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A371699 The smallest composite number which divides the concatenation of its descending ordered prime factors, with repetition, when written in base n.

%I A371699 #30 Apr 13 2024 23:17:23
%S A371699 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,
%T A371699 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,
%U A371699 25,4,9,4,238,4,1131,4,9,4,25,4,555,4,9,4,14405,4,12207
%N A371699 The smallest composite number which divides the concatenation of its descending ordered prime factors, with repetition, when written in base n.
%C A371699 Conjecture: a(n) <= A371641(n) for n >= 2.
%H A371699 Chai Wah Wu, <a href="/A371699/b371699.txt">Table of n, a(n) for n = 2..10000</a>
%F A371699 If p is prime, then a(p*(m+2)-1) <= p^2 for all m >= 0.
%F A371699 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.
%F A371699 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, ...
%F A371699 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.
%F A371699 Suppose n>2 is even. 2 concatenated twice in base n is 2(n+1) which is not divisible by 4.
%F A371699 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.
%F A371699 This implies that if n = 6*k+2 for some k > 0, then a(n) = A371641(n) = 9.
%e A371699 a(2) = 42 since 42 = 7*3*2 = 111_2 * 11_2 * 10 _2 and 42 divides 126 = 1111110_2.
%e A371699 a(10) = 378 since 278 = 7*3*3*3*2 and 278 divides 73332.
%o A371699 (Python)
%o A371699 from itertools import count
%o A371699 from sympy import factorint, integer_log
%o A371699 def A371699(n):
%o A371699     for m in count(4):
%o A371699         f = factorint(m)
%o A371699         if sum(f.values()) > 1:
%o A371699             c = 0
%o A371699             for p in sorted(f,reverse=True):
%o A371699                 a = pow(n,integer_log(p,n)[0]+1,m)
%o A371699                 for _ in range(f[p]):
%o A371699                     c = (c*a+p)%m
%o A371699             if not c:
%o A371699                 return m
%Y A371699 Cf. A020639, A027746, A259047, A322843, A248915, A371641, A371821.
%K A371699 nonn,base
%O A371699 2,1
%A A371699 _Chai Wah Wu_, Apr 12 2024