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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).

%I A347892 #28 Feb 10 2022 20:06:25
%S A347892 36,72,180,216,252,360,396,432,468,504,612,684,792,828,864,936,1044,
%T A347892 1080,1116,1224,1260,1296,1332,1368,1476,1512,1548,1656,1692,1908,
%U A347892 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
%N 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).
%C A347892 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.
%H A347892 Antti Karttunen, <a href="/A347892/b347892.txt">Table of n, a(n) for n = 1..20000</a>
%t A347892 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 *)
%o A347892 (PARI)
%o A347892 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); }
%o A347892 isA347892(n) = (!issquarefree(n) && !(A003968(n)%n));
%Y A347892 Intersection of A013929 and A348499. Subsequence of A338539.
%Y A347892 Cf. A003968, A007814, A007949, A056170.
%K A347892 nonn
%O A347892 1,1
%A A347892 _Antti Karttunen_, Oct 29 2021