A181373 Least m>0 such that prime(n) divides S(m)=A007908(m)=123...m and all numbers obtained by cyclic permutations of its digits; 0 if no such m exists.

0, 2, 0, 100, 106, 120, 196, 102, 542, 400, 181, 21, 216, 372, 10446, 127, 10086, 616, 399, 1703, 196, 2009, 118, 12350, 516, 416, 13244, 884, 15462, 15146, 106, 1006942, 10762, 10814, 11634, 5808, 12408, 576, 30076, 4996, 25290, 1015092, 1108, 26874, 24036, 5994

Offset: 1

Author

M. F. Hasler and Marco Ripà, Jan 27 2011

Keywords

• nonn
• base

Comments

The first three primes 2, 3 and 5 are particular cases, cf. examples. It happens that all other primes < 47 are in A180346 (and therefore have a(n) < 1000). P=37 is the only one among them with a(n) < 100 (but m=123 is another possibility for this prime).

Conjecture: a(n) > 0 for n <> 1 and n <> 3. - Chai Wah Wu, Oct 06 2023

Least m>0 such that prime(n) divides both A007908(m) and 10^A058183(m)-1; or 0 if no such m exists. - Chai Wah Wu, Oct 07 2023

Examples

			For prime(1)=2, no such m can exist (consider e.g. the initial 1 is permuted to the end), therefore a(1)=0.
For prime(2)=3, we have S(2)=12 and the permutation 21 both divisible by 3, thus a(2)=2. (There are many m for which the divisibility property is satisfied; it is equivalent to 1+...+m=0 (mod 3), or equivalently the sum of all these digits is divisible by 3. Therefore, the permutations do not need to be checked.)
For prime(3)=5, similar to prime(1)=2, no such m can exist.
For prime(4)=7, it turns out the m=100 is the least possibility, i.e., 123...99100 and the permutations 234...991001, 345...9910012, ... 100123...99, (00)123...991, (0)123...9910 are all divisible by 7.
		

Links

• Chai Wah Wu, Table of n, a(n) for n = 1..4042

Crossrefs

Cf. A000040, A007908, A058183, A180346 (see references there), A181373.

Programs

PARI

A181373(p,LIM=999,MIN=1)={ p=prime(p); p!=2 & p!=5 & for(n=MIN,LIM, my(S=eval(concat(vector(n,i,Str(i)))),L=#Str(S)-1); S%p & next; for(k=1,L, (S=[1,10^L]*divrem(S,10)) % p & next(2)); return(n)) } /* highly unoptimized code, for illustration purpose */

Python

from sympy import prime
def A181373(n):
    s, p, l = '', prime(n), 0
    for m in range(1,10**6):
        u = str(m)
        s += u
        l += len(u)
        t = s
        if not int(t) % p:
            for i in range(l-1):
                t = t[1:]+t[0]
                if int(t) % p:
                    break
            else:
                return m
    else:
        return 'search limit reached.' # Chai Wah Wu, Nov 12 2015

Python

from itertools import count
from sympy import prime
def A181373(n):
    if n == 1 or n == 3: return 0
    p, c, q, a, b = prime(n), 0, 1, 10, 10
    for m in count(1):
        if m >= b:
            a = 10*a%p
            b *= 10
        c = (c*a + m) % p
        q = q*a % p
        if not (c or (q-1)%p):
            return m # Chai Wah Wu, Oct 07 2023

Formula

A007908( A181373(n) ) = 0 (mod A000040(n)).

Extensions

a(15)-a(31) from Chai Wah Wu, Nov 12 2015

a(32)-a(46) from Chai Wah Wu, Oct 06 2023