A082662 Numbers k such that the odd part of k is less than sqrt(2k).

1, 2, 4, 6, 8, 12, 16, 20, 24, 28, 32, 40, 48, 56, 64, 72, 80, 88, 96, 104, 112, 120, 128, 144, 160, 176, 192, 208, 224, 240, 256, 272, 288, 304, 320, 336, 352, 368, 384, 400, 416, 432, 448, 464, 480, 496, 512, 544, 576, 608, 640, 672, 704, 736, 768, 800

Offset: 1

Author

Naohiro Nomoto, May 18 2003

Keywords

• easy
• nonn
• nice

Comments

Theorem: The following eight definitions are equivalent.

(P1) Numbers k such that the odd part of k (A000265(k)) is < sqrt(2k).

(P1) is the new definition, repeated here for convenience. Note that this is not the same as saying A000265(k) < A172471(k), since A172471(k) = floor(sqrt(2*k)).

(P2) Numbers k such that the odd divisors of k are < sqrt(2k).

(P2) and (P1) are obviously equivalent.

(P3) The numbers 1, S_0, S_1, S_2, ..., where

S_m = { 2^(m+1)*(2^m+i) : i = 0 .. 3*2^m - 1 }.

So S_0 = {2,4,6}, S_1 = {8,12,16,20,24,28}, S_2 = {32,40,48,...,120}, S_3 = {128,144,...,496}, ...

The proof that (P3) and (P1) are the same sequence is not difficult and will be added later. (P3) is equivalent to a formula stated without proof (it may have been only an empirical observation) in the original version of this entry.

(P4) Numbers k such that the odd part of k is <= A003056(k).

That is, the odd part of k is <= floor((sqrt(1+8*n)-1)/2). It is more difficult to show this is equivalent to (P1), but it is true.

(P5) Numbers k such that the odd divisors of k are <= A003056(k).

(P5) and (P4) are obviously equivalent.

(P6) Numbers k such that A001227(k) = A082647(k).

(P6) was the original definition. In words, it says that the number of odd divisors of k is equal to the number of ways to write k as a sum of an odd number of consecutive positive integers, or equivalently as a sum of d consecutive positive integers for some d dividing k. To show that (P6) is equivalent to (P1) one makes use of the Hirschhorn-Hirschhorn article.

(P7) Numbers k such that the odd part of k is <= the sum of divisors of the even part.

(P7) was contributed by Jaycob Coleman, Jun 21 2014. To show (P7) is equivalent to (P1), write k as 2^m*s where s is odd. Equality holds if and only if k is an even perfect number.

(P8) Numbers k such that A000265(k) <= A000203(A006519(k)) or also such that A000265(k) <= A038712(k).

(P8) was contributed by Michel Marcus, Aug 14 2014. It is a restatement of (P7).

(End of theorem)

A further equivalent property, (P9), follows at once from (P4). This was conjectured by Omar E. Pol, Apr 18 2017

(P9) These are the numbers k such that the sequence of successive widths in the symmetric representation of sigma(k) is unimodal.

Yet another equivalent property:

(P10) Numbers k >= 1 such if k = i + (i+1) + (i+2) + ... + (i+j-1) for some i >= 1 and j >= 1 then j is odd [Caballero, 2019]. - Michel Marcus, Jan 16 2020

This is a subsequence of A005153. - Jaycob Coleman, Jun 21 2014

The complement of this sequence is A281005. - Omar E. Pol, Apr 18 2017

Subsequence of A174973. - Omar E. Pol, Feb 01 2021

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• José Manuel Rodríguez Caballero, Integers Which Cannot Be Partitioned Into an Even Number of Consecutive Parts, INTEGERS, Volume 19 (2019), #A20.
• M. D. Hirschhorn and P. M. Hirschhorn, Partitions into Consecutive Parts, Mathematics Magazine: 2003, Volume 76, Number 4, pp. 306-308.
• Hsien-Kuei Hwang, Svante Janson, and Tsung-Hsi Tsai, Identities and periodic oscillations of divide-and-conquer recurrences splitting at half, arXiv:2210.10968 [cs.DS], 2022, p. 38.
• Eric Weisstein's World of Mathematics, Even Part
• Eric Weisstein's World of Mathematics, Odd Part

Crossrefs

Cf. A281005 (complement).

Cf. A000196, A000203, A000265, A001227, A003056, A005153, A006519, A038712, A053644, A082647, A116882, A172471, A174973, A237593, A279387.

Programs

Mathematica

cnt[n_] := DivisorSum[n, Boole[OddQ[#] && #>Sqrt[2n]]&]; Select[Range[800], cnt[#]==0&] (* Jean-François Alcover, Feb 16 2017 *)

PARI

isok(n) = my(q = sqrt(2*n)); (sumdiv(n, d, (d%2) && (d < q)) == sumdiv(n, d, d%2)); \\ Michel Marcus, Jul 04 2014

Formula

G.f. = 1 + (1/(1-x)^2) * Sum_{m >= 0} (2^(m+1)*x^(3*2^m-2) * ( x^(3*2^m)*(2^(m+2)*(x-1)-x) - 2^m*(x-1) + x ) ). (This follows from (P3).) :w

- N. J. A. Sloane, Feb 02 2021

a(n+1) = a(n) + A053644(A000196(2*a(n))). - Peter Munn, Oct 03 2023

Extensions

Edited by N. J. A. Sloane, Jan 28 2021: Replaced original indirect definition by simple direct definition; rearranged comments; provided proofs (not yet included here) that the various definitions are equivalent