1,4

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.

a(n) is the number of 0s at the end of n when n is written in base 2.

a(n) = A063787(n) - A000120(n) [From Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 04 2009]

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.

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

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]

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)

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;

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)

(PARI) a(n)=valuation(n, 2)

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]

Sequence in context: A093057 A065334 A162590 this_sequence A083280 A060689 A053119

Adjacent sequences: A007811 A007812 A007813 this_sequence A007815 A007816 A007817

nonn,nice,easy

John Tromp (tromp(AT)math.uwaterloo.ca)