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A335879
a(n) = A332215(A335882(n)).
1
15, 5, 30, 63, 255, 10, 60, 13, 126, 2047, 510, 20, 120, 26, 252, 4094, 262143, 11, 1020, 4194303, 40, 240, 52, 504, 8188, 61, 524286, 22, 2040, 8388606, 80, 480, 104, 1008, 16376, 122, 1048572, 140737488355327, 44, 4080, 59, 4503599627370495, 16777212, 160, 960, 208, 2016, 32752, 244, 2097144, 253, 281474976710654, 2417851639229258349412351
OFFSET
1,1
COMMENTS
For all n, a(n) <> A335882(n). Proof: We need to consider only the odd terms, because for n > 1, A332215(2^k * n) = 2^k * A332215(n). The odd terms of A335882 are either primes or semiprimes whose both factors are Mersenne primes, terms of A144482.
(A) If A335882(n) is a prime, then a(n) = A332215(A335882(n)) is a term of A000225 (of the form 2^k - 1, a binary repunit), while primes in A335882 are certainly not of that form, as all Mersenne primes (A000668) are on a different row in array A335430 (on row 1, A335431).
(B) For any semiprime k in A335882, there is only one non-leading zero in the binary representation of A332215(k). On the other hand, a product of two Mersenne primes always contains more than one non-leading zero in its base-2 representation: for three times a Mersenne prime, there are two such zeros, as explained in A279389, and products of two Mersenne primes > 3 are always of the form 8k+1, with at least two zeros immediately left of the least significant 1-bit.
FORMULA
a(n) = A332215(A335882(n)).
For all n >= 1, A007814(a(n)) = A007814(A335882(n)).
KEYWORD
nonn
AUTHOR
Antti Karttunen, Jul 10 2020
STATUS
approved