A169858 -id:A169858 - cp's OEIS frontend

A177834 Opmanis's sequence: a(n) is the smallest integer k such that k or one of its nonzero substrings (regarded as an integer) is divisible by every integer in the range 1 through n.

1, 2, 6, 12, 45, 54, 56, 56, 245, 504, 1440, 1440, 5044, 5044, 10456, 10569, 11704, 11704, 11704, 13608, 13608, 13608, 26460, 26460, 198007, 258064, 264600, 264600, 475440, 475440, 1754608, 1754608, 2258064, 2258064, 2646004, 2646004, 2992520

Offset: 1

Martins Opmanis, May 14 2010

Comment from N. J. A. Sloane, May 28 2010: (Start)
The factorizations of the initial terms are:
1, 2, 2*3, 2^2*3, 3^2*5, 2*3^3, 2^3*7, 2^3*7, 5*7^2, 2^3*3^2*7, 2^5*3^2*5, 2^5*3^2*5, 2^2*13*97, 2^2*13*97, 2^3*1307, 3*13*271, 2^3*7*11*19,
2^3*7*11*19, 2^3*7*11*19, 2^3*3^5*7, 2^3*3^5*7, 2^3*3^5*7, 2^2*3^3*5*7^2, 2^2*3^3*5*7^2, 23*8609, 2^4*127^2, 2^3*3^3*5^2*7^2, 2^3*3^3*5^2*7^2, 2^4*3*5*7*283,
2^4*3*5*7*283, 2^4*109663, 2^4*109663, 2^4*3^3*5227, 2^4*3^3*5227, 2^2*139*4759, 2^2*139*4759, 2^3*5*79*947, ...
The name "Opmanis's sequence" is due to N. J. A. Sloane, not the author. (End)

			a(8)=56 because 56 is divisible by 1,2,4,7,8; 5 is divisible by 5; 6 is divisible by 3 and 6. Therefore the set {1,2,3,4,5,6,7,8} is covered by the divisors. 56 is the smallest number with this property.

Robert Gerbicz, Table of n, a(n) for n = 1..102

Cf. A003418 (a weak upper bound), A169819, A169858, A178544.

k = 1; lst = {}; mx = 0; f[n_] := Block[{a, d, id = IntegerDigits@ n}, a = Complement[ Union[ FromDigits /@ Flatten[ Table[ Partition[ id, k, 1], {k, Length@ id}], 1]], {0}]; d = Union[ Flatten[ Divisors /@ a]]; Complement[ Range@ 100, d][[1]] - 1]; While[k < 3000000, a = f@k; If[a > mx, Print[{a, k}]; AppendTo[lst, k]; mx = a]; k++ ] (* Zak Seidov & Robert G. Wilson v, May 30 2010 *)

def substrings(n): # returns set of nonzero substrings of n
    s = str(n)
    ss = (s[i:j] for i in range(len(s)) for j in range(i+1, len(s)+1))
    return set(int(sij) for sij in ss) - {0}
def a(n, startk=1):
    k = startk
    while True:
        subsk = substrings(k)
        if all(any(kij%m == 0 for kij in subsk) for m in range(1, n+1)):
            return k
        k += 1
def afind():
    n, an = 1, 1
    while True:
        n, an = n+1, a(n, startk=an)
        print(an, end=", ")
afind() # Michael S. Branicky, Jan 22 2022

Edited by N. J. A. Sloane, May 28 2010

a(1)-a(37) confirmed by Zak Seidov, May 28 2010

A175516 a(n) = smallest positive even integer k such that k or one of its left substrings is divisible by any integer from {1..n}.

Original entry on oeis.org

2, 2, 6, 12, 60, 60, 252, 504, 504, 504, 7260, 7260, 10296, 11760, 11760, 11760, 56160, 56160, 198016, 198016, 1176480, 1323008, 2992500, 6240366, 13442580, 13442580, 37536408, 37536408, 90725804, 90725804, 319800096, 319800096, 319800096, 800640126, 2201169600, 2201169600, 4656965040, 5250966084, 5250966084

Offset: 1

Zak Seidov, Jun 03 2010

Hugo van der Sanden, A169858: C source code to calculate terms (has an option to compute A175516).

Cf. A169858.

from itertools import count, islice
def agen(): # generator of terms
    n = 1
    for k in count(2, 2):
        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(3, 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

Corrected and extended by Hugo van der Sanden, Jun 01 2010

cp's OEIS Frontend

A177834 Opmanis's sequence: a(n) is the smallest integer k such that k or one of its nonzero substrings (regarded as an integer) is divisible by every integer in the range 1 through n.

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