cp's OEIS Frontend

A257065 Number of length 3 1..(n+1) arrays with every leading partial sum divisible by 2 or 3.

%I A257065 #8 Dec 20 2018 15:44:33
%S A257065 2,6,18,27,64,81,141,200,293,343,512,578,776,954,1208,1331,1728,1875,
%T A257065 2291,2652,3147,3375,4096,4356,5070,5678,6494,6859,8000,8405,9497,
%U A257065 10416,11633,12167,13824,14406,15956,17250,18948,19683,21952,22743,24831,26564
%N A257065 Number of length 3 1..(n+1) arrays with every leading partial sum divisible by 2 or 3.
%H A257065 R. H. Hardin, <a href="/A257065/b257065.txt">Table of n, a(n) for n = 1..210</a>
%F A257065 Empirical: a(n) = a(n-1) + 3*a(n-6) - 3*a(n-7) - 3*a(n-12) + 3*a(n-13) + a(n-18) - a(n-19).
%F A257065 Empirical for n mod 6 = 0: a(n) = (8/27)*n^3 + (4/9)*n^2 + (1/6)*n
%F A257065 Empirical for n mod 6 = 1: a(n) = (8/27)*n^3 + (2/3)*n^2 + (17/18)*n + (5/54)
%F A257065 Empirical for n mod 6 = 2: a(n) = (8/27)*n^3 + (2/3)*n^2 + (7/9)*n - (16/27)
%F A257065 Empirical for n mod 6 = 3: a(n) = (8/27)*n^3 + (8/9)*n^2 + (1/2)*n + (1/2)
%F A257065 Empirical for n mod 6 = 4: a(n) = (8/27)*n^3 + (4/9)*n^2 + (2/9)*n + (1/27)
%F A257065 Empirical for n mod 6 = 5: a(n) = (8/27)*n^3 + (8/9)*n^2 + (8/9)*n + (8/27).
%F A257065 Empirical g.f.: x*(2 + 4*x + 12*x^2 + 9*x^3 + 37*x^4 + 17*x^5 + 54*x^6 + 47*x^7 + 57*x^8 + 23*x^9 + 58*x^10 + 15*x^11 + 24*x^12 + 13*x^13 + 11*x^14 + x^16) / ((1 - x)^4*(1 + x)^3*(1 - x + x^2)^3*(1 + x + x^2)^3). - _Colin Barker_, Dec 20 2018
%e A257065 Some solutions for n=4:
%e A257065 ..3....2....2....4....2....4....2....3....3....4....2....3....4....3....4....2
%e A257065 ..5....1....1....2....4....5....4....3....5....5....4....5....4....1....4....1
%e A257065 ..4....5....1....2....3....3....2....3....1....1....4....2....1....5....2....3
%Y A257065 Row 3 of A257062.
%K A257065 nonn
%O A257065 1,1
%A A257065 _R. H. Hardin_, Apr 15 2015