# Greetings from The On-Line Encyclopedia of Integer Sequences! http://oeis.org/ Search: id:a377119 Showing 1-1 of 1 %I A377119 #10 Oct 31 2024 01:46:40 %S A377119 0,1,1,6,12,42,108,324,900,2592,7344,20952,59616,169776,483408, %T A377119 1376352,3919104,11158560,31772736,90465984,257587776,733432320, %U A377119 2088322560,5946120576,16930529280,48206658816,137259899136,390823128576,1112799347712,3168498166272 %N A377119 a(n) = coefficient of 2^(2/3) in the expansion of (2^(1/3) + 2^(2/3))^n. %C A377119 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 2,2,4,5,2,5 and these periods: %C A377119 p = 2: (2,1) %C A377119 p = 3: (7, 1) %C A377119 p = 5: (21, 5, 13, 1) %C A377119 p = 7: (15, 17, 20, 43, 1) %C A377119 p = 11: (9, 1) %C A377119 p = 13: (102, 5, 17, 15, 1) %C A377119 See A377109 for a guide to related sequences. %H A377119 Index entries for linear recurrences with constant coefficients, signature (0,6,6). %F A377119 a(n) = 6*a(n-2) + 6*a(n-3) for n>=1, with a(0)=0, a(1)=1, a(3)=1. %F A377119 G.f.: (x (1 + x))/(1 - 6 x^2 - 6 x^3). %e A377119 ((2^(1/3) + 2^(2/3)))^3 = 4 + 2*2^(1/3) + 2^(2/3), so a(3) = 1. %t A377119 (* Program 1 generates sequences A377117-A377119. *) %t A377119 tbl = Table[Expand[(2^(1/3) + 2^(2/3))^n], {n, 0, 30}]; %t A377119 u = MapApply[{#1/#2, #2} /. {1, #} -> {{1}, {#}} &, %t A377119 Map[({#1, #1 /. _^_ -> 1} &), Map[(Apply[List, #1] &), tbl]]]; %t A377119 {s1, s2, s3} = Transpose[(PadRight[#1, 3] &) /@ Last /@ u][[1 ;; 3]]; %t A377119 s3 (* _Peter J. C. Moses_, Oct 16 2024 *) %t A377119 (* Program 2 generates (a(n)) for n>=1. *) %t A377119 LinearRecurrence[{0,6,6}, {0,1,1}, 15]. %Y A377119 Cf. A377109, A377117, A377118. %K A377119 nonn,easy %O A377119 0,4 %A A377119 _Clark Kimberling_, Oct 26 2024 # Content is available under The OEIS End-User License Agreement: http://oeis.org/LICENSE