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A102566
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a(n) = {minimal k such that f^k(prime(n)) = 1} where f(m) = (m+1)/2^r, 2^r is the highest power of two dividing m+1.
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0
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2, 1, 2, 1, 2, 2, 4, 3, 2, 2, 1, 4, 4, 3, 2, 3, 2, 2, 5, 4, 5, 3, 4, 4, 5, 4, 3, 3, 3, 4, 1, 6, 6, 5, 5, 4, 4, 5, 4, 4, 4, 4, 2, 6, 5, 4, 4, 2, 4, 4, 4, 2, 4, 2, 8, 6, 6, 5, 6, 6, 5, 6, 5, 4, 5, 4, 5, 6, 4, 4, 6, 4, 3, 4, 3, 2, 6, 5, 6, 5, 5, 5, 3, 5, 3, 3, 6, 5, 4, 3, 4, 2, 3, 3, 3, 2, 2, 8, 7, 6, 7, 6, 6, 6, 5
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OFFSET
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1,1
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COMMENTS
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LINKS
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FORMULA
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EXAMPLE
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f(f(f(f(17)))) = 1, prime(7) = 17, so a(7) = 4.
prime(16) = 53 = (2*27-1) = (2*(2^2*7-1)-1) = (2*(2^2*(2^3*1-1)-1)-1), has 3 levels, so a(16) = 3.
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PROG
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(PARI) f(n) = (n+1)/2^(valuation(n+1, 2));
a(n) = {my(k = 1, p = prime(n)); while((q=f(p)) != 1, k++; p = q); k; } \\ Michel Marcus, Nov 20 2016
(PARI) a(n) = my(p=prime(n)); 2 + logint(p, 2) - hammingweight(p); \\ Kevin Ryde, Nov 06 2023
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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