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a(13) = 4 because 13 = (1101) -> (1+1+0+1 = 11) -> (1+1 = 10) -> (1+0 = 1) = 1. (Three iterations were required to reach 1).)
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a(13) = 4 because 13 = (1101) -> (1+1+0+1 = 11) -> (1+1 = 10) -> (1+0 = 1) = 1. (Three iterations were required to reach 1).
Description corrected by Antti Karttunen, Jul 09 2017
Write n in binary; repeatedly sum the "digits" until reaching 1; a(n) = 1 + number of steps required.
<a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
m a(1) = 1; for n > 1, a(n) = "1 + a(A000120(n)), where A000120 gives the number of occurrences of binary digit 1 in n"; a(n) = 1 (if n = 1); a(binary representation of n) = 1 + a(m) otherwise.
a(n) = 1 + A180094(n). - Antti Karttunen, Jul 09 2017
One more than A180094.
Antti Karttunen, <a href="/A078627/b078627.txt">Table of n, a(n) for n = 1..8192</a>
(PARI) A078627(n) = { my(k=1); while(n>1, n = hammingweight(n); k += 1); (k); }; \\ Antti Karttunen, Jul 09 2017
Cf. A000120.
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Table[Length[NestWhileList[Total[IntegerDigits[#, 2]]&, n, #>1&]], {n, 110}] (* From _Harvey P. Dale, _, Oct 10 2011 *)
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