A056538 -id:A056538 - cp's OEIS frontend

A304254 Irregular triangle read by rows: row n lists divisors of n^2 in decreasing order.

1, 4, 2, 1, 9, 3, 1, 16, 8, 4, 2, 1, 25, 5, 1, 36, 18, 12, 9, 6, 4, 3, 2, 1, 49, 7, 1, 64, 32, 16, 8, 4, 2, 1, 81, 27, 9, 3, 1, 100, 50, 25, 20, 10, 5, 4, 2, 1, 121, 11, 1, 144, 72, 48, 36, 24, 18, 16, 12, 9, 8, 6, 4, 3, 2, 1, 169, 13, 1, 196, 98, 49, 28, 14, 7, 4, 2, 1

Offset: 1

Kritsada Moomuang, May 09 2018

			Triangle begins:
  1;
  4, 2, 1;
  9, 3, 1;
  16, 8, 4, 2, 1;
  25, 5, 1;
  36, 18, 12, 9, 6, 4, 3, 2, 1;
  49, 7, 1;
  64, 32, 16, 8, 4, 2, 1;
  81, 27, 9, 3, 1;
  100, 50, 25, 20, 10, 5, 4, 2, 1;
  121, 11, 1;
  144, 72, 48, 36, 24, 18, 16, 12, 9, 8, 6, 4, 3, 2, 1;
  169, 13, 1;
  196, 98, 49, 28, 14, 7, 4, 2, 1;
  225, 75, 45, 25, 15, 9, 5, 3, 1;
  256, 128, 64, 32, 16, 8, 4, 2, 1;
  289, 17, 1;
  324, 162, 108, 81, 54, 36, 27, 18, 12, 9, 6, 4, 3, 2, 1;
  361, 19, 1;
  400, 200, 100, 80, 50, 40, 25, 20, 16, 10, 8, 5, 4, 2, 1;

Amiram Eldar, Table of n, a(n) for n = 1..11958 (rows 1..600 flattened)

Cf. A048691 (row lengths), A065764 (row sums), A056538.

Table[Reverse[Divisors[n^2]], {n, 1, 14}] // Flatten (* Amiram Eldar, May 02 2025 *)

row(n) = Vecrev(divisors(n^2)); \\ Amiram Eldar, May 02 2025

A306303 Numbers that appear in the concatenation in decreasing order of their aliquot parts.

Original entry on oeis.org

95, 110, 220, 312, 330, 371, 432, 440, 510, 525, 550, 624, 660, 714, 770, 864, 880, 936, 990, 995, 1111, 2480, 4100, 5115, 5125, 5134, 5135, 5625, 6120, 6432, 6794, 8100, 8145, 9225, 9864, 9950, 9995, 10250, 11100, 11110, 12864, 15375, 16200, 20500, 22200, 22220

Offset: 1

Paolo P. Lava, Feb 05 2019

The first terms that appear in both the concatenations in increasing and decreasing order are 510, 550, 624, 714, 1111, 5625, 50100, 51020, 53125, 56250, 67134, ...

			Aliquot parts of 95 are 1, 5, 19 and their concatenation in decreasing order is 1951 where 95 appears.

Paolo P. Lava, Table of n, a(n) for n = 1..250

Cf. A056538, A248323.

with(numtheory): P:=proc(n) local a,b,k;
a:=sort([op(divisors(n) minus {n})]); b:=a[nops(a)];
for k from nops(a) by -1 to 2 do b:=b*10^(ilog10(a[k-1])+1)+a[k-1]; od;
if searchtext(convert(n,string),convert(b,string))>0
then n; fi; end: seq(P(i),i=2..22220);

Select[Range[23000],SequenceCount[Flatten[IntegerDigits/@Reverse[Most[Divisors[ #]]]],IntegerDigits[ #]]>0&] (* Harvey P. Dale, Nov 20 2023 *)

A309400 Irregular triangle read by rows in which row n lists in reverse order the partitions of n into equal parts.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 1, 3, 1, 1, 1, 1, 2, 2, 4, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 6, 1, 1, 1, 1, 1, 1, 1, 7, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 4, 4, 8, 1, 1, 1, 1, 1, 1, 1, 1, 1, 3, 3, 3, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 5, 5, 10, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11

Offset: 1

Omar E. Pol, Nov 30 2019

The number of parts in row n equals sigma(n) = A000203(n), the sum of the divisors of n. More generally, the number of parts congruent to 0 (mod m) in row m*n equals sigma(n).
The number of parts greater than 1 in row n equals A001065(n), the sum of the aliquot parts of n.
The number of parts greater than 1 and less than n in row n equals A048050(n), the sum of divisors of n except for 1 and n.
The number of partitions in row n equals A000005(n), the number of divisors of n.
The number of partitions in row n with an odd number of parts equals A001227(n).
The sum of odd parts in row n equals the sum of parts of the partitions in row n that have an odd number of parts, and equals the sum of all parts in the partitions of n into consecutive parts, and equals A245579(n) = n*A001227(n).
The sum of row n equals n*A000005(n) = A038040(n).
Records in row n give the n-th row of A027750.
First n rows contain A000217(n) 1's.
The number of k's in row n is A126988(n,k).
The number of odd parts in row n is A002131(n).
The k-th block in row n has A056538(n,k) parts.
Column 1 gives A000012.
Right border gives A000027.

			Triangle begins:
[1];
[1,1], [2];
[1,1,1], [3];
[1,1,1,1], [2,2], [4];
[1,1,1,1,1], [5];
[1,1,1,1,1,1], [2,2,2], [3,3], [6];
[1,1,1,1,1,1,1], [7];
[1,1,1,1,1,1,1,1], [2,2,2,2], [4,4], [8];
[1,1,1,1,1,1,1,1,1], [3,3,3], [9];
[1,1,1,1,1,1,1,1,1,1], [2,2,2,2,2], [5,5], [10];
[1,1,1,1,1,1,1,1,1,1,1], [11];
[1,1,1,1,1,1,1,1,1,1,1,1], [2,2,2,2,2,2], [3,3,3,3], [4,4,4], [6,6], [12];
[1,1,1,1,1,1,1,1,1,1,1,1,1], [13];
[1,1,1,1,1,1,1,1,1,1,1,1,1,1], [2,2,2,2,2,2,2], [7,7], [14];
[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1], [3,3,3,3,3], [5,5,5], [15];
[1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1], [2,2,2,2,2,2,2,2], [4,4,4,4], [8,8], [16];
...

Mirror of A244051.

Cf. A000005, A000012, A000027, A000203, A000217, A001065, A001227, A002131, A027750, A038040, A048050, A056538, A126988, A237593, A245579, A299765, A328365.

A319135 Irregular triangle read by rows in which row n lists the positive divisors of n that are <= sqrt(n) in decreasing order.

Original entry on oeis.org

1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 1, 2, 1, 1, 3, 2, 1, 1, 2, 1, 3, 1, 4, 2, 1, 1, 3, 2, 1, 1, 4, 2, 1, 3, 1, 2, 1, 1, 4, 3, 2, 1, 5, 1, 2, 1, 3, 1, 4, 2, 1, 1, 5, 3, 2, 1, 1, 4, 2, 1, 3, 1, 2, 1, 5, 1, 6, 4, 3, 2, 1, 1, 2, 1, 3, 1, 5, 4, 2, 1, 1, 6, 3, 2, 1

Offset: 1

Charles Kusniec, Feb 27 2021

Reversing rows gives A161906.

			Triangle begins:
1
1
1
2 1
1
2 1
1
2 1
3 1
2 1
1
3 2 1
1
2 1
3 1
4 2 1

Cf. A038548 (row widths), A033676 (first column), A333750 (second column), A161906 (rows reversed), A027750, A056538, A061017, A340791, A340792.

row(n) = Vecrev(select(x->(x<=sqrt(n)), divisors(n))); \\ Jinyuan Wang, Mar 13 2021

a(n) * A161908(n) = A340792(n)
A161906(n) * A340791(n) = A340792(n)
A027750(n) * A056538(n) = A061017(n)

More terms from Jinyuan Wang, Mar 13 2021

A383963 Irregular triangle read by rows: T(n,k) is the sum of the k-th pair of conjugate divisors of n. If n is a square then the central term in the row n is equal to 2*sqrt(n), with n >= 1, 1 <= k <= A000005(n).

Original entry on oeis.org

2, 3, 3, 4, 4, 5, 4, 5, 6, 6, 7, 5, 5, 7, 8, 8, 9, 6, 6, 9, 10, 6, 10, 11, 7, 7, 11, 12, 12, 13, 8, 7, 7, 8, 13, 14, 14, 15, 9, 9, 15, 16, 8, 8, 16, 17, 10, 8, 10, 17, 18, 18, 19, 11, 9, 9, 11, 19, 20, 20, 21, 12, 9, 9, 12, 21, 22, 10, 10, 22, 23, 13, 13, 23, 24, 24, 25, 14, 11, 10, 10, 11, 14, 25

Offset: 1

Omar E. Pol, Jun 17 2025

Row n is a palindromic composition of A074400(n) = 2*A000203(n).

Shares infinitely many rows with the virtual sequence 2*A237270.

			Triangle begins:
   n |   Row n
   1 |   2;
   2 |   3,  3;
   3 |   4,  4;
   4 |   5,  4,  5;
   5 |   6,  6;
   6 |   7,  5,  5,  7;
   7 |   8,  8;
   8 |   9,  6,  6,  9;
   9 |  10,  6, 10;
  10 |  11,  7,  7, 11;
  11 |  12, 12;
  12 |  13,  8,  7,  7,  8, 13;
  13 |  14, 14;
  14 |  15,  9,  9, 15;
  15 |  16,  8,  8, 16;
  16 |  17, 10,  8, 10, 17;
  ...
For n = 8 the divisors of 8 are [1, 2, 4, 8] and the sums of the conjugate divisors are respectively [1 + 8 = 9], [2 + 4 = 6], [4 + 2 = 6], [8 + 1 = 9], so the 8th row is [9, 6, 6, 9].
For n = 9 the divisors of 9 are [1, 3, 9] and the sums of the conjugate divisors are respectively [1 + 9 = 10], [3 + 3 = 6], [9 + 1 = 10], so the 9th row is [10, 6, 10]. Since 9 is a square then the central term in the row is equal to 2*sqrt(9) = 2*3 = 6. Also in this case the 9th row is the same as the 9th row of the virtual sequence 2*A237270 because the 9th row of A237270 is [5, 3, 5].

Row lengths give A000005.
Row sums give A074400 = 2*A000203.
Column 1 gives A000027 except the 1, the same for the right border.
Cf. A027750, A056538, A237270, A272025.

row[n_] := Module[{d = Divisors[n]}, d + Reverse[d]]; Array[row, 24] // Flatten (* Amiram Eldar, Jun 18 2025 *)

row(n) = my(d=divisors(n)); vector(#d, k, d[k]+n/d[k]); \\ Michel Marcus, Jun 18 2025

T(n,k) = A027750(n,k) + A056538(n,k).

cp's OEIS Frontend

A304254 Irregular triangle read by rows: row n lists divisors of n^2 in decreasing order.

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A306303 Numbers that appear in the concatenation in decreasing order of their aliquot parts.

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A309400 Irregular triangle read by rows in which row n lists in reverse order the partitions of n into equal parts.

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A319135 Irregular triangle read by rows in which row n lists the positive divisors of n that are <= sqrt(n) in decreasing order.

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A383963 Irregular triangle read by rows: T(n,k) is the sum of the k-th pair of conjugate divisors of n. If n is a square then the central term in the row n is equal to 2*sqrt(n), with n >= 1, 1 <= k <= A000005(n).

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