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 6,5,14,4,4,8 and these periods:
p = 2: (1, 2, 2, 1, 3, 3)
p = 3: (1, 4, 3, 8, 8)
p = 5: (1, 5, 4, 1, 1, 6, 6, 6, 6, 6, 6, 2, 4, 6)
p = 7: (1, 16, 1, 18)
p = 11: (1, 30, 1, 32)
p = 13: (1, 10, 3, 17, 10, 1, 28, 14)
See A377109 for a guide to related sequences. (End)
LINKS
G. C. Greubel, Table of n, a(n) for n = 0..1000 (terms 0..200 from Vincenzo Librandi)
FORMULA
From G. C. Greubel, Apr 10 2018: (Start)
Empirical: a(n) = 4*a(n-1) + 4*a(n-2) - 16*a(n-3) + 8*a(n-4).
Empirical: G.f.: 2*x^2*(1-x)/(1 - 4*x - 4*x^2 + 16*x^3 - 8*x^4). (End)
EXAMPLE
a(3) = 6, 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/2] - j - 1 - k) 3^j Multinomial[2 k + n - 2 Floor[n/2], 2 j + 1, 2 Floor[n/2] - 2 k - 1 - 2 j], {j, 0, Floor[n/2] - k - 1}], {k, 0, Floor[n/2] - 1}]; Table[a[n], {n, 0, 25}]
a[n_] := Coefficient[ Expand[(1 + Sqrt[2] + Sqrt[3])^n], Sqrt[6]]; Table[a[n], {n, 0, 25}] (* Jean-François Alcover, Jan 08 2013 *)
CROSSREFS
KEYWORD
nonn
AUTHOR
Mateusz Szymański, Dec 28 2012
EXTENSIONS
Keyword tabl removed by Michel Marcus, Apr 11 2018
Edited by Clark Kimberling, Oct 23 2024
STATUS
approved