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Sum of the binary weights of the lengths of 1-runs in base-2 representation of n: a(n) = A000523(A286575(n)).
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%I #22 Feb 04 2022 14:36:50

%S 0,1,1,1,1,2,1,2,1,2,2,2,1,2,2,1,1,2,2,2,2,3,2,3,1,2,2,2,2,3,1,2,1,2,

%T 2,2,2,3,2,3,2,3,3,3,2,3,3,2,1,2,2,2,2,3,2,3,2,3,3,3,1,2,2,2,1,2,2,2,

%U 2,3,2,3,2,3,3,3,2,3,3,2,2,3,3,3,3,4,3,4,2,3,3,3,3,4,2,3,1,2,2,2,2,3,2,3,2,3,3,3,2,3,3,2,2,3,3,3,3,4,3,4,1

%N Sum of the binary weights of the lengths of 1-runs in base-2 representation of n: a(n) = A000523(A286575(n)).

%C a(0) = 0 (an empty sum).

%H Antti Karttunen, <a href="/A286574/b286574.txt">Table of n, a(n) for n = 0..65536</a>

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

%F a(n) = A000523(A286575(n)). [Log_2 of run-length transform of A001316.]

%F a(n) = A064547(A005940(1+n)).

%e For n = 27, "11011" in binary, there are two 1-runs, both of length 2, thus a(27) = A000120(2) + A000120(2) = 1 + 1 = 2.

%e For n = 29, "11101" in binary, there are two 1-runs, of lengths 1 and 3, thus a(29) = A000120(1) + A000120(3) = 1 + 2 = 3.

%e For n = 61, "111101" in binary, there are two 1-runs, of lengths 1 and 4, thus a(61) = A000120(1) + A000120(4) = 1 + 1 = 2.

%o (Scheme) (define (A286574 n) (A000523 (A286575 n)))

%o (Python)

%o from sympy import factorint, prime, log

%o import math

%o def wt(n): return bin(n).count("1")

%o def a037445(n):

%o f=factorint(n)

%o return 2**sum([wt(f[i]) for i in f])

%o def A(n): return n - 2**int(math.floor(log(n, 2)))

%o def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))

%o def a286575(n): return a037445(b(n))

%o def a(n): return int(math.floor(log(a286575(n), 2))) # _Indranil Ghosh_, May 30 2017

%o (Python)

%o # uses RLT function from A278159

%o def A286574(n): return len(bin(RLT(n,lambda m: 2**(bin(m).count('1')))))-3 # _Chai Wah Wu_, Feb 04 2022

%Y Cf. A000120, A000523, A001316, A005940, A064547, A286575.

%K nonn,base

%O 0,6

%A _Antti Karttunen_, May 28 2017