

A048298


a(n) = n if n=2^i, i=0,1,2,3,...; else = 0.


13



0, 1, 2, 0, 4, 0, 0, 0, 8, 0, 0, 0, 0, 0, 0, 0, 16, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 32, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 64, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
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OFFSET

0,3


COMMENTS

Expand x/(x1) = Sum_{n >= 0} 1/x^n as Sum a(n) / (1+x^n).
Nimbinomial transform of the natural numbers. If {t(n)} is the Nimbinomial transform of {a(n)}, then t(n)=(S^n)a(0), where Sf(n) denotes the Nimsum of f(n) and f(n+1); and S^n=S(S^(n1)).  John W. Layman, Mar 06 2001


REFERENCES

J.P. Allouche and J. Shallit, The ring of kregular sequences, II, Theoret. Computer Sci., 307 (2003), 329.


LINKS

Table of n, a(n) for n=0..102.
J.P. Allouche and J. Shallit, The Ring of kregular Sequences, II
Daniele A. Gewurz and Francesca Merola, Sequences realized as Parker vectors ..., J. Integer Seqs., Vol. 6, 2003.


FORMULA

Multiplicative with a(2^e)=2^e and a(p^e)=0 for p > 2.  Vladeta Jovovic, Jan 27 2002
Inverse mod 2 binomial transform of n. a(n)=sum{k=0..n, (1)^A010060(nk)*mod(C(n, k), 2)*k}  Paul Barry, Jan 03 2005
Dirichlet g.f.: 2^s/(2^s2).  Ralf Stephan, Jun 17 2007
For n>=1, we have a recursion Sum{dn}(1)^(1+(n/d))a(d)=1. [From Vladimir Shevelev, Jun 09 2009]
for n>=1 there is the recurrence n=sum(k=1,n,a(k)*g(n/k)) where g(x)=floor(x)2*floor(x/2) [From Benoit Cloitre, Nov 11 2010]


EXAMPLE

If n=1 we have a(n)=1; if n=p is prime, then (1)^(p+1)+a(p)=1, thus a(2)=2, and a(p)=0,if p>2. [From Vladimir Shevelev, Jun 09 2009]


CROSSREFS

A kind of inverse to A048272. Cf. A060147.
This is Guy Steele's sequence GS(5, 1) (see A135416).
Sequence in context: A104774 A087263 A099894 * A123565 A246160 A081120
Adjacent sequences: A048295 A048296 A048297 * A048299 A048300 A048301


KEYWORD

easy,nonn,mult


AUTHOR

Adam Kertesz (adamkertesz(AT)worldnet.att.net)


EXTENSIONS

More terms from Keiko L. Noble (s1180624(AT)cedarville.edu).


STATUS

approved



