A347892 Nonsquarefree numbers k such that A003968(k) is a multiple of k, where A003968 is multiplicative with a(p^e) = p*(p+1)^(e-1).

36, 72, 180, 216, 252, 360, 396, 432, 468, 504, 612, 684, 792, 828, 864, 936, 1044, 1080, 1116, 1224, 1260, 1296, 1332, 1368, 1476, 1512, 1548, 1656, 1692, 1908, 1980, 2088, 2124, 2160, 2196, 2232, 2340, 2376, 2412, 2520, 2556, 2592, 2628, 2664, 2772, 2808, 2844, 2952, 2988, 3024, 3060, 3096, 3204, 3276, 3384, 3420

Offset: 1

Author

Antti Karttunen, Oct 29 2021

Keywords

• nonn

Comments

All terms k are multiples of 36 and A056170(k) = 2. Proof: Imagine k had a non-unitary prime divisor p^e, with p > 3 and e > 1. Then p^e divides A003968(k) only if k has also another non-unitary prime divisor q^h (with h > 1), such that p divides (q+1), which implies that q > p. But then q^h divides A003968(k) only if there is yet another non-unitary prime divisor r^i, such that r > q (and i > 1), and so on, which is clearly impossible by reductio ad infinitum. Therefore we should consider only the cases p=2 and p=3, because they are only primes that can occur as non-unitary prime factors in k, and at least either of them must occur with exponent larger than one, because every k is nonsquarefree. Let e = A007814(k) and h = A007949(k), so that 2^e and 3^h are the highest powers of 2 and 3 that divide k. Because A003968 changes "extra" 2's to 3's and extra 3's to 4's, it must follow that e >= h > e/2. Therefore, if e >= 2 (k is a multiple of 4), h must be at least 2. On the other hand, if h >= 2, then e also must be at least 2. In other words, if k is a multiple of 4, it must then also be a multiple of 9, and vice versa, thus k is a multiple of 36 and k has exactly two non-unitary prime divisors (2^e and 3^h, with e, h > 1), therefore this is a subsequence of A338539.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..20000

Crossrefs

Intersection of A013929 and A348499. Subsequence of A338539.

Cf. A003968, A007814, A007949, A056170.

Programs

Mathematica

f[p_, e_] := p*(p + 1)^(e - 1); s[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[3500], ! SquareFreeQ[#] && Divisible[s[#], #] &] (* Amiram Eldar, Oct 29 2021 *)

PARI

A003968(n) = {my(f=factor(n)); for (i=1, #f~, p= f[i, 1]; f[i, 1] = p*(p+1)^(f[i, 2]-1); f[i, 2] = 1); factorback(f); }
isA347892(n) = (!issquarefree(n) && !(A003968(n)%n));