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David A. Corneth, <a href="/A330687/b330687.txt">Table of n, a(n) for n = 1..1004</a> (terms <= 10^100; first 294 terms from Antti Karttunen, terms <= 10^100)

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David A. Corneth, <a href="/A330687/b330687_1.txt">Table of n, a(n) for n = 1..1004</a> (first 294 terms from Antti Karttunen, terms <= 10^100)

editing

approved

Each term is a perfect square. Proof: A050377(n) is multiplicative with a(p^e) = A018819(e) and A018819(2k) = A018819(2k+1) and this sequence considers just records so we only need exponents of the form 2k, ; i.e., terms are squares.

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More specifically: Let S(n) be the list, possibly with duplicates , of exponents occurring in the prime factorizations of terms with the sum of exponents in the prime factorization <= n.

For example, S(8) is found from the following terms: 4, 16, 64, 144, 256, 576 and 1296 as the exponents in the prime factorization are (2), (4), (6), (4, 2), (8), (6, 2), (4, 4). The sums of each of these exponents per term is <= 8. There are 10 exponents listed. Of these 10 there are 5 of them that are divisible by 4. Therefore R(8) = 5/10.

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It seems that most exponents of a(n) are divisible by 4. (End)

More specifically: Let S(n) be the list, possibly with duplicates of exponents occurring in the prime factorizations of terms with the sum of exponents in the prime factorization <= n.

Let R(n) = |{x : x==4, S(n)}| / |S(n)|.

For example, S(8) is found from the following terms: 4, 16, 64, 144, 256, 576 and 1296 as the exponents in the prime factorization are (2), (4), (6), (4, 2), (8), (6, 2), (4, 4). The sums of each of these exponents per term is <= 8. There are 10 exponents listed. Of these 10 there are 5 of them divisible by 4. Therefore R(8) = 5/10.

Then it seems that R(n) tends to some value > 0.8 as n grows. (End)