6, 1, 1, 3, 1, 4, 2, 6)
p = 5: (6, 6, 6, 2, 4, 6, 1, 5, 4, 1, 1, 6, 6, 6)
p = 7: (9, 18, 9)
p = 11: (43, 10, 11)
p = 13: (21, 7, 28, 7, 21)
See A377109 for a guide to related sequences. (End)
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).
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
EXAMPLE
a(3) = 14 because (1+sqrt(2)+sqrt(3))^3 = 16 + 14*sqrt(2) + 12*sqrt(3) + 6*sqrt(6).
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 *)
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Mateusz Szymański, Dec 28 2012
EXTENSIONS
Edited by Clark Kimberling, Oct 20 2024
STATUS
approved