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

Author

Martins Opmanis, May 14 2010

Keywords

• nonn
• base
• nice

Comments

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)

Examples

			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.
		

Links

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

Crossrefs

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

Programs

Mathematica

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 *)

Python

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

Extensions

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

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