A200156 Number of 0..n arrays x(0..4) of 5 elements with zero 4th difference.
2, 15, 40, 103, 202, 381, 636, 1033, 1550, 2287, 3212, 4451, 5946, 7869, 10140, 12969, 16238, 20211, 24744, 30147, 36222, 43349, 51296, 60493, 70646, 82267, 95016, 109467, 125206, 142897, 162076, 183477, 206546, 232123, 259596, 289879, 322262
Offset: 1
Keywords
Examples
Some solutions for n=6: ..6....4....1....0....2....6....2....2....4....6....6....0....6....6....2....4 ..4....5....1....6....0....1....2....3....6....2....5....6....6....5....3....4 ..1....3....1....6....2....0....0....2....3....0....2....6....4....4....2....5 ..0....1....2....4....4....2....0....1....0....0....1....3....3....4....2....5 ..4....2....5....4....2....6....6....2....2....2....6....0....6....6....6....2
Links
- R. H. Hardin, Table of n, a(n) for n = 1..200
Crossrefs
Cf. A200154.
Formula
Empirical: a(n) = a(n-1) +a(n-2) +a(n-4) -3*a(n-5) -a(n-6) +a(n-8) +3*a(n-9) -a(n-10) -a(n-12) -a(n-13) +a(n-14).
Empirical g.f.: x*(2 + 13*x + 23*x^2 + 48*x^3 + 57*x^4 + 67*x^5 + 60*x^6 + 48*x^7 + 26*x^8 + 11*x^9 + x^10 - x^12 + x^13) / ((1 - x)^5*(1 + x)^3*(1 + x^2)^2*(1 + x + x^2)). - Colin Barker, May 17 2018
Comments