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 1-3 of 3 results.

A176206 Irregular triangle T(n,k) (n >= 1, k >= 1) read by rows: row n has length A000070(n-1) and every column k gives the positive integers.

Original entry on oeis.org

1, 2, 1, 3, 2, 1, 1, 4, 3, 2, 2, 1, 1, 1, 5, 4, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1, 6, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 7, 6, 5, 5, 4, 4, 4, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 8, 7, 6, 6, 5, 5, 5, 4, 4, 4, 4, 4, 3, 3, 3, 3, 3, 3, 3
Offset: 1

Views

Author

Alford Arnold, Apr 11 2010

Keywords

Comments

The original definition was: An irregular table: Row n begins with n, counts down to 1 and repeats the intermediate numbers as often as given by the partition numbers.
Row n contains a decreasing sequence where n-k is repeated A000041(k) times, k = 0..n-1.
From Omar E. Pol, Nov 23 2020: (Start)
Row n lists in nonincreasing order the first A000070(n-1) terms of A336811.
In other words: row n lists in nonincreasing order the terms from the first n rows of triangle A336811.
Conjecture: all divisors of all terms in row n are also all parts of all partitions of n.
For more information see the example and A336811 which contains the most elementary conjecture about the correspondence divisors/partitions.
Row sums give A014153.
A338156 lists the divisors of every term of this sequence.
The n-th row of A340581 lists in nonincreasing order the terms of the first n rows of this triangle.
For a regular triangle with the same row sums see A141157. (End)
From Omar E. Pol, Jul 31 2021: (Start)
The number of k's in row n is equal to A000041(n-k), 1 <= k <= n.
The number of terms >= k in row n is equal to A000070(n-k), 1 <= k <= n.
The number of k's in the first n rows (or in the first A014153(n-1) terms of the sequence) is equal to A000070(n-k), 1 <= k <= n.
The number of terms >= k in the first n rows (or in the first A014153(n-1) terms of the sequence) is equal to A014153(n-k), 1 <= k <= n. (End)

Examples

			Triangle begins:
  1;
  2, 1;
  3, 2, 1, 1;
  4, 3, 2, 2, 1, 1, 1;
  5, 4, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1;
  6, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1;
  7, 6, 5, 5, 4, 4, 4, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, ...
  ... Extended by _Omar E. Pol_, Nov 23 2020
From _Omar E. Pol_, Jan 25 2020: (Start)
For n = 5, by definition the length of row 5 is A000070(5-1) = A000070(4) = 12, so the row 5 of triangle has 12 terms. Since every column lists the positive integers A000027 so the row 5 is [5, 4, 3, 3, 2, 2, 2, 1, 1, 1, 1, 1].
Then we have that the divisors of the numbers of the 5th row are:
.
5th row of triangle -----> 5  4  3  3  2  2  2  1  1  1  1  1
                           1  2  1  1  1  1  1
                              1
.
There are twelve 1's, four 2's, two 3's, one 4 and one 5.
In total there are 12 + 4 + 2 + 1 + 1 = 20 divisors.
On the other hand the partitions of 5 are as shown below:
.
.      5
.      3  2
.      4  1
.      2  2  1
.      3  1  1
.      2  1  1  1
.      1  1  1  1  1
.
There are twelve 1's, four 2's, two 3's, one 4 and one 5, as shown also in the 5th row of triangle A066633.
In total there are 12 + 4 + 2 + 1 + 1 = A006128(5) = 20 parts.
Finally in accordance with the conjecture we can see that all divisors of all numbers in the 5th row of the triangle are the same positive integers as all parts of all partitions of 5. (End)

Links

Crossrefs

Cf. A000027 (columns), A000070 (row lengths), A338156 (divisors), A340061 (mirror).
Cf. A000041, A006128, A014153, A027750, A066633, A141157, A176207, A336811, A338156, A340581.

Programs

  • Mathematica
    Table[Flatten[Table[ConstantArray[n-k,PartitionsP[k]],{k,0,n-1}]],{n,10}] (* Paolo Xausa, May 30 2022 *)

Extensions

New name, changed offset, edited and more terms from Omar E. Pol, Nov 22 2020

A284870 Expansion of Sum_{i>=1} i*x^i/(1 - x) * Product_{j=1..i} 1/(1 - x^j).

Original entry on oeis.org

0, 1, 4, 10, 22, 42, 77, 131, 217, 345, 537, 812, 1211, 1767, 2547, 3615, 5078, 7043, 9687, 13185, 17815, 23867, 31766, 41972, 55146, 71997, 93519, 120813, 155358, 198811, 253374, 321509, 406436, 511802, 642264, 803140, 1001154, 1243966, 1541167, 1903754, 2345300, 2881404, 3531195, 4316632, 5264444, 6405389
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 04 2017

Keywords

Comments

Total number of parts in all partitions of all positive integers <= n.
Sum of largest parts of all partitions of all positive integers <= n.
From Omar E. Pol, Feb 16 2021: (Start)
Apart from initial zero this is as follows:
Convolution of A341062 and A014153.
Convolution of A000005 and A000070.
Convolution of nonzero terms of A006218 and A000041.
a(n) is also the total number of divisors of all terms in the n-th row of triangle A340581. These divisors are also all parts of all partitions of all positive integers <= n. (End)

Examples

			a(4) = 22 because we have 1 = 1, 2 = 2, 1 + 1 = 2, 3 = 3, 2 + 1 = 3, 1 + 1 + 1 = 3, 4 = 4, 3 + 1 = 4, 2 + 2 = 4, 2 + 1 + 1 = 4 and 1 + 1 + 1 + 1 = 4 therefore 1 + 1 + 2 + 1 + 2 + 3 + 1 + 2 + 2 + 3 + 4 = 22 (total number of parts) or 1 + 2 + 1 + 3 + 2 + 1 + 4 + 3 + 2 + 2 + 1 = 22 (sum of largest parts).

Links

Crossrefs

Cf. A000070, A006128, A015724, A026905, A124920, A138137, A182738, A299779.
Cf. A000041, A006218, A014153, A340581, A341062.

Programs

  • Maple
    b:= proc(n, i) option remember; `if`(n=0 or i=1, [1, n],
      b(n, i-1) +(p-> p+[0, p[1]])(b(n-i, min(n-i, i))))
    end:
a:= proc(n) a(n):= `if`(n<1, 0, a(n-1)+b(n$2)[2]) end:
seq(a(n), n=0..45);  # Alois P. Heinz, Feb 16 2021
  • Mathematica
    nmax = 45; CoefficientList[Series[Sum[i x^i /(1 - x) Product[1/(1 - x^j), {j, 1, i}], {i, 1, nmax}], {x, 0, nmax}], x]
nmax = 45; CoefficientList[Series[1/(1 - x) Sum[x^i /(1 - x^i), {i, 1, nmax}] Product[1/(1 - x^j), {j, 1, nmax}], {x, 0, nmax}], x]
Accumulate[Table[Sum[DivisorSigma[0, k] PartitionsP[n - k], {k, 1, n}], {n, 0, 45}]]

Formula

G.f.: Sum_{i>=1} i*x^i/(1 - x) * Product_{j=1..i} 1/(1 - x^j).
G.f.: (1/(1 - x)) * Sum_{i>=1} x^i/(1 - x^i) * Product_{j>=1} 1/(1 - x^j).
a(n) = Sum_{k=0..n} A006128(k).
a(n) = A124920(n+1) - 1.
a(n) = Sum_{k=1..n} k * A299779(n,k). - Alois P. Heinz, May 14 2018

A345116 Irregular triangle T(n,k) read by rows in which row n has length the n-th triangular number A000217(n) and every column k lists the positive integers A000027, n >= 1, k >= 1.

Original entry on oeis.org

1, 2, 1, 1, 3, 2, 2, 1, 1, 1, 4, 3, 3, 2, 2, 2, 1, 1, 1, 1, 5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1, 6, 5, 5, 4, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 7, 6, 6, 5, 5, 5, 4, 4, 4, 4, 3, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 8, 7, 7, 6, 6, 6, 5, 5, 5, 5
Offset: 1

Views

Author

Omar E. Pol, Jun 08 2021

Keywords

Comments

Row n lists the terms of the n-th row of A333516 in nonincreasing order.
The sum of the divisors of the terms of the n-th row of the triangle is equal to A175254(n), equaling the volume of the stepped pyramid with n levels described in A245092.

Examples

			Triangle begins:
1;
2, 1, 1;
3, 2, 2, 1, 1, 1;
4, 3, 3, 2, 2, 2, 1, 1, 1, 1;
5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1, 1;
6, 5, 5, 4, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1;
...
For n = 6 the divisors of the terms of the 6th row of triangle are:
6, 5, 5, 4, 4, 4, 3, 3, 3, 3, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1;
3, 1, 1, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1;
2,       1, 1, 1;
1;
The sum of these divisors is equal to A175254(6) = 82, equaling the volume of the stepped pyramid with six levels described in A245092.

Crossrefs

Mirror of A110730.
Row lengths gives A000217, n >= 1.
Row sums gives A000292, n >= 1.
Every column gives A000027.
Cf. A000203, A175254, A237593, A245092, A333516, A340581.
Showing 1-3 of 3 results.

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