A330889 -id:A330889 - cp's OEIS frontend

A285914 Irregular triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists k's interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

1, 1, 1, 2, 1, 0, 1, 2, 1, 0, 3, 1, 2, 0, 1, 0, 0, 1, 2, 3, 1, 0, 0, 4, 1, 2, 0, 0, 1, 0, 3, 0, 1, 2, 0, 0, 1, 0, 0, 4, 1, 2, 3, 0, 5, 1, 0, 0, 0, 0, 1, 2, 0, 0, 0, 1, 0, 3, 4, 0, 1, 2, 0, 0, 0, 1, 0, 0, 0, 5, 1, 2, 3, 0, 0, 6, 1, 0, 0, 4, 0, 0, 1, 2, 0, 0, 0, 0, 1, 0, 3, 0, 0, 0, 1, 2, 0, 0, 5, 0, 1, 0, 0, 4, 0, 0, 1, 2, 3, 0, 0, 6

Offset: 1

Omar E. Pol, Apr 28 2017

Conjecture 1: T(n,k) is the number of parts in the partition of n into k consecutive parts, if T(n,k) > 0.
Conjecture 2: row sums give A204217, which should be also the total number of parts in all partitions of n into consecutive parts.
(The conjectures are true. See Joerg Arndt's proof in the Links section.) - Omar E. Pol, Jun 14 2017
From Omar E. Pol, May 05 2020: (Start)
Theorem: Let T(n,k) be an irregular triangle read by rows in which column k lists k's interleaved with k-1 zeros, and the first element of column k is in the row that is the k-th (m+2)-gonal number, with n >= 1, k >= 1, m >= 0. T(n,k) is also the number of parts in the partition of n into k consecutive parts that differ by m, including n as a valid partition. Hence the sum of row n gives the total number of parts in all partitions of n into consecutive parts that differ by m.
About the above theorem, this is the case for m = 1. For m = 0 see the triangle A127093, in which row sums give A000203. For m = 2 see the triangle A330466, in which row sums give A066839 (conjectured). For m = 3 see the triangle A330888, in which row sums give A330889.
Note that there are infinitely many triangles of this kind, with m >= 0. Also, every triangle can be represented with a diagram of overlapping curves, in which every column of triangle is represented by a periodic curve. (End)

			Triangle begins (rows 1..28):
1;
1;
1,  2;
1,  0;
1,  2;
1,  0,  3;
1,  2,  0;
1,  0,  0;
1,  2,  3;
1,  0,  0,  4;
1,  2,  0,  0;
1,  0,  3,  0;
1,  2,  0,  0;
1,  0,  0,  4;
1,  2,  3,  0,  5;
1,  0,  0,  0,  0;
1,  2,  0,  0,  0;
1,  0,  3,  4,  0;
1,  2,  0,  0,  0;
1,  0,  0,  0,  5;
1,  2,  3,  0,  0,  6;
1,  0,  0,  4,  0,  0;
1,  2,  0,  0,  0,  0;
1,  0,  3,  0,  0,  0;
1,  2,  0,  0,  5,  0;
1,  0,  0,  4,  0,  0;
1,  2,  3,  0,  0,  6;
1,  0,  0,  0,  0,  0,  7;
...
In accordance with the conjectures, for n = 15 there are four partitions of 15 into consecutive parts: [15], [8, 7], [6, 5, 4] and [5, 4, 3, 2, 1]. These partitions are formed by 1, 2, 3 and 5 consecutive parts respectively, so the 15th row of the triangle is [1, 2, 3, 0, 5].
Illustration of initial terms:
Row                                                         _
1                                                         _|1|
2                                                       _|1 _|
3                                                     _|1  |2|
4                                                   _|1   _|0|
5                                                 _|1    |2 _|
6                                               _|1     _|0|3|
7                                             _|1      |2  |0|
8                                           _|1       _|0 _|0|
9                                         _|1        |2  |3 _|
10                                      _|1         _|0  |0|4|
11                                    _|1          |2   _|0|0|
12                                  _|1           _|0  |3  |0|
13                                _|1            |2    |0 _|0|
14                              _|1             _|0   _|0|4 _|
15                            _|1              |2    |3  |0|5|
16                          _|1               _|0    |0  |0|0|
17                        _|1                |2     _|0 _|0|0|
18                      _|1                 _|0    |3  |4  |0|
19                    _|1                  |2      |0  |0 _|0|
20                  _|1                   _|0     _|0  |0|5 _|
21                _|1                    |2      |3   _|0|0|6|
22              _|1                     _|0      |0  |4  |0|0|
23            _|1                      |2       _|0  |0  |0|0|
24          _|1                       _|0      |3    |0 _|0|0|
25        _|1                        |2        |0   _|0|5  |0|
26      _|1                         _|0       _|0  |4  |0 _|0|
27    _|1                          |2        |3    |0  |0|6 _|
28   |1                            |0        |0    |0  |0|0|7|
...
Note that the k's are placed exactly below the k-th horizontal line segment of every row.
The above structure is related to the triangle A237591, also to the left-hand part of the triangle A237593, and also to the left-hand part of the front view of the pyramid described in A245092.

Joerg Arndt, Proof of the conjectures of A204217 and A285914, SeqFan Mailing Lists, Jun 03 2017.

Row n has length A003056(n).
Column k starts in row A000217(k).
The number of positive terms in row n is A001227(n), the number of partitions of n into consecutive parts.
Cf. A000203, A066839, A196020, A204217, A235791, A236104, A237048, A237591, A237593, A245092, A261699, A285898, A262626, A330889.
Triangles of the same family are A127093, this sequence, A330466, A330888.

With[{nn = 6}, Table[Boole[If[EvenQ@ k, Mod[(n - k/2), k] == 0, Mod[n, k] == 0]] k, {n, nn (nn + 3)/2}, {k, Floor[((Sqrt[8 n + 1] - 1)/2)]}]] // Flatten (* Michael De Vlieger, Jun 15 2017, after Python by Indranil Ghosh *)

t(n, k) = if (k % 2, (n % k) == 0, ((n - k/2) % k) == 0); \\ A237048
tabf(nn) = {for (n=1, nn, for (k=1, floor((sqrt(1+8*n)-1)/2), print1(k*t(n, k), ", "); ); print(); ); } \\ Michel Marcus, Nov 04 2019

from sympy import sqrt
import math
def a237048(n, k):
    return int(n%k == 0) if k%2 else int(((n - k//2)%k) == 0)
def T(n, k): return k*a237048(n, k)
for n in range(1, 29): print([T(n, k) for k in range(1, int(math.floor((sqrt(8*n + 1) - 1)/2)) + 1)]) # Indranil Ghosh, Apr 30 2017

T(n,k) = k*A237048(n,k).

A334949 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 6.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 4, 3, 1, 6, 1, 3, 4, 3, 1, 6, 1, 3, 4, 3, 1, 6, 1, 3, 4, 7, 1, 6, 1, 7, 4, 3, 1, 10, 1, 3, 4, 7, 1, 6, 1, 7, 4, 3, 1, 10, 1, 3, 4, 7, 6, 6, 1, 7, 4, 8, 1, 10, 1, 3, 9, 7, 1, 6, 1, 12, 4, 3, 1, 10, 6, 3, 4, 7, 1, 11, 1, 7, 4

Offset: 1

Omar E. Pol, May 27 2020

nonn

The one-part partition n = n is included in the count.

For the relation to the octagonal numbers see also A334947.

			For n = 24 there are three partitions of 24 into consecutive parts that differ by 6, including 24 as a valid partition. They are [24], [15, 9] and [14, 8, 2]. There are 1, 2 and 3 parts respectively, hence the total number of parts is 1 + 2 + 3 = 6, so a(24) = 6.

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Row sums of A334947.
Column k=6 of A334466.
Sequences of the same family whose consecutive parts differs by k are: A000203 (k=0), A204217 (k=1), A066839 (k=2), A330889 (k=3), A334464 (k=4), A334732 (k=5), this sequence (k=6).
Cf. A000567, A334946, A334947, A334948, A334953.

nmax = 100;
CoefficientList[Sum[n x^(n(3n-2)-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* Jean-François Alcover, Nov 30 2020 *)
Table[Sum[If[n > 3*k*(k-1), k, 0], {k, Divisors[n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 22 2024 *)

my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(3*k-2))/(1-x^k))) \\ Seiichi Manyama, Dec 04 2020

G.f.: Sum_{n>=1} n*x^(n*(3*n-2))/(1-x^n). (For proof, see A330889. - Omar E. Pol, Nov 22 2020)

Sum_{k=1..n} a(k) ~ 2 * n^(3/2) / 3^(3/2). - Vaclav Kotesovec, Oct 23 2024

A334466 Square array read by antidiagonals upwards: T(n,k) is the total number of parts in all partitions of n into consecutive parts that differ by k, with n >= 1, k >= 0.

Original entry on oeis.org

1, 3, 1, 4, 1, 1, 7, 3, 1, 1, 6, 1, 1, 1, 1, 12, 3, 3, 1, 1, 1, 8, 4, 1, 1, 1, 1, 1, 15, 3, 3, 3, 1, 1, 1, 1, 13, 1, 1, 1, 1, 1, 1, 1, 1, 18, 6, 3, 3, 3, 1, 1, 1, 1, 1, 12, 5, 4, 1, 1, 1, 1, 1, 1, 1, 1, 28, 3, 3, 3, 3, 3, 1, 1, 1, 1, 1, 1, 14, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 24, 3, 6, 3, 3, 3, 3, 1

Offset: 1

Omar E. Pol, May 01 2020

The one-part partition n = n is included in the count.
The column k is related to (k+2)-gonal numbers, assuming that 2-gonals are the nonnegative numbers, 3-gonals are the triangular numbers, 4-gonals are the squares, 5-gonals are the pentagonal numbers, and so on.
Note that the number of parts for T(n,0) = A000203(n), equaling the sum of the divisors of n.
For fixed k>0, Sum_{j=1..n} T(j,k) ~ 2^(3/2) * n^(3/2) / (3*sqrt(k)). - Vaclav Kotesovec, Oct 23 2024

			Square array starts:
   n\k|   0  1  2  3  4  5  6  7  8  9 10 11 12
   ---+---------------------------------------------
   1  |   1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   2  |   3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   3  |   4, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   4  |   7, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   5  |   6, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, ...
   6  |  12, 4, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, ...
   7  |   8, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, 1, ...
   8  |  15, 1, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, 1, ...
   9  |  13, 6, 4, 3, 1, 3, 1, 3, 1, 1, 1, 1, 1, ...
  10  |  18, 5. 3. 1. 3. 1, 3, 1, 3, 1, 1, 1, 1, ...
  11  |  12, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 1, 1, ...
  12  |  28, 4, 6, 4, 3, 1, 3, 1, 3, 1, 3, 1, 1, ...
  ...
For n = 9 we have that:
For k = 0 the partitions of 9 into consecutive parts that differ by 0 (or simply: the partitions of 9 into equal parts) are [9], [3,3,3], [1,1,1,1,1,1,1,1,1]. In total there are 13 parts, so T(9,0) = 13.
For k = 1 the partitions of 9 into consecutive parts that differ by 1 (or simply: the partitions of 9 into consecutive parts) are [9], [5,4], [4,3,2]. In total there are six parts, so T(9,1) = 6.
For k = 2 the partitions of 9 into consecutive parts that differ by 2 are [9], [5, 3, 1]. In total there are four parts, so T(9,2) = 4.

Columns k: A000203 (k=0), A204217 (k=1), A066839 (k=2), A330889 (k=3), A334464 (k=4), A334732 (k=5), A334949 (k=6), A377300 (k=7), A377301 (k=8).
Triangles whose row sums give the column k: A127093 (k=0), A285914 (k=1), A330466 (k=2) (conjectured), A330888 (k=3), A334462 (k=4), A334540 (k=5), A339947 (k=6).
Sequences of number of partitions related to column k: A000005 (k=0), A001227 (k=1), A038548 (k=2), A117277 (k=3), A334461 (k=4), A334541 (k=5), A334948 (k=6).
Tables of partitions related to column k: A010766 (k=0), A286001 (k=1), A332266 (k=2), A334945 (k=3), A334618 (k=4).
Polygonal numbers related to column k: A001477 (k=0), A000217 (k=1), A000290 (k=2), A000326 (k=3), A000384 (k=4), A000566 (k=5), A000567 (k=6).
Cf. A051735, A237048, A303300, A330887, A334460, A334465, A334946.
Cf. A038040, A245579, A060872, A334467, A327262, A334733, A334953.
Cf. A057145, A323345.

nmax = 14;
col[k_] := col[k] = CoefficientList[Sum[n x^(n(k n - k + 2)/2)/(1 - x^n), {n, 1, nmax}] + O[x]^(nmax+1), x];
T[n_, k_] := col[k][[n+1]];
Table[T[n-k, k], {n, 1, nmax}, {k, 0, n-1}] // Flatten (* Jean-François Alcover, Nov 30 2020 *)

The g.f. for column k is Sum_{n>=1} n*x^(n*(k*n-k+2)/2)/(1-x^n). (For proof, see A330889. - N. J. A. Sloane, Nov 21 2020)

A334464 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 4.

Original entry on oeis.org

1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 4, 3, 1, 6, 1, 3, 4, 3, 1, 6, 1, 3, 4, 7, 1, 6, 1, 7, 4, 3, 1, 10, 1, 3, 4, 7, 1, 6, 1, 7, 9, 3, 1, 10, 1, 8, 4, 7, 1, 6, 6, 7, 4, 3, 1, 15, 1, 3, 4, 7, 6, 12, 1, 7, 4, 8, 1, 16, 1, 3, 9, 7, 1, 12, 1, 12, 4, 3, 1, 16, 6, 3, 4, 7, 1, 17, 8, 7, 4

Offset: 1

Omar E. Pol, May 05 2020

nonn

The one-part partition n = n is included in the count.

For the relation to hexagonal numbers see also A334462.

			For n = 28 there are three partitions of 28 into consecutive parts that differ by 4, including 28 as a valid partition. They are [28], [16, 12] and [13, 9, 5, 1]. The number of parts of these partitions are 1, 2, 4 respectively. The total number of parts is 1 + 2 + 4 = 7, so a(28) = 7.

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Row sums of A334462.
Column k=4 of A334466.
Cf. A000384.
Sequences of the same family whose consecutive parts differs by k are: A000203 (k=0), A204217 (k=1), A066839 (k=2), A330889 (k=3), this sequence (k=4), A334732 (k=5), A334949 (k=6).
Cf. A334461.

nmax = 100;
CoefficientList[Sum[n x^(n(2n-1)-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* Jean-François Alcover, Nov 30 2020 *)
Table[Sum[If[n > 2*k*(k-1), k, 0], {k, Divisors[n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 22 2024 *)

my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(2*k-1))/(1-x^k))) \\ Seiichi Manyama, Dec 04 2020

G.f.: Sum_{n>=1} n*x^(n*(2*n-1))/(1-x^n). (For proof, see A330889. - N. J. A. Sloane, Nov 21 2020)

Sum_{k=1..n} a(k) ~ sqrt(2) * n^(3/2) / 3. - Vaclav Kotesovec, Oct 23 2024

A334732 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 5.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 1, 3, 4, 3, 1, 6, 1, 3, 4, 3, 1, 6, 1, 3, 4, 3, 1, 6, 5, 3, 4, 3, 5, 6, 1, 3, 8, 3, 1, 6, 5, 3, 4, 3, 5, 6, 1, 3, 8, 8, 1, 6, 5, 3, 9, 3, 5, 6, 1, 8, 8, 3, 1, 6, 10, 3, 4, 3, 5, 11, 1, 3, 8, 3, 6, 12, 5, 3, 4, 8, 5, 12, 1, 3, 13, 3, 1, 12

Offset: 1

Omar E. Pol, May 09 2020

nonn

The one-part partition n = n is included in the count.

For the relation to heptagonal numbers see also A334540.

			For n = 27 there are three partitions of 27 into consecutive parts that differ by 5, including 27 as a valid partition. They are [27], [16, 11] and [14, 9, 4]. The number of parts of these partitions are 1, 2, 3 respectively and the total number of parts is 1 + 2 + 3 = 6, so the a(27) = 6.

Seiichi Manyama, Table of n, a(n) for n = 1..10000

Row sums of A334540.
Column k=5 of A334466.
Sequences of the same family whose consecutive parts differs by k are: A000203 (k=0), A204217 (k=1), A066839 (k=2), A330889 (k=3), A334464 (k=4), this sequence (k=5), A334949 (k=6).
Cf. A000566, A334465, A334540, A334541, A334733.

A334732 := proc(n)
    local a,k;
    a := 0 ;
    for k from 1 do
        if n>= A000566(k) then
            a := a+A334540(n,k);
        else
            return a;
        end if;
    end do:
end proc:
seq(A334732(n),n=1..120) ; # R. J. Mathar, Oct 02 2020

nmax = 100;
CoefficientList[Sum[n x^(n(5n-3)/2-1)/(1-x^n), {n, 1, nmax}]+O[x]^nmax, x] (* Jean-François Alcover, Nov 30 2020 *)
Table[Sum[If[n > 5*k*(k-1)/2 && IntegerQ[n/k - 5*(k-1)/2], k, 0], {k, Divisors[2*n]}], {n, 1, 100}] (* Vaclav Kotesovec, Oct 23 2024 *)

my(N=66, x='x+O('x^N)); Vec(sum(k=1, N, k*x^(k*(5*k-3)/2)/(1-x^k))) \\ Seiichi Manyama, Dec 04 2020

G.f.: Sum_{n>=1} n*x^(n*(5*n-3)/2)/(1-x^n) (for proof see A330889).

Sum_{k=1..n} a(k) ~ 2^(3/2) * n^(3/2) / (3*sqrt(5)). - Vaclav Kotesovec, Oct 23 2024

More terms from R. J. Mathar, Oct 02 2020

A334463 a(n) is the sum of all parts of all partitions of n into consecutive parts that differ by 3.

Original entry on oeis.org

1, 2, 3, 4, 10, 6, 14, 8, 18, 10, 22, 24, 26, 14, 45, 16, 34, 36, 38, 20, 63, 44, 46, 48, 50, 52, 81, 28, 58, 90, 62, 32, 99, 68, 105, 72, 74, 76, 117, 80, 82, 126, 86, 44, 180, 92, 94, 96, 98, 150, 204, 52, 106, 162, 165, 56, 228, 116, 118, 180, 122, 124, 252, 64, 195, 198, 134, 68, 276, 280, 142, 144

Offset: 1

Omar E. Pol, May 05 2020

nonn

The one-part partition n = n is included in the count.

			For n = 21 there are three partitions of 21 into consecutive parts that differ by 3, including 21 as a valid partition. They are [21], [12, 9] and [10, 7, 4]. The sum of the parts is [21] + [12 + 9] + [10 + 7 + 4] = 63, the same as 3*21 = 63, since there are three of these partitions of 21, so a(21) = 63.

Cf. A038040, A245579, A330887, A330888, A330889.

Sequences of the same family where the parts differs by k are: A038040 (k=0), A245579 (k=1), A060872 (k=2), this sequence (k=3), A327262 (k=4).

a(n) = n*A117277(n).

A334945 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists successive blocks of k consecutive integers that differ by 3, where the m-th block starts with m, m >= 1, and the first element of column k is in the row that is the k-th pentagonal number (A000326).

Original entry on oeis.org

1, 2, 3, 4, 5, 1, 6, 4, 7, 2, 8, 5, 9, 3, 10, 6, 11, 4, 12, 7, 1, 13, 5, 4, 14, 8, 7, 15, 6, 2, 16, 9, 5, 17, 7, 8, 18, 10, 3, 19, 8, 6, 20, 11, 9, 21, 9, 4, 22, 12, 7, 1, 23, 10, 10, 4, 24, 13, 5, 7, 25, 11, 8, 10, 26, 14, 11, 2, 27, 12, 6, 5, 28, 15, 9, 8, 29, 13, 12, 11, 30, 16, 7, 3

Offset: 1

Omar E. Pol, May 27 2020

This triangle can be interpreted as a table of partitions into consecutive parts that differ by 3 (see the Example section).

Also, every triangle of this family has the property that starting from row n the sum of k positive and consecutive terms in the column k is equal to n. - Omar E. Pol, Dec 18 2020

			Triangle begins:
   1;
   2;
   3;
   4;
   5,  1;
   6,  4;
   7,  2;
   8,  5;
   9,  3;
  10,  6;
  11,  4;
  12,  7,  1;
  13,  5,  4;
  14,  8,  7;
  15,  6,  2;
  16,  9,  5;
  17,  7,  8;
  18, 10,  3;
  19,  8,  6;
  20, 11,  9;
  21,  9,  4;
  22, 12,  7,  1;
...
Figures A..G show the location (in the columns of the table) of the partitions of n = 1..7 (respectively) into consecutive parts that differ by 3:
.   -----------------------------------------------------
Fig:   A     B     C     D       E        F        G
.   -----------------------------------------------------
. n:   1     2     3     4       5        6        7
Row -----------------------------------------------------
1   | [1];|  1; |  1; |  1; |  1;     |  1;   |  1;     |
2   |     | [2];|  2; |  2; |  2;     |  2;   |  2;     |
3   |     |     | [3];|  3; |  3;     |  3;   |  3;     |
4   |     |     |     | [4];|  4;     |  4;   |  4;     |
5   |     |     |     |     | [5],[1];|  5, 1;|  5,  1; |
6   |     |     |     |     |  6, [4];| [6],4;|  6,  4; |
7   |     |     |     |     |         |       | [7],[2];|
8   |     |     |     |     |         |       |  8, [5];|
.   -----------------------------------------------------
Figure G: for n = 7 the partitions of 7 into consecutive parts that differ by 3 (but with the parts in increasing order) are [7] and [2, 5]. These partitions have one part and two parts respectively. On the other hand  we can find the mentioned partitions in the columns 1 and 2 of this table, starting at the row 7.
.
Illustration of initial terms arranged into a triangular structure:
.                                                           _
.                                                         _|1|
.                                                       _|2  |
.                                                     _|3    |
.                                                   _|4     _|
.                                                 _|5      |1|
.                                               _|6       _|4|
.                                             _|7        |2  |
.                                           _|8         _|5  |
.                                         _|9          |3    |
.                                       _|10          _|6    |
.                                     _|11           |4     _|
.                                   _|12            _|7    |1|
.                                 _|13             |5      |4|
.                               _|14              _|8     _|7|
.                             _|15               |6      |2  |
.                           _|16                _|9      |5  |
.                         _|17                 |7       _|8  |
.                       _|18                  _|10     |3    |
.                     _|19                   |8        |6    |
.                   _|20                    _|11      _|9    |
.                 _|21                     |9        |4     _|
.                |22                       |12       |7    |1|
...
The number of horizontal line segments in the n-th row of the diagram equals A117277(n), the number of partitions of n into consecutive parts that differ by 3.

Tables of the same family where the consecutive parts differ by d are A010766 (d=0), A286001 (d=1), A332266 (d=2), this sequence (d=3), A334618(d=4).

Cf. A000326, A117277, A330887, A330888, A330889, A334463.

A338730 Generating function Sum_{n >= 1} a(n)x^n = Sum_{k>=1} kx^(k(3k+1)/2)/(1-x^k).

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 3, 1, 6, 1, 3, 4, 3, 1, 6, 1, 3, 4, 3, 5, 6, 1, 3, 8, 3, 1, 6, 5, 3, 4, 3, 5, 6, 6, 3, 8, 3, 1, 11, 5, 3, 4, 3, 10, 6, 1, 3, 8, 8, 1, 12, 5, 3, 9, 3, 5, 12, 1, 8, 8, 3, 1, 12, 10, 3, 4, 3, 5, 17, 1, 10, 8, 3, 6, 12, 5, 3, 11, 8, 5, 12, 1, 3

Offset: 0

N. J. A. Sloane, Dec 02 2020

nonn

The OEIS contains many very similar sequences, but this one was missing.

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Cf. A001227, A081757, A117277, A330889, A338731, A338732.

my(N=66, x='x+O('x^N)); concat([0, 0], Vec(sum(k=1, N, k*x^(k*(3*k+1)/2)/(1-x^k)))) \\ Seiichi Manyama, Dec 03 2020

A338731 Generating function Sum_{n >= 0} a(n)x^n = Sum_{k>=1} x^(k(3*k+1)/2)/(1-x^k).

Original entry on oeis.org

0, 0, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 3, 2, 1, 3, 2, 2, 2, 2, 2, 3, 2, 2, 3, 2, 1, 4, 2, 2, 2, 2, 3, 3, 1, 2, 3, 3, 1, 4, 2, 2, 3, 2, 2, 4, 1, 3, 3, 2, 1, 4, 3, 2, 2, 2, 2, 5, 1, 3, 3, 2, 2, 4, 2, 2, 3, 3, 2, 4, 1, 2, 4, 3, 1

Offset: 0

N. J. A. Sloane, Dec 02 2020

nonn

The OEIS contains many very similar sequences, but this one was missing.

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Cf. A001227, A081757, A117277, A330889, A338730, A338732.

my(N=66, x='x+O('x^N)); concat([0, 0], Vec(sum(k=1, N, x^(k*(3*k+1)/2)/(1-x^k)))) \\ Seiichi Manyama, Dec 03 2020

A338732 Generating function Sum_{n >= 0} a(n)x^n = Sum_{k>=1} x^(k(3*k+1))/(1-x^k).

Original entry on oeis.org

0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 3, 1, 3, 1, 3, 2, 2, 1, 4, 1, 2, 2, 3, 1, 3, 1, 3, 2, 2, 1, 4, 1, 2, 2, 3, 1, 3, 1, 4, 2, 2, 1, 4, 2, 2, 2, 3, 1, 4, 1, 3

Offset: 0

N. J. A. Sloane, Dec 03 2020

nonn

The OEIS contains many very similar sequences, but this one was missing.

Seiichi Manyama, Table of n, a(n) for n = 0..10000

Cf. A001227, A081757, A117277, A330889, A338730, A338731.

my(N=66, x='x+O('x^N)); concat([0, 0, 0, 0], Vec(sum(k=1, N, x^(k*(3*k+1))/(1-x^k)))) \\ Seiichi Manyama, Dec 04 2020

cp's OEIS Frontend

A285914 Irregular triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists k's interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

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A334949 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 6.

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A334466 Square array read by antidiagonals upwards: T(n,k) is the total number of parts in all partitions of n into consecutive parts that differ by k, with n >= 1, k >= 0.

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A334464 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 4.

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A334732 a(n) is the total number of parts in all partitions of n into consecutive parts that differ by 5.

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A334463 a(n) is the sum of all parts of all partitions of n into consecutive parts that differ by 3.

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A334945 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists successive blocks of k consecutive integers that differ by 3, where the m-th block starts with m, m >= 1, and the first element of column k is in the row that is the k-th pentagonal number (A000326).

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A338730 Generating function Sum_{n >= 1} a(n)*x^n = Sum_{k>=1} k*x^(k*(3*k+1)/2)/(1-x^k).

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A338730 Generating function Sum_{n >= 1} a(n)x^n = Sum_{k>=1} kx^(k(3k+1)/2)/(1-x^k).

A338731 Generating function Sum_{n >= 0} a(n)x^n = Sum_{k>=1} x^(k(3*k+1)/2)/(1-x^k).

A338732 Generating function Sum_{n >= 0} a(n)x^n = Sum_{k>=1} x^(k(3*k+1))/(1-x^k).