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A033264
Number of blocks of {1,0} in the binary expansion of n.
18
0, 1, 0, 1, 1, 1, 0, 1, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 2, 2, 2, 2, 1, 1, 1, 2, 1, 2, 2, 2, 1, 1, 1, 2, 1, 1, 1, 1, 0, 1, 1, 2, 1, 2, 2, 2, 1, 2, 2, 3, 2, 2, 2, 2, 1, 2, 2, 3, 2, 3, 3, 3, 2, 2, 2, 3, 2, 2, 2, 2, 1, 1, 1, 2, 1, 2, 2, 2
OFFSET
1,10
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 down-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
G.f.: 1/(1-x) * Sum_(k>=0, t^2/(1+t)/(1+t^2), t=x^2^k). - Ralf Stephan, Sep 10 2003
a(n) = A069010(n) - (n mod 2). - Ralf Stephan, Sep 10 2003
a(4n) = a(4n+1) = a(2n), a(4n+2) = a(n)+1, a(4n+3) = a(n). - Ralf Stephan, Aug 20 2003
a(n) = A087116(n) for n > 0, since strings of 0's alternate with strings of 1's, which end in (1,0). - Jonathan Sondow, Jan 17 2016
Sum_{n>=1} a(n)/(n*(n+1)) = Pi/4 - log(2)/2 (A196521) (Allouche and Shallit, 1990). - Amiram Eldar, Jun 01 2021
MAPLE
f:= proc(n) option remember; local k;
k:= n mod 4;
if k = 2 then procname((n-2)/4) + 1
elif k = 3 then procname((n-3)/4)
else procname((n-k)/2)
fi
end proc:
f(1):= 0: f(0):= q:
seq(f(i), i=1..100); # Robert Israel, Aug 31 2015
MATHEMATICA
Table[Count[Partition[IntegerDigits[n, 2], 2, 1], {1, 0}], {n, 102}] (* Michael De Vlieger, Aug 31 2015, after Robert G. Wilson v at A014081 *)
Table[SequenceCount[IntegerDigits[n, 2], {1, 0}], {n, 110}] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jan 26 2017 *)
PROG
(Haskell)
a033264 = f 0 . a030308_row where
f c [] = c
f c (0 : 1 : bs) = f (c + 1) bs
f c (_ : bs) = f c bs
-- Reinhard Zumkeller, Feb 20 2014, Jun 17 2012
(PARI)
a(n) = { hammingweight(bitand(n>>1, bitneg(n))) }; \\ Gheorghe Coserea, Aug 30 2015
CROSSREFS
a(n) = A005811(n) - ceiling(A005811(n)/2) = A005811(n) - A069010(n).
Equals (A072219(n+1)-1)/2.
Cf. also A175047, A030308.
Essentially the same as A087116.
Sequence in context: A047988 A037818 A087116 * A258045 A239302 A256983
KEYWORD
nonn,base,easy
STATUS
approved