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%I #22 Nov 14 2017 08:19:50
%S 1,3,9,11,33,35,41,43,129,131,137,139,161,163,169,171,513,515,521,523,
%T 545,547,553,555,641,643,649,651,673,675,681,683,2049,2051,2057,2059,
%U 2081,2083,2089,2091,2177,2179,2185,2187,2209,2211,2217,2219,2561,2563
%N 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).
%C Theorem. An odd number is in the sequence iff in its binary expansion all digits 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)).
%C 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 no 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. Find the conditions of the solvability of this equation in B and give, if it is possible, other examples. [_Vladimir Shevelev_, Oct 21 2008]
%C 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. [_Vladimir Shevelev_, Oct 24 2008]
%H Klaus Brockhaus, <a href="/A145812/b145812.txt">Table of n, a(n) for n=1..8192</a> [From _Klaus Brockhaus_, Nov 01 2008]
%H V. Shevelev, <a href="http://arxiv.org/abs/0811.0290">On unique additive representations of positive integers and some close problems</a>, arXiv:0811.0290 [math.NT]
%F If f(x) = Sum_{n>=1} x^a(n), abs(x) < 1, then f(x)*f(x^2) = x^3/(1-x^2).
%F To get a(n), write n-1 as Sum b_j 2^j, then a(n) = 1 + 2Sum b_j 2^(2j). Conversely, 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. [_Vladimir Shevelev_, Oct 25 2008]
%F 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)). [_Vladimir Shevelev_, Nov 04 2008]
%F a(n) = 2*A000695(n-1) + 1. [_Vladimir Shevelev_, Nov 07 2008]
%F From _Robert Israel_, Aug 20 2017: (Start)
%F a(2n-1) = 4*a(n)-3.
%F a(2n) = 4*a(n)-1.
%F G.f. g(z) satisfies g(z) = (4 + 4/z) g(z^2) - (z^2 + 3 z)/(1-z^2). (End)
%e 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 and thus 2(521-1) = 1040 = 2^10 + 2^4, then the equation a(x)=521 has the unique solution x = 2^4 + 2^1 + 1 = 19. [_Vladimir Shevelev_, Oct 25 2008]
%p filter:= proc(n) local L; L:= convert(n,base,2);
%p andmap(i -> L[2*i+1]=0,[$1..(nops(L)-1)/2])
%p end proc:
%p select(filter, [seq(i,i=1..10000,2)]); # _Robert Israel_, Aug 20 2017
%t a[1]=1; a[2]=3; a[n_] := a[n] = If[OddQ[n], 4*a[(n+1)/2]-3, 4*a[n/2]-1];
%t Array[a, 50] (* _Jean-François Alcover_, Nov 14 2017, after _Robert Israel_ *)
%o (PARI) isok(n) = {if (! (n % 2), return (0)); b = binary(n); forstep (i = #b, 1, -1, rpos = #b - i + 1; if ((rpos > 1) && (rpos % 2) && b[i], return (0));); return (1);} \\ _Michel Marcus_, Jan 19 2014
%o (Haskell)
%o a145812 n = a145812_list !! (n-1)
%o a145812_list = filter f [1, 3 ..] where
%o f v = v == 0 || even w && f w where w = v `div` 4
%o -- _Reinhard Zumkeller_, Mar 13 2014
%Y Cf. A000695.
%K nonn,look
%O 1,2
%A _Vladimir Shevelev_, Oct 20 2008
%E Extended beyond a(16) by _Klaus Brockhaus_, Oct 22 2008
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