A189323 Number of nondecreasing arrangements of n+2 numbers in 0..6 with the last equal to 6 and each after the second equal to the sum of one or two of the preceding four.
10, 18, 36, 64, 110, 179, 275, 393, 528, 676, 836, 1008, 1192, 1388, 1596, 1816, 2048, 2292, 2548, 2816, 3096, 3388, 3692, 4008, 4336, 4676, 5028, 5392, 5768, 6156, 6556, 6968, 7392, 7828, 8276, 8736, 9208, 9692, 10188, 10696, 11216, 11748, 12292, 12848
Offset: 1
Keywords
Examples
Some solutions for n=3: ..1....1....2....2....1....3....1....2....2....0....1....1....3....2....1....1 ..5....6....3....4....4....6....5....2....4....3....3....3....3....3....5....3 ..5....6....3....4....5....6....5....4....4....3....4....3....3....3....6....3 ..5....6....6....4....5....6....6....6....6....3....5....4....6....3....6....6 ..6....6....6....6....6....6....6....6....6....6....6....6....6....6....6....6
Links
- R. H. Hardin, Table of n, a(n) for n = 1..200
Crossrefs
Cf. A189326.
Formula
Empirical: a(n) = 6*n^2 + 34*n - 264 for n>8.
Empirical g.f.: x*(10 - 12*x + 12*x^2 + 8*x^4 + 5*x^5 + 4*x^6 - 5*x^7 - 5*x^8 - 4*x^9 - x^10) / (1 - x)^3. - Colin Barker, May 02 2018
Comments