A169858 - OEIS

A169858

Smallest integer k such that k or one of its left substrings (or prefixes, regarded as an integer) is divisible by any integer from {1,2,...,n}.

1, 2, 6, 12, 45, 60, 245, 245, 504, 504, 5049, 5049, 10296, 11760, 11760, 11760, 56160, 56160, 198016, 198016, 1008159, 1323008, 2340849, 6240366, 13442580, 13442580, 37536408, 37536408, 75432065, 75432065, 180092645, 319800096, 319800096, 800640126, 2201169600, 2201169600, 3780487275, 5250966084, 5250966084, 6832425609, 36960308625, 36960308625, 62244072512, 62244072512, 62244072512, 62244072512, 372960042489, 372960042489

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OFFSET

1,2

LINKS

Table of n, a(n) for n=1..48.

Hugo van der Sanden A169858: C source code to calculate terms.

FORMULA

a(n) = min m: forall d in {1..n}: exists k in {0..log_10(m)}: d | floor(m / 10^k).

a(n) <= A003418(n). - Michael S. Branicky, Jun 09 2023

EXAMPLE

a(5) = 45 as the left substrings of 45 are {4, 45} and for every d in {1,2,...,n} = {1, 2, 3, 4, 5} there is a left substring of 45 such that d | 45. That is: 1 | 4, 2 | 4, 3 | 45, 4 | 4, 5 | 45. - David A. Corneth, Jun 09 2023

PROG

(Python)

from itertools import count, islice

def agen(): # generator of terms

n = 1

for k in count(1):

s = str(k)

prefixes = [int(s[:i+1]) for i in range(len(s))]

if all(any(ki%m == 0 for ki in prefixes) for m in range(1, n+1)):

yield k; n += 1

while any(ki%n == 0 for ki in prefixes):

yield k; n += 1