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.

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A003056 n appears n+1 times. Also the array A(n,k) = n+k (n >= 0, k >= 0) read by antidiagonals. Also inverse of triangular numbers.

Original entry on oeis.org

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

Views

Author

Keywords

Comments

Also triangle read by rows: T(n,k), n>=0, k>=0, in which n appears n+1 times in row n. - Omar E. Pol, Jul 15 2012
The PARI functions t1, t2 can be used to read a triangular array T(n,k) (n >= 0, 0 <= k <= n-1) by rows from left to right: n -> T(t1(n), t2(n)). - Michael Somos, Aug 23 2002
Number of terms in partition of n with greatest number of distinct terms. - Amarnath Murthy, May 20 2001
Summation table for (x+y) = (0+0),(0+1),(1+0),(0+2),(1+1),(2+0), ...
Also the number of triangular numbers less than or equal to n, not counting 0 as triangular. - Robert G. Wilson v, Oct 21 2005
Permutation of A116939: a(n) = A116939(A116941(n)), a(A116942(n)) = A116939(n). - Reinhard Zumkeller, Feb 27 2006
Maximal size of partitions of n into distinct parts, see A000009. - Reinhard Zumkeller, Jun 13 2009
Also number of digits of A000462(n). - Reinhard Zumkeller, Mar 27 2011
Also the maximum number of 1's contained in the list of hook-lengths of a partition of n. E.g., a(4)=2 because hooks of partitions of n=4 comprise {4,3,2,1}, {4,2,1,1}, {3,2,2,1}, {4,1,2,1}, {4,3,2,1} where the number of 1's in each is 1,2,1,2,1. Hence the maximum is 2. - T. Amdeberhan, Jun 03 2012
Fan, Yang, and Yu (2012) prove a conjecture of Amdeberhan on the generating function of a(n). - Jonathan Sondow, Dec 17 2012
Also the number of partitions of n into distinct parts p such that max(p) - min(p) <= length(p). - Clark Kimberling, Apr 18 2014
Also the maximum number of occurrences of any single value among the previous terms. - Ivan Neretin, Sep 20 2015
Where records occur gives A000217. - Omar E. Pol, Nov 05 2015
Also number of peaks in the largest Dyck path of the symmetric representation of sigma(n), n >= 1. Cf. A237593. - Omar E. Pol, Dec 19 2016

Examples

			G.f. = x + x^2 + 2*x^3 + 2*x^4 + 2*x^5 + 3*x^6 + 3*x^7 + 3*x^8 + 3*x^9 + 4*x^10 + ...
As triangle, the sequence starts
  0;
  1, 1;
  2, 2, 2;
  3, 3, 3, 3;
  4, 4, 4, 4, 4;
  5, 5, 5, 5, 5, 5;
  6, 6, 6, 6, 6, 6, 6;
  7, 7, 7, 7, 7, 7, 7, 7;
  8, 8, 8, 8, 8, 8, 8, 8, 8;
  ...
		

Crossrefs

a(n) = A002024(n+1)-1.
Cf. A000196, A000217, A000462, A001227, A001462, A001614, A004247 (multiplication table), A006463 (partial sums), A016655, A050600, A050602, A048645, A122797, A131507, A238005.
Partial sums of A073424.

Programs

  • Haskell
    a003056 = floor . (/ 2) . (subtract 1) .
                      sqrt . (+ 1) . (* 8) . fromIntegral
    a003056_row n = replicate (n + 1) n
    a003056_tabl = map a003056_row [0..]
    a003056_list = concat $ a003056_tabl
    -- Reinhard Zumkeller, Aug 02 2014, Oct 17 2010
    
  • Magma
    [Floor((Sqrt(1+8*n)-1)/2): n in [0..80]]; // Vincenzo Librandi, Oct 23 2011
    
  • Maple
    A003056 := (n,k) -> n: # Peter Luschny, Oct 29 2011
    a := [ 0 ]: for i from 1 to 15 do for j from 1 to i+1 do a := [ op(a),i ]; od: od: a;
    A003056 := proc(n)
        floor((sqrt(1+8*n)-1)/2) ;
    end proc: # R. J. Mathar, Jul 10 2015
  • Mathematica
    f[n_] := Floor[(Sqrt[1 + 8n] - 1)/2]; Table[ f[n], {n, 0, 87}] (* Robert G. Wilson v, Oct 21 2005 *)
    Table[x, {x, 0, 13}, {y, 0, x}] // Flatten
    T[ n_, k_] := If[ n >= k >= 0, n, 0]; (* Michael Somos, Dec 22 2016 *)
    Flatten[Table[PadRight[{},n+1,n],{n,0,12}]] (* Harvey P. Dale, Jul 03 2021 *)
  • PARI
    A003056(n)=(sqrtint(8*n+1)-1)\2  \\ M. F. Hasler, Oct 08 2011
    
  • PARI
    t1(n)=floor(-1/2+sqrt(2+2*n)) /* A003056 */
    
  • PARI
    t2(n)=n-binomial(floor(1/2+sqrt(2+2*n)),2) /* A002262 */
    
  • Python
    from math import isqrt
    def A003056(n): return (k:=isqrt(m:=n+1<<1))+int((m<<2)>(k<<2)*(k+1)+1)-1 # Chai Wah Wu, Jul 26 2022

Formula

a(n) = floor((sqrt(1+8*n)-1)/2). - Antti Karttunen
a(n) = floor(-1/2 + sqrt(2*n+b)) with 1/4 <= b < 9/4 or a(n) = floor((sqrt(8*n+b)-1)/2) with 1 <= b < 9. - Michael A. Childers (childers_moof(AT)yahoo.com), Nov 11 2001
a(n) = f(n,0) with f(n,k) = k if n <= k, otherwise f(n-k-1, k+1). - Reinhard Zumkeller, May 23 2009
a(n) = 2*n + 1 - A001614(n+1) = n + 1 - A122797(n+1). - Reinhard Zumkeller, Feb 12 2012
a(n) = k if k*(k+1)/2 <= n < (k+1)*(k+2)/2. - Jonathan Sondow, Dec 17 2012
G.f.: (1-x)^(-1)*Sum_{n>=1} x^(n*(n+1)/2) = (Theta_2(0,x^(1/2)) - 2*x^(1/8))/(2*x^(1/8)*(1-x)) where Theta_2 is a Jacobi Theta function. - Robert Israel, May 21 2015
a(n) = floor((A000196(1+8*n)-1)/2). - Pontus von Brömssen, Dec 10 2018
a(n+1) = a(n-a(n)) + 1, a(0) = 0. - Rok Cestnik, Dec 29 2020
a(n) = A001227(n) + A238005(n), n >= 1. - Omar E. Pol, Sep 30 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = log(2)/2 (cf. A016655). - Amiram Eldar, Sep 24 2023
G.f. as array: (x + y - 2*x*y)/((1 - x)^2*(1 - y)^2). - Stefano Spezia, Dec 20 2023 [corrected by Stefano Spezia, Apr 22 2024]

Extensions

Definition clarified by N. J. A. Sloane, Dec 08 2020

A237270 Triangle read by rows in which row n lists the parts of the symmetric representation of sigma(n).

Original entry on oeis.org

1, 3, 2, 2, 7, 3, 3, 12, 4, 4, 15, 5, 3, 5, 9, 9, 6, 6, 28, 7, 7, 12, 12, 8, 8, 8, 31, 9, 9, 39, 10, 10, 42, 11, 5, 5, 11, 18, 18, 12, 12, 60, 13, 5, 13, 21, 21, 14, 6, 6, 14, 56, 15, 15, 72, 16, 16, 63, 17, 7, 7, 17, 27, 27, 18, 12, 18, 91, 19, 19, 30, 30, 20, 8, 8, 20, 90
Offset: 1

Views

Author

Omar E. Pol, Feb 19 2014

Keywords

Comments

T(n,k) is the number of cells in the k-th region of the n-th set of regions in a diagram of the symmetry of sigma(n), see example.
Row n is a palindromic composition of sigma(n).
Row sums give A000203.
Row n has length A237271(n).
In the row 2n-1 of triangle both the first term and the last term are equal to n.
If n is an odd prime then row n is [m, m], where m = (1 + n)/2.
The connection with A196020 is as follows: A196020 --> A236104 --> A235791 --> A237591 --> A237593 --> A239660 --> this sequence.
For the boundary segments in an octant see A237591.
For the boundary segments in a quadrant see A237593.
For the boundary segments in the spiral see also A239660.
For the parts in every quadrant of the spiral see A239931, A239932, A239933, A239934.
We can find the spiral on the terraces of the stepped pyramid described in A244050. - Omar E. Pol, Dec 07 2016
T(n,k) is also the area of the k-th terrace, from left to right, at the n-th level, starting from the top, of the stepped pyramid described in A245092 (see Links section). - Omar E. Pol, Aug 14 2018

Examples

			Illustration of the first 27 terms as regions (or parts) of a spiral constructed with the first 15.5 rows of A239660:
.
.                  _ _ _ _ _ _ _ _
.                 |  _ _ _ _ _ _ _|_ _ _ _ _ _ _ 7
.                 | |             |_ _ _ _ _ _ _|
.             12 _| |                           |
.               |_ _|  _ _ _ _ _ _              |_ _
.         12 _ _|     |  _ _ _ _ _|_ _ _ _ _ 5      |_
.      _ _ _| |    9 _| |         |_ _ _ _ _|         |
.     |  _ _ _|  9 _|_ _|                   |_ _ 3    |_ _ _ 7
.     | |      _ _| |      _ _ _ _          |_  |         | |
.     | |     |  _ _| 12 _|  _ _ _|_ _ _ 3    |_|_ _ 5    | |
.     | |     | |      _|   |     |_ _ _|         | |     | |
.     | |     | |     |  _ _|           |_ _ 3    | |     | |
.     | |     | |     | |    3 _ _        | |     | |     | |
.     | |     | |     | |     |  _|_ 1    | |     | |     | |
.    _|_|    _|_|    _|_|    _|_| |_|    _|_|    _|_|    _|_|    _
.   | |     | |     | |     | |         | |     | |     | |     | |
.   | |     | |     | |     |_|_ _     _| |     | |     | |     | |
.   | |     | |     | |    2  |_ _|_ _|  _|     | |     | |     | |
.   | |     | |     |_|_     2    |_ _ _|7   _ _| |     | |     | |
.   | |     | |    4    |_                 _|  _ _|     | |     | |
.   | |     |_|_ _        |_ _ _ _        |  _|    _ _ _| |     | |
.   | |    6      |_      |_ _ _ _|_ _ _ _| | 15 _|    _ _|     | |
.   |_|_ _ _        |_   4        |_ _ _ _ _|  _|     |    _ _ _| |
.  8      | |_ _      |                       |      _|   |  _ _ _|
.         |_    |     |_ _ _ _ _ _            |  _ _|28  _| |
.           |_  |_    |_ _ _ _ _ _|_ _ _ _ _ _| |      _|  _|
.          8  |_ _|  6            |_ _ _ _ _ _ _|  _ _|  _|
.                 |                               |  _ _|  31
.                 |_ _ _ _ _ _ _ _                | |
.                 |_ _ _ _ _ _ _ _|_ _ _ _ _ _ _ _| |
.                8                |_ _ _ _ _ _ _ _ _|
.
.
[For two other drawings of the spiral see the links. - _N. J. A. Sloane_, Nov 16 2020]
If the sequence does not contain negative terms then its terms can be represented in a quadrant. For the construction of the diagram we use the symmetric Dyck paths of A237593 as shown below:
---------------------------------------------------------------
Triangle         Diagram of the symmetry of sigma (n = 1..24)
---------------------------------------------------------------
.              _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
1;            |_| | | | | | | | | | | | | | | | | | | | | | | |
3;            |_ _|_| | | | | | | | | | | | | | | | | | | | | |
2, 2;         |_ _|  _|_| | | | | | | | | | | | | | | | | | | |
7;            |_ _ _|    _|_| | | | | | | | | | | | | | | | | |
3, 3;         |_ _ _|  _|  _ _|_| | | | | | | | | | | | | | | |
12;           |_ _ _ _|  _| |  _ _|_| | | | | | | | | | | | | |
4, 4;         |_ _ _ _| |_ _|_|    _ _|_| | | | | | | | | | | |
15;           |_ _ _ _ _|  _|     |  _ _ _|_| | | | | | | | | |
5, 3, 5;      |_ _ _ _ _| |      _|_| |  _ _ _|_| | | | | | | |
9, 9;         |_ _ _ _ _ _|  _ _|    _| |    _ _ _|_| | | | | |
6, 6;         |_ _ _ _ _ _| |  _|  _|  _|   |  _ _ _ _|_| | | |
28;           |_ _ _ _ _ _ _| |_ _|  _|  _ _| | |  _ _ _ _|_| |
7, 7;         |_ _ _ _ _ _ _| |  _ _|  _|    _| | |    _ _ _ _|
12, 12;       |_ _ _ _ _ _ _ _| |     |     |  _|_|   |* * * *
8, 8, 8;      |_ _ _ _ _ _ _ _| |  _ _|  _ _|_|       |* * * *
31;           |_ _ _ _ _ _ _ _ _| |  _ _|  _|      _ _|* * * *
9, 9;         |_ _ _ _ _ _ _ _ _| | |_ _ _|      _|* * * * * *
39;           |_ _ _ _ _ _ _ _ _ _| |  _ _|    _|* * * * * * *
10, 10;       |_ _ _ _ _ _ _ _ _ _| | |       |* * * * * * * *
42;           |_ _ _ _ _ _ _ _ _ _ _| |  _ _ _|* * * * * * * *
11, 5, 5, 11; |_ _ _ _ _ _ _ _ _ _ _| | |* * * * * * * * * * *
18, 18;       |_ _ _ _ _ _ _ _ _ _ _ _| |* * * * * * * * * * *
12, 12;       |_ _ _ _ _ _ _ _ _ _ _ _| |* * * * * * * * * * *
60;           |_ _ _ _ _ _ _ _ _ _ _ _ _|* * * * * * * * * * *
...
The total number of cells in the first n set of symmetric regions of the diagram equals A024916(n), the sum of all divisors of all positive integers <= n, hence the total number of cells in the n-th set of symmetric regions of the diagram equals sigma(n) = A000203(n).
For n = 9 the 9th row of A237593 is [5, 2, 2, 2, 2, 5] and the 8th row of A237593 is [5, 2, 1, 1, 2, 5] therefore between both symmetric Dyck paths there are three regions (or parts) of sizes [5, 3, 5], so row 9 is [5, 3, 5].
The sum of divisors of 9 is 1 + 3 + 9 = A000203(9) = 13. On the other hand the sum of the parts of the symmetric representation of sigma(9) is 5 + 3 + 5 = 13, equaling the sum of divisors of 9.
For n = 24 the 24th row of A237593 is [13, 4, 3, 2, 1, 1, 1, 1, 2, 3, 4, 13] and the 23rd row of A237593 is [12, 5, 2, 2, 1, 1, 1, 1, 2, 2, 5, 12] therefore between both symmetric Dyck paths there are only one region (or part) of size 60, so row 24 is 60.
The sum of divisors of 24 is 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = A000203(24) = 60. On the other hand the sum of the parts of the symmetric representation of sigma(24) is 60, equaling the sum of divisors of 24.
Note that the number of *'s in the diagram is 24^2 - A024916(24) = 576 - 491 = A004125(24) = 85.
From _Omar E. Pol_, Nov 22 2020: (Start)
Also consider the infinite double-staircases diagram defined in A335616 (see the theorem).
For n = 15 the diagram with first 15 levels looks like this:
.
Level                         "Double-staircases" diagram
.                                          _
1                                        _|1|_
2                                      _|1 _ 1|_
3                                    _|1  |1|  1|_
4                                  _|1   _| |_   1|_
5                                _|1    |1 _ 1|    1|_
6                              _|1     _| |1| |_     1|_
7                            _|1      |1  | |  1|      1|_
8                          _|1       _|  _| |_  |_       1|_
9                        _|1        |1  |1 _ 1|  1|        1|_
10                     _|1         _|   | |1| |   |_         1|_
11                   _|1          |1   _| | | |_   1|          1|_
12                 _|1           _|   |1  | |  1|   |_           1|_
13               _|1            |1    |  _| |_  |    1|            1|_
14             _|1             _|    _| |1 _ 1| |_    |_             1|_
15            |1              |1    |1  | |1| |  1|    1|              1|
.
Starting from A196020 and after the algorithm described in A280850 and A296508 applied to the above diagram we have a new diagram as shown below:
.
Level                             "Ziggurat" diagram
.                                          _
6                                         |1|
7                            _            | |            _
8                          _|1|          _| |_          |1|_
9                        _|1  |         |1   1|         |  1|_
10                     _|1    |         |     |         |    1|_
11                   _|1      |        _|     |_        |      1|_
12                 _|1        |       |1       1|       |        1|_
13               _|1          |       |         |       |          1|_
14             _|1            |      _|    _    |_      |            1|_
15            |1              |     |1    |1|    1|     |              1|
.
The 15th row
of A249351 :  [1,1,1,1,1,1,1,1,0,0,0,1,1,1,2,1,1,1,0,0,0,1,1,1,1,1,1,1,1]
The 15th row
of triangle:  [              8,            8,            8              ]
The 15th row
of A296508:   [              8,      7,    1,    0,      8              ]
The 15th row
of A280851    [              8,      7,    1,            8              ]
.
More generally, for n >= 1, it appears there is the same correspondence between the original diagram of the symmetric representation of sigma(n) and the "Ziggurat" diagram of n.
For the definition of subparts see A239387 and also A296508, A280851. (End)
		

Crossrefs

Programs

  • Mathematica
    T[n_,k_] := Ceiling[(n + 1)/k - (k + 1)/2] (* from A235791 *)
    path[n_] := Module[{c = Floor[(Sqrt[8n + 1] - 1)/2], h, r, d, rd, k, p = {{0, n}}}, h = Map[T[n, #] - T[n, # + 1] &, Range[c]]; r = Join[h, Reverse[h]]; d = Flatten[Table[{{1, 0}, {0, -1}}, {c}], 1];
    rd = Transpose[{r, d}]; For[k = 1, k <= 2c, k++, p = Join[p, Map[Last[p] + rd[[k, 2]] * # &, Range[rd[[k, 1]]]]]]; p]
    segments[n_] := SplitBy[Map[Min, Drop[Drop[path[n], 1], -1] - path[n - 1]], # == 0 &]
    a237270[n_] := Select[Map[Apply[Plus, #] &, segments[n]], # != 0 &]
    Flatten[Map[a237270, Range[40]]] (* data *)
    (* Hartmut F. W. Hoft, Jun 23 2014 *)

Formula

T(n, k) = (A384149(n, k) + A384149(n, m+1-k))/2, where m = A237271(n) is the row length. (conjectured) - Peter Munn, Jun 01 2025

Extensions

Drawing of the spiral extended by Omar E. Pol, Nov 22 2020

A237591 Irregular triangle read by rows: T(n,k) is the difference between the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the total number of partitions of all positive integers <= n into exactly k+1 consecutive parts (n>=1, 1<=k<=A003056(n)).

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Feb 22 2014

Keywords

Comments

The original name was: Triangle read by rows: T(n,k) = A235791(n,k) - A235791(n,k+1), assuming that the virtual right border of triangle A235791 is A000004.
T(n,k) is also the length of the k-th segment in a zig-zag path on the first quadrant of the square grid, connecting the point (n, 0) with the point (m, m), starting with a segment in vertical direction, where m <= n.
Conjecture: the area of the polygon defined by the x-axis, this zig-zag path and the diagonal [(0, 0), (m, m)], is equal to A024916(n)/2, one half of the sum of all divisors of all positive integers <= n. Therefore the reflected polygon, which is adjacent to the y-axis, with the zig-zag path connecting the point (0, n) with the point (m, m), has the same property. And so on for each octant in the four quadrants.
For the representation of A024916 and A000203 we use two octants, for example: the first octant and the second octant, or the 6th octant and the 7th octant, etc., see A237593.
At least up to n = 128, two zig-zag paths never cross (checked by hand).
The finite sequence formed by the n-th row of triangle together with its mirror row gives the n-th row of triangle A237593.
The connection between A196020 and A237271 is as follows: A196020 --> A236104 --> A235791 --> this sequence --> A237593 --> A239660 --> A237270 --> A237271.
Comments from Franklin T. Adams-Watters on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014. (Start)
The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple.
You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric.
Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n-1; and the number of squares added is sigma(n).
Finally, look at the connected areas enclosed by n but not by n-1; the size of these areas is the symmetric representation of sigma. (End)
From Hartmut F. W. Hoft, Apr 07 2014: (Start)
The row sum is A235791(n,1) - A235791(n,floor((sqrt(8n+1)-1)/2)+1) = n - 0.
Mathematica function has been written to check the conjecture as well as non-crossing zig-zag paths (Dyck paths rotated by 90 degrees) up through n=30000 (same applies to A237593). (End)
The n-th zig-zag path ending at the point (m, m), where m = A240542(n). - Omar E. Pol, Apr 16 2014
From Omar E. Pol, Aug 23 2015: (Start)
n is an odd prime if and only if T(n,2) = 1 + T(n-1,2) and T(n,k) = T(n-1,k) for the rest of the values of k.
The elements of the n-th row of triangle together with the elements of the n-th row of triangle A261350 give the n-th row of triangle A237593.
T(n,k) is also the area (or the number of cells) of the k-th vertical side at the n-th level (starting from the top) in the left hand part of the front view of the stepped pyramid described in A245092, see Example section.
(End)
From Omar E. Pol, Nov 19 2015: (Start)
T(n,k) is also the number of cells between the k-th and the (k+1)st line segments (from left to right) in the n-th row of the diagram as shown in Example section.
Note that the number of horizontal line segments in the n-th row of the diagram equals A001227(n), the number of odd divisors of n. (End)
Conjecture: the values f(n,k) in the n-th row of the triangle are either 1 or 2 for all k with ceiling((sqrt(4*n+1)-1)/2) <= k <= floor((sqrt(8*n+1)-1)/2) = r(n), the length of the n-th row, though the lower bound need not be minimal; tested through 2500000. See also A285356. - Hartmut F. W. Hoft, Apr 17 2017
Conjecture: T(n,k) is the difference between the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the total number of partitions of all positive integers <= n into exactly k+1 consecutive parts. - Omar E. Pol, Apr 30 2017
From Omar E. Pol, Aug 31 2021: (Start)
It appears that T(n,2)/T(n,1) converges to 1/3.
It appears that T(n,3)/T(n,2) converges to 1/2.
It appears that T(n,4)/T(n,3) converges to 3/5.
It appears that T(n,5)/T(n,4) converges to 2/3. (End)
In other words: T(n,k) is the length of the k-th line segment of the largest Dyck path of the symmetric representation of sigma(n). - Omar E. Pol, Sep 08 2021

Examples

			Triangle begins:
   1;
   2;
   2, 1;
   3, 1;
   3, 2;
   4, 1, 1;
   4, 2, 1;
   5, 2, 1;
   5, 2, 2;
   6, 2, 1, 1;
   6, 3, 1, 1;
   7, 2, 2, 1;
   7, 3, 2, 1;
   8, 3, 1, 2;
   8, 3, 2, 1, 1;
   9, 3, 2, 1, 1;
   9, 4, 2, 1, 1;
  10, 3, 2, 2, 1;
  10, 4, 2, 2, 1;
  11, 4, 2, 1, 2;
  11, 4, 3, 1, 1, 1;
  12, 4, 2, 2, 1, 1;
  12, 5, 2, 2, 1, 1;
  13, 4, 3, 2, 1, 1;
  13, 5, 3, 1, 2, 1;
  14, 5, 2, 2, 2, 1;
  14, 5, 3, 2, 1, 2;
  15, 5, 3, 2, 1, 1, 1;
  ...
For n = 10 the 10th row of triangle A235791 is [10, 4, 2, 1] so row 10 is [6, 2, 1, 1].
From _Omar E. Pol_, Aug 23 2015: (Start)
Illustration of initial terms:
  Row                                                         _
   1                                                        _|1|
   2                                                      _|2 _|
   3                                                    _|2  |1|
   4                                                  _|3   _|1|
   5                                                _|3    |2 _|
   6                                              _|4     _|1|1|
   7                                            _|4      |2  |1|
   8                                          _|5       _|2 _|1|
   9                                        _|5        |2  |2 _|
  10                                      _|6         _|2  |1|1|
  11                                    _|6          |3   _|1|1|
  12                                  _|7           _|2  |2  |1|
  13                                _|7            |3    |2 _|1|
  14                              _|8             _|3   _|1|2 _|
  15                            _|8              |3    |2  |1|1|
  16                          _|9               _|3    |2  |1|1|
  17                        _|9                |4     _|2 _|1|1|
  18                      _|10                _|3    |2  |2  |1|
  19                    _|10                 |4      |2  |2 _|1|
  20                  _|11                  _|4     _|2  |1|2 _|
  21                _|11                   |4      |3   _|1|1|1|
  22              _|12                    _|4      |2  |2  |1|1|
  23            _|12                     |5       _|2  |2  |1|1|
  24          _|13                      _|4      |3    |2 _|1|1|
  25        _|13                       |5        |3   _|1|2  |1|
  26      _|14                        _|5       _|2  |2  |2 _|1|
  27    _|14                         |5        |3    |2  |1|2 _|
  28   |15                           |5        |3    |2  |1|1|1|
  ...
Also the diagram represents the left part of the front view of the pyramid described in A245092. For the other half front view see A261350. For more information about the pyramid and the symmetric representation of sigma see A237593. (End)
From _Omar E. Pol_, Sep 08 2021: (Start)
For n = 12 the symmetric representation of sigma(12) in the fourth quadrant is as shown below:
.                           _
                           | |
                           | |
                           | |
                           | |
                           | |
                      _ _ _| |
                    _|    _ _|
                  _|     |
                 |      _|
                 |  _ _|1
      _ _ _ _ _ _| |  2
     |_ _ _ _ _ _ _|2
            7
.
The lengths of the successive line segments from the first vertex to the central vertex of the largest Dyck path are [7, 2, 2, 1] respectively, the same as the 12th row of triangle. (End)
		

Crossrefs

Row n has length A003056(n) hence column k starts in row A000217(k).
Row sums give A000027.
Column 1 is A008619, n >= 1.
Right border gives A042974.

Programs

  • Mathematica
    row[n_]:= Floor[(Sqrt[8*n+1] -1)/2];  f[n_,k_]:= Ceiling[(n+1)/k-(k+1)/2] - Ceiling[(n+1)/(k+1)-(k+2)/2];
    Table[f[n,k],{n,1,50},{k,1,row[n]}]//Flatten
    (* Hartmut F. W. Hoft, Apr 08 2014 *)
  • PARI
    row235791(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i);
    row(n) = {my(orow = concat(row235791(n), 0)); vector(#orow -1, i, orow[i] - orow[i+1]);} \\ Michel Marcus, Mar 27 2014
    
  • Python
    from sympy import sqrt
    import math
    def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2)) - int(math.ceil((n + 1)/(k + 1) - (k + 2)/2))
    for n in range(1, 29): print([T(n, k) for k in range(1, int((sqrt(8*n + 1) - 1)/2) + 1)]) # Indranil Ghosh, Apr 30 2017

Formula

T(n,k) = ceiling((n+1)/k - (k+1)/2) - ceiling((n+1)/(k+1) - (k+2)/2), for 1 <= n and 1 <= k <= floor((sqrt(8n+1)-1)/2). - Hartmut F. W. Hoft, Apr 07 2014

Extensions

3 more rows added by Omar E. Pol, Aug 23 2015
New name from a comment dated Apr 30 2017. - Omar E. Pol, Jun 18 2023

A000096 a(n) = n*(n+3)/2.

Original entry on oeis.org

0, 2, 5, 9, 14, 20, 27, 35, 44, 54, 65, 77, 90, 104, 119, 135, 152, 170, 189, 209, 230, 252, 275, 299, 324, 350, 377, 405, 434, 464, 495, 527, 560, 594, 629, 665, 702, 740, 779, 819, 860, 902, 945, 989, 1034, 1080, 1127, 1175, 1224, 1274, 1325, 1377, 1430, 1484, 1539, 1595, 1652, 1710, 1769
Offset: 0

Views

Author

Keywords

Comments

For n >= 1, a(n) is the maximal number of pieces that can be obtained by cutting an annulus with n cuts. See illustration. - Robert G. Wilson v
n(n-3)/2 (n >= 3) is the number of diagonals of an n-gon. - Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr)
n(n-3)/2 (n >= 4) is the degree of the third-smallest irreducible presentation of the symmetric group S_n (cf. James and Kerber, Appendix 1).
a(n) is also the multiplicity of the eigenvalue (-2) of the triangle graph Delta(n+1). (See p. 19 in Biggs.) - Felix Goldberg (felixg(AT)tx.technion.ac.il), Nov 25 2001
For n > 3, a(n-3) = dimension of the traveling salesman polytope T(n). - Benoit Cloitre, Aug 18 2002
Also counts quasi-dominoes (quasi-2-ominoes) on an n X n board. Cf. A094170-A094172. - Jon Wild, May 07 2004
Coefficient of x^2 in (1 + x + 2*x^2)^n. - Michael Somos, May 26 2004
a(n) is the number of "prime" n-dimensional polyominoes. A "prime" n-polyomino cannot be formed by connecting any other n-polyominoes except for the n-monomino and the n-monomino is not prime. E.g., for n=1, the 1-monomino is the line of length 1 and the only "prime" 1-polyominoes are the lines of length 2 and 3. This refers to "free" n-dimensional polyominoes, i.e., that can be rotated along any axis. - Bryan Jacobs (bryanjj(AT)gmail.com), Apr 01 2005
Solutions to the quadratic equation q(m, r) = (-3 +- sqrt(9 + 8(m - r))) / 2, where m - r is included in a(n). Let t(m) = the triangular number (A000217) less than some number k and r = k - t(m). If k is neither prime nor a power of two and m - r is included in A000096, then m - q(m, r) will produce a value that shares a divisor with k. - Andrew S. Plewe, Jun 18 2005
Sum_{k=2..n+1} 4/(k*(k+1)*(k-1)) = ((n+3)*n)/((n+2)*(n+1)). Numerator(Sum_{k=2..n+1} 4/(k*(k+1)*(k-1))) = (n+3)*n/2. - Alexander Adamchuk, Apr 11 2006
Number of rooted trees with n+3 nodes of valence 1, no nodes of valence 2 and exactly two other nodes. I.e., number of planted trees with n+2 leaves and exactly two branch points. - Theo Johnson-Freyd (theojf(AT)berkeley.edu), Jun 10 2007
If X is an n-set and Y a fixed 2-subset of X then a(n-2) is equal to the number of (n-2)-subsets of X intersecting Y. - Milan Janjic, Jul 30 2007
For n >= 1, a(n) is the number of distinct shuffles of the identity permutation on n+1 letters with the identity permutation on 2 letters (12). - Camillia Smith Barnes, Oct 04 2008
If s(n) is a sequence defined as s(1) = x, s(n) = kn + s(n-1) + p for n > 1, then s(n) = a(n-1)*k + (n-1)*p + x. - Gary Detlefs, Mar 04 2010
The only primes are a(1) = 2 and a(2) = 5. - Reinhard Zumkeller, Jul 18 2011
a(n) = m such that the (m+1)-th triangular number minus the m-th triangular number is the (n+1)-th triangular number: (m+1)(m+2)/2 - m(m+1)/2 = (n+1)(n+2)/2. - Zak Seidov, Jan 22 2012
For n >= 1, number of different values that Sum_{k=1..n} c(k)*k can take where the c(k) are 0 or 1. - Joerg Arndt, Jun 24 2012
On an n X n chessboard (n >= 2), the number of possible checkmate positions in the case of king and rook versus a lone king is 0, 16, 40, 72, 112, 160, 216, 280, 352, ..., which is 8*a(n-2). For a 4 X 4 board the number is 40. The number of positions possible was counted including all mirror images and rotations for all four sides of the board. - Jose Abutal, Nov 19 2013
If k = a(i-1) or k = a(i+1) and n = k + a(i), then C(n, k-1), C(n, k), C(n, k+1) are three consecutive binomial coefficients in arithmetic progression and these are all the solutions. There are no four consecutive binomial coefficients in arithmetic progression. - Michael Somos, Nov 11 2015
a(n-1) is also the number of independent components of a symmetric traceless tensor of rank 2 and dimension n >= 1. - Wolfdieter Lang, Dec 10 2015
Numbers k such that 8k + 9 is a square. - Juri-Stepan Gerasimov, Apr 05 2016
Let phi_(D,rho) be the average value of a generic degree D monic polynomial f when evaluated at the roots of the rho-th derivative of f, expressed as a polynomial in the averaged symmetric polynomials in the roots of f. [See the Wojnar et al. link] The "last" term of phi_(D,rho) is a multiple of the product of all roots of f; the coefficient is expressible as a polynomial h_D(N) in N:=D-rho. These polynomials are of the form h_D(N)= ((-1)^D/(D-1)!)*(D-N)*N^chi*g_D(N) where chi = (1 if D is odd, 0 if D is even) and g_D(N) is a monic polynomial of degree (D-2-chi). Then a(n) are the negated coefficients of the next to the highest order term in the polynomials N^chi*g_D(N), starting at D=3. - Gregory Gerard Wojnar, Jul 19 2017
For n >= 2, a(n) is the number of summations required to solve the linear regression of n variables (n-1 independent variables and 1 dependent variable). - Felipe Pedraza-Oropeza, Dec 07 2017
For n >= 2, a(n) is the number of sums required to solve the linear regression of n variables: 5 for two variables (sums of X, Y, X^2, Y^2, X*Y), 9 for 3 variables (sums of X1, X2, Y1, X1^2, X1*X2, X1*Y, X2^2, X2*Y, Y^2), and so on. - Felipe Pedraza-Oropeza, Jan 11 2018
a(n) is the area of a triangle with vertices at (n, n+1), ((n+1)*(n+2)/2, (n+2)*(n+3)/2), ((n+2)^2, (n+3)^2). - J. M. Bergot, Jan 25 2018
Number of terms less than 10^k: 1, 4, 13, 44, 140, 446, 1413, 4471, 14141, 44720, 141420, 447213, ... - Muniru A Asiru, Jan 25 2018
a(n) is also the number of irredundant sets in the (n+1)-path complement graph for n > 2. - Eric W. Weisstein, Apr 11 2018
a(n) is also the largest number k such that the largest Dyck path of the symmetric representation of sigma(k) has exactly n peaks, n >= 1. (Cf. A237593.) - Omar E. Pol, Sep 04 2018
For n > 0, a(n) is the number of facets of associahedra. Cf. A033282 and A126216 and their refinements A111785 and A133437 for related combinatorial and analytic constructs. See p. 40 of Hanson and Sha for a relation to projective spaces and string theory. - Tom Copeland, Jan 03 2021
For n > 0, a(n) is the number of bipartite graphs with 2n or 2n+1 edges, no isolated vertices, and a stable set of cardinality 2. - Christian Barrientos, Jun 13 2022
For n >= 2, a(n-2) is the number of permutations in S_n which are the product of two different transpositions of adjacent points. - Zbigniew Wojciechowski, Mar 31 2023
a(n) represents the optimal stop-number to achieve the highest running score for the Greedy Pig game with an (n-1)-sided die with a loss on a 1. The total at which one should stop is a(s-1), e.g. for a 6-sided die, one should pass the die at 20. See Sparks and Haran. - Nicholas Stefan Georgescu, Jun 09 2024

Examples

			G.f. = 2*x + 5*x^2 + 9*x^3 + 14*x^4 + 20*x^5 + 27*x^6 + 35*x^7 + 44*x^8 + 54*x^9 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), Table 22.7, p. 797.
  • Norman Biggs, Algebraic Graph Theory, 2nd ed. Cambridge University Press, 1993.
  • G. James and A. Kerber, The Representation Theory of the Symmetric Group, Encyclopedia of Maths. and its Appls., Vol. 16, Addison-Wesley, 1981, Reading, MA, U.S.A.
  • D. G. Kendall et al., Shape and Shape Theory, Wiley, 1999; see p. 4.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Complement of A007401. Column 2 of A145324. Column of triangle A014473, first skew subdiagonal of A033282, a diagonal of A079508.
Occurs as a diagonal in A074079/A074080, i.e., A074079(n+3, n) = A000096(n-1) for all n >= 2. Also A074092(n) = 2^n * A000096(n-1) after n >= 2.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488.
Similar sequences are listed in A316466.

Programs

Formula

G.f.: A(x) = x*(2-x)/(1-x)^3. a(n) = binomial(n+1, n-1) + binomial(n, n-1).
Connection with triangular numbers: a(n) = A000217(n+1) - 1.
a(n) = a(n-1) + n + 1. - Bryan Jacobs (bryanjj(AT)gmail.com), Apr 01 2005
a(n) = 2*t(n) - t(n-1) where t() are the triangular numbers, e.g., a(5) = 2*t(5) - t(4) = 2*15 - 10 = 20. - Jon Perry, Jul 23 2003
a(-3-n) = a(n). - Michael Somos, May 26 2004
2*a(n) = A008778(n) - A105163(n). - Creighton Dement, Apr 15 2005
a(n) = C(3+n, 2) - C(3+n, 1). - Zerinvary Lajos, Dec 09 2005
a(n) = A067550(n+1) / A067550(n). - Alexander Adamchuk, May 20 2006
a(n) = A126890(n,1) for n > 0. - Reinhard Zumkeller, Dec 30 2006
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Paul Curtz, Jan 02 2008
Starting (2, 5, 9, 14, ...) = binomial transform of (2, 3, 1, 0, 0, 0, ...). - Gary W. Adamson, Jul 03 2008
For n >= 0, a(n+2) = b(n+1) - b(n), where b(n) is the sequence A005586. - K.V.Iyer, Apr 27 2009
A002262(a(n)) = n. - Reinhard Zumkeller, May 20 2009
Let A be the Toeplitz matrix of order n defined by: A[i,i-1]=-1, A[i,j]=Catalan(j-i), (i<=j), and A[i,j]=0, otherwise. Then, for n>=1, a(n-1)=coeff(charpoly(A,x),x^(n-2)). - Milan Janjic, Jul 08 2010
a(n) = Sum_{k=1..n} (k+1)!/k!. - Gary Detlefs, Aug 03 2010
a(n) = n(n+1)/2 + n = A000217(n) + n. - Zak Seidov, Jan 22 2012
E.g.f.: F(x) = 1/2*x*exp(x)*(x+4) satisfies the differential equation F''(x) - 2*F'(x) + F(x) = exp(x). - Peter Bala, Mar 14 2012
a(n) = binomial(n+3, 2) - (n+3). - Robert G. Wilson v, Mar 15 2012
a(n) = A181971(n+1, 2) for n > 0. - Reinhard Zumkeller, Jul 09 2012
a(n) = A214292(n+2, 1). - Reinhard Zumkeller, Jul 12 2012
G.f.: -U(0) where U(k) = 1 - 1/((1-x)^2 - x*(1-x)^4/(x*(1-x)^2 - 1/U(k+1))); (continued fraction, 3-step). - Sergei N. Gladkovskii, Sep 27 2012
A023532(a(n)) = 0. - Reinhard Zumkeller, Dec 04 2012
a(n) = A014132(n,n) for n > 0. - Reinhard Zumkeller, Dec 12 2012
a(n-1) = (1/n!)*Sum_{j=0..n} binomial(n,j)*(-1)^(n-j)*j^n*(j-1). - Vladimir Kruchinin, Jun 06 2013
a(n) = 2n - floor(n/2) + floor(n^2/2). - Wesley Ivan Hurt, Jun 15 2013
a(n) = Sum_{i=2..n+1} i. - Wesley Ivan Hurt, Jun 28 2013
Sum_{n>0} 1/a(n) = 11/9. - Enrique Pérez Herrero, Nov 26 2013
a(n) = Sum_{i=1..n} (n - i + 2). - Wesley Ivan Hurt, Mar 31 2014
A023531(a(n)) = 1. - Reinhard Zumkeller, Feb 14 2015
For n > 0: a(n) = A101881(2*n-1). - Reinhard Zumkeller, Feb 20 2015
a(n) + a(n-1) = A008865(n+1) for all n in Z. - Michael Somos, Nov 11 2015
a(n+1) = A127672(4+n, n), n >= 0, where A127672 gives the coefficients of the Chebyshev C polynomials. See the Abramowitz-Stegun reference. - Wolfdieter Lang, Dec 10 2015
a(n) = (n+1)^2 - A000124(n). - Anton Zakharov, Jun 29 2016
Dirichlet g.f.: (zeta(s-2) + 3*zeta(s-1))/2. - Ilya Gutkovskiy, Jun 30 2016
a(n) = 2*A000290(n+3) - 3*A000217(n+3). - J. M. Bergot, Apr 04 2018
a(n) = Stirling2(n+2, n+1) - 1. - Peter Luschny, Jan 05 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 4*log(2)/3 - 5/9. - Amiram Eldar, Jan 10 2021
From Amiram Eldar, Jan 20 2021: (Start)
Product_{n>=1} (1 + 1/a(n)) = 3.
Product_{n>=1} (1 - 1/a(n)) = 3*cos(sqrt(17)*Pi/2)/(4*Pi). (End)
Product_{n>=0} a(4*n+1)*a(4*n+4)/(a(4*n+2)*a(4*n+3)) = Pi/6. - Michael Jodl, Apr 05 2025

A236104 Triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k copies of the positive squares in nondecreasing order, and the first element of column k is in row k(k+1)/2.

Original entry on oeis.org

1, 4, 9, 1, 16, 1, 25, 4, 36, 4, 1, 49, 9, 1, 64, 9, 1, 81, 16, 4, 100, 16, 4, 1, 121, 25, 4, 1, 144, 25, 9, 1, 169, 36, 9, 1, 196, 36, 9, 4, 225, 49, 16, 4, 1, 256, 49, 16, 4, 1, 289, 64, 16, 4, 1, 324, 64, 25, 9, 1, 361, 81, 25, 9, 1, 400, 81, 25, 9, 4
Offset: 1

Views

Author

Omar E. Pol, Jan 23 2014

Keywords

Comments

These are the squares of the entries of the triangle in A235791: T(n,k) = (A235791(n,k))^2.
Row n has length A003056(n) hence the first element of column k is in row A000217(k).
Columns 1-3 (including the initial zeros) are A000290, A008794, A211547.
Also column k lists the partial sums of the k-th column of triangle A196020 which gives an identity for sigma.
Since all the elements of this sequence are squares, we can draw an illustration of the alternating sum of row n step by step, and a symmetric diagram for A000203, A024916, A004125; see example.
For more information about the diagram see A237593.

Examples

			Triangle begins:
    1;
    4;
    9,   1;
   16,   1;
   25,   4;
   36,   4,   1;
   49,   9,   1;
   64,   9,   1;
   81,  16,   4;
  100,  16,   4,   1;
  121,  25,   4,   1;
  144,  25,   9,   1;
  169,  36,   9,   1;
  196,  36,   9,   4;
  225,  49,  16,   4,   1;
  256,  49,  16,   4,   1;
  289,  64,  16,   4,   1;
  324,  64,  25,   9,   1;
  361,  81,  25,   9,   1;
  400,  81,  25,   9,   4;
  441, 100,  36,   9,   4,   1;
  484, 100,  36,  16,   4,   1;
  529, 121,  36,  16,   4,   1;
  576, 121,  49,  16,   4,   1;
  ...
For n = 6 the sum of all divisors of all positive integers <= 6 is [1] + [1+2] + [1+3] + [1+2+4] + [1+5] + [1+2+3+6] = 1 + 3 + 4 + 7 + 6 + 12 = 33. On the other hand the 6th row of triangle is 36, 4, 1, therefore the alternating row sum is 36 - 4 + 1 = 33, equaling the sum of all divisors of all positive integers <= 6.
Illustration of the alternating sum of the 6th row as the area of a polygon (or the number of cells), step by step, in the fourth quadrant:
.     _ _ _ _ _ _       _ _ _ _ _ _       _ _ _ _ _ _
.    |           |     |           |     |           |
.    |           |     |           |     |           |
.    |           |     |           |     |           |
.    |           |     |        _ _|     |          _|
.    |           |     |       |         |        _|
.    |_ _ _ _ _ _|     |_ _ _ _|         |_ _ _ _|
.
.          36           36 - 4 = 32     36 - 4 + 1 = 33
.
Then using this method we can draw a symmetric diagram for A000203, A024916, A004125, as shown below:
--------------------------------------------------
n     A000203  A024916            Diagram
--------------------------------------------------
.                         _ _ _ _ _ _ _ _ _ _ _ _
1        1        1      |_| | | | | | | | | | | |
2        3        4      |_ _|_| | | | | | | | | |
3        4        8      |_ _|  _|_| | | | | | | |
4        7       15      |_ _ _|    _|_| | | | | |
5        6       21      |_ _ _|  _|  _ _|_| | | |
6       12       33      |_ _ _ _|  _| |  _ _|_| |
7        8       41      |_ _ _ _| |_ _|_|    _ _|
8       15       56      |_ _ _ _ _|  _|     |* *
9       13       69      |_ _ _ _ _| |      _|* *
10      18       87      |_ _ _ _ _ _|  _ _|* * *
11      12       99      |_ _ _ _ _ _| |* * * * *
12      28      127      |_ _ _ _ _ _ _|* * * * *
.
The total number of cells in the first n set of symmetric regions of the diagram equals A024916(n). It appears that the total number of cells in the n-th set of symmetric regions of the diagram equals sigma(n) = A000203(n). Example: for n = 12 the 12th row of triangle is 144, 25, 9, 1, hence the alternating sums is 144 - 25 + 9 - 1 = 127. On the other hand we have that A000290(12) - A004125(12) = 144 - 17 = A024916(12) = 127, equaling the total number of cells in the diagram after 12 stages. The number of cells in the 12th set of symmetric regions of the diagram is sigma(12) = A000203(12) = 28. Note that in this case there is only one region. Finally, the number of *'s is A004125(12) = 17.
Note that the diagram is also the top view of the stepped pyramid described in A245092. - _Omar E. Pol_, Feb 12 2018
		

Crossrefs

Programs

  • Mathematica
    Table[Ceiling[(n + 1)/k - (k + 1)/2]^2, {n, 20}, {k, Floor[(Sqrt[8 n + 1] - 1)/2]}] // Flatten (* Michael De Vlieger, Feb 10 2018, after Hartmut F. W. Hoft at A235791 *)
  • Python
    from sympy import sqrt
    import math
    def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2))
    for n in range(1, 21): print([T(n, k)**2 for k in range(1, int(math.floor((sqrt(8*n + 1) - 1)/2)) + 1)]) # Indranil Ghosh, Apr 25 2017

Formula

Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A024916(n). [Although this was stated as a fact, as far as I can tell, no proof was known. However, Don Reble has recently found a proof, which will be added here soon. - N. J. A. Sloane, Nov 23 2020]
A000203(n) = Sum_{k=1..A003056(n)} (-1)^(k-1) * (T(n,k) - T(n-1,k)), assuming that T(k*(k+1)/2-1,k) = 0. - Omar E. Pol, Oct 10 2018

A235791 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists k copies of every positive integer in nondecreasing order, and the first element of column k is in row k(k+1)/2.

Original entry on oeis.org

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

Views

Author

Omar E. Pol, Jan 23 2014

Keywords

Comments

The alternating sum of the squares of the elements of the n-th row equals the sum of all divisors of all positive integers <= n, i.e., Sum_{k=1..A003056(n)} (-1)^(k-1)*(T(n,k))^2 = A024916(n).
Row n has length A003056(n) hence the first element of column k is in row A000217(k).
For more information see A236104.
The sum of row n gives A060831(n), the sum of the number of odd divisors of all positive integers <= n. - Omar E. Pol, Mar 01 2014. [An equivalent assertion is that the sum of row n of A237048 is the number of odd divisors of n, and this was proved by Hartmut F. W. Hoft in a comment in A237048. - N. J. A. Sloane, Dec 07 2020]
Comments from Franklin T. Adams-Watters on sequences related to the "symmetric representation of sigma" in A235791 and related sequences, Mar 31 2014: (Start)
The place to start is with A235791, which is very simple. Then go to A237591, also very simple, and A237593, still very simple.
You then need to interpret the rows of A237593 as Dyck paths. This interpretation is in terms of run lengths, so 2,1,1,2 means up twice, down once, up once, and down twice. Because the rows of A237593 are symmetric and of even length, this path will always be symmetric.
Now the surprising fact is that the areas enclosed by the Dyck path for n (laid on its side) always includes the area enclosed for n-1; and the number of squares added is sigma(n).
Finally, look at the connected areas enclosed by n but not by n-1; the size of these areas is the symmetric representation of sigma. (End)
From Hartmut F. W. Hoft, Apr 07 2014: (Start)
Mathematica function has been written to check the first property up to n = 20000.
T(n,(sqrt(8n+1)-1)/2+1) = 0 for all n >= 1, which is useful for formulas for A237591 and A237593. (End)
Alternating row sums give A240542. - Omar E. Pol, Apr 16 2014
Conjecture: T(n,k) is also the total number of partitions of all positive integers <= n into exactly k consecutive parts, i.e., the partial column sum of A285898, or in accordance with the triangles of the same family: the partial column sum of A237048. - Omar E. Pol, Apr 28 2017, Nov 24 2020
The above conjecture is true. The proof will be added soon (it uses the generating function for the columns). - N. J. A. Sloane, Nov 24 2020
T(n,k) is also the total length of all line segments between the k-th vertex and the central vertex of the largest Dyck path of the symmetric representation of sigma(n). In other words: T(n,k) is the sum of the last (A003056(n)-k+1) terms of the n-th row of A237591. - Omar E. Pol, Sep 07 2021
T(n,k) is also the Manhattan distance between the k-th vertex and the central vertex of the Dyck path described in the n-th row of the triangle A237593. - Omar E. Pol, Jan 11 2023

Examples

			Triangle begins:
   1;
   2;
   3,  1;
   4,  1;
   5,  2;
   6,  2,  1;
   7,  3,  1;
   8,  3,  1;
   9,  4,  2;
  10,  4,  2,  1;
  11,  5,  2,  1;
  12,  5,  3,  1;
  13,  6,  3,  1;
  14,  6,  3,  2;
  15,  7,  4,  2,  1;
  16,  7,  4,  2,  1;
  17,  8,  4,  2,  1;
  18,  8,  5,  3,  1;
  19,  9,  5,  3,  1;
  20,  9,  5,  3,  2;
  21, 10,  6,  3,  2,  1;
  22, 10,  6,  4,  2,  1;
  23, 11,  6,  4,  2,  1;
  24, 11,  7,  4,  2,  1;
  25, 12,  7,  4,  3,  1;
  26, 12,  7,  5,  3,  1;
  27, 13,  8,  5,  3,  2;
  28, 13,  8,  5,  3,  2,  1;
  ...
For n = 10 the 10th row of triangle is 10, 4, 2, 1, so we have that 10^2 - 4^2 + 2^2 - 1^2 = 100 - 16 + 4 - 1 = 87, the same as A024916(10) = 87, the sum of all divisors of all positive integers <= 10.
From _Omar E. Pol_, Nov 19 2015: (Start)
Illustration of initial terms in the third quadrant:
.                                                            y
Row                                                         _|
1                                                         _|1|
2                                                       _|2 _|
3                                                     _|3  |1|
4                                                   _|4   _|1|
5                                                 _|5    |2 _|
6                                               _|6     _|2|1|
7                                             _|7      |3  |1|
8                                           _|8       _|3 _|1|
9                                         _|9        |4  |2 _|
10                                      _|10        _|4  |2|1|
11                                    _|11         |5   _|2|1|
12                                  _|12          _|5  |3  |1|
13                                _|13           |6    |3 _|1|
14                              _|14            _|6   _|3|2 _|
15                            _|15             |7    |4  |2|1|
16                          _|16              _|7    |4  |2|1|
17                        _|17               |8     _|4 _|2|1|
18                      _|18                _|8    |5  |3  |1|
19                    _|19                 |9      |5  |3 _|1|
20                  _|20                  _|9     _|5  |3|2 _|
21                _|21                   |10     |6   _|3|2|1|
22              _|22                    _|10     |6  |4  |2|1|
23            _|23                     |11      _|6  |4  |2|1|
24          _|24                      _|11     |7    |4 _|2|1|
25        _|25                       |12       |7   _|4|3  |1|
26      _|26                        _|12      _|7  |5  |3 _|1|
27    _|27                         |13       |8    |5  |3|2 _|
28   |28                           |13       |8    |5  |3|2|1|
...
T(n,k) is also the number of cells between the k-th vertical line segment (from left to right) and the y-axis in the n-th row of the structure.
Note that the number of horizontal line segments in the n-th row of the structure equals A001227(n), the number of odd divisors of n.
Also the diagram represents the left part of the front view of the pyramid described in A245092. (End)
For more information about the diagram see A286001. - _Omar E. Pol_, Dec 19 2020
From _Omar E. Pol_, Sep 08 2021: (Start)
For n = 12 the symmetric representation of sigma(12) in the fourth quadrant is as shown below:
                            _
                           | |
                           | |
                           | |
                           | |
                           | |
                      _ _ _| |
                    _|    _ _|
                  _|     |
                 |      _|
                 |  _ _|
      _ _ _ _ _ _| |3   1
     |_ _ _ _ _ _ _|
    12              5
.
For n = 12 and k = 1 the total length of all line segments between the first vertex and the central vertex of the largest Dyck path is equal to 12, so T(12,1) = 12.
For n = 12 and k = 2 the total length of all line segments between the second vertex and the central vertex of the largest Dyck path is equal to 5, so T(12,2) = 5.
For n = 12 and k = 3 the total length of all line segments between the third vertex and the central vertex of the largest Dyck path is equal to 3, so T(12,3) = 3.
For n = 12 and k = 4 the total length of all line segments between the fourth vertex and the central vertex of the largest Dyck path is equal to 1, so T(12,4) = 1.
Hence the 12th row of triangle is [12, 5, 3, 1]. (End)
		

Crossrefs

Columns 1..3: A000027, A008619, A008620.
Operations on rows: A003056 (number of terms), A237591 (differences between terms), A060831 (sums), A339577 (products), A240542 (alternating sums), A236104 (squares), A339576 (sums of squares), A024916 (alternating sums of squares), A237048 (differences between rows), A042974 (right border).

Programs

  • Mathematica
    row[n_] := Floor[(Sqrt[8*n + 1] - 1)/2]; f[n_, k_] := Ceiling[(n + 1)/k - (k + 1)/2]; Table[f[n, k], {n, 1, 150}, {k, 1, row[n]}] // Flatten (* Hartmut F. W. Hoft, Apr 07 2014 *)
  • PARI
    row(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i); \\ Michel Marcus, Mar 27 2014
    
  • Python
    from sympy import sqrt
    import math
    def T(n, k): return int(math.ceil((n + 1)/k - (k + 1)/2))
    for n in range(1, 21): print([T(n, k) for k in range(1, int(math.floor((sqrt(8*n + 1) - 1)/2)) + 1)]) # Indranil Ghosh, Apr 25 2017

Formula

T(n,k) = ceiling((n+1)/k - (k+1)/2) for 1 <= n, 1 <= k <= floor((sqrt(8n+1)-1)/2) = A003056(n). - Hartmut F. W. Hoft, Apr 07 2014
G.f. for column k (k >= 1): x^(k*(k+1)/2)/( (1-x)*(1-x^k) ). - N. J. A. Sloane, Nov 24 2020
T(n,k) = Sum_{j=1..n} A237048(j,k). - Omar E. Pol, May 18 2017
T(n,k) = sqrt(A236104(n,k)). - Omar E. Pol, Feb 14 2018
Sigma(n) = Sum_{k=1..A003056(n)} (-1)^(k-1) * (T(n,k)^2 - T(n-1,k)^2), assuming that T(k*(k+1)/2-1,k) = 0. - Omar E. Pol, Oct 10 2018
a(s(n,k)) = T(n,k), n >= 1, 1 <= k <= r = floor((sqrt(8*n + 1) - 1)/2), where s(n,k) = r*n - r*(r+1)*(r+2)/6 + k translates position (row n, column k) in the triangle of this sequence to its position in the sequence. - Hartmut F. W. Hoft, Feb 24 2021

A196020 Irregular triangle read by rows: T(n,k), n >= 1, k >= 1, in which column k lists the odd numbers interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.

Original entry on oeis.org

1, 3, 5, 1, 7, 0, 9, 3, 11, 0, 1, 13, 5, 0, 15, 0, 0, 17, 7, 3, 19, 0, 0, 1, 21, 9, 0, 0, 23, 0, 5, 0, 25, 11, 0, 0, 27, 0, 0, 3, 29, 13, 7, 0, 1, 31, 0, 0, 0, 0, 33, 15, 0, 0, 0, 35, 0, 9, 5, 0, 37, 17, 0, 0, 0, 39, 0, 0, 0, 3, 41, 19, 11, 0, 0, 1, 43, 0, 0, 7, 0, 0, 45, 21, 0, 0, 0, 0, 47, 0, 13, 0, 0, 0
Offset: 1

Views

Author

Omar E. Pol, Feb 02 2013

Keywords

Comments

Gives an identity for sigma(n): alternating sum of row n equals the sum of divisors of n. For proof see Max Alekseyev link.
Row n has length A003056(n) hence column k starts in row A000217(k).
The number of positive terms in row n is A001227(n), the number of odd divisors of n.
If n = 2^j then the only positive integer in row n is T(n,1) = 2^(j+1) - 1.
If n is an odd prime then the only two positive integers in row n are T(n,1) = 2n - 1 and T(n,2) = n - 2.
If T(n,k) = 3 then T(n+1,k+1) = 1, the first element of the column k+1.
The partial sums of column k give the column k of A236104.
The connection with the symmetric representation of sigma is as follows: A236104 --> A235791 --> A237591 --> A237593 --> A239660 --> A237270.
Alternating sum of row n equals the number of units cubes that protrude from the n-th level of the stepped pyramid described in A245092. - Omar E. Pol, Oct 28 2015
Conjecture: T(n,k) is the difference between the square of the total number of partitions of all positive integers <= n into exactly k consecutive parts, and the square of the total number of partitions of all positive integers < n into exactly k consecutive parts. - Omar E. Pol, Feb 14 2018
From Omar E. Pol, Nov 24 2020: (Start)
T(n,k) is also the number of steps in the first n levels of the k-th double-staircase that has at least one step in the n-th level of the "Double- staircases" diagram, otherwise T(n,k) = 0, (see the Example section).
For the connection with A280851 see also the algorithm of A280850 and the conjecture of A296508. (End)
The number of zeros in the n-th row equals A238005(n). - Omar E. Pol, Sep 11 2021
Apart from the alternating row sums and the sum of divisors function A000203 another connection with Euler's pentagonal theorem is that in the irregular triangle of A238442 the k-th column starts in the row that is the k-th generalized pentagonal number A001318(k) while here the k-th column starts in the row that is the k-th generalized hexagonal number A000217(k). Both A001318 and A000217 are successive members of the same family: the generalized polygonal numbers. - Omar E. Pol, Sep 23 2021
Other triangle with the same row lengths and alternating row sums equals sigma(n) is A252117. - Omar E. Pol, May 03 2022

Examples

			Triangle begins:
   1;
   3;
   5,  1;
   7,  0;
   9,  3;
  11,  0,  1;
  13,  5,  0;
  15,  0,  0;
  17,  7,  3;
  19,  0,  0,  1;
  21,  9,  0,  0;
  23,  0,  5,  0;
  25, 11,  0,  0;
  27,  0,  0,  3;
  29, 13,  7,  0,  1;
  31,  0,  0,  0,  0;
  33, 15,  0,  0,  0;
  35,  0,  9,  5,  0;
  37, 17,  0,  0,  0;
  39,  0,  0,  0,  3;
  41, 19, 11,  0,  0,  1;
  43,  0,  0,  7,  0,  0;
  45, 21,  0,  0,  0,  0;
  47,  0, 13,  0,  0,  0;
  49, 23,  0,  0,  5,  0;
  51,  0,  0,  9,  0,  0;
  53, 25, 15,  0,  0,  3;
  55,  0,  0,  0,  0,  0,  1;
  ...
For n = 15 the divisors of 15 are 1, 3, 5, 15, so the sum of divisors of 15 is 1 + 3 + 5 + 15 = 24. On the other hand, the 15th row of the triangle is 29, 13, 7, 0, 1, so the alternating row sum is 29 - 13 + 7 - 0 + 1 = 24, equaling the sum of divisors of 15.
If n is even then the alternating sum of the n-th row is simpler to evaluate than the sum of divisors of n. For example the sum of divisors of 24 is 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 = 60, and the alternating sum of the 24th row of triangle is 47 - 0 + 13 - 0 + 0 - 0 = 60.
From _Omar E. Pol_, Nov 24 2020: (Start)
For an illustration of the rows of triangle consider the infinite "double-staircases" diagram defined in A335616 (see also the theorem there).
For n = 15 the diagram with first 15 levels looks like this:
.
Level                         "Double-staircases" diagram
.                                          _
1                                        _|1|_
2                                      _|1 _ 1|_
3                                    _|1  |1|  1|_
4                                  _|1   _| |_   1|_
5                                _|1    |1 _ 1|    1|_
6                              _|1     _| |1| |_     1|_
7                            _|1      |1  | |  1|      1|_
8                          _|1       _|  _| |_  |_       1|_
9                        _|1        |1  |1 _ 1|  1|        1|_
10                     _|1         _|   | |1| |   |_         1|_
11                   _|1          |1   _| | | |_   1|          1|_
12                 _|1           _|   |1  | |  1|   |_           1|_
13               _|1            |1    |  _| |_  |    1|            1|_
14             _|1             _|    _| |1 _ 1| |_    |_             1|_
15            |1              |1    |1  | |1| |  1|    1|              1|
.
The first largest double-staircase has 29 horizontal steps, the second double-staircase has 13 steps, the third double-staircase has 7 steps, and the fifth double-staircases has only one step. Note that the fourth double-staircase does not count because it does not have horizontal steps in the 15th level, so the 15th row of triangle is [29, 13, 7, 0, 1].
For a connection with the "Ziggurat" diagram and the parts and subparts of the symmetric representation of sigma(15) see also A237270. (End)
		

Crossrefs

Programs

  • Maple
    T_row := proc(n) local T;
    T := (n, k) -> if modp(n-k/2, k) = 0 and n >= k*(k+1)/2 then 2*n/k-k else 0 fi;
    seq(T(n,k), k=1..floor((sqrt(8*n+1)-1)/2)) end:
    seq(print(T_row(n)),n=1..24); # Peter Luschny, Oct 27 2015
  • Mathematica
    T[n_, k_] := If[Mod[n - k*(k+1)/2, k] == 0 ,2*n/k - k, 0]
    row[n_] := Floor[(Sqrt[8n+1]-1)/2]
    line[n_] := Map[T[n, #]&, Range[row[n]]]
    a196020[m_, n_] := Map[line, Range[m, n]]
    Flatten[a196020[1,22]] (* data *)
    (* Hartmut F. W. Hoft, Oct 26 2015 *)
    A196020row = Function[n,Table[If[Divisible[Numerator[n-k/2],k] && CoprimeQ[ Denominator[n- k/2], k],2*n/k-k,0],{k,1,Floor[(Sqrt[8 n+1]-1)/2]}]]
    Flatten[Table[A196020row[n], {n,1,24}]] (* Peter Luschny, Oct 28 2015 *)
  • Sage
    def T(n,k):
        q = (2*n-k)/2
        b = k.divides(q.numerator()) and gcd(k,q.denominator()) == 1
        return 2*n/k - k if b else 0
    for n in (1..24): [T(n, k) for k in (1..floor((sqrt(8*n+1)-1)/2))] # Peter Luschny, Oct 28 2015

Formula

A000203(n) = Sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k).
T(n,k) = 2*A211343(n,k) - 1, if A211343(n,k) >= 1 otherwise T(n,k) = 0.
If n==k/2 (mod k) and n>=k(k+1)/2, then T(n,k) = 2*n/k - k; otherwise T(n,k) = 0. - Max Alekseyev, Nov 18 2013
T(n,k) = A236104(n,k) - A236104(n-1,k), assuming that A236104(k*(k+1)/2-1,k) = 0. - Omar E. Pol, Oct 14 2018
T(n,k) = A237048(n,k)*A338721(n,k). - Omar E. Pol, Feb 22 2022

A237271 Number of parts in the symmetric representation of sigma(n).

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 2, 1, 3, 2, 2, 1, 2, 2, 3, 1, 2, 1, 2, 1, 4, 2, 2, 1, 3, 2, 4, 1, 2, 1, 2, 1, 4, 2, 3, 1, 2, 2, 4, 1, 2, 1, 2, 2, 3, 2, 2, 1, 3, 3, 4, 2, 2, 1, 4, 1, 4, 2, 2, 1, 2, 2, 5, 1, 4, 1, 2, 2, 4, 3, 2, 1, 2, 2, 4, 2, 3, 2, 2, 1, 5, 2, 2, 1, 4, 2, 4, 1, 2, 1
Offset: 1

Views

Author

Omar E. Pol, Feb 25 2014

Keywords

Comments

The diagram of the symmetry of sigma has been via A196020 --> A236104 --> A235791 --> A237591 --> A237593.
For more information see A237270.
a(n) is also the number of terraces at n-th level (starting from the top) of the stepped pyramid described in A245092. - Omar E. Pol, Apr 20 2016
a(n) is also the number of subparts in the first layer of the symmetric representation of sigma(n). For the definion of "subpart" see A279387. - Omar E. Pol, Dec 08 2016
Note that the number of subparts in the symmetric representation of sigma(n) equals A001227(n), the number of odd divisors of n. (See the second example). - Omar E. Pol, Dec 20 2016
From Hartmut F. W. Hoft, Dec 26 2016: (Start)
Using odd prime number 3, observe that the 1's in the 3^k-th row of the irregular triangle of A237048 are at index positions
3^0 < 2*3^0 < 3^1 < 2*3^1 < ... < 2*3^((k-1)/2) < 3^(k/2) < ...
the last being 2*3^((k-1)/2) when k is odd and 3^(k/2) when k is even. Since odd and even index positions alternate, each pair (3^i, 2*3^i) specifies one part in the symmetric representation with a center part present when k is even. A straightforward count establishes that the symmetric representation of 3^k, k>=0, has k+1 parts. Since this argument is valid for any odd prime, every positive integer occurs infinitely many times in the sequence. (End)
a(n) = number of runs of consecutive nonzero terms in row n of A262045. - N. J. A. Sloane, Jan 18 2021
Indices of odd terms give A071562. Indices of even terms give A071561. - Omar E. Pol, Feb 01 2021
a(n) is also the number of prisms in the three-dimensional version of the symmetric representation of k*sigma(n) where k is the height of the prisms, with k >= 1. - Omar E. Pol, Jul 01 2021
With a(1) = 0; a(n) is also the number of parts in the symmetric representation of A001065(n), the sum of aliquot parts of n. - Omar E. Pol, Aug 04 2021
The parity of this sequence is also the characteristic function of numbers that have middle divisors. - Omar E. Pol, Sep 30 2021
a(n) is also the number of polycubes in the 3D-version of the ziggurat of order n described in A347186. - Omar E. Pol, Jun 11 2024
Conjecture 1: a(n) is the number of odd divisors of n except the "e" odd divisors described in A005279. Thus a(n) is the length of the n-th row of A379288. - Omar E. Pol, Dec 21 2024
The conjecture 1 was checked up n = 10000 by Amiram Eldar. - Omar E. Pol, Dec 22 2024
The conjecture 1 is true. For a proof see A379288. - Hartmut F. W. Hoft, Jan 21 2025
From Omar E. Pol, Jul 31 2025: (Start)
Conjecture 2: a(n) is the number of 2-dense sublists of divisors of n.
We call "2-dense sublists of divisors of n" to the maximal sublists of divisors of n whose terms increase by a factor of at most 2.
In a 2-dense sublist of divisors of n the terms are in increasing order and two adjacent terms are the same two adjacent terms in the list of divisors of n.
Example: for n = 10 the list of divisors of 10 is [1, 2, 5, 10]. There are two 2-dense sublists of divisors of 10, they are [1, 2], [5, 10], so a(10) = 2.
The conjecture 2 is essentially the same as the second conjecture in the Comments of A384149. See also Peter Munn's formula in A237270.
The indices where a(n) = 1 give A174973 (2-dense numbers). See the proof there. (End)
Conjecture 3: a(n) is the number of divisors p of n such that p is greater than twice the adjacent previous divisor of n. The divisors p give the n-th row of A379288. - Omar E. Pol, Aug 02 2025

Examples

			Illustration of initial terms (n = 1..12):
---------------------------------------------------------
n   A000203  A237270    a(n)            Diagram
---------------------------------------------------------
.                               _ _ _ _ _ _ _ _ _ _ _ _
1       1      1         1     |_| | | | | | | | | | | |
2       3      3         1     |_ _|_| | | | | | | | | |
3       4      2+2       2     |_ _|  _|_| | | | | | | |
4       7      7         1     |_ _ _|    _|_| | | | | |
5       6      3+3       2     |_ _ _|  _|  _ _|_| | | |
6      12      12        1     |_ _ _ _|  _| |  _ _|_| |
7       8      4+4       2     |_ _ _ _| |_ _|_|    _ _|
8      15      15        1     |_ _ _ _ _|  _|     |
9      13      5+3+5     3     |_ _ _ _ _| |      _|
10     18      9+9       2     |_ _ _ _ _ _|  _ _|
11     12      6+6       2     |_ _ _ _ _ _| |
12     28      28        1     |_ _ _ _ _ _ _|
...
For n = 9 the sum of divisors of 9 is 1+3+9 = A000203(9) = 13. On the other hand the 9th set of symmetric regions of the diagram is formed by three regions (or parts) with 5, 3 and 5 cells, so the total number of cells is 5+3+5 = 13, equaling the sum of divisors of 9. There are three parts: [5, 3, 5], so a(9) = 3.
From _Omar E. Pol_, Dec 21 2016: (Start)
Illustration of the diagram of subparts (n = 1..12):
---------------------------------------------------------
n   A000203  A279391  A001227           Diagram
---------------------------------------------------------
.                               _ _ _ _ _ _ _ _ _ _ _ _
1       1      1         1     |_| | | | | | | | | | | |
2       3      3         1     |_ _|_| | | | | | | | | |
3       4      2+2       2     |_ _|  _|_| | | | | | | |
4       7      7         1     |_ _ _|  _ _|_| | | | | |
5       6      3+3       2     |_ _ _| |_|  _ _|_| | | |
6      12      11+1      2     |_ _ _ _|  _| |  _ _|_| |
7       8      4+4       2     |_ _ _ _| |_ _|_|  _ _ _|
8      15      15        1     |_ _ _ _ _|  _|  _| |
9      13      5+3+5     3     |_ _ _ _ _| |  _|  _|
10     18      9+9       2     |_ _ _ _ _ _| |_ _|
11     12      6+6       2     |_ _ _ _ _ _| |
12     28      23+5      2     |_ _ _ _ _ _ _|
...
For n = 6 the symmetric representation of sigma(6) has two subparts: [11, 1], so A000203(6) = 12 and A001227(6) = 2.
For n = 12 the symmetric representation of sigma(12) has two subparts: [23, 5], so A000203(12) = 28 and A001227(12) = 2. (End)
From _Hartmut F. W. Hoft_, Dec 26 2016: (Start)
Two examples of the general argument in the Comments section:
Rows 27 in A237048 and A249223 (4 parts)
i:  1  2 3 4 5 6 7 8 9 . . 12
27: 1  1 1 0 0 1                           1's in A237048 for odd divisors
    1 27 3     9                           odd divisors represented
27: 1  0 1 1 1 0 0 1 1 1 0 1               blocks forming parts in A249223
Rows 81 in A237048 and A249223 (5 parts)
i:  1  2 3 4 5 6 7 8 9 . . 12. . . 16. . . 20. . . 24
81: 1  1 1 0 0 1 0 0 1 0 0 0                          1's in A237048 f.o.d
    1 81 3    27     9                                odd div. represented
81: 1  0 1 1 1 0 0 0 1 1 1 1 1 1 1 1 0 0 0 1 1 1 0 1  blocks fp in A249223
(End)
		

Crossrefs

Programs

  • Mathematica
    a237271[n_] := Length[a237270[n]] (* code defined in A237270 *)
    Map[a237271, Range[90]] (* data *)
    (* Hartmut F. W. Hoft, Jun 23 2014 *)
    a[n_] := Module[{d = Partition[Divisors[n], 2, 1]}, 1 + Count[d, ?(OddQ[#[[2]]] && #[[2]] >= 2*#[[1]] &)]]; Array[a, 100] (* _Amiram Eldar,  Dec 22 2024 *)
  • PARI
    fill(vcells, hga, hgb) = {ic = 1; for (i=1, #hgb, if (hga[i] < hgb[i], for (j=hga[i], hgb[i]-1, cell = vector(4); cell[1] = i - 1; cell[2] = j; vcells[ic] = cell; ic ++;););); vcells;}
    findfree(vcells) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[3] == 0) && (vcelli[4] == 0), return (i));); return (0);}
    findxy(vcells, x, y) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[1]==x) && (vcelli[2]==y) && (vcelli[3] == 0) && (vcelli[4] == 0), return (i));); return (0);}
    findtodo(vcells, iz) = {for (i=1, #vcells, vcelli = vcells[i]; if ((vcelli[3] == iz) && (vcelli[4] == 0), return (i)); ); return (0);}
    zcount(vcells) = {nbz = 0; for (i=1, #vcells, nbz = max(nbz, vcells[i][3]);); nbz;}
    docell(vcells, ic, iz) = {x = vcells[ic][1]; y = vcells[ic][2]; if (icdo = findxy(vcells, x-1, y), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x+1, y), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x, y-1), vcells[icdo][3] = iz); if (icdo = findxy(vcells, x, y+1), vcells[icdo][3] = iz); vcells[ic][4] = 1; vcells;}
    docells(vcells, ic, iz) = {vcells[ic][3] = iz; while (ic, vcells = docell(vcells, ic, iz); ic = findtodo(vcells, iz);); vcells;}
    nbzb(n, hga, hgb) = {vcells = vector(sigma(n)); vcells = fill(vcells, hga, hgb); iz = 1; while (ic = findfree(vcells), vcells = docells(vcells, ic, iz); iz++;); zcount(vcells);}
    lista(nn) = {hga = concat(heights(row237593(0), 0), 0); for (n=1, nn, hgb = heights(row237593(n), n); nbz = nbzb(n, hga, hgb); print1(nbz, ", "); hga = concat(hgb, 0););} \\ with heights() also defined in A237593; \\ Michel Marcus, Mar 28 2014
    
  • Python
    from sympy import divisors
    def a(n: int) -> int:
        divs = list(divisors(n))
        d = [divs[i:i+2] for i in range(len(divs) - 1)]
        s = sum(1 for pair in d if len(pair) == 2 and pair[1] % 2 == 1 and pair[1] >= 2 * pair[0])
        return s + 1
    print([a(n) for n in range(1, 80)])  # Peter Luschny, Aug 05 2025

Formula

a(n) = A001227(n) - A239657(n). - Omar E. Pol, Mar 23 2014
a(p^k) = k + 1, where p is an odd prime and k >= 0. - Hartmut F. W. Hoft, Dec 26 2016
Theorem: a(n) <= number of odd divisors of n (cf. A001227). The differences are in A239657. - N. J. A. Sloane, Jan 19 2021
a(n) = A340846(n) - A340833(n) + 1 (Euler's formula). - Omar E. Pol, Feb 01 2021
a(n) = A000005(n) - A243982(n). - Omar E. Pol, Aug 02 2025

A245092 The even numbers (A005843) and the values of sigma function (A000203) interleaved.

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 6, 7, 8, 6, 10, 12, 12, 8, 14, 15, 16, 13, 18, 18, 20, 12, 22, 28, 24, 14, 26, 24, 28, 24, 30, 31, 32, 18, 34, 39, 36, 20, 38, 42, 40, 32, 42, 36, 44, 24, 46, 60, 48, 31, 50, 42, 52, 40, 54, 56, 56, 30, 58, 72, 60, 32, 62, 63, 64, 48
Offset: 0

Views

Author

Omar E. Pol, Jul 15 2014

Keywords

Comments

Consider an irregular stepped pyramid with n steps. The base of the pyramid is equal to the symmetric representation of A024916(n), the sum of all divisors of all positive integers <= n. Two of the faces of the pyramid are the same as the representation of the n-th triangular numbers as a staircase. The total area of the pyramid is equal to 2*A024916(n) + A046092(n). The volume is equal to A175254(n). By definition a(2n-1) is A000203(n), the sum of divisors of n. Starting from the top a(2n-1) is also the total area of the horizontal part of the n-th step of the pyramid. By definition, a(2n) = A005843(n) = 2n. Starting from the top, a(2n) is also the total area of the irregular vertical part of the n-th step of the pyramid.
On the other hand the sequence also has a symmetric representation in two dimensions, see Example.
From Omar E. Pol, Dec 31 2016: (Start)
We can find the pyramid after the following sequences: A196020 --> A236104 --> A235791 --> A237591 --> A237593.
The structure of this infinite pyramid arises after the 90-degree-zig-zag folding of the diagram of the isosceles triangle A237593 (see the links).
The terraces at the m-th level of the pyramid are also the parts of the symmetric representation of sigma(m), m >= 1, hence the sum of the areas of the terraces at the m-th level equals A000203(m).
Note that the stepped pyramid is also one of the 3D-quadrants of the stepped pyramid described in A244050.
For more information about the pyramid see A237593 and all its related sequences. (End)

Examples

			Illustration of initial terms:
----------------------------------------------------------------------
a(n)                             Diagram
----------------------------------------------------------------------
0    _
1   |_|\ _
2    \ _| |\ _
3     |_ _| | |\ _
4      \ _ _|_| | |\ _
4       |_ _|  _| | | |\ _
6        \ _ _|  _| | | | |\ _
7         |_ _ _|  _|_| | | | |\ _
8          \ _ _ _|  _ _| | | | | |\ _
6           |_ _ _| |    _| | | | | | |\ _
10           \ _ _ _|  _|  _|_| | | | | | |\ _
12            |_ _ _ _|  _|  _ _| | | | | | | |\ _
12             \ _ _ _ _|  _|  _ _| | | | | | | | |\ _
8               |_ _ _ _| |  _|  _ _|_| | | | | | | | |\ _
14               \ _ _ _ _| |  _| |  _ _| | | | | | | | | |\ _
15                |_ _ _ _ _| |_ _| |  _ _| | | | | | | | | | |\ _
16                 \ _ _ _ _ _|  _ _|_|  _ _|_| | | | | | | | | | |\
13                  |_ _ _ _ _| |  _|  _|  _ _ _| | | | | | | | | | |
18                   \ _ _ _ _ _| |  _|  _|    _ _| | | | | | | | | |
18                    |_ _ _ _ _ _| |  _|     |  _ _|_| | | | | | | |
20                     \ _ _ _ _ _ _| |      _| |  _ _ _| | | | | | |
12                      |_ _ _ _ _ _| |  _ _|  _| |  _ _ _| | | | | |
22                       \ _ _ _ _ _ _| |  _ _|  _|_|  _ _ _|_| | | |
28                        |_ _ _ _ _ _ _| |  _ _|  _ _| |  _ _ _| | |
24                         \ _ _ _ _ _ _ _| |  _| |    _| |  _ _ _| |
14                          |_ _ _ _ _ _ _| | |  _|  _|  _| |  _ _ _|
26                           \ _ _ _ _ _ _ _| | |_ _|  _|  _| |
24                            |_ _ _ _ _ _ _ _| |  _ _|  _|  _|
28                             \ _ _ _ _ _ _ _ _| |  _ _|  _|
24                              |_ _ _ _ _ _ _ _| | |  _ _|
30                               \ _ _ _ _ _ _ _ _| | |
31                                |_ _ _ _ _ _ _ _ _| |
32                                 \ _ _ _ _ _ _ _ _ _|
...
a(n) is the total area of the n-th set of symmetric regions in the diagram.
.
From _Omar E. Pol_, Aug 21 2015: (Start)
The above structure contains a hidden pattern, simpler, as shown below:
Level                              _ _
1                                _| | |_
2                              _|  _|_  |_
3                            _|   | | |   |_
4                          _|    _| | |_    |_
5                        _|     |  _|_  |     |_
6                      _|      _| | | | |_      |_
7                    _|       |   | | |   |       |_
8                  _|        _|  _| | |_  |_        |_
9                _|         |   |  _|_  |   |         |_
10             _|          _|   | | | | |   |_          |_
11           _|           |    _| | | | |_    |           |_
12         _|            _|   |   | | |   |   |_            |_
13       _|             |     |  _| | |_  |     |             |_
14     _|              _|    _| |  _|_  | |_    |_              |_
15   _|               |     |   | | | | |   |     |               |_
16  |                 |     |   | | | | |   |     |                 |
...
The symmetric pattern emerges from the front view of the stepped pyramid.
Note that starting from this diagram A000203 is obtained as follows:
In the pyramid the area of the k-th vertical region in the n-th level on the front view is equal to A237593(n,k), and the sum of all areas of the vertical regions in the n-th level on the front view is equal to 2n.
The area of the k-th horizontal region in the n-th level is equal to A237270(n,k), and the sum of all areas of the horizontal regions in the n-th level is equal to sigma(n) = A000203(n). (End)
From _Omar E. Pol_, Dec 31 2016: (Start)
Illustration of the top view of the pyramid with 16 levels:
.
n   A000203    A237270    _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
1      1   =      1      |_| | | | | | | | | | | | | | | |
2      3   =      3      |_ _|_| | | | | | | | | | | | | |
3      4   =    2 + 2    |_ _|  _|_| | | | | | | | | | | |
4      7   =      7      |_ _ _|    _|_| | | | | | | | | |
5      6   =    3 + 3    |_ _ _|  _|  _ _|_| | | | | | | |
6     12   =     12      |_ _ _ _|  _| |  _ _|_| | | | | |
7      8   =    4 + 4    |_ _ _ _| |_ _|_|    _ _|_| | | |
8     15   =     15      |_ _ _ _ _|  _|     |  _ _ _|_| |
9     13   =  5 + 3 + 5  |_ _ _ _ _| |      _|_| |  _ _ _|
10    18   =    9 + 9    |_ _ _ _ _ _|  _ _|    _| |
11    12   =    6 + 6    |_ _ _ _ _ _| |  _|  _|  _|
12    28   =     28      |_ _ _ _ _ _ _| |_ _|  _|
13    14   =    7 + 7    |_ _ _ _ _ _ _| |  _ _|
14    24   =   12 + 12   |_ _ _ _ _ _ _ _| |
15    24   =  8 + 8 + 8  |_ _ _ _ _ _ _ _| |
16    31   =     31      |_ _ _ _ _ _ _ _ _|
... (End)
		

Crossrefs

Programs

  • Mathematica
    Table[If[EvenQ@ n, n, DivisorSigma[1, (n + 1)/2]], {n, 0, 65}] (* or *)
    Transpose@ {Range[0, #, 2], DivisorSigma[1, #] & /@ Range[#/2 + 1]} &@ 65 // Flatten (* Michael De Vlieger, Dec 31 2016 *)
    With[{nn=70},Riffle[Range[0,nn,2],DivisorSigma[1,Range[nn/2]]]] (* Harvey P. Dale, Aug 05 2024 *)

Formula

a(2*n-1) + a(2n) = A224880(n).

A024916 a(n) = Sum_{k=1..n} k*floor(n/k); also Sum_{k=1..n} sigma(k) where sigma(n) = sum of divisors of n (A000203).

Original entry on oeis.org

1, 4, 8, 15, 21, 33, 41, 56, 69, 87, 99, 127, 141, 165, 189, 220, 238, 277, 297, 339, 371, 407, 431, 491, 522, 564, 604, 660, 690, 762, 794, 857, 905, 959, 1007, 1098, 1136, 1196, 1252, 1342, 1384, 1480, 1524, 1608, 1686, 1758, 1806, 1930, 1987, 2080, 2152
Offset: 1

Views

Author

Keywords

Comments

Row sums of triangle A128489. E.g., a(5) = 15 = (10 + 3 + 1 + 1), sum of row 4 terms of triangle A128489. - Gary W. Adamson, Jun 03 2007
Row sums of triangle A134867. - Gary W. Adamson, Nov 14 2007
a(10^4) = 82256014, a(10^5) = 8224740835, a(10^6) = 822468118437, a(10^7) = 82246711794796; see A072692. - M. F. Hasler, Nov 22 2007
Equals row sums of triangle A158905. - Gary W. Adamson, Mar 29 2009
n is prime if and only if a(n) - a(n-1) - 1 = n. - Omar E. Pol, Dec 31 2012
Also the alternating row sums of A236104. - Omar E. Pol, Jul 21 2014
a(n) is also the total number of parts in all partitions of the positive integers <= n into equal parts. - Omar E. Pol, Apr 30 2017
a(n) is also the total area of the terraces of the stepped pyramid with n levels described in A245092. - Omar E. Pol, Nov 04 2017
a(n) is also the area under the Dyck path described in the n-th row of A237593 (see example). - Omar E. Pol, Sep 17 2018
From Omar E. Pol, Feb 17 2020: (Start)
Convolution of A340793 and A000027.
Convolved with A340793 gives A000385. (End)
a(n) is also the number of cubic cells (or cubes) in the n-th level starting from the top of the stepped pyramid described in A245092. - Omar E. Pol, Jan 12 2022

Examples

			From _Omar E. Pol_, Aug 20 2021: (Start)
For n = 6 the sum of all divisors of the first six positive integers is [1] + [1 + 2] + [1 + 3] + [1 + 2 + 4] + [1 + 5] + [1 + 2 + 3 + 6] = 1 + 3 + 4 + 7 + 6 + 12 = 33, so a(6) = 33.
On the other hand the area under the Dyck path of the 6th diagram as shown below is equal to 33, so a(6) = 33.
Illustration of initial terms:                        _ _ _ _
                                        _ _ _        |       |_
                            _ _ _      |     |       |         |_
                  _ _      |     |_    |     |_ _    |           |
          _ _    |   |_    |       |   |         |   |           |
    _    |   |   |     |   |       |   |         |   |           |
   |_|   |_ _|   |_ _ _|   |_ _ _ _|   |_ _ _ _ _|   |_ _ _ _ _ _|
.
    1      4        8          15           21             33         (End)
		

References

  • Hardy and Wright, "An introduction to the theory of numbers", Oxford University Press, fifth edition, p. 266.

Crossrefs

Programs

  • Haskell
    a024916 n = sum $ map (\k -> k * div n k) [1..n]
    -- Reinhard Zumkeller, Apr 20 2015
    
  • Magma
    [(&+[DivisorSigma(1, k): k in [1..n]]): n in [1..60]]; // G. C. Greubel, Mar 15 2019
    
  • Maple
    A024916 := proc(n)
        add(numtheory[sigma](k),k=0..n) ;
    end proc: # Zerinvary Lajos, Jan 11 2009
    # second Maple program:
    a:= proc(n) option remember; `if`(n=0, 0,
          numtheory[sigma](n)+a(n-1))
        end:
    seq(a(n), n=1..100);  # Alois P. Heinz, Sep 12 2019
  • Mathematica
    Table[Plus @@ Flatten[Divisors[Range[n]]], {n, 50}] (* Alonso del Arte, Mar 06 2006 *)
    Table[Sum[n - Mod[n, m], {m, n}], {n, 50}] (* Roger L. Bagula and Gary W. Adamson, Oct 06 2006 *)
    a[n_] := Sum[DivisorSigma[1, k], {k, n}]; Table[a[n], {n, 51}] (* Jean-François Alcover, Dec 16 2011 *)
    Accumulate[DivisorSigma[1,Range[60]]] (* Harvey P. Dale, Mar 13 2014 *)
  • PARI
    A024916(n)=sum(k=1,n,n\k*k) \\ M. F. Hasler, Nov 22 2007
    
  • PARI
    A024916(z) = { my(s,u,d,n,a,p); s = z*z; u = sqrtint(z); p = 2; for(d=1, u, n = z\d - z\(d+1); if(n<=1, p=d; break(), a = z%d; s -= (2*a+(n-1)*d)*n/2); ); u = z\p; for(d=2, u, s -= z%d); return(s); } \\ See the link for a nicely formatted version. - P. L. Patodia (pannalal(AT)usa.net), Jan 11 2008
    
  • PARI
    A024916(n)={my(s=0,d=1,q=n);while(dPeter Polm, Aug 18 2014
    
  • PARI
    A024916(n)={ my(s=n^2, r=sqrtint(n), nd=n, D); for(d=1, r, (1>=D=nd-nd=n\(d+1)) && (r=d-1) && break; s -= n%d*D+(D-1)*D\2*d); s - sum(d=2, n\(r+1), n%d)} \\ Slightly optimized version of Patodia's code. - M. F. Hasler, Apr 18 2015
    (C#) See Polm link.
    
  • Python
    def A024916(n): return sum(k*(n//k) for k in range(1,n+1)) # Chai Wah Wu, Dec 17 2021
    
  • Python
    from math import isqrt
    def A024916(n): return (-(s:=isqrt(n))**2*(s+1) + sum((q:=n//k)*((k<<1)+q+1) for k in range(1,s+1)))>>1 # Chai Wah Wu, Oct 21 2023
  • Sage
    [sum(sigma(k) for k in (1..n)) for n in (1..60)] # G. C. Greubel, Mar 15 2019
    

Formula

From Benoit Cloitre, Apr 28 2002: (Start)
a(n) = n^2 - A004125(n).
Asymptotically a(n) = n^2*Pi^2/12 + O(n*log(n)). (End)
G.f.: (1/(1-x))*Sum_{k>=1} x^k/(1-x^k)^2. - Benoit Cloitre, Apr 23 2003
a(n) = Sum_{m=1..n} (n - (n mod m)). - Roger L. Bagula and Gary W. Adamson, Oct 06 2006
a(n) = n^2*Pi^2/12 + O(n*log(n)^(2/3)) [Walfisz]. - Charles R Greathouse IV, Jun 19 2012
a(n) = A000217(n) + A153485(n). - Omar E. Pol, Jan 28 2014
a(n) = A000292(n) - A076664(n), n > 0. - Omar E. Pol, Feb 11 2014
a(n) = A078471(n) + A271342(n). - Omar E. Pol, Apr 08 2016
a(n) = (1/2)*(A222548(n) + A006218(n)). - Ridouane Oudra, Aug 03 2019
From Greg Dresden, Feb 23 2020: (Start)
a(n) = A092406(n) + 8, n>3.
a(n) = A160664(n) - 1, n>0. (End)
a(2*n) = A326123(n) + A326124(n). - Vaclav Kotesovec, Aug 18 2021
a(n) = Sum_{k=1..n} k * A010766(n,k). - Georg Fischer, Mar 04 2022
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