

A145818


Odd positive integers a(n) such that for every integer m==3(mod 4) there exists a unique representation of the form m=a(l)+2a(s), but there are no such representations for m==1(mod 4)


11



1, 5, 17, 21, 65, 69, 81, 85, 257, 261, 273, 277, 321, 325, 337, 341, 1025, 1029, 1041, 1045, 1089, 1093, 1105, 1109, 1281, 1285, 1297, 1301, 1345, 1349, 1361, 1365, 4097, 4101, 4113, 4117, 4161, 4165, 4177, 4181, 4353, 4357, 4369, 4373, 4417, 4421, 4433
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OFFSET

1,2


COMMENTS

Theorem. A positive odd number is in the sequence iff in its binary expansion all bits in the kth position from the end, for k=2, 4, 6, ..., are zeros. For example, 337, 341 have binary expansions 101010001, 101010101. Thus both of them are in the sequence. If A(x) is the counting function of a(n)<=x, then A(x)=O(sqrt(x))and Omega(sqrt(x)). If f(x)=sum_{n>=1}x^a(n), abs(x)<1, then f(x)*f(x^2)=x^3/(1x^4); a(n)=2A145812(n)1.
Every positive odd integer m==3 (mod 2^(2r)) is a unique sum of the form a(2^(r1)*(s1)+1)+a(2^(r1)*(t1)+1),r=1,2,..., while other odd integers are not expressible in such form (see also comment to A145812). [From Vladimir Shevelev, Oct 21 2008]
To get the decomposition of m=4k+3 as sum a(l)+2a(s), write m2 as Sum b_j 2^j, then a(s) = 1 + Sum_{j odd} b_j 2^(j1). For example, if m=55, then we have: 53=2^0+2^2+2^4+2^5. Thus a(l)=1+2^4 =17 and the required decomposition is: 55=a(l)+2*17,such that a(l)=21. We see that l=4,s=3, i.e. "index coordinates" of 55 are (4,3). Thus we have a onetoone map of positive integers of the form 4k+3 to the positive lattice points on the plane. [From Vladimir Shevelev, Oct 26 2008]


LINKS

Klaus Brockhaus, Table of n, a(n) for n=1..8192 [From Klaus Brockhaus, Nov 01 2008]


CROSSREFS

A145812
Sequence in context: A191210 A191143 A032376 * A029986 A076275 A316307
Adjacent sequences: A145815 A145816 A145817 * A145819 A145820 A145821


KEYWORD

nonn


AUTHOR

Vladimir Shevelev, Oct 20 2008, Oct 21 2008


EXTENSIONS

Extended beyond a(16). Klaus Brockhaus, Oct 22 2008


STATUS

approved



