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.

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

A359279 Irregular triangle T(n,k) (n>=1, k>=1) read by rows in which the length of row n equals the partition number A000041(n-1) and every column k gives the positive triangular numbers A000217.

Original entry on oeis.org

1, 3, 6, 1, 10, 3, 1, 15, 6, 3, 1, 1, 21, 10, 6, 3, 3, 1, 1, 28, 15, 10, 6, 6, 3, 3, 1, 1, 1, 1, 36, 21, 15, 10, 10, 6, 6, 3, 3, 3, 3, 1, 1, 1, 1, 45, 28, 21, 15, 15, 10, 10, 6, 6, 6, 6, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 55, 36, 28, 21, 21, 15, 15, 10, 10, 10, 10, 6, 6, 6, 6, 3, 3, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1, 1
Offset: 1

Views

Author

Omar E. Pol, Dec 23 2022

Keywords

Comments

All divisors of the largest partition into consecutive parts of all terms in row n are also all parts of all partitions of n.

Examples

			Triangle begins:
   1;
   3;
   6,  1;
  10,  3,  1;
  15,  6,  3,  1,  1;
  21, 10,  6,  3,  3,  1,  1;
  28, 15, 10,  6,  6,  3,  3, 1, 1, 1, 1;
  36, 21, 15, 10, 10,  6,  6, 3, 3, 3, 3, 1, 1, 1, 1;
  45, 28, 21, 15, 15, 10, 10, 6, 6, 6, 6, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 1;
...
From _Omar E. Pol_, Feb 28 2023: (Start)
For n = 4 the fourth row is [10, 3, 1]. The largest partition into consecutive parts of every term are respectively [4, 3, 2, 1], [2, 1], [1]. The divisors of these parts are [(1, 2, 4), (1, 3), (1, 2), (1)], [(1, 2), (1)], [1]. These 12 divisors are also all parts of all partitions of 4. They are  [(4), (2, 2), (3, 1), (2, 1, 1), (1, 1, 1, 1)]. (End)

Links

Crossrefs

Row sums give A014153 (convolution of A000041 and A000027).
This sequence has the same row sums as A176206, A299779 and A359350.
Cf. A000217, A336811, A336812.

Programs

  • Mathematica
    A359279[rowmax_]:=Table[Flatten[Table[ConstantArray[(n-m)(n-m+1)/2,PartitionsP[m]-PartitionsP[m-1]],{m,0,n-1}]],{n,rowmax}];
A359279[10] (* Generates 10 rows *) (* Paolo Xausa, Mar 06 2023 *)
  • PARI
    A359279(rowmax)=vector(rowmax,n,concat(vector(n,m,vector(numbpart(m-1)-numbpart(m-2),i,(n-m+1)*(n-m+2)/2))));
A359279(10) \\ Generates 10 rows - Paolo Xausa, Mar 06 2023

Formula

T(n,k) = A000217(A336811(n,k)).

A359350 Irregular triangle T(n,k) (n >= 1, k >= 1) read by rows: row n is constructed by replacing A336811(n,k) with the largest partition into consecutive parts of A000217(A336811(n,k)).

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Dec 27 2022

Keywords

Comments

All divisors of all terms in row n are also all parts of all partitions of n.
The terms of row n listed in nonincreasing order give the n-th row of A176206.
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.
Row n is constructed replacing A336811(n,k) with the largest partition into consecutive parts of A359279(n,k).

Examples

			Triangle begins:
  1;
  2, 1;
  3, 2, 1, 1;
  4, 3, 2, 1, 2, 1, 1;
  5, 4, 3, 2, 1, 3, 2, 1, 2, 1, 1, 1;
  6, 5, 4, 3, 2, 1, 4, 3, 2, 1, 3, 2, 1, 2, 1, 2, 1, 1, 1;
  ...
Or also the triangle begins:
  [1];
  [2, 1];
  [3, 2, 1],          [1];
  [4, 3, 2, 1],       [2, 1],       [1];
  [5, 4, 3, 2, 1],    [3, 2, 1],    [2, 1],    [1],    [1];
  [6, 5, 4, 3, 2, 1], [4, 3, 2, 1], [3, 2, 1], [2, 1], [2, 1], [1], [1];
  ...
For n = 3 the third row is [3, 2, 1, 1]. The divisors of these terms are [1, 3], [1, 2], [1], [1]. These six divisors are also all parts of all partitions of 3. They are [3], [2, 1], [1, 1, 1].

Links

Crossrefs

Row sums give A014153 (convolution of A000041 and A000027).
Row lengths give A000070.
Row n has A000041(n-1) blocks.
This triangle has the same row sums as A176206, A299779 and A359279.
Cf. A000217, A336811, A336812, A338156, A359280.

Programs

  • Mathematica
    A359350row[n_]:=Flatten[Table[ConstantArray[Range[n-m,1,-1],PartitionsP[m]-PartitionsP[m-1]],{m,0,n-1}]];Array[A359350row,10] (* Paolo Xausa, Sep 01 2023 *)
Showing 1-3 of 3 results.

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