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

1,2

Comments

Column 5 of A212868.

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

Example

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

Crossrefs

Sequence in context: A315080 A315081 A075343 * A102176 A315082 A315083

Adjacent sequences: A212862 A212863 A212864 * A212866 A212867 A212868

Keyword

nonn

Author

R. H. Hardin, May 29 2012

Status

approved