cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-4 of 4 results.

A279102 Numbers n having three parts in the symmetric representation of sigma(n).

Original entry on oeis.org

9, 15, 25, 35, 45, 49, 50, 70, 77, 91, 98, 110, 121, 130, 135, 143, 154, 169, 170, 182, 187, 190, 209, 221, 225, 238, 242, 247, 266, 286, 289, 299, 315, 322, 323, 338, 350, 361, 374, 391, 405, 418, 437, 442, 484, 493, 494, 506, 527, 529, 550, 551, 572, 578, 589, 598, 638, 646, 650, 667, 675, 676, 682
Offset: 1

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Author

Hartmut F. W. Hoft, Dec 06 2016

Keywords

Comments

Let n = 2^m * q with m >= 0 and q odd, let row(n) = floor(sqrt(8*n+1) - 1)/2), and let 1 = d_1 < ... < d_h <= row(n) < d_(h+1) < ... < d_k = q be the k odd divisors of n.
The symmetric representation of sigma(n) consists of 3 parts precisely when there is a unique i, 1 <= i < h, such that 2^(m+1) * d_i < d_(i+1) and d_h <= row(n) < 2^(m+1) * d_h.
This property of the odd divisors of n is equivalent to the n-th row of the irregular triangle of A249223 consisting of a block of positive numbers, followed by a block of zeros, followed in turn by a block of positive numbers, i.e., determining the first part and the left half of the center part of the symmetric representation of sigma(n), resulting in 3 parts.
Let n be the product of two primes p and q satisfying 2 < p < q < 2*p. Then n satisfies the property above so that the odd numbers in A087718 form a subsequence.

Examples

			a(4) = 35 = 5*7 is in the sequence since 1 < 2 < 5 < row(35) = 7 < 10;
a(8) = 70 = 2*5*7 is in the sequence since 1 < 4 < 5 < row(70) = 11 < 20;
140 = 4*5*7 is not in the sequence since 1 < 5 < 7 < 8 < row(140) = 16 < 20;
a(506) = 5950 = 2*25*7*17 is in the sequence since 1*4 < 5 is the only pair of odd divisors 1 < 5 < 7 < 17 < 25 < 35 < 85 < row(5950) = 108 satisfying the property (see A251820).
		

Crossrefs

Column 3 of A240062.

Programs

  • Mathematica
    (* support functions are defined in A237048 and A262045 *)
    segmentsSigma[n_] := Length[Select[SplitBy[a262045[n], #!=0&], First[#]!=0&]]
    a279102[m_, n_] := Select[Range[m, n], segmentsSigma[#]==3&]
    a279102[1, 700] (* sequence data *)
    (* An equivalent, but slower computation is based on A237271 *)
    a279102[m_, n_] := Select[Range[m, n], a237271[#]==3&]
    a279102[1,700] (* sequence data *)

A265999 Numbers k such that in the symmetric representation of sigma(k) all parts are of the same size.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 22, 23, 24, 26, 28, 29, 30, 31, 32, 34, 36, 37, 38, 40, 41, 42, 43, 44, 46, 47, 48, 52, 53, 54, 56, 58, 59, 60, 61, 62, 64, 66, 67, 68, 71, 72, 73, 74, 76, 78, 79, 80, 82, 83, 84, 86, 88, 89, 90, 92, 94, 96, 97, 100
Offset: 1

Views

Author

Omar E. Pol, Dec 19 2015

Keywords

Comments

All powers of 2, all prime numbers and all even perfect numbers are members of this sequence.
For more information about the symmetric representation of sigma see A237270 and A237593.
Sequence A174973: the symmetric representation of sigma, SRS(A174973(n)) consisting of 1 part, and sequence A239929: SRS(A239929(n)) consisting of 2 parts, are proper subsequences. Sequence A251820: SRS(A251820(n)) consisting of 3 equal parts, contains the only other known members 15 and 5950 of this sequence. No number m with SRS(m) consisting of 4 or more equal parts is known. - Hartmut F. W. Hoft, Jan 11 2025

Examples

			9 is not in the sequence because the parts of the symmetric representation of sigma(9) = 13 are [5, 3, 5].
10 is in the sequence because the parts of the symmetric representation of sigma(10) = 18 are [9, 9].
SRS(15) = { 8, 8, 8 } and SRS(5950) = { 4464, 4464, 4464 }. - _Hartmut F. W. Hoft_, Jan 11 2025
		

Crossrefs

Programs

  • Mathematica
    (* Function partsSRS[ ] is defined in A377654 *)
    a265999[n_] := Select[Range[n], Length[Union[partsSRS[#]]]==1&]
    a265999[100] (* Hartmut F. W. Hoft, Jan 11 2025 *)

A266000 Numbers k such that the symmetric representation of sigma(k) has at least two parts of distinct size.

Original entry on oeis.org

9, 21, 25, 27, 33, 35, 39, 45, 49, 50, 51, 55, 57, 63, 65, 69, 70, 75, 77, 81, 85, 87, 91, 93, 95, 98, 99, 105, 110, 111, 115, 117, 119, 121, 123, 125, 129, 130, 133, 135, 141, 143, 145, 147, 153, 154, 155, 159, 161, 165, 169, 170, 171, 175, 177, 182, 183, 185, 187, 189, 190, 195
Offset: 1

Views

Author

Omar E. Pol, Dec 19 2015

Keywords

Comments

In other words: numbers k such that the symmetric representation of sigma(k) has at least two parts with distinct number of cells.
For more information about the symmetric representation of sigma see A237270 and A237593.
When the symmetric representation of sigma of m, SRS(m), consists of 2n-1 or 2n parts, n>=1, then at most n parts can be of distinct sizes. For the published terms in A239663, SRS(A239663(n)) consists of n parts representing ceiling(n/2) parts of distinct sizes, n>=1. Only two numbers m are known, 15 and 5950 in A251820, for which SRS(m) consists of n parts of less than ceiling(n/2) distinct sizes. - Hartmut F. W. Hoft, Jan 11 2025

Examples

			The symmetric representation of sigma(9) = 13 in the first quadrant looks like this:
y
.
._ _ _ _ _ 5
|_ _ _ _ _|
.         |_ _ 3
.         |_  |
.           |_|_ _ 5
.               | |
.               | |
.               | |
.               | |
. . . . . . . . |_| . . x
.
There are three parts: 5 + 3 + 5 = 13, so 9 is in the sequence because the structure contains at least two parts of distinct size.
From _Hartmut F. W. Hoft_, Jan 11 2025: (Start)
SRS(a(1)) = SRS(A239663(3)) = SRS(9) = { 5, 3, 5 } is the smallest with 2 parts of distinct sizes.
SRS(a(14)) = SRS(A239663(5)) = SRS(63) = { 32, 12, 16, 12, 32 } is the smallest with 3 parts of distinct sizes.
SRS(a(127)) = SRS(A239663(7)) = SRS(357) = { 179, 61, 29, 38, 29, 61, 179 } is the smallest with 4 parts of distinct sizes. (End)
		

Crossrefs

Programs

  • Mathematica
    (* Function partsSRS[ ] is defined in A377654 *)
    a266000[n_] := Select[Range[n], Length[Union[partsSRS[#]]]>=2&]
    a266000[200] (* Hartmut F. W. Hoft, Jan 11 2025 *)

Extensions

Extended from a(37) to a(62) by Hartmut F. W. Hoft, Jan 11 2025

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.

Original entry on oeis.org

15, 207, 1023, 2975, 5950, 19359, 147455, 294910, 1207359, 5017599, 2170814463
Offset: 1

Views

Author

Hartmut F. W. Hoft, May 04 2023

Keywords

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

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 *)
Showing 1-4 of 4 results.