A056520 -id:A056520 - cp's OEIS frontend

A368045 Triangle read by rows. T(n, k) = (k(k + 1)(2k + 1) + n(n + 1)(2n + 1)) / 6.

0, 1, 2, 5, 6, 10, 14, 15, 19, 28, 30, 31, 35, 44, 60, 55, 56, 60, 69, 85, 110, 91, 92, 96, 105, 121, 146, 182, 140, 141, 145, 154, 170, 195, 231, 280, 204, 205, 209, 218, 234, 259, 295, 344, 408, 285, 286, 290, 299, 315, 340, 376, 425, 489, 570

Offset: 0

Peter Luschny, Dec 09 2023

Consider a sequence-to-triangle transformation a -> T, where a is a 0-based sequence and T a regular (0, 0)-based triangular array. The transformation is recursively defined, starting with T(0, 0) = 0, and T(n, n) = a(n) + T(n, n - 1) for n > 0. For k <> n let T(n, k) = a(n) + T(n-1, k).
If a(n) = 1, then T = A051162; if a(n) = n, then T = A367964 (generalizing the triangular numbers); if a(n) = n^2, then T is this triangle.
In the multiplicative form of the transformation, T(0, 0) is set to 1, and the operation '+' is replaced by '*'. For instance, a(n) = 2 is then mapped to T = A368043 and a(n) = n to A143216.

			Triangle T(n, k) starts:
  [0] [  0]
  [1] [  1,   2]
  [2] [  5,   6,  10]
  [3] [ 14,  15,  19,  28]
  [4] [ 30,  31,  35,  44,  60]
  [5] [ 55,  56,  60,  69,  85, 110]
  [6] [ 91,  92,  96, 105, 121, 146, 182]
  [7] [140, 141, 145, 154, 170, 195, 231, 280]
  [8] [204, 205, 209, 218, 234, 259, 295, 344, 408]
  [9] [285, 286, 290, 299, 315, 340, 376, 425, 489, 570]

Cf. A000330 (T(n,0)), A056520 (T(n,1)), A005900 (T(n-1,n)), A006331 (T(n,n)), A094952 (T(2*n,n)), A368046 (row sums), A368047 (alternating row sums).
Cf. A051162 (transform of n^0), A367964 (transform of n^1), this sequence (transform of n^2).
Cf. A368043, A143216.

Module[{n=1},NestList[Append[#+n^2,Last[#]+2(n++^2)]&,{0},10]] (* or *)
Table[(k(k+1)(2k+1)+n(n+1)(2n+1))/6,{n,0,10},{k,0,n}] (* Paolo Xausa, Dec 10 2023 *)

from functools import cache
@cache
def Trow(n: int) -> list[int]:
    if n == 0: return [0]
    row = Trow(n - 1) + [0]
    for k in range(n): row[k] += n * n
    row[n] = row[n - 1] + n * n
    return row
print([k for n in range(10) for k in Trow(n)])

T(n, k) = A000330(k) + A000330(n).

A180174 Triangle read by rows of the numbers C(n,k) of k-subsets of a quadratically populated n-multiset M.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 3, 5, 7, 9, 10, 10, 10, 10, 10, 9, 7, 5, 3, 1, 1, 4, 9, 16, 25, 35, 45, 55, 65, 75, 84, 91, 96, 99, 100, 100, 100, 99, 96, 91, 84, 75, 65, 55, 45, 35, 25, 16, 9, 4, 1, 1, 5, 14, 30, 55, 90, 135, 190, 255, 330, 414, 505, 601, 700, 800, 900, 1000, 1099

Offset: 0

Thomas Wieder, Aug 15 2010

The multiplicity m(i) of the i-th element with 1 <= i <= n is m(i)=i^2.
Thus M=[1,2,2,2,2,...,i^2 x i,...,n^2 x n].
Row sum is equal to A028361.
Column for k=2 is equal to AA000096.
Column for k=3 is equal to AA005581.
Column for k=4 is equal to AA005582.
The number of coefficients C(n,k) for given n is equal to A056520.

			For n=4 one has M=[1, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4].
For k=7 we have 55 subsets from M:
[1, 2, 2, 3, 3, 4, 4], [1, 2, 3, 3, 4, 4, 4], [1, 2, 3, 3, 3, 4, 4],
[1, 2, 2, 3, 4, 4, 4], [1, 2, 2, 3, 3, 3, 4], [1, 2, 2, 2, 3, 4, 4],
[1, 2, 2, 2, 3, 3, 4], [2, 2, 3, 3, 4, 4, 4], [2, 2, 3, 3, 3, 4, 4],
[2, 2, 2, 3, 3, 4, 4], [1, 2, 2, 2, 3, 3, 3], [1, 2, 2, 2, 4, 4, 4],
[1, 3, 3, 3, 4, 4, 4], [2, 3, 3, 3, 4, 4, 4], [2, 2, 2, 3, 4, 4, 4],
[2, 2, 2, 3, 3, 3, 4], [1, 2, 3, 4, 4, 4, 4], [1, 2, 3, 3, 3, 3, 4],
[1, 2, 2, 2, 2, 3, 4], [1, 2, 2, 3, 3, 3, 3], [1, 2, 2, 2, 2, 3, 3],
[1, 2, 2, 4, 4, 4, 4], [1, 2, 2, 2, 2, 4, 4], [1, 3, 3, 4, 4, 4, 4],
[1, 3, 3, 3, 3, 4, 4], [2, 3, 3, 4, 4, 4, 4], [2, 3, 3, 3, 3, 4, 4],
[2, 2, 3, 4, 4, 4, 4], [2, 2, 3, 3, 3, 3, 4], [2, 2, 2, 2, 3, 4, 4],
[2, 2, 2, 2, 3, 3, 4], [2, 2, 2, 3, 3, 3, 3], [2, 2, 2, 2, 3, 3, 3],
[2, 2, 2, 4, 4, 4, 4], [2, 2, 2, 2, 4, 4, 4], [3, 3, 3, 4, 4, 4, 4],
[3, 3, 3, 3, 4, 4, 4], [1, 2, 3, 3, 3, 3, 3], [1, 2, 4, 4, 4, 4, 4],
[1, 3, 4, 4, 4, 4, 4], [1, 3, 3, 3, 3, 3, 4], [2, 3, 4, 4, 4, 4, 4],
[2, 3, 3, 3, 3, 3, 4], [2, 2, 3, 3, 3, 3, 3], [2, 2, 4, 4, 4, 4, 4],
[3, 3, 4, 4, 4, 4, 4], [3, 3, 3, 3, 3, 4, 4], [1, 3, 3, 3, 3, 3, 3],
[1, 4, 4, 4, 4, 4, 4], [2, 3, 3, 3, 3, 3, 3], [2, 4, 4, 4, 4, 4, 4],
[3, 4, 4, 4, 4, 4, 4], [3, 3, 3, 3, 3, 3, 4], [3, 3, 3, 3, 3, 3, 3],
[4, 4, 4, 4, 4, 4, 4].

Cf. A007318, A008302, A028361, A056520, A000096, A005581, A005582.

with(combinat)
kend := 4;
Liste := NULL;
for k from 0 to kend do
Liste := Liste, `$`(k, k^2)
end do;
Liste := [Liste];
for k from 0 to 2^(kend+1)-1 do
Teilergebnis[k] := choose(Liste, k)
end do;
seq(nops(Teilergebnis[k]), k = 0 .. 2^(kend+1)-1)
' Excel VBA
Sub A180174()
Dim n As Long, nend As Long, k As Long, kk As Long, length_row As Long, length_sum As Long
Dim ATable(10, -1000 To 1000) As Double, Summe As Double
Dim offset_row As Integer, offset_column As Integer
Worksheets("Tabelle2").Select
Cells.Select
Selection.ClearContents
Range("A1").Select
offset_row = 1
offset_column = 1
nend = 7
ATable(0, 0) = 1
Cells(0 + offset_row, 0 + offset_column) = 1
For n = 1 To nend
length_row = n * (n + 1) * (2 * n + 1) / 6
length_sum = n ^ 2 + 1
For k = 0 To length_row / 2
Summe = 0
For kk = k - length_sum + 1 To k
Summe = Summe + ATable(n - 1, kk)
Next kk
ATable(n, k) = Summe
Cells(n + offset_row, k + offset_column) = ATable(n, k)
ATable(n, length_row - k) = Summe
Cells(n + offset_row, length_row - k + 0 + offset_column) = ATable(n, k)
Next k
Next n
End Sub

C(0,0) = 0.
C(n,k) = sum_{j=(k-LS+1)}^{k} C(n-1,j).
for n > 0 and k=1,...,LR with LS = n^2+1 and LR = n*(n+1)*(2*n+1)/6.
C(n,k) = C(n,LR-k).

A214651 Count down from n to 1, n times.

Original entry on oeis.org

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

Offset: 1

Arkadiusz Wesolowski, Jul 24 2012

This sequence contains every positive integer infinitely often.

This is a fractal sequence. Striking out the first instance of every term produces 1, 2, 1, 2, 1, 3, 2, 1, 3, 2, 1, 3, ..., which is the same as the original sequence, as far as it goes.

			1;
2, 1, 2, 1;
3, 2, 1, 3, 2, 1, 3, 2, 1;
...

Arkadiusz Wesolowski, Rows n = 1..14 of triangle, flattened
Eric Weisstein's World of Mathematics, Positive Integer
Wikipedia, Fractal sequence

Cf. A056520 (locations of new values), A060432 (locations of 1's).

Cf. A000290 (row lengths), A002411 (row sums), A036740 (row products).

f[n_] := Table[Range[n, 1, -1], {n}]; Flatten@Array[f, 6] (* Wesolowski *)
Flatten[Table[Table[Range[n, 1, -1], {n}], {n, 6}]] (* Alonso del Arte, Jul 24 2012 *)

A215147 For n odd, a(n) = 1^2+2^2+3^2+...+n^2; for n even, a(n) = (1^2+2^2+3^2+...+n^2)+1.

Original entry on oeis.org

1, 2, 5, 6, 14, 15, 30, 31, 55, 56, 91, 92, 140, 141, 204, 205, 285, 286, 385, 386, 506, 507, 650, 651, 819, 820, 1015, 1016, 1240, 1241, 1496, 1497, 1785, 1786, 2109, 2110, 2470, 2471, 2870, 2871, 3311, 3312, 3795, 3796, 4324, 4325, 4900, 4901, 5525, 5526

Offset: 1

Kritsana Sokhuma, Aug 04 2012

Square pyramidal numbers when n is odd.

An interleaving of A000330 and A056520. - Michel Marcus, Aug 07 2013

Paolo Xausa, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,3,-3,-3,3,1,-1).

Cf. A000330, A056520.

for i from 1 to 100 do a(2*i-1):=sum('k^2','k'=1..i);
a(2*i):=a(2*i-1)+1; end do;

LinearRecurrence[{1, 3, -3, -3, 3, 1, -1}, {1, 2, 5, 6, 14, 15, 30}, 50] (* or *)
Riffle[#, #+1] & [Accumulate[Range[25]^2]] (* Paolo Xausa, Feb 22 2024 *)

From Colin Barker, Nov 16 2012: (Start)
a(n) = (6*(5+3*(-1)^n)+(13-9*(-1)^n)*n-3*(-3+(-1)^n)*n^2+2*n^3)/48.
G.f.: -x*(x^6-x^5-2*x^4+2*x^3-x-1)/((x-1)^4*(x+1)^3). (End)

More terms from Paolo Xausa, Feb 22 2024

A263689 a(n) = (2n^6 - 6n^5 + 5*n^4 - n^2 + 12)/12.

Original entry on oeis.org

1, 1, 2, 34, 277, 1301, 4426, 12202, 29009, 61777, 120826, 220826, 381877, 630709, 1002002, 1539826, 2299201, 3347777, 4767634, 6657202, 9133301, 12333301, 16417402, 21571034, 28007377, 35970001, 45735626, 57617002, 71965909, 89176277, 109687426, 133987426, 162616577, 196171009, 235306402, 280741826

Offset: 0

Ilya Gutkovskiy, Nov 20 2015

			a(0) = 1,
a(1) = 0^5 + 1 = 1,
a(2) = 1^5 + 1 = 2,
a(3) = 2^5 + 2 = 34,
a(4) = 3^5 + 34 = 227,
a(5) = 4^5 + 227 = 1301, etc.

Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Cf. A000124, A000539, A000584, A056520, A154323.

Table[(1/12) (12 + (-1 + n)^2 n^2 (-1 + 2 (-1 + n) n)), {n, 0, 35}]

first(m)=vector(m,n,n--;(2*n^6 - 6*n^5 + 5*n^4 - n^2 + 12)/12) \\ Anders Hellström, Nov 20 2015

G.f.: (1 - 6*x + 16*x^2 + 6*x^3 + 81*x^4 + 20*x^5 + 2*x^6)/(1 - x)^7.
a(n + 1) = a(n) + n^5, a(0) = 1.
a(n + 1) - a(n) = A000584(n).
a(n + 1) = A000539(n) + 1.
Sum_{n>0} 1/(a(n + 1) - a(n)) = zeta(5) = 1.036927755...

A267691 a(n) = (n + 1)(6n^4 - 21n^3 + 31n^2 - 31*n + 30)/30.

Original entry on oeis.org

1, 1, 2, 18, 99, 355, 980, 2276, 4677, 8773, 15334, 25334, 39975, 60711, 89272, 127688, 178313, 243849, 327370, 432346, 562667, 722667, 917148, 1151404, 1431245, 1763021, 2153646, 2610622, 3142063, 3756719, 4464000, 5274000, 6197521, 7246097, 8432018, 9768354

Offset: 0

Ilya Gutkovskiy, Jan 19 2016

			a(0) = 1,
a(1) = 1 + 0^4 = 1,
a(2) = 1 + 1^4 = 2,
a(3) = 2 + 2^4 = 18,
a(4) = 18+ 3^4 = 99, etc.

Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1)

Essentially the same as A000538.
Cf. A000124, A000583, A008514, A056520, A154323, A263689.
Cf. A013662 (zeta(4)).

[(n+1)*(6*n^4-21*n^3+31*n^2-31*n+30)/30: n in [0..35]]; // Vincenzo Librandi, Jan 20 2016

Table[(n + 1) (6 n^4 - 21 n^3 + 31 n^2 - 31 n + 30)/30, {n, 0, 30}]
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 1, 2, 18, 99, 355}, 40] (* Vincenzo Librandi, Jan 20 2016 *)

a(n)=(n+1)*(6*n^4-21*n^3+31*n^2-31*n+30)/30 \\ Charles R Greathouse IV, Jan 19 2016

Vec((1-5*x+11*x^2+x^3+16*x^4)/(x-1)^6 + O(x^100)) \\ Altug Alkan, Jan 19 2016

G.f.: (1 - 5*x + 11*x^2 + x^3 + 16*x^4)/(1 - x)^6.
a(n + 1) = a(n) + n^4.
a(n + 1) = A000538(n) + 1.
a(n + 2) - a(n) = A008514(n).
Sum_{n>=0} 1/a(n) = 2.570450909491318975...
Sum_{n>=1} 1/(a(n + 1) - a(n)) = zeta(4) = Pi^4/90.

A338819 The entries in the rows of the n X n identity matrix, multiplied by the size of the matrix (n).

Original entry on oeis.org

1, 2, 0, 0, 2, 3, 0, 0, 0, 3, 0, 0, 0, 3, 4, 0, 0, 0, 0, 4, 0, 0, 0, 0, 4, 0, 0, 0, 0, 4, 5, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 5, 0, 0, 0, 0, 0, 5, 6, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 6, 0, 0, 0, 0, 0, 0, 6, 7, 0

Offset: 1

Julia Zimmerman, Nov 10 2020

			For every identity matrix of size n starting with n=1, append n*(each entry of each row of the matrix), e.g., n=1 -> 1, n=2 -> 2,0,0,2, so the first 5 terms of the sequence are 1,2,0,0,2.

Cf. A191747 (with 1's), A191748 (locations of nonzero terms), A056520 (start location of each matrix).

import numpy as np
def n_id_sequence(iterations):
    sequence = []
    for i in range(1,iterations+1):
        matrix = i*(np.identity(i))
        for row in matrix:
            for entry in row:
                sequence.append(int(entry))
    return sequence

A386319 Triangle read by rows where row n is the start, corner and end vertex numbers of a triangular spiral with n sides on a triangular grid, starting from 1 and working inwards (0 <= k <= n).

Original entry on oeis.org

1, 1, 1, 1, 2, 3, 1, 3, 5, 6, 1, 4, 7, 9, 10, 1, 5, 9, 12, 14, 15, 1, 6, 11, 15, 18, 20, 21, 1, 7, 13, 18, 22, 25, 27, 28, 1, 8, 15, 21, 26, 30, 33, 35, 36, 1, 9, 17, 24, 30, 35, 39, 42, 44, 45, 1, 10, 19, 27, 34, 40, 45, 49, 52, 54, 55, 1, 11, 21, 30, 38, 45, 51, 56, 60, 63, 65, 66, 1, 12, 23, 33, 42, 50, 57, 63, 68, 72, 75, 77, 78

Offset: 0

Binay Krishna Maity, Jul 18 2025

The first 2 sides are length n-1 so that T(n,1) = 1 + (n-1) and T(n,2) = 1 + 2*(n-1) and then the side lengths decrease by 1 each time as it spirals in (ending at triangular number A000217(n) when n>=1).
These sides mesh to fill the triangle as they go inwards, and can also be thought of going outwards tracing out the sides of the triangle.
The resulting vertex numbers are 1 together with row n of A141419.
Row n=1 is taken as a side of length 0 so the start and end numbers are both 1 (which is not really a spiral but is consistent with the formula and two points 1,2 would be even less like a triangle filled by a spiral).

			Triangle begins:
--------------------------------------
   n\k  0   1   2   3   4   5   6   7
--------------------------------------
   0|   1;
   1|   1,  1;
   2|   1,  2,  3;
   3|   1,  3,  5,  6;
   4|   1,  4,  7,  9, 10;
   5|   1,  5,  9, 12, 14, 15;
   6|   1,  6, 11, 15, 18, 20, 21;
   7|   1,  7, 13, 18, 22, 25, 27, 28;
  ...
For n = 2 the spiral is 2 sides of length 1 so row [1, 2, 3],
   1 --- 2
       /
     3
For n = 4 the spiral is:
   1  2  3  4
    9  10  5
      8  6
        7
The start, corner and end vertices are [1, 4, 7, 9, 10].

Columns: A000012 (k=0), A000027 (k=1), A144396 (k=3).
Cf. A179865(n+1) (main diagonal), A056520 (row sums).
Cf. A000217, A141419.

T[n_,k_]:=If[k==0,1,k(2n-k+1)/2];Table[T[n,k],{n,0,12},{k,0,n}]//Flatten (* James C. McMahon, Jul 31 2025 *)

T(n,0) = 1.

T(n,k) = k*(2*n - k + 1)/2 for k >= 1.

cp's OEIS Frontend

A368045 Triangle read by rows. T(n, k) = (k*(k + 1)*(2*k + 1) + n*(n + 1)*(2*n + 1)) / 6.

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A180174 Triangle read by rows of the numbers C(n,k) of k-subsets of a quadratically populated n-multiset M.

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A214651 Count down from n to 1, n times.

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A215147 For n odd, a(n) = 1^2+2^2+3^2+...+n^2; for n even, a(n) = (1^2+2^2+3^2+...+n^2)+1.

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A263689 a(n) = (2*n^6 - 6*n^5 + 5*n^4 - n^2 + 12)/12.

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A267691 a(n) = (n + 1)*(6*n^4 - 21*n^3 + 31*n^2 - 31*n + 30)/30.

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Magma

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A338819 The entries in the rows of the n X n identity matrix, multiplied by the size of the matrix (n).

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A386319 Triangle read by rows where row n is the start, corner and end vertex numbers of a triangular spiral with n sides on a triangular grid, starting from 1 and working inwards (0 <= k <= n).

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A368045 Triangle read by rows. T(n, k) = (k(k + 1)(2k + 1) + n(n + 1)(2n + 1)) / 6.

A263689 a(n) = (2n^6 - 6n^5 + 5*n^4 - n^2 + 12)/12.

A267691 a(n) = (n + 1)(6n^4 - 21n^3 + 31n^2 - 31*n + 30)/30.