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A037800
Number of occurrences of 01 in the binary expansion of n.
15
0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 0, 1, 1, 1, 1, 2, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 1, 1, 2, 2, 2, 1, 2, 1, 1, 1, 2, 2, 2, 2, 3, 2, 2, 1, 2, 2, 2, 1, 2, 1, 1, 0, 1, 1, 1, 1, 2, 1
OFFSET
0,22
COMMENTS
Number of i such that d(i)>d(i-1), where Sum{d(i)*2^i: i=0,1,...,m} is base 2 representation of n.
This is the base-2 up-variation sequence; see A297330. - Clark Kimberling, Jan 18 2017
LINKS
Jean-Paul Allouche and Jeffrey Shallit, Sums of digits and the Hurwitz zeta function, in: K. Nagasaka and E. Fouvry (eds.), Analytic Number Theory, Lecture Notes in Mathematics, Vol. 1434, Springer, Berlin, Heidelberg, 1990, pp. 19-30.
Eric Weisstein's World of Mathematics, Digit Block.
FORMULA
a(2n) = a(n), a(2n+1) = a(n) + [n is even]. - Ralf Stephan, Aug 21 2003
G.f.: 1/(1-x) * Sum_{k>=0} t^5/(1+t)/(1+t^2) where t=x^2^k. - Ralf Stephan, Sep 10 2003
a(n) = A069010(n) - 1, n>0. - Ralf Stephan, Sep 10 2003
Sum_{n>=1} a(n)/(n*(n+1)) = log(2)/2 + Pi/4 - 1 = A231902 - 1 (Allouche and Shallit, 1990). - Amiram Eldar, Jun 01 2021
MATHEMATICA
Table[SequenceCount[IntegerDigits[n, 2], {0, 1}], {n, 0, 120}] (* Harvey P. Dale, Aug 10 2023 *)
PROG
(Haskell)
a037800 = f 0 . a030308_row where
f c [_] = c
f c (1 : 0 : bs) = f (c + 1) bs
f c (_ : bs) = f c bs
-- Reinhard Zumkeller, Feb 20 2014
(PARI)
a(n) = { if(n == 0, 0, -1 + hammingweight(bitnegimply(n, n>>1))) }; \\ Gheorghe Coserea, Aug 31 2015
KEYWORD
nonn,base,easy
STATUS
approved