A141157 -id:A141157 - cp's OEIS frontend

A014153 Expansion of 1/((1-x)^2*Product_{k>=1} (1-x^k)).

1, 3, 7, 14, 26, 45, 75, 120, 187, 284, 423, 618, 890, 1263, 1771, 2455, 3370, 4582, 6179, 8266, 10980, 14486, 18994, 24757, 32095, 41391, 53123, 67865, 86325, 109350, 137979, 173450, 217270, 271233, 337506, 418662, 517795, 638565, 785350, 963320, 1178628

Offset: 0

N. J. A. Sloane

nonn

Number of partitions of n with three kinds of 1. E.g., a(2)=7 because we have 2, 1+1, 1+1', 1+1", 1'+1', 1'+1", 1"+1". - Emeric Deutsch, Mar 22 2005
Partial sums of the partial sums of the partition numbers A000041. Partial sums of A000070. Euler transform of 3,1,1,1,...
Also sum of parts, counted without multiplicity, in all partitions of n, offset 1. Also Sum phi(p), where the sum is taken over all parts p of all partitions of n, offset 1. - Vladeta Jovovic, Mar 26 2005
Equals row sums of triangle A141157. - Gary W. Adamson, Jun 12 2008
A014153 convolved with A010815 = (1, 2, 3, ...). n-th partial sum sequence of A000041 convolved with A010815 = (n-1)-th column of Pascal's triangle, starting (1, n, ...). - Gary W. Adamson, Nov 09 2008
From Omar E. Pol, May 25 2012: (Start)
a(n) is also the sum of all parts of the (n+1)st column of a version of the section model of partitions in which every section has its parts aligned to the right margin (cf. A210953, A210970, A135010).
Rows of triangle A210952 converge to this sequence. (End)
Using the above result (see Jovovic's comment) of Jovovic and Mertens's theorem on the average order of the phi function, we can obtain the estimate a(n-1) = (6/Pi^2)*n*p(n) + O(log(n)*A006128(n)), where p(n) is the partition function A000041(n). It can be shown that A006128(n) = O(sqrt(n)*log(n)*p(n)), so we have the asymptotic result a(n) ~ (6/Pi^2)*n*p(n). - Peter Bala, Dec 23 2013
a(n-2) is the number of partitions of 2n or 2n-1 with palindromicity 2; that is, partitions that can be listed in palindromic order except for a central sequence of two distinct parts. - Gregory L. Simay, Nov 01 2015
Convolution of A000041 and A000027. - Omar E. Pol, Jun 17 2021
Convolution of A002865 and the positive terms of A000217. Partial sums give A014160. - Omar E. Pol, Mar 01 2023

Alois P. Heinz and Vaclav Kotesovec, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Alois P. Heinz)
Mircea Merca and Maxie D. Schmidt, The partition function p(n) in terms of the classical Möbius function, arXiv:2310.13658 [math.CO], 2023.

Cf. A000041, A000070, A116930, A141157.
Cf. A010815. - Gary W. Adamson, Nov 09 2008
Column k=3 of A292508.
Cf. A000217, A002865, A014160.

m:=45; R:=PowerSeriesRing(Integers(), m); Coefficients(R!( 1/((1-x)^2*(&*[1-x^k: k in [1..50]])) )); // G. C. Greubel, Oct 15 2018

with(numtheory):
a:= proc(n) option remember;
      `if`(n=0, 1, add((2+sigma(j)) *a(n-j), j=1..n)/n)
    end:
seq(a(n), n=0..40);  # Alois P. Heinz, Feb 13 2012

a[n_] := a[n] = If[n == 0, 1, Sum[(2+DivisorSigma[1, j])*a[n-j], {j, 1, n}]/n]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Mar 03 2014, after Alois P. Heinz *)
Table[Sum[(n-k)*PartitionsP[k],{k,0,n}],{n,1,50}] (* Vaclav Kotesovec, Jun 23 2015 *)
t[n_, k_] := Sum[StirlingS1[n, j]* Binomial[i + j - 1, i]* PartitionsP[k - n - i], {j, 0, n}, {i, 0, k - n}]; Print@ Table[t[n, k], {k, 10}, {n, 0, k - 1}]; Table[t[2, k], {k, 3, 43}] (* George Beck, May 25 2016 *)

x='x+O('x^45); Vec(1/((1-x)^2*prod(k=1,50, 1-x^k))) \\ G. C. Greubel, Oct 15 2018

Let t(n_, k_) = Sum_{i = 0..k} Sum_{j = 0..n} s(n, j)*C(i,  j)*p(k - n - i), where s(n, j) are Stirling numbers of the first kind, C(i, j) are the number of compositions of i distinct objects into j parts, and p is the integer partition function. Then a(k) = t(2, k+2) (conjectured). The formula for t(n, k) is the same as at A126442 except that there the Stirling numbers are of the second kind. - George Beck, May 21 2016
a(n) = (n+1)*A000070(n+1) - A182738(n+1). - Vaclav Kotesovec, Nov 04 2016
a(n) ~ exp(sqrt(2*n/3)*Pi)*sqrt(3)/(2*Pi^2) * (1 + 23*Pi/(24*sqrt(6*n))). - Vaclav Kotesovec, Nov 04 2016

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

Alford Arnold, Apr 11 2010

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)

			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)

Paolo Xausa, Table of n, a(n) for n = 1..10980 (rows 1..21 of triangle, flattened)

Cf. A000027 (columns), A000070 (row lengths), A338156 (divisors), A340061 (mirror).

Cf. A000041, A006128, A014153, A027750, A066633, A141157, A176207, A336811, A338156, A340581.

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

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

cp's OEIS Frontend

A014153 Expansion of 1/((1-x)^2*Product_{k>=1} (1-x^k)).

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