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A297155
a(1) = a(2) = 0, after which, a(n) = 1+a(n/2) if n is of the form 4k+2, otherwise a(n) = a(A252463(n)).
5
0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1, 0, 2, 0, 0, 1, 1, 1, 1, 0, 1, 1, 1, 0, 2, 0, 1, 1, 1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 2, 0, 1, 1, 0, 1, 2, 0, 1, 1, 2, 0, 1, 0, 1, 1, 1, 1, 2, 0, 1, 0, 1, 0, 2, 1, 1, 1, 1, 0, 2, 1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 2, 0, 1, 2
OFFSET
1,30
COMMENTS
Consider the binary tree illustrated in A005940: If we start from any vertex containing n, computing successive iterations of A252463 until 1 is reached, a(n) gives the number of the numbers of the form 4k+2 (with k >= 1) encountered on the path (i.e., excluding 2 from the count but including the starting n if it is of the form 4k+2).
FORMULA
a(n) = A252464(n) - A297113(n).
a(n) = A037800(A156552(n)).
a(n) = A001221(n) - 1 for all n > 1. - Velin Yanev, Mar 26 2019
PROG
(PARI)
A064989(n) = {my(f); f = factor(n); if((n>1 && f[1, 1]==2), f[1, 2] = 0); for (i=1, #f~, f[i, 1] = precprime(f[i, 1]-1)); factorback(f)};
A297155(n) = if(n<=2, 0, if(n%2, A297155(A064989(n)), (2==(n%4))+A297155(n/2)));
(Scheme, with memoization-macro definec) (definec (A297155 n) (cond ((<= n 2) 0) ((= 2 (modulo n 4)) (+ 1 (A297155 (/ n 2)))) (else (A297155 (A252463 n)))))
KEYWORD
nonn
AUTHOR
Antti Karttunen, Dec 27 2017
STATUS
approved