A371641 The smallest composite number which divides the concatenation of its ascending ordered prime factors, with repetition, when written in base n.

85329, 4, 224675, 4, 1140391, 4, 9, 4, 28749, 4, 841, 4, 9, 4, 239571, 4, 343, 4, 9, 4, 231, 4, 25, 4, 9, 4, 315, 4, 343, 4, 9, 4, 25, 4, 9761637601, 4, 9, 4, 4234329, 4, 715, 4, 9, 4, 609, 4, 49, 4, 9, 4, 195, 4, 25, 4, 9, 4, 1015, 4, 76729, 4, 9, 4, 25, 4, 14332171

Offset: 2

Author

Scott R. Shannon, Mar 30 2024

Keywords

• nonn
• base

Comments

The number 4 = 2*2 in any base b = 3 + 2*n, n >= 0, will always divide the concatenation of its prime divisors as 4 = "2"_b + "2"_b = "22"_b = 2*(3 + 2*n) + 2 = 8 + 4*n, which is divisible by 4.

The number 9 = 3*3 in any base b = 8 + 6*n, n >= 0, will always divide the concatenation of its prime divisors as 9 = "3"_b + "3"_b = "33"_b = 3*(8 + 6*n) + 3 = 27 + 18*n, which is divisible by 9.

Theorem: if p is prime, then a(p*(m+2)-1) <= p^2 for all m >= 0. Proof: If p is prime and base b = p*(m+2)-1 for some m >= 0, then b > p and p expressed in base b = "p"b and thus "pp"_b = p*(b+1) = p^2*(m+2), i.e., divisible by p^2. - _Chai Wah Wu, Apr 01 2024

a(36) <= 9761637601. a(40) = 4234329. By the theorem above if n+1 is composite, then a(n) <= p^2 where p = A020639(n+1) is the smallest prime factor of n+1. The first few terms where the inequality is strict are: a(406) = 105, a(766) = 105, a(988) = 195, a(1036) = 105, a(1072) = 231, ... - Chai Wah Wu, Apr 11 2024

From Chai Wah Wu, Apr 12 2024: (Start)

Theorem: If n is even, then a(n) >= 9 is odd.

Proof: This is true for n = 2. Let n > 2 be even. Let m be even.

If m=2^k for k>1, then 2 concatenated k times in base n is 2*(n^(k-1)+...+n+1) = 2*(n^k-1)/(n-1).

Since n^k-1 is odd, m does not divide 2*(n^k-1)/(n-1). If m has an odd prime divisor then concatenating the primes in base n will result in an odd number that is not divisible by m.

Finally, a(n) >= 9 since the first odd composite number is 9. (End)

Examples

			a(2) = 85329 as 85329 = 3_10 * 3_10 * 19_10 * 499_10 = 11_2 * 11_2 * 10011_2 * 111110011_2 = "111110011111110011"_2 = 255987_10 which is divisible by 85329.
a(10) = 28749 as 28749 =  3_10 * 7_10 * 37_10 * 37_10 = "373737"_10 = 373737_10 which is divisible by 28749. See also A259047.
		

Links

• Michael S. Branicky, Table of n, a(n) for n = 2..129

Crossrefs

Cf. A020639, A027746, A259047, A322843, A248915.

Programs

PARI

has(F,n)=my(f=F[2],t); for(i=1,#f~, my(p=f[i,1],d=#digits(p,n),D=n^d); for(j=1,f[i,2], t=D*t+p)); t%F[1]==0
a(k,lim=10^6,startAt=4)=forfactored(n=startAt,lim, if(vecsum(n[2][,2])>1 && has(n,k), return(n[1]))); a(k,2*lim,lim+1) \\ Charles R Greathouse IV, Apr 11 2024

Python

from itertools import count
from sympy.ntheory import digits
from sympy import factorint, isprime
def fromdigits(d, b):
    n = 0
    for di in d: n *= b; n += di
    return n
def a(n):
    for k in count(4):
        if isprime(k): continue
        sf = []
        for p, e in factorint(k).items():
            sf.extend(e*digits(p, n)[1:])
        if fromdigits(sf, n)%k == 0:
            return k
print([a(n) for n in range(2, 6)]) # Michael S. Branicky, Apr 01 2024

Python

from itertools import count
from sympy import factorint, integer_log
def A371641(n):
    for m in count(4):
        f = factorint(m)
        if sum(f.values()) > 1:
            c = 0
            for p in sorted(f):
                a = pow(n,integer_log(p,n)[0]+1,m)
                for _ in range(f[p]):
                    c = (c*a+p)%m
            if not c:
                return m # Chai Wah Wu, Apr 11 2024

Extensions

a(36) and beyond from Michael S. Branicky, Apr 27 2024