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

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
print(list(islice(agen(), 20))) # Michael S. Branicky, Jun 09 2023

Crossrefs

Cf. A177834, A175516, A003418.

Sequence in context: A261467 A180070 A177834 * A292132 A208147 A369330

Adjacent sequences: A169855 A169856 A169857 * A169859 A169860 A169861

Keyword

nonn,base

Author

Hugo van der Sanden, Jun 01 2010

Extensions

Corrected and extended by Hugo van der Sanden, Jun 04 2010 (errors reported by Zak Seidov).

Status

approved