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) + 2*a(s), but there are no such representations for m == 1 (mod 4).

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

Offset: 1

Author

Vladimir Shevelev, Oct 20 2008, Oct 21 2008

Keywords

• nonn

Comments

Theorem. A positive odd number is in the sequence iff in its binary expansion all bits in the k-th 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/(1-x^4); a(n) = 2*A145812(n) - 1.

Every positive odd integer m == 3 (mod 2^(2r)) is a unique sum of the form a(2^(r-1)*(s-1)+1) + a(2^(r-1)*(t-1)+1), r=1,2,..., while other odd integers are not expressible in such form (see also comment to A145812). - Vladimir Shevelev, Oct 21 2008

To get the decomposition of m=4k+3 as the sum a(l)+2*a(s), write m-2 as Sum b_j 2^j, then a(s) = 1 + Sum_{j odd} b_j 2^(j-1). 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 one-to-one map of positive integers of the form 4k+3 to the positive lattice points on the plane. - Vladimir Shevelev, Oct 26 2008

Odd terms of A000695 (the Moser-de Bruijn sequence: sums of distinct powers of 4). - Jon E. Schoenfield, Mar 18 2021

Links

• Klaus Brockhaus, Table of n, a(n) for n = 1..8192

Crossrefs

Cf. A000695, A145812.

Programs

Mathematica

aQ[n_] := OddQ[n] && Module[{d = Reverse[IntegerDigits[n, 2]]}, Length[d] < 2 || Max[d[[2;; -1;; 2]]] == 0]; Select[Range[4500], aQ] (* Amiram Eldar, Dec 15 2018 *)

PARI

isok(n) = {if (n % 2, my(rb = Vecrev(binary(n)), brb = vector(#rb\2, k, rb[2*k])); (#brb == 0) || vecmax(brb) == 0, 0);} \\ Michel Marcus, Dec 15 2018

Extensions

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