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Numbers k such that A051027(k) = Product_{p^e||k} A051027(p^e) = A353802(n). Here each p^e is the maximal prime power divisor of k, and A051027(k) = sigma(sigma(k)). Numbers at which points A051027 appears to be multiplicative.

Proof that this interpretation is equal to the main definition:

(1) If none of sigma(p_1^e_1), ..., sigma(p_k^e_k) share prime factors, then A051027(k) = sigma(sigma(p_1^e_1) * ... * sigma(p_k^e_k)) = A051027(p_1^e_1) * ... * A051027(p_k^e_k), by multiplicativity of sigma.

(2) On the other hand, if say, gcd(sigma(p_i^e_i), sigma(p_j^e_j)) = c > 1 for some distinct i, j, then that c has at least one prime factor q, with product t = sigma(p_1^e_1) * ... * sigma(p_k^e_k) having a divisor of the form q^v (where v = valuation(t,q)), and the same prime factor q occurs as a divisor in more than one of the sigma(p_i^e_j), in the form q^k, with the exponents summing to v, then it is impossible to form sigma(q^v) = (1 + q + q^2 + ... + q^v) as a product of some sigma(q^k_1) * ... * sigma(q^k_z), i.e., as a product of (1 + q + ... + q^k_1) * ... * (1 + q + ... + q^k_z), with v = k_1 + ... + k_z, because such a product is always larger than (1 + q + ... + q^v). And if there are more such cases of "split primes", then each of them brings its own share to this monotonic inequivalence, thus Product_{p^e|n} A051027(p^e) = A353802(n) >= A051027(n), for all n.

From Antti Karttunen, May 07 2022: (Start)

Also numbers k such that A062401(k) = phi(sigma(n)) = Product_{p^e||k} A062401(p^e) = A353752(n).

Proof that also this interpretation is equal to the main definition:

(1) like in (1) above, if none of sigma(p_1^e_1), ..., sigma(p_k^e_k) share prime factors, then by the multiplicativity of phi.

(2) On the other hand, if say, gcd(sigma(p_i^e_i), sigma(p_j^e_j)) = c > 1 for some distinct i, j, then that c must have a prime factor q occurring in both sigma(p_i^e_i) and sigma(p_j^e_j), with say q^x being the highest power of q in the former, and q^y in the latter. Then phi(q^x)*phi(q^y) < phi(q^(x+y)), i.e. here the inequivalence acts to the opposite direction than with sigma(sigma(...)), so we have A353752(n) <= A062401(n) for all n.

(End)

From Antti Karttunen, May 22 2022: (Start)

All even perfect numbers (even terms of A000396) are included in this sequence. In general, for any perfect number n in this sequence, map k -> A026741(sigma(k)) induces on its unitary prime power divisors (p^e||n) a permutation that is a single cycle, mapping each one of them to the next larger one, except that the largest is mapped to the smallest one. Therefore, for a hypothetical odd perfect number n = x*a*b*c*d*e*f*g*h to be included in this sequence, where x is Euler's special factor of the form (4k+1)^(4h+1), and a .. h are even powers of odd primes (of which there are at least eight distinct ones, see P. P. Nielsen reference in A228058), further constraints are imposed on it: (1) that h < x < 2*a (here assuming that a, b, c, ..., g, h have already been sorted by their size, thus we have a < b < ... < g < h < x < 2*a, and (2), that we must also have sigma(a) = b, sigma(b) = c, ..., sigma(f) = g, sigma(h) = x, and sigma(x) = 2*a. Note that of the 400 initial terms of A008848, only its second term 81 is a prime power, so empirically this seems highly unlikely to ever happen.

(End)

These are the square roots of squares in A006532. - M. F. Hasler, Oct 23 2010

Also n such that the squarefree part of n (A007913) equals the squarefree part of sigma(n), A355928.

Also n such that abundancy of n, sigma(n)/n is a rational square. - Michel Marcus, Oct 06 2013

See A230043, resp. A230538, for n whose abundancy is a rational cube, resp. fourth power. - M. F. Hasler, Nov 02 2013

1, 5286126, 45025305, 54742416, ... are also hexagonal and their sum of divisors too. - Michel Marcus, Apr 20 2014

Conjecture: There are no 1's after the initial term. Remark: If there were some k = x^2 > 1, for which a(x) = 1, then sigma(k) would be a divisor of A003961(k). In other words, d = A350073(k) = A064989(sigma(k)) would be a divisor of k. Then, if that divisor were also a unitary divisor [with gcd(d,k/d) = 1], it would need to satisfy the equation sigma(k) = sigma(d) * sigma(k/d) = sigma(A064989(sigma(k))) * sigma(k/A064989(sigma(k))), because sigma is a multiplicative function. (Minor correction by Antti Karttunen, Jul 11 2023)

Note that if d = A064989(sigma(k)) were a unitary divisor of a square k, then sigma(k) would also be a square, the cases which are quite rare (see A008848 and A336547). Also compare to A349756. - Antti Karttunen, Jul 24 2022

Contains exactly odd terms of A008848.

Odd n is a term if and only if both n and sigma(n) are perfect squares.

The sum of divisors of a square is always odd, therefore these numbers have the property that x=n^2, y=sigma(x) and z=sigma(y) are all three odd.

This is the subsequence of odd terms of A008847.

The next term, if it exists, is > 10^11. - Donovan Johnson, Aug 24 2012

a(4), if it exists, satisfies sigma(a(4)) > 10^36. - Hiroaki Yamanouchi, Sep 10 2014

If n belongs to this sequence, it may have at most two distinct prime divisors. If n=p^k, then sigma(p^k) = (p^(k+1)-1)/(p-1) = r^2 for some prime r. For k=1, it trivially has the only solution n=3; while for k>1 it is a partial case of the Nagell-Ljunggren equation and has the only prime solution r=11 (see Bennett-Levin 2015) corresponding to n=3^4=81. If n=p^k*q^m, then sigma(n) = (p^(k+1)-1)/(p-1) * (q^(m+1)-1)/(q-1) = r^2 for some prime r, implying that (p^(k+1)-1)/(p-1) = (q^(m+1)-1)/(q-1) = r. Here k+1 and m+1 must be odd distinct primes. The Goormaghtigh conjecture would imply that its only solution is n=400 with (p,k,q,m)=(5,2,2,4). - Max Alekseyev, Apr 24 2015

This sequence is infinite since it contains all the numbers of the form 4^(k^2). - Giovanni Resta, May 28 2016