%I #23 Feb 16 2019 21:12:53
%S 1,2,3,4,5,6,7,8,9,10,11,12,13,14,16,17,18,19,20,22,23,24,25,26,27,28,
%T 29,31,32,34,36,37,38,40,41,43,44,46,47,48,49,50,52,53,54,56,58,59,61,
%U 62,64,67,68,71,72,73,74,76,79,80,81,82,83,86,87,88,89,92,94,96,97,98,100,101,103,104,106,107,108,109,112,113,116,118,121
%N Numbers n such that A324108(n) = A324054(n-1).
%C Numbers n such that A324054(n-1) is equal to A324108(n), which is a multiplicative function with A324108(p^e) = A324054((p^e)-1).
%C Prime powers (A000961) is a subsequence by definition.
%C Also A070776 is a subsequence. This follows because for every n of the form 2^i * p^j (where p is an odd prime, and i >= 0, j >= 0), we have A324108(2^i * p^j) = A324054(2^i - 1)*A324054(p^j - 1) = sigma(A005940(2^i)) * sigma(A005940(p^j)). Because A005940(1) = 1, and A005940(2n) = 2*A005940(n), the powers of two are among the fixed points of A005940 (cf. A029747), thus the left half of product is sigma(2^i), while on the other hand, we know that A005940(p^j) is odd (because A005940 also preserves parity), and thus the whole product is equal to sigma(2^i * A005940(p^j)) = sigma(A005940(2^i * p^j)) = A324054((2^i * p^j)-1).
%C See subsequence A324111 for less regular solutions.
%H Antti Karttunen, <a href="/A324109/b324109.txt">Table of n, a(n) for n = 1..10001</a>
%o (PARI)
%o A324054(n) = { my(p=2,mp=p*p,m=1); while(n, if(!(n%2), p=nextprime(1+p); mp = p*p, if(3==(n%4),mp *= p,m *= (mp-1)/(p-1))); n>>=1); (m); };
%o A324108(n) = { my(f=factor(n)); prod(i=1, #f~, A324054((f[i,1]^f[i,2])-1)); };
%o isA324109(n) = (A324054(n-1)==A324108(n));
%o for(n=1,121,if(isA324109(n), print1(n,", ")));
%Y Union of A070776 and A324111.
%Y Cf. A000961 (a subsequence), A029747, A324054, A324107, A324108, A324110 (complement).
%K nonn
%O 1,2
%A _Antti Karttunen_ and _David A. Corneth_, Feb 15 2019