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

2,1

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)

Links

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

Example

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.

Prog

(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
(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

Crossrefs

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

Sequence in context: A194777 A375263 A031671 * A371821 A010083 A235968

Adjacent sequences: A371638 A371639 A371640 * A371642 A371643 A371644

Keyword

nonn,base

Author

Scott R. Shannon, Mar 30 2024

Extensions

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

Status

approved