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0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0, 0, 1, 4, 0, 2, 1, 0, 0, 0, 1, 0, 0, 0, 3, 0, 0, 0, 1, 0, 0, 0, 1, 2, 0, 0, 1, 8, 4, 0, 1, 0, 2, 0, 1, 0, 0, 0, 1, 0, 0, 2, 2, 0, 0, 0, 1, 0, 0, 0, 3, 0, 0, 4, 1, 0, 0, 0, 1, 2, 0, 0, 1, 0, 0, 0, 1, 0, 2, 0, 1, 0, 0, 0, 1, 0, 8, 2, 5, 0, 0, 0, 1, 0
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OFFSET
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1,9
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COMMENTS
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Conjecture: Each k >= 0 occurs for the first time at A334110(k) = A019565(k)^2. Note that each k must occur first time on square n, because of the identity a(n) = a(A008833(n)). However, is there any reason to exclude squares with prime exponents > 2 from the candidates? See also comments in A334204.
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LINKS
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FORMULA
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Alternatively, additive with a(prime(i)^(2^k)) = 2^(i-1) * A329697(prime(k+1)), a(m*n) = a(m)+a(n) if A059895(m,n) = 1. - Peter Munn, May 04 2020
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MATHEMATICA
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Map[-1 + Length@ NestWhileList[# - #/FactorInteger[#][[-1, 1]] &, #, # != 2^IntegerExponent[#, 2] &] &, Array[If[# == 1, 1, Times @@ Flatten@ Map[Function[{p, e}, Map[Prime[Log2@ # + 1]^(2^(PrimePi@ p - 1)) &, DeleteCases[NumberExpand[e, 2], 0]]] @@ # &, FactorInteger[#]]] &, 105] ] (* Michael De Vlieger, May 26 2020 *)
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PROG
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(PARI)
A019565(n) = factorback(vecextract(primes(logint(n+!n, 2)+1), n));
A329697(n) = if(!bitand(n, n-1), 0, 1+A329697(n-(n/vecmax(factor(n)[, 1]))));
A334109(n) = { my(f=factor(n), pis=apply(primepi, f[, 1]), es=f[, 2]); sum(k=1, #f~, (2^(pis[k]-1))*A329697(A019565(es[k]))); };
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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