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Search: id:A007814
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| A007814 |
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Exponent of highest power of 2 dividing n (the binary carry sequence). |
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+0 175
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| 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 6, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 5, 0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0
(list; graph; listen)
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OFFSET
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1,4
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COMMENT
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This sequence is an exception to my usual rule that when every other term of a sequence is 0 then those 0's should be omitted. In this case we would get A001511. - N. J. A. Sloane (njas(AT)research.att.com).
To construct the sequence: start with 0,1, concatenate to get 0,1,0,1. Add + 1 to last term gives 0,1,0,2. Concatenate those 4 terms to get 0,1,0,2,0,1,0,2. Add + 1 to last term etc. - Benoit Cloitre (benoit7848c(AT)orange.fr), Mar 06 2003
a(n) = A091090(n-1) + A036987(n-1) - 1.
Fixed point of the morphism 0->01, 1->02, 2->03, 3->04, ..., n->0(n+1), ..., starting from a(1) = 0. - DELEHAM Philippe (kolotoko(AT)wanadoo.fr), Mar 15 2004
a(n) is also the number of times to repeat a step on an even number in the hailstone sequence referenced in the Collatz conjecture. - Alex T. Flood (whiteangelsgrace(AT)gmail.com), Sep 22 2006
Let F(n) be the n-th Fermat number (A000215). Then F(a(r-1)) divides F(n)+2^k for r=mod(k,2^n) and r != 1. - T. D. Noe (noe(AT)sspectra.com), Jul 12 2007
A007814(n) = A001511(n) - 1. [From Ivor C. Quence (Ivan(AT)email_address.too), May 02 2009]
a(n) is the number of 0s at the end of n when n is written in base 2. [From Ivor C. Quence (Ivan(AT)email_address.too), May 02 2009]
a(n) = A063787(n) - A000120(n) [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 04 2009]
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REFERENCES
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K. Atanassov, On the 37-th and the 38-th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 2, 83-85.
K. Atanassov, On Some of Smarandache's Problems, American Research Press, 1999, 16-21.
F. Smarandache, Only Problems, not Solutions!, Xiquan Publ., Phoenix-Chicago, 1993.
P. M. B. Vitanyi, An optimal simulation of counter machines, SIAM J. Comput, 14:1(1985), 1-33.
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LINKS
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T. D. Noe, Table of n, a(n) for n=1..10000
K. Atanassov, On Some of Smarandache's Problems
M. Hassani, Equations and inequalities involving v_p(n!), J. Inequ. Pure Appl. Math. 6 (2005) vol. 2, #29
M. L. Perez et al., eds., Smarandache Notions Journal
V. Shevelev, Several results on sequences which are similar to the positive integers [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), Apr 15 2009]
F. Smarandache, Only Problems, Not Solutions!.
R. Stephan, Some divide-and-conquer sequences ...
R. Stephan, Table of generating functions
Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.
Eric Weisstein's World of Mathematics, Link to a section of The World of Mathematics.
Eric Weisstein's World of Mathematics, Binary
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FORMULA
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a(n) = if n is odd then 0 else 1 + a(n/2). - Reinhard Zumkeller (reinhard.zumkeller(AT)gmail.com), Aug 11 2001
Sum(k=1, n, a(k))=n-A000120(n) - Benoit Cloitre (benoit7848c(AT)orange.fr), Oct 19 2002
G.f.: A(x) = Sum(k=1, infinity, x^(2^k)/(1-x^(2^k))). - Ralf Stephan (ralf(AT)ark.in-berlin.de), Apr 10 2002
The sequence is invariant under the following two transformations: increment every element by one (1, 2, 1, 3, 1, 2, 1, 4, ..), put a zero in front and between adjacent elements (0, 1, 0, 2, 0, 1, 0, 3, 0, 1, 0, 2, 0, 1, 0, 4, ..). The intermediate result is A001511. - Ralf Hinze (ralf(AT)informatik.uni-bonn.de), Aug 26 2003
G.f. A(x) satisfies A(x) = A(x^2) + x^2/(1-x^2). A(x) = B(x^2) = B(x) - x/(1-x), where B(x) is the g.f. for A001151. - Frank Adams-Watters (FrankTAW(AT)Netscape.net), Feb 09 2006
Totally additive with a(p) = 1 if p = 2, 0 otherwise.
Dirichlet g.f.: zeta(s)/(2^s-1). - Ralf Stephan, Jun 17 2007
Define 0 <= k <= 2^n - 1; binary: k = b(0) + 2.b(1) + 4.b(2) + ... + 2^(n-1).b(n-1); where b(x) are 0 or 1 for 0 <= x <= n - 1; Define c(x) = 1 - b(x) for 0 <= x <= n - 1; Then: a007814(k) = c(0) + c(0).c(1) + c(0).c(1).c(2) + ... + c(0).c(1)...c(n-1); a007814(k+1) = b(0) + b(0).b(1) + b(0).b(1).b(2) + ... + b(0).b(1)...b(n-1) - Arie Werksma (werksma(AT)tiscali.nl), May 10 2008
a(n) = floor(A002487(n - 1) / A002487(n)). [From Reikku Kulon (reikku(AT)gmail.com), Oct 05 2008]
Sum(k=1,n, (-1)^A000120(n-k)*a(k))=(-1)^(A000120(n)-1)*(A000120(n)-A000035(n)). [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), Mar 17 2009]
a(A001147(n)+A057077(n-1))=a(2n) [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), Mar 21 2009]
For n>=1, a(A004760(n+1))=a(n). [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), Apr 15 2009]
2^(a(n))=A006519(n). [From Philippe DELEHAM (kolotoko(AT)wanadoo.fr), Apr 22 2009]
a(C(n,k))=A000120(k)+A000120(n-k)-A000120(n). [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), Jul 19 2009]
a(n!)=n-A000120(n). [From Vladimir Shevelev (shevelev(AT)bgu.ac.il), Jul 20 2009]
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EXAMPLE
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2^3 divides 24, so a(24)=3.
Contribution from Omar E. Pol (info(AT)polprimos.com), Jun 12 2009: (Start)
Triangle begins:
0;
1,0;
2,0,1,0;
3,0,1,0,2,0,1,0;
4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0;
5,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0;
6,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,4,0,1,0,2,0,1,0,3,0,1,0,2,0,1,0,5,0,1,0,2,...
(End)
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MAPLE
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ord:=proc(n) local i, j; if n=0 then RETURN(0); fi; i:=0; j:=n; while j mod 2 <> 1 do i:=i+1; j:=j/2; od: i; end;
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MATHEMATICA
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Table[IntegerExponent[n, 2], {n, 64}] (From Eric Weisstein)
p=2; Array[ If[ Mod[ #, p ]==0, Select[ FactorInteger[ # ], Function[ q, q[ [ 1 ] ]==p ], 1 ][ [ 1, 2 ] ], 0 ]&, 96 ]
DigitCount[BitXor[x, x - 1], 2, 1] - 1; a different version based on the same concept: Floor[Log[2, BitXor[x, x - 1]]] (from Jaume Simon Gispert (jaume(AT)nuem.com), Aug 29 2004)
Nest[Join[ # , ReplacePart[ # , Length[ # ] -> Last[ # ] + 1]] &, {0, 1}, 5] (from N. J. Gunther, May 23 2009)
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PROGRAM
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(PARI) a(n)=valuation(n, 2)
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CROSSREFS
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A053398(1, n)
a(n) = A001511[n]-1, column/row 1 of table A050602. Cf. A006519, A001511.
a(2n)=A050603(2n).
First differences of A011371. Bisection of A050605 and |A088705|.
Cf. A122840, A122841, A007949, A112765.
See A050603 and A136480 for a(n)+a(n+1).
This is Guy Steele's sequence GS(1, 4) (see A135416).
Cf. A002487 [From Reikku Kulon (reikku(AT)gmail.com), Oct 05 2008]
A063787 [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 04 2009]
Cf. A000079. [From Omar E. Pol (info(AT)polprimos.com), Jun 12 2009]
Adjacent sequences: A007811 A007812 A007813 this_sequence A007815 A007816 A007817
Sequence in context: A093057 A065334 A162590 this_sequence A083280 A060689 A053119
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KEYWORD
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nonn,nice,easy
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AUTHOR
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John Tromp (tromp(AT)math.uwaterloo.ca)
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