A362809 Numbers k for which the area of the first part of the symmetric representation of sigma(k) equals sigma(k)/3 and its width is 1.

15, 207, 1023, 2975, 5950, 19359, 147455, 294910, 1207359, 5017599, 2170814463

Offset: 1

Author

Hartmut F. W. Hoft, May 04 2023

Keywords

• nonn
• hard
• more

Comments

The symmetric representation of sigma(k), SRS(k), of every term k in this sequence consists of at least 3 parts, with a(1) = 15 and a(5) = 5950 being the only ones among the first 11 terms for which the SRS consists of exactly 3 parts. A251820 is a subsequence. a(12) > 5*10^9.

Suppose that a(n) = 2^i * q, i >= 0 and q odd. Because the first part of SRS(a(n)) has width 1, the smallest prime factor p of q satisfies p > 2^(i+1) -- see the locations of 1's in the triangle of A237048 and computation of widths in A249223. The area of the first part of SRS(a(n)) is (2^(i+1) - 1) * (q+1)/2 = (a(n) + 2^i) * (2^(i+1) - 1) / 2^(i+1) since the first part has 2^(i+1) - 1 legs (see the formula in A237591). Therefore, when 2^i * q is in the sequence then 2^j * q, for 0 <= j <= i is also. Sigma(a(n)) = A068156(i+1) * (a(n) + 2^i) / 2^(i+1).

Conjecture: The area of the first part of SRS(n) being equal to sigma(n)/3 implies that the first part has width 1.

This is true for all odd a(n) since their first part consists of a single leg of width 1. It also holds for even numbers through 10^7.

Observation: Consider the known 11 terms. Apart from 5950 and 294910 the rest are also in A063906. Question: Is A063906 a subsequence? - Omar E. Pol, Jul 09 2023

Answer: A063906 consists of the odd terms of this sequence since the first part of the symmetric representation of sigma for odd a(n) equals (a(n)+1)/2 which is equivalent to a(n) = 2*sigma(a(n))/3 - 1, i.e., a(n) is in A063906. - Hartmut F. W. Hoft, Jul 10 2023

A063906(10) = 58946212863 is a term here. - Omar E. Pol, Jul 12 2023

Examples

			15 belongs to the sequence since SRS(15) consists of the parts {8, 8, 8} of maximum widths {1, 2, 1} and sigma(15) = 24.
294910 belongs to the sequence since SRS(294910) consists of the 5 parts {221184, 109440, 2304, 109440, 221184} of maximum widths {1, 3, 2, 3, 1}, with 109440 + 2304 + 109440 = 211184 and sigma(294910) = 3 * 211184 = 663552.
From _Omar E. Pol_, Jul 07 2023: (Start)
Illustration of a(1) = 15.
The 14th row of triangle A237593 is [8, 3, 1, 2, 2, 1, 3, 8] and the 15th row of the same triangle is [8, 3, 2, 1, 1, 1, 1, 2, 3, 8] so the diagram of the symmetric representation of sigma(15) in the fourth quadrant is constructed as shown below:
.                                _
.                               | |
.                               | |
.                               | |
.                               | |
.                               | |
.                               | |
.                               | |
.                          _ _ _|_|
.                      _ _| |      8
.                     |    _|
.                    _|  _|
.                   |_ _|  8
.                   |
.    _ _ _ _ _ _ _ _|
.   |_ _ _ _ _ _ _ _|
.                    8
.
The area of the first part (or polygon) of the diagram equals sigma(15)/3 = 24/3 = 8 and its width is 1 so 15 is in the sequence. (End)
		

Crossrefs

Cf. A068156, A235791, A237048, A237591, A237593, A249223, A251820.

Cf. A063906.

Programs

Mathematica

(* substitute code suggested by Andrey Zabolotskiy *)
cd[n_, k_] := Boole[Divisible[n, k]]
a237048[s_, j_] := If[OddQ[j], cd[s, j], cd[s-j/2, j]]
firstZeroQ[s_, a_] := Sum[(-1)^(j+1)a237048[s, j], {j, a}]==0
evenPart[n_] := 2^IntegerExponent[n, 2]
a362809[{m_, n_}] := Module[{a, b}, Select[Range[m, n], (a=evenPart[#]; b=(2a-1)/(2a); DivisorSigma[1, #]==3b(#+a)&&firstZeroQ[#, 2a])&]]
a362809[{1, 2170814463}] (* a(11) has a long computation time *)