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A145812
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Odd positive integers a(n) such that for every odd integer m>1 there exists a unique representation of m as a sum of the form a(l)+2a(s)
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17
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1, 3, 9, 11, 33, 35, 41, 43, 129, 131, 137, 139, 161, 163, 169, 171, 513, 515, 521, 523, 545, 547, 553, 555, 641, 643, 649, 651, 673, 675, 681, 683, 2049, 2051, 2057, 2059, 2081, 2083, 2089, 2091, 2177, 2179, 2185, 2187, 2209, 2211, 2217, 2219, 2561, 2563
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OFFSET
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1,2
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COMMENTS
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Theorem. An odd number is in the sequence iff in its binary expansion all didits on k-th positions from the end, k=3, 5, 7, ..., are zeros. For example, 169, 171 have binary expansions 10101001, 10101011. 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))
Every positive odd integer m==3 (mod 2^(2r-1)) is a unique sum of the form a(2^(r-1)*(s-1)+1)+2a(2^(r-1)*(t-1)+1),r=1,2,..., while the other odd integers are not expressible in such a form (see also the comment to A145818). Problem: Let B be the set of analytical functions f(x),f(0)=0,abs(x) > 1, with (0,1)-Taylor coefficients. Let F(x) be in B. It could be proved that the equation f(x)*f(x^2)=F(x) has not more than one solution in B.In case of F(x)=x^3/(1-x^(2^r)), r=1,2,...,it has one solution, which leads to A145812, A145818, A145849, A145850, etc. To find the conditions of the solvability of this equation in B and give, if it is possible, other examples. [From Vladimir Shevelev, Oct 21 2008]
To get the decomposition of odd m as sum a(l)+2a(s), write m-2 as Sum b_j 2^j, then a(l) = 1 + Sum_{j odd} b_j 2^j. This algorithm follows from our comment to A088442 and the algorithm of calculation of A088442(n). For example, if m=81, then we have: 79=1*2^0+1*2^1+1*2^2+1*2^3+1*2^6. Thus a(l)=1+(1*2^1+1*2^3) =11 and the required decomposition is: 81=11+2*35,such that a(s)=35. We see that l=4,s=6, i.e. "index coordinates" of 81 are (4,6). Thus we have a one-to-one map of odd integers > 1 to the positive lattice points in the plane. [From Vladimir Shevelev, Oct 24 2008]
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REFERENCES
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V. Shevelev,On unique additive representations of positive integers and some close problems, http:// arxiv.org/abs/0811.0290 [From Vladimir Shevelev, Nov 04 2008]
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LINKS
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Klaus Brockhaus, Table of n, a(n) for n=1..8192 [From Klaus Brockhaus, Nov 01 2008]
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FORMULA
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If f(x)=sum_{n>=1}x^a(n), abs(x)<1, then f(x)*f(x^2)=x^3/(1-x^2).
To get a(n), write n-1 as Sum b_j 2^j, then a(n) = 1 + 2Sum b_j 2^(2j). Conversly, if an odd number m==r(mod 4),r=1 or 3, has the form, which is indicated in the theorem, i.e. 2m-2r=Sum{j>=1}b_j 2^(2j), b_j=0 or 1, then the only solution of the equation a(x)=m is x=Sum{j>=1}b_j 2^(j-1)+(r+1)/2. [From Vladimir Shevelev, Oct 25 2008]
For n>=2, a(n)=a(n-1)+(4^(t+1)+2)/3, where t>=0 is such that n-1==2^t (mod 2^(t+1)). [From Vladimir Shevelev, Nov 04 2008]
a(n)=2A000695(n-1)+1 [From Vladimir Shevelev, Nov 07 2008]
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EXAMPLE
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If n=13, we have n-1=12=2^3+2^2. Therefore a(13)=1+2(4^3+4^2)=161. If m=521 such that 2(521-1)=1040=2^10+2^4, then the equation a(x)=521 has the only solution x=2^4+2^1+1=19. [From Vladimir Shevelev, Oct 25 2008]
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CROSSREFS
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Cf. A000695.
Sequence in context: A003597 A018705 A018381 * A220944 A200879 A027895
Adjacent sequences: A145809 A145810 A145811 * A145813 A145814 A145815
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KEYWORD
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nonn
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AUTHOR
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Vladimir Shevelev, Oct 20 2008
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EXTENSIONS
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Extended beyond a(16) by Klaus Brockhaus, Oct 22 2008
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STATUS
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approved
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