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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), so we can apply the multiplicativity of sigma (in reverse), 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).

See subsequence A324111 for more "irregular" less regular solutions.

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 (See also cf. A029747), thus the left half of product is sigma(2^i) = 2^(i+1) - 1, , while on the other hand, we know that A005940(p^j) is odd (because A005940 also preserves parity), so we can apply the multiplicativity of sigma, (in reverse), and thus the whole product is (2^(i+i)-1)*A005940(p^j), which is equal to sigma(2^i * A005940(p^j)) = sigma(A005940(2^i * p^j)) = A324054((2^i * p^j)-1).

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)). Two's Because A005940(1) = 1, and A005940(2n) = 2*A005940(n), the powers of two are among the fixed points of A005940 (because A005940(2n) = 2*A005940(n). See also A029747), thus the left half of product is sigma(2^i) = 2^(i+1) - 1. On , while on the other hand, we know that A005940(p^j) is odd (because A005940 also preserves parity), thus so we can apply the multiplicativity of sigma, and thus the whole product is (2^(i+i)-1)*A005940(p^j), which is equal to A324054((2^i * p^j)-1) = sigma(2^i * A005940(2^i * p^j)) = sigma(2^i * A005940(2^i * p^j)) = A324054((2^i * p^j)-1).

Subsequence See subsequence A324111 gives the for more "irregular" solutions.

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), and as A324054(2^i - 1) = A000203sigma(A005940(2^i)) = A000203(2^i) = 2^(i+1) - 1. Let's say k = A324054(p^j - 1) = A000203* sigma(A005940(p^j)). On Two's powers are among the other hand, A324054((2^i*p^j) - 1) = A000203(fixed points of A005940(2^i * p^j)) = A000203 (2^i * because A005940(p^j)2n) = A000203(2^i) * A005940(pn). See also A029747), thus the left half of product is sigma(2^ji) = (2^(i+1) - 1)*k [. On the other hand, we know that A00590A005940(p^j) is odd, (because A005940 preserves parity), thus we can apply the multiplicativity of sigma], and thus the whole product is (2^(i+i)-1)*A005940(p^j), which is equal to A324054((2^i * p^j)-1) = sigma(A005940(2^i * p^j)) = sigma(2^i * A005940(p^j)).

Sequence Subsequence A324111 gives the more "irregular " solutions.

Cf. A000961 (a subsequence), A029747, A324054, A324107, A324108, A324110 (complement).

Also A070776 is a subsequence, 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), and as A324054(2^j i - 1) = A000203(A005940(2^ji)) = A000203(2^ji) = 2^(ji+1) - 1 and . Let's say k = A324054(p^j - 1) = A000203(A005940(p^j)). On the other hand, A324054((2^i*p^j ) - 1) = A000203(A005940(2^i * p^j)) = A000203(2^i * A005940(p^j)) = A000203(2^i) * A005940(p^j) = (2^(i+1)-1)*k, [we know that A00590(p^j) is odd, thus we can apply the multiplicativity of sigma].

and on Sequence A324111 gives the other hand,more irregular solutions.

A324054((2^i*p^j) - 1) = A000203(A005940(2^i * p^j)) =

A324054(p^j - 1) = . Sequence A324111 gives the more irregular solutions.

Also A070776 is a subsequence, 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), and as A324054(2^j - 1) = A000203(A005940(2^j)) = A000203(2^j) = 2^(j+1) - 1. and A324054(p^j - 1) = A000203(A005940(p^j)) = k,

and on the other hand,

A324054((2^i*p^j) - 1) = A000203(A005940(2^i * p^j)) =

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), is equal to A324054(n-1).

Also A070776 is a subsequence, 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). , and as A324054(2^j - 1) = A000203(A005940(2^j)) = A000203(2^j) = 2^(j+1) - 1.

Numbers n such that A324108(n) which is a multiplicative function with A324108(p^e) = A324054((p^e)-1), is equal to A324054(n-1).

Prime powers (A000961) is a subsequence by definition.

Also A070776 is a subsequence, 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) = . Sequence A324111 gives the more irregular solutions. A324054(2^j - 1) = A000203(A005940(2^j)) =

A324054(p^j - 1) = . Sequence A324111 gives the more irregular solutions.