A189328 Number of nondecreasing arrangements of 5 numbers in 0..n with the last equal to n and each after the second equal to the sum of one or two of the preceding four.
2, 8, 11, 20, 21, 36, 31, 49, 42, 63, 51, 79, 60, 93, 72, 105, 80, 125, 89, 133, 104, 149, 109, 168, 117, 178, 135, 190, 138, 213, 147, 219, 166, 234, 166, 257, 176, 263, 197, 274, 196, 303, 205, 304, 227, 319, 225, 346, 234, 347, 259, 360, 254, 392, 262, 389, 290, 404
Offset: 1
Keywords
Examples
All solutions for n=3: ..1....1....1....1....0....1....3....0....2....1....1 ..2....2....3....2....1....1....3....3....3....1....1 ..2....2....3....3....1....2....3....3....3....2....1 ..3....2....3....3....2....3....3....3....3....2....2 ..3....3....3....3....3....3....3....3....3....3....3
Links
- R. H. Hardin, Table of n, a(n) for n = 1..200
Crossrefs
Cf. A189326.
Formula
Empirical: a(n) = -3*a(n-1) -5*a(n-2) -5*a(n-3) -2*a(n-4) +3*a(n-5) +8*a(n-6) +10*a(n-7) +8*a(n-8) +3*a(n-9) -2*a(n-10) -5*a(n-11) -5*a(n-12) -3*a(n-13) -a(n-14).
Empirical g.f.: x*(2 + 14*x + 45*x^2 + 103*x^3 + 180*x^4 + 264*x^5 + 326*x^6 + 350*x^7 + 322*x^8 + 258*x^9 + 173*x^10 + 97*x^11 + 40*x^12 + 11*x^13) / ((1 - x)^2*(1 + x)^2*(1 + x^2)*(1 + x + x^2)^2*(1 + x + x^2 + x^3 + x^4)). - Colin Barker, May 02 2018
Comments