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.
30, 190, 860, 2640, 6730, 14730, 29060, 52900, 90390, 146610, 228000, 342120, 498030, 706270, 979100, 1330440, 1776250, 2334330, 3024740, 3869740, 4893990, 6124530, 7591200, 9326400, 11365470, 13746670, 16511420, 19704240, 23373130
Offset: 1
Keywords
Examples
Some solutions for n=6: 2 4 0 0 6 6 0 6 4 0 1 4 2 4 2 5 3 3 2 6 5 0 1 3 6 0 3 6 1 3 4 1 4 6 0 6 4 0 5 6 0 6 4 0 5 3 6 2 3 0 3 1 2 4 5 0 0 0 4 5 6 1 0 5 5 0 2 2 0 6 6 5 3 1 5 6 5 2 2 1
Links
- R. H. Hardin, Table of n, a(n) for n = 1..210
Crossrefs
Row 1 of A249466.
Formula
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).
Also a polynomial of degree 5 plus a constant pseudonomial with period 12:
Empirical for n mod 12 = 0: a(n) = n^5 + (15/4)*n^4 + (25/3)*n^3 + (15/2)*n^2 + 4*n
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)
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)
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)
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)
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)
Empirical for n mod 12 = 6: a(n) = n^5 + (15/4)*n^4 + (25/3)*n^3 + (15/2)*n^2 + 4*n
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)
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)
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)
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)
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).
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