A342034 a(n) is the number of numbers k with n digits where k has digits in nondecreasing order and satisfies k < (product of digits of k) * (sum of digits of k).

8, 41, 140, 367, 789, 1432, 2276, 3280, 4326, 5350, 6254, 7009, 7588, 7970, 8175, 8210, 8120, 7923, 7633, 7272, 6877, 6445, 6013, 5555, 5122, 4693, 4298, 3901, 3534, 3189, 2872, 2562, 2285, 2029, 1789, 1576, 1376, 1194, 1037, 893, 759, 654, 548, 454, 384, 315, 254, 210, 168, 127, 97, 79, 56, 39, 31, 21, 12, 8, 4

Offset: 1

Author

David A. Corneth, Mar 05 2021

Keywords

• nonn
• easy
• base

Comments

As A066310 is finite there exists m such that a(n) = 0 for all n > m.

a(n) = 0 for n >= 85 since 9^n*9n <= 10^(n-1) for n >= 85. This may occur as early as n = 60, as 9^n*9n <= 10^n-1 for n >= 60. But a(59) > 0 since 10^59-1 < 9^59*9*59. - Michael S. Branicky, Mar 05 2021

Examples

			a(1) = 8 as there are 8 one-digit numbers k as described in name. Those are {2, 3, 4, 5, 6, 7, 8, 9}.
		

Links

• David A. Corneth, PARI program

Crossrefs

Cf. A009994, A066310.

Programs

PARI

See PARI link

Python

from math import prod
from itertools import combinations_with_replacement as cwr
def c(digs): return int("".join(map(str, digs))) < prod(digs) * sum(digs)
def a(n): return sum(1 for u in cwr(range(1, 10), n) if c(u))
print([a(n) for n in range(1, 16)]) # Michael S. Branicky, Mar 05 2021

Extensions

a(27)-a(41) from Michael S. Branicky, Mar 05 2021