A328361 -id:A328361 - cp's OEIS frontend

A126988 Triangle read by rows: T(n,k) = n/k if k is a divisor of n; T(n,k) = 0 if k is not a divisor of n (1 <= k <= n).

1, 2, 1, 3, 0, 1, 4, 2, 0, 1, 5, 0, 0, 0, 1, 6, 3, 2, 0, 0, 1, 7, 0, 0, 0, 0, 0, 1, 8, 4, 0, 2, 0, 0, 0, 1, 9, 0, 3, 0, 0, 0, 0, 0, 1, 10, 5, 0, 0, 2, 0, 0, 0, 0, 1, 11, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 12, 6, 4, 3, 0, 2, 0, 0, 0, 0, 0, 1

Offset: 1

Gary W. Adamson, Dec 31 2006

Row sums = A000203, sigma(n).
k-th column (k=0,1,2,...) is (1,2,3,...) interspersed with n consecutive zeros starting after the "1".
The nonzero entries of row n are the divisors of n in decreasing order. - Emeric Deutsch, Jan 17 2007
Alternating row sums give A000593. - Omar E. Pol, Feb 11 2018
T(n,k) is the number of k's in the partitions of n into equal parts. - Omar E. Pol, Nov 25 2019

			First few rows of the triangle are:
   1;
   2, 1;
   3, 0, 1;
   4, 2, 0, 1;
   5, 0, 0, 0, 1;
   6, 3, 2, 0, 0, 1;
   7, 0, 0, 0, 0, 0, 1;
   8, 4, 0, 2, 0, 0, 0, 1;
   9, 0, 3, 0, 0, 0, 0, 0, 1;
  10, 5, 0, 0, 2, 0, 0, 0, 0, 1;
  ...
sigma(12) = A000203(n) = 28.
sigma(12) = 28, from 12th row = (12 + 6 + 4 + 3 + 2 + 1), deleting the zeros, from left to right.
For n = 6 the partitions of 6 into equal parts are [6], [3,3], [2,2,2], [1,1,1,1,1,1], so the number of k's are [6, 3, 2, 0, 0, 1] respectively, equaling the 6th row of triangle. - _Omar E. Pol_, Nov 25 2019

David Wells, "Prime Numbers, the Most Mysterious Figures in Math", John Wiley & Sons, Inc, 2005, Appendix B.

Reinhard Zumkeller, Rows n = 1..125 of triangle, flattened

Cf. A000203, A008284, A127446, A244051, A328361.

a126988 n k = a126988_tabl !! (n-1) !! (k-1)
a126988_row n = a126988_tabl !! (n-1)
a126988_tabl = zipWith (zipWith (*)) a010766_tabl a051731_tabl
-- Reinhard Zumkeller, Jan 20 2014

[[(n mod k) eq 0 select n/k else 0: k in [1..n]]: n in [1..12]]; // G. C. Greubel, May 29 2019

A126988:=proc(n,k) if type(n/k, integer)=true then n/k else 0 fi end: for n from 1 to 12 do seq(A126988(n,k),k=1..n) od; # yields sequence in triangular form - Emeric Deutsch, Jan 17 2007

Table[If[Mod[n, m]==0, n/m, 0], {n,1,12}, {m,1,n}]//Flatten (* Roger L. Bagula, Sep 06 2008, simplified by Franklin T. Adams-Watters, Aug 24 2011 *)

{T(n,k) = if(n%k==0, n/k, 0)}; \\ G. C. Greubel, May 29 2019

def T(n, k):
    if (n%k==0): return n/k
    else: return 0
[[T(n, k) for k in (1..n)] for n in (1..12)] # G. C. Greubel, May 29 2019

From Emeric Deutsch, Jan 17 2007: (Start)
G.f. of column k: z^k/(1-z^k)^2 (k=1,2,...).
G.f.: G(t,z) = Sum_{k>=1} t^k*z^k/(1-z^k)^2. (End)
G.f.: F(x,z) = log(1/(Product_{n >= 1} (1 - x*z^n))) = Sum_{n >= 1} (x*z)^n/(n*(1 - z^n)) = x*z + (2*x + x^2)*z^2/2 + (3*x + x^3)*z^3/3 + .... Note, exp(F(x,z)) is a g.f. for A008284 (with an additional term T(0,0) equal to 1). - Peter Bala, Jan 13 2015
T(n,k) = A010766(n,k)*A051731(n,k), k=1..n. - Reinhard Zumkeller, Jan 20 2014

Edited by N. J. A. Sloane, Jan 24 2007

Comment from Emeric Deutsch made name by Franklin T. Adams-Watters, Aug 24 2011

A204217 G.f.: Sum_{n>=1} n * x^(n*(n+1)/2) / (1 - x^n).

Original entry on oeis.org

1, 1, 3, 1, 3, 4, 3, 1, 6, 5, 3, 4, 3, 5, 11, 1, 3, 8, 3, 6, 12, 5, 3, 4, 8, 5, 12, 8, 3, 13, 3, 1, 12, 5, 15, 12, 3, 5, 12, 6, 3, 15, 3, 9, 26, 5, 3, 4, 10, 10, 12, 9, 3, 17, 18, 8, 12, 5, 3, 17, 3, 5, 28, 1, 18, 19, 3, 9, 12, 17, 3, 13, 3, 5, 27, 9, 21, 20, 3, 6, 21, 5, 3, 19, 18, 5, 12, 12, 3

Offset: 1

Paul D. Hanna, Jan 12 2012

Conjecture: a(n) is the total number of parts in all partitions of n into consecutive parts. - Omar E. Pol, Apr 23 2017
Conjecture: row sums of A285914. - Omar E. Pol, Apr 30 2017
(The conjectures are true. See Joerg Arndt's proof in the Links section.) - Omar E. Pol, Jun 14 2017
Row lengths of A299765. - Omar E. Pol, Jul 23 2018
From Omar E. Pol, Oct 20 2019: (Start)
Row sums of A328361.
a(n) = 3 iff n is an odd prime. (End)

			G.f.: A(x) = x + x^2 + 3*x^3 + x^4 + 3*x^5 + 4*x^6 + 3*x^7 + x^8 + ...
follows by expanding A(x) = x/(1-x) + 2*x^3/(1-x^2) + 3*x^6/(1-x^3) + 4*x^10/(1-x^4) + ...
Also, by a Ramanujan identity:
A(x)*Theta4(x)^2 = x*(1-x)/(1+x)^2 - 2*x^3*(1-x^2)/(1+x^2)^2 + 3*x^6*(1-x^3)/(1+x^3)^2 - 4*x^10*(1-x^4)/(1+x^4)^2 + 5*x^15*(1-x^5)/(1+x^5)^2 + ...
For n = 15 there are four partitions of 15 into consecutive parts: [15], [8, 7], [6, 5, 4] and [5, 4, 3, 2, 1]. The total number of parts is 11, so a(15) = 11. - _Omar E. Pol_, Apr 23 2017
From _Omar E. Pol_, Nov 30 2020: (Start)
Illustration of initial terms:
                         Diagram
n   a(n)                       _
1     1                      _|1
2     1                    _|1 _
3     3                  _|1  |2
4     1                _|1   _|
5     3              _|1    |2 _
6     4            _|1     _| |3
7     3          _|1      |2  |
8     1        _|1       _|  _|
9     6      _|1        |2  |3 _
10    5     |1          |   | |4
...
a(n) is the total length of all vertical line segments that are below and that share one vertex with the horizontal line segments that are in the n-th level of the diagram. For more information about the diagram see A286001 and A237593. (End)

David A. Corneth, Table of n, a(n) for n = 1..10000 (First 1024 terms from Paul D. Hanna)
Joerg Arndt, Proof of the conjectures of A204217, SeqFan Mailing List, Jun 03 2017.
Michael H. Mertens, Ken Ono and Larry Rolen, Mock modular Eisenstein series with Nebentypus, arXiv:1906.07410 [math.NT], 2019, p. 4, 9.

Cf. A204218, A237593, A285914, A286001, A299765, A328361, A357618.

terms = 1024; Sum[n*x^(n*(n+1)/2)/(1-x^n), {n, 1, Ceiling[Sqrt[2*terms]]}] + O[x]^(terms+1) // CoefficientList[#, x]& // Rest (* Jean-François Alcover, Jun 04 2017 *)
Table[Sum[If[n > k*(k-1)/2 && IntegerQ[n/k - (k-1)/2], k, 0], {k, Divisors[2*n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 23 2024 *)

{a(n)=polcoeff(sum(m=1,n,m*x^(m*(m+1)/2)/(1-x^m+x*O(x^n))),n)}

{a(n)=local(Theta4=1+2*sum(m=1,sqrtint(n+1),(-x)^(m^2))+x*O(x^n));polcoeff(1/Theta4^2*sum(m=1,sqrtint(2*n+1),(-1)^(m-1)*m*x^(m*(m+1)/2)*(1-x^m)/(1+x^m+x*O(x^n))^2),n)}

a(n) = {nb = 0; forpart(v=n, nbp = #v; if ((#Set(v)==#v) && (v[nbp] - v[1] == #v-1), nb += #v); ); nb; } \\ Michel Marcus, Apr 23 2017

a(n) = {my(i=2, t=1); n--; while(n>0, t += (i*(n%i==0)); n-=i; i++); t} \\ David A. Corneth, Apr 28 2017

a(k) = 1 iff k = 2^n for n>=0.
G.f.: (1/Theta4(x)^2) * Sum_{n>=1} (-1)^(n-1)* n*x^(n*(n+1)/2) * (1 - x^n)/(1 + x^n)^2 where Theta4(x) = 1 + 2*Sum_{n>=1} (-x)^(n^2), due to an identity of Ramanujan.
Conjecture: a(n) = Sum_{k=1..n} k*A285898(n,k). - R. J. Mathar, Apr 30 2017
Conjecture: a(n) = Sum_{k=1..A003056(n)} k*A237048(n,k). - Omar E. Pol, Apr 30 2017
(The conjectures are true. See Joerg Arndt's proof in the Links section.) - Omar E. Pol, Jun 14 2017
Sum_{k=1..n} a(k) ~ 2^(3/2) * n^(3/2) / 3. - Vaclav Kotesovec, Oct 23 2024

A328365 Irregular triangle read by rows, T(n,k), n >= 1, k >= 1, in which row n lists in reverse order the partitions of n into consecutive parts.

Original entry on oeis.org

1, 2, 1, 2, 3, 4, 2, 3, 5, 1, 2, 3, 6, 3, 4, 7, 8, 2, 3, 4, 4, 5, 9, 1, 2, 3, 4, 10, 5, 6, 11, 3, 4, 5, 12, 6, 7, 13, 2, 3, 4, 5, 14, 1, 2, 3, 4, 5, 4, 5, 6, 7, 8, 15, 16, 8, 9, 17, 3, 4, 5, 6, 5, 6, 7, 18, 9, 10, 19, 2, 3, 4, 5, 6, 20, 1, 2, 3, 4, 5, 6, 6, 7, 8, 10, 11, 21, 4, 5, 6, 7, 22, 11, 12, 23, 7, 8, 9, 24

Offset: 1

Omar E. Pol, Oct 22 2019

For m >= 0, row 2^m consists of just one element (2^m). - Paolo Xausa, May 24 2025

			Triangle begins:
  [1];
  [2];
  [1, 2], [3];
  [4];
  [2, 3], [5];
  [1, 2, 3], [6];
  [3, 4], [7];
  [8];
  [2, 3, 4], [4, 5], [9];
  [1, 2, 3, 4], [10];
  [5, 6], [11];
  [3, 4, 5], [12];
  [6, 7], [13];
  [2, 3, 4, 5], [14];
  [1, 2, 3, 4, 5], [4, 5, 6], [7, 8], [15];
  [16];
  [8, 9], [17];
  [3, 4, 5, 6], [5, 6, 7], [18];
  [9, 10], [19];
  [2, 3, 4, 5, 6], [20];
  [1, 2, 3, 4, 5, 6], [6, 7, 8], [10, 11], [21];
  [4, 5, 6, 7], [22];
  [11, 12], [23];
  [7, 8, 9], [24];
  [3, 4, 5, 6, 7], [12, 13], [25];
  [5, 6, 7, 8], [26];
  [2, 3, 4, 5, 6, 7], [8, 9, 10], [13, 14], [27];
  [1, 2, 3, 4, 5, 6, 7], [28];
  ...
For n = 9 there are three partitions of 9 into consecutive parts, they are [9], [5, 4], [4, 3, 2], so the 9th row of triangle is [2, 3, 4], [4, 5], [9].
Note that in the below diagram the number of horizontal line segments in the n-th row equals A001227(n), the number of partitions of n into consecutive parts, so we can find the partitions of n into consecutive parts as follows: consider the vertical blocks of numbers that start exactly in the n-th row of the diagram, for example: for n = 15 consider the vertical blocks of numbers that start exactly in the 15th row. They are [1, 2, 3, 4, 5], [4, 5, 6], [7, 8], [15], equaling the 15th row of the above triangle.
Row        _
  1       |1|_
  2       |_ 2|_
  3       |1|  3|_
  4       |2|_   4|_
  5       |_ 2|    5|_
  6       |1|3|_     6|_
  7       |2|  3|      7|_
  8       |3|_ 4|_       8|_
  9       |_ 2|  4|        9|_
  10      |1|3|  5|_        10|_
  11      |2|4|_   5|         11|_
  12      |3|  3|  6|_          12|_
  13      |4|_ 4|    6|           13|_
  14      |_ 2|5|_   7|_            14|_
  15      |1|3|  4|    7|             15|_
  16      |2|4|  5|    8|_              16|_
  17      |3|5|_ 6|_     8|               17|_
  18      |4|  3|  5|    9|_                18|_
  19      |5|_ 4|  6|      9|                 19|_
  20      |_ 2|5|  7|_    10|_                  20|_
  21      |1|3|6|_   6|     10|                   21|_
  22      |2|4|  4|  7|     11|_                    22|_
  23      |3|5|  5|  8|_      11|                     23|_
  24      |4|6|_ 6|    7|     12|_                      24|_
  25      |5|  3|7|_   8|       12|                       25|_
  26      |6|_ 4|  5|  9|_      13|_                        26|_
  27      |_ 2|5|  6|    8|       13|                         27|_
  28      |1|3|6|  7|    9|       14|                           28|
  ...
The diagram is infinite. For more information about the diagram see A286001.
For an amazing connection with sum of divisors function (A000203) see A237593.

Paolo Xausa, Table of n, a(n) for n = 1..10350 (rows 1..500 of triangle, flattened)

Mirror of A299765.
Row n has length A204217(n).
Row sums give A245579.
Column 1 gives A118235.
Right border gives A000027.
Records give A000027.
Where records occur gives A285899.
The number of partitions into consecutive parts in row n is A001227(n).
For tables of partitions into consecutive parts see A286000 and A286001.
Cf. A000203, A026792, A235791, A237048, A237591, A237593, A245092, A285914, A286013, A288529, A288772, A288773, A288774, A328361, A328362.

Table[With[{h = Floor[n/2] - Boole[EvenQ@ n]},Append[Array[Which[Total@ # == n, #, Total@ Most@ # == n, Most[#], True, Nothing] &@ NestWhile[Append[#, #[[-1]] + 1] &, {#}, Total@ # <= n &, 1, h - # + 1] &, h], {n}]], {n, 24}] // Flatten (* Michael De Vlieger, Oct 22 2019 *)

A285899 Total number of parts in all partitions of all positive integers <= n into consecutive parts.

Original entry on oeis.org

1, 2, 5, 6, 9, 13, 16, 17, 23, 28, 31, 35, 38, 43, 54, 55, 58, 66, 69, 75, 87, 92, 95, 99, 107, 112, 124, 132, 135, 148, 151, 152, 164, 169, 184, 196, 199, 204, 216, 222, 225, 240, 243, 252, 278, 283, 286, 290, 300, 310, 322, 331, 334, 351, 369, 377, 389, 394, 397, 414, 417, 422, 450, 451, 469, 488, 491, 500, 512, 529

Offset: 1

Omar E. Pol, May 02 2017

nonn

Partial sums of A204217.
Sum of first n rows of the triangle A285914.
Where records occur in A328365. - Omar E. Pol, Oct 22 2019
Row sums of A328368. - Omar E. Pol, Nov 04 2019

			For n = 15 there are four partitions of 15 into consecutive parts: [15], [8, 7], [6, 5, 4] and [5, 4, 3, 2, 1]. The total number of parts in these four partitions is 11, and a(14) = 43, so a(15) = 43 + 11 = 54.

Cf. A001227, A196020, A204217, A235791, A237048, A237591, A237593, A245092, A285900, A285914, A299765, A328361, A328362, A328365, A328368.

A328362 Triangle read by rows: T(n,k) is the sum of all parts k in all partitions of n into consecutive parts, (1 <= k <= n).

Original entry on oeis.org

1, 0, 2, 1, 2, 3, 0, 0, 0, 4, 0, 2, 3, 0, 5, 1, 2, 3, 0, 0, 6, 0, 0, 3, 4, 0, 0, 7, 0, 0, 0, 0, 0, 0, 0, 8, 0, 2, 3, 8, 5, 0, 0, 0, 9, 1, 2, 3, 4, 0, 0, 0, 0, 0, 10, 0, 0, 0, 0, 5, 6, 0, 0, 0, 0, 11, 0, 0, 3, 4, 5, 0, 0, 0, 0, 0, 0, 12, 0, 0, 0, 0, 0, 6, 7, 0, 0, 0, 0, 0, 13, 0, 2, 3, 4, 5, 0, 0, 0, 0, 0, 0, 0, 0, 14

Offset: 1

Omar E. Pol, Oct 20 2019

Iff n is a power of 2 (A000079) then row n lists n - 1 zeros together with n.

Iff n is an odd prime (A065091) then row n lists (n - 3)/2 zeros, (n - 1)/2, (n + 1)/2, (n - 3)/2 zeros, n.

			Triangle begins:
1;
0, 2;
1, 2, 3;
0, 0, 0, 4;
0, 2, 3, 0, 5;
1, 2, 3, 0, 0, 6;
0, 0, 3, 4, 0, 0, 7;
0, 0, 0, 0, 0, 0, 0, 8;
0, 2, 3, 8, 5, 0, 0, 0, 9;
1, 2, 3, 4, 0, 0, 0, 0, 0, 10;
0, 0, 0, 0, 5, 6, 0, 0, 0,  0, 11;
0, 0, 3, 4, 5, 0, 0, 0, 0,  0,  0, 12;
0, 0, 0, 0, 0, 6, 7, 0, 0,  0,  0,  0, 13;
0, 2, 3, 4, 5, 0, 0, 0, 0,  0,  0,  0,  0, 14;
1, 2, 3, 8,10, 6, 7, 8, 0,  0,  0,  0,  0,  0, 15;
0, 0, 0, 0, 0, 0, 0, 0, 0,  0,  0,  0,  0,  0,  0, 16;
...
For n = 9 there are three partitions of 9 into consecutive parts, they are [9], [5, 4], [4, 3, 2], so the 9th row of triangle is [0, 2, 3, 8, 5, 0, 0, 0, 9].

Row sums give A245579.
Column 1 gives A010054, n => 1.
Leading diagonal gives A000027.
Cf. A000079, A001227, A065091, A138785, A204217, A237048, A237593, A266531, A285898, A285899, A285900, A285914, A286000, A286001, A299765, A328361.

T(n,k) = k*A328361(n,k).

A328368 Irregular triangle read by rows: T(n,k) is the total number of parts in all partitions of all positive integers <= n into k consecutive parts.

Original entry on oeis.org

1, 2, 3, 2, 4, 2, 5, 4, 6, 4, 3, 7, 6, 3, 8, 6, 3, 9, 8, 6, 10, 8, 6, 4, 11, 10, 6, 4, 12, 10, 9, 4, 13, 12, 9, 4, 14, 12, 9, 8, 15, 14, 12, 8, 5, 16, 14, 12, 8, 5, 17, 16, 12, 8, 5, 18, 16, 15, 12, 5, 19, 18, 15, 12, 5, 20, 18, 15, 12, 10, 21, 20, 18, 12, 10, 6, 22, 20, 18, 16, 10, 6, 23, 22, 18, 16, 10, 6

Offset: 1

Omar E. Pol, Nov 02 2019

Column k lists k times every nonzero multiple of k in nondecreasing order.

Column k lists the partial sums of the k-th column of triangle A285914.

			Triangle begins:
   1;
   2;
   3,  2;
   4,  2;
   5,  4;
   6,  4,  3;
   7,  6,  3;
   8,  6,  3;
   9,  8,  6;
  10,  8,  6,  4;
  11, 10,  6,  4;
  12, 10,  9,  4;
  13, 12,  9,  4;
  14, 12,  9,  8;
  15, 14, 12,  8,  5;
  16, 14, 12,  8,  5;
  17, 16, 12,  8,  5;
  18, 16, 15, 12,  5;
  19, 18, 15, 12,  5;
  20, 18, 15, 12, 10;
  21, 20, 18, 12, 10,  6;
  22, 20, 18, 16, 10,  6;
  23, 22, 18, 16, 10,  6;
  24, 22, 21, 16, 10,  6;
  25, 24, 21, 16, 15,  6;
  26, 24, 21, 20, 15,  6;
  27, 26, 24, 20, 15, 12;
  28, 26, 24, 20, 15, 12, 7;
...

Row sums give A285899.
Row n has length A003056(n).
Column 1 gives A000027.
Column k starts with k in the row A000217(k).
Cf.  A052928, A196020, A204217, A211343, A235791, A236104, A235791, A237048, A237591, A237593, A245579, A262612, A285900, A285914, A285891, A286000, A286001, A286013, A299765, A328361, A328365, A328371.

tt(n, k) = k*(if (k % 2, (n % k) == 0, ((n - k/2) % k) == 0)); \\ A285891
t(n, k) = sum(j=k*(k+1)/2, n, tt(j, k));
tabf(nn) = {for (n=1, nn, for (k=1, floor((sqrt(1+8*n)-1)/2), print1(t(n, k), ", "); ); print(); ); } \\ Michel Marcus, Nov 04 2019

A329321 a(n) is the total number of odd parts in all partitions of n into consecutive parts.

Original entry on oeis.org

1, 0, 2, 0, 2, 2, 2, 0, 3, 2, 2, 2, 2, 2, 6, 0, 2, 4, 2, 2, 6, 2, 2, 2, 5, 2, 6, 4, 2, 6, 2, 0, 6, 2, 8, 6, 2, 2, 6, 2, 2, 8, 2, 4, 14, 2, 2, 2, 5, 4, 6, 4, 2, 8, 10, 4, 6, 2, 2, 8, 2, 2, 14, 0, 10, 10, 2, 4, 6, 8, 2, 6, 2, 2, 14, 4, 10, 10, 2, 2, 11, 2, 2, 10, 10, 2, 6, 6, 2, 16

Offset: 1

Omar E. Pol, Nov 10 2019

nonn

a(n) = 0 if and only if n is an even power of 2.

			For n = 15 there are four partitions of 15 into consecutive part, they are [15], [8, 7], [6, 5, 4], [5, 4, 3, 2, 1]. In total there are six odd parts, they are [15, 7, 5, 5, 3, 1], so a(15) = 6.

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A204217 (total number of parts).

Cf. A000079, A001227, A002131, A237048, A245579, A285898, A285914, A286000, A286001, A299765, A328361, A328365, A329322.

A329321(n) = { my(i=2, t=(n%2)); n--; while(n>0, if(!(n%i), t += (((n/i)%2)+i)\2); n-=i; i++); t }; \\ (After David A. Corneth's program for A204217) - Antti Karttunen, Dec 09 2021

a(n) = A204217(n) - A329322(n).

A329322 a(n) is the total number of even parts in all partitions of n into consecutive parts.

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 1, 3, 3, 1, 2, 1, 3, 5, 1, 1, 4, 1, 4, 6, 3, 1, 2, 3, 3, 6, 4, 1, 7, 1, 1, 6, 3, 7, 6, 1, 3, 6, 4, 1, 7, 1, 5, 12, 3, 1, 2, 5, 6, 6, 5, 1, 9, 8, 4, 6, 3, 1, 9, 1, 3, 14, 1, 8, 9, 1, 5, 6, 9, 1, 7, 1, 3, 13, 5, 11, 10, 1, 4, 10, 3, 1, 9, 8, 3, 6, 6, 1, 18

Offset: 1

Omar E. Pol, Nov 10 2019

nonn

			For n = 15 there are four partitions of 15 into consecutive part, they are [15], [8, 7], [6, 5, 4], [5, 4, 3, 2, 1]. In total there are five even parts, they are [8, 6, 4, 4, 2], so a(15) = 5.

Antti Karttunen, Table of n, a(n) for n = 1..20000

Cf. A204217 (total number of parts).

Cf. A001227, A237048, A245579, A285898, A285914, A286000, A286001, A299765, A328361, A328365, A329321.

A329322(n) = { my(i=2, t=!(n%2)); n--; while(n>0, if(!(n%i), t += (!((n/i)%2)+i)\2); n-=i; i++); t }; \\ (After David A. Corneth's program for A204217) - Antti Karttunen, Dec 09 2021

a(n) = A204217(n) - A329321(n).

A329255 Irregular triangle read by rows: T(n,k) is greatest positive integer <= n that have a partition into k consecutive parts, 1 <= k <= A003056(n), n >= 1.

Original entry on oeis.org

1, 2, 3, 3, 4, 3, 5, 5, 6, 5, 6, 7, 7, 6, 8, 7, 6, 9, 9, 9, 10, 9, 9, 10, 11, 11, 9, 10, 12, 11, 12, 10, 13, 13, 12, 10, 14, 13, 12, 14, 15, 15, 15, 14, 15, 16, 15, 15, 14, 15, 17, 17, 15, 14, 15, 18, 17, 18, 18, 15, 19, 19, 18, 18, 15, 20, 19, 18, 18, 20, 21, 21, 21, 18, 20, 21, 22, 21, 21, 22, 20, 21

Offset: 1

Omar E. Pol, Nov 09 2019

T(n,k) is also the positive integer whose partition into k consecutive parts is associated to the k-th vertex, from left to right, of the largest Dyck path of the symmetric representation of sigma(n). For more information see A237593.

Also this triangle can be constructed replacing every zero of triangle A285891 with the previous positive integer from the same column.

			Triangle begins:
   1;
   2;
   3,  3;
   4,  3;
   5,  5;
   6,  5,  6;
   7,  7,  6;
   8,  7,  6;
   9,  9,  9;
  10,  9,  9, 10;
  11, 11,  9, 10;
  12, 11, 12, 10;
  13, 13, 12, 10;
  14, 13, 12, 14;
  15, 15, 15, 14, 15;
  16, 15, 15, 14, 15;
  17, 17, 15, 14, 15;
  18, 17, 18, 18, 15;
  19, 19, 18, 18, 15;
  20, 19, 18, 18, 20;
  21, 21, 21, 18, 20, 21;
  22, 21, 21, 22, 20, 21;
  23, 23, 21, 22, 20, 21;
  24, 23, 24, 22, 20, 21;
  25, 25, 24, 22, 25, 21;
  26, 25, 24, 26, 25, 21;
  27, 27, 27, 26, 25, 27;
  28, 27, 27, 26, 25, 27, 28;
...

Column k stars with A000217(k) in the row A000217(k).
Row n has length A003056(n).
Cf. A196020, A204217, A211343, A235791, A236104, A235791, A237048, A237591, A237593, A285900, A285914, A285891, A286000, A286001, A286013, A299765, A328361, A328365, A328371.

cp's OEIS Frontend

A126988 Triangle read by rows: T(n,k) = n/k if k is a divisor of n; T(n,k) = 0 if k is not a divisor of n (1 <= k <= n).

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A204217 G.f.: Sum_{n>=1} n * x^(n*(n+1)/2) / (1 - x^n).

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A328365 Irregular triangle read by rows, T(n,k), n >= 1, k >= 1, in which row n lists in reverse order the partitions of n into consecutive parts.

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A285899 Total number of parts in all partitions of all positive integers <= n into consecutive parts.

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A328362 Triangle read by rows: T(n,k) is the sum of all parts k in all partitions of n into consecutive parts, (1 <= k <= n).

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A328368 Irregular triangle read by rows: T(n,k) is the total number of parts in all partitions of all positive integers <= n into k consecutive parts.

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PARI

A329321 a(n) is the total number of odd parts in all partitions of n into consecutive parts.

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A329322 a(n) is the total number of even parts in all partitions of n into consecutive parts.