This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A249467 #10 Nov 09 2018 16:28:14 %S A249467 30,190,860,2640,6730,14730,29060,52900,90390,146610,228000,342120, %T A249467 498030,706270,979100,1330440,1776250,2334330,3024740,3869740,4893990, %U A249467 6124530,7591200,9326400,11365470,13746670,16511420,19704240,23373130 %N A249467 Number of length 1+4 0..n arrays with no five consecutive terms having four times any element equal to the sum of the remaining four. %H A249467 R. H. Hardin, <a href="/A249467/b249467.txt">Table of n, a(n) for n = 1..210</a> %F A249467 Empirical: a(n) = 4*a(n-1) - 6*a(n-2) + 5*a(n-3) - 4*a(n-4) + 2*a(n-5) + 2*a(n-6) - 4*a(n-7) + 5*a(n-8) - 6*a(n-9) + 4*a(n-10) - a(n-11). %F A249467 Also a polynomial of degree 5 plus a constant pseudonomial with period 12: %F A249467 Empirical for n mod 12 = 0: a(n) = n^5 + (15/4)*n^4 + (25/3)*n^3 + (15/2)*n^2 + 4*n %F A249467 Empirical for n mod 12 = 1: a(n) = n^5 + (15/4)*n^4 + (25/3)*n^3 + (15/2)*n^2 + 4*n + (65/12) %F A249467 Empirical for n mod 12 = 2: a(n) = n^5 + (15/4)*n^4 + (25/3)*n^3 + (15/2)*n^2 + 4*n - (20/3) %F A249467 Empirical for n mod 12 = 3: a(n) = n^5 + (15/4)*n^4 + (25/3)*n^3 + (15/2)*n^2 + 4*n + (35/4) %F A249467 Empirical for n mod 12 = 4: a(n) = n^5 + (15/4)*n^4 + (25/3)*n^3 + (15/2)*n^2 + 4*n - (40/3) %F A249467 Empirical for n mod 12 = 5: a(n) = n^5 + (15/4)*n^4 + (25/3)*n^3 + (15/2)*n^2 + 4*n + (145/12) %F A249467 Empirical for n mod 12 = 6: a(n) = n^5 + (15/4)*n^4 + (25/3)*n^3 + (15/2)*n^2 + 4*n %F A249467 Empirical for n mod 12 = 7: a(n) = n^5 + (15/4)*n^4 + (25/3)*n^3 + (15/2)*n^2 + 4*n - (55/12) %F A249467 Empirical for n mod 12 = 8: a(n) = n^5 + (15/4)*n^4 + (25/3)*n^3 + (15/2)*n^2 + 4*n - (20/3) %F A249467 Empirical for n mod 12 = 9: a(n) = n^5 + (15/4)*n^4 + (25/3)*n^3 + (15/2)*n^2 + 4*n + (75/4) %F A249467 Empirical for n mod 12 = 10: a(n) = n^5 + (15/4)*n^4 + (25/3)*n^3 + (15/2)*n^2 + 4*n - (40/3) %F A249467 Empirical for n mod 12 = 11: a(n) = n^5 + (15/4)*n^4 + (25/3)*n^3 + (15/2)*n^2 + 4*n + (25/12). %F A249467 Empirical g.f.: 10*x*(3 + 7*x + 28*x^2 + 19*x^3 + 50*x^4 + 5*x^5 + 32*x^6 - 3*x^7 + 3*x^8) / ((1 - x)^6*(1 + x)*(1 + x^2)*(1 + x + x^2)). - _Colin Barker_, Nov 09 2018 %e A249467 Some solutions for n=6: %e A249467 2 4 0 0 6 6 0 6 4 0 1 4 2 4 2 5 %e A249467 3 3 2 6 5 0 1 3 6 0 3 6 1 3 4 1 %e A249467 4 6 0 6 4 0 5 6 0 6 4 0 5 3 6 2 %e A249467 3 0 3 1 2 4 5 0 0 0 4 5 6 1 0 5 %e A249467 5 0 2 2 0 6 6 5 3 1 5 6 5 2 2 1 %Y A249467 Row 1 of A249466. %K A249467 nonn %O A249467 1,1 %A A249467 _R. H. Hardin_, Oct 29 2014