A371699 The smallest composite number which divides the concatenation of its descending ordered prime factors, with repetition, when written in base n.
42, 4, 42, 4, 374, 4, 9, 4, 378, 4, 609, 4, 9, 4, 3525, 4, 343, 4, 9, 4, 70, 4, 25, 4, 9, 4, 195, 4, 343, 4, 9, 4, 25, 4, 130, 4, 9, 4, 366, 4, 3562, 4, 9, 4, 42, 4, 49, 4, 9, 4, 474, 4, 25, 4, 9, 4, 238, 4, 1131, 4, 9, 4, 25, 4, 555, 4, 9, 4, 14405, 4, 12207
Offset: 2
Examples
a(2) = 42 since 42 = 7*3*2 = 111_2 * 11_2 * 10 _2 and 42 divides 126 = 1111110_2. a(10) = 378 since 278 = 7*3*3*3*2 and 278 divides 73332.
Links
- Chai Wah Wu, Table of n, a(n) for n = 2..10000
Programs
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Python
from itertools import count from sympy import factorint, integer_log def A371699(n): for m in count(4): f = factorint(m) if sum(f.values()) > 1: c = 0 for p in sorted(f,reverse=True): 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
Formula
If p is prime, then a(p*(m+2)-1) <= p^2 for all m >= 0.
If n+1 is composite, then a(n) <= q^2, where q = A020639(n+1) is the smallest prime factor of n+1. This implies that if n > 2 is odd, then a(n) = A371641(n) = 4.
The first few terms n where n+1 is composite and a(n) < A020639(n+1)^2 are a(288) = 70, a(298) = 42, a(340) = 42, a(360) = 182, ...
If n is even, then a(n) >= 9. This is true as it is easy to verify that a(n) cannot be equal to 4, 6 or 8 in this case.
Suppose n>2 is even. 2 concatenated twice in base n is 2(n+1) which is not divisible by 4.
Next, 3 concatenated by 2 is 3*n+2 which is not divisible by 6. Finally 2 concatenated 3 times is 2(n^3-1)/(n-1) which is not divisible by 8 since n^3-1 is odd.
This implies that if n = 6*k+2 for some k > 0, then a(n) = A371641(n) = 9.
Comments