A212865 Number of nondecreasing sequences of n 1..5 integers with no element dividing the sequence sum.
0, 5, 9, 15, 22, 32, 40, 59, 74, 97, 124, 159, 188, 229, 260, 301, 347, 415, 477, 559, 630, 715, 801, 897, 987, 1106, 1214, 1342, 1471, 1623, 1760, 1934, 2099, 2287, 2475, 2683, 2878, 3116, 3334, 3581, 3832, 4115, 4377, 4681, 4968, 5283, 5605, 5965, 6310, 6707
Offset: 1
Keywords
Examples
Some solutions for n=8: ..2....3....2....2....3....2....2....2....3....2....4....2....3....4....2....2 ..2....3....2....3....3....4....5....3....3....2....4....2....3....5....3....2 ..2....3....2....3....3....4....5....4....4....3....4....3....3....5....3....2 ..2....3....3....3....3....4....5....4....4....3....4....3....3....5....3....4 ..2....3....5....3....3....4....5....4....5....4....4....3....3....5....4....4 ..3....4....5....3....3....4....5....4....5....5....4....4....3....5....4....5 ..3....5....5....3....3....4....5....5....5....5....4....4....4....5....5....5 ..3....5....5....3....4....5....5....5....5....5....5....4....4....5....5....5
Links
- R. H. Hardin, Table of n, a(n) for n = 1..210
Formula
Empirical: a(n) = 2*a(n-1) -2*a(n-3) +a(n-5) +2*a(n-6) -2*a(n-7) -2*a(n-8) +2*a(n-9) +a(n-10) -a(n-12) -2*a(n-13) +2*a(n-14) +2*a(n-15) -2*a(n-16) -a(n-17) +2*a(n-19) +a(n-20) -4*a(n-21) +a(n-22) +2*a(n-23) -a(n-25) -2*a(n-26) +2*a(n-27) +2*a(n-28) -2*a(n-29) -a(n-30) +a(n-32) +2*a(n-33) -2*a(n-34) -2*a(n-35) +2*a(n-36) +a(n-37) -2*a(n-39) +2*a(n-41) -a(n-42).
If the above empirical recurrence by R. H. Hardin is correct, then the denominator of the g.f. (that determines the above recurrence) equals (1-x)^2*(1-x^2)*(1-x^12)*(1-x^15)*(1-x^20)/((1-x^4)*(1-x^5)). - Petros Hadjicostas, Sep 09 2019
Comments