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
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;
...
Cf.
A000196,
A000217,
A000462,
A001227,
A001462,
A001614,
A004247 (multiplication table),
A006463 (partial sums),
A016655,
A050600,
A050602,
A048645,
A122797,
A131507,
A238005.
-
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
-
[Floor((Sqrt(1+8*n)-1)/2): n in [0..80]]; // Vincenzo Librandi, Oct 23 2011
-
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
-
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 *)
-
A003056(n)=(sqrtint(8*n+1)-1)\2 \\ M. F. Hasler, Oct 08 2011
-
t1(n)=floor(-1/2+sqrt(2+2*n)) /* A003056 */
-
t2(n)=n-binomial(floor(1/2+sqrt(2+2*n)),2) /* A002262 */
-
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
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
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)
- Robert Price, Table of n, a(n) for n = 1..15542 (rows n = 1..5000, flattened)
- Hartmut F. W. Hoft, Sample visual documentation for Mathematica code
- Michel Marcus, A colored version of the symmetric representation of sigma(n), multipage, n = 1..85
- Omar E. Pol, An infinite stepped pyramid (A237593, A237270, A262626)
- Omar E. Pol, Perspective view of the stepped pyramid (16 levels)
- Omar E. Pol, Perspective view of the stepped pyramid into four quadrants (11 levels). This is formed by combing four copies of the pyramid back-to-back (cf. A244050).
- N. J. A. Sloane, Another drawing of the spiral
- N. J. A. Sloane, Spiral showing only the outer boundary
- Index entries for sequences related to sigma(n)
Cf.
A000203,
A004125,
A023196,
A024916,
A153485,
A196020,
A221529,
A231347,
A235791,
A235796,
A236104,
A236112,
A236540,
A237046,
A237048,
A237271,
A237590,
A237591,
A237593,
A239050,
A239660,
A239663,
A239665,
A239931,
A239932,
A239933,
A239934,
A240020,
A240062,
A244050,
A245092,
A249351,
A262626,
A280850,
A280851,
A296508,
A335616,
A340035,
A384149.
-
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 *)
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
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)
Cf.
A000203,
A001227,
A024916,
A196020,
A235791,
A236104,
A237048,
A237270,
A237271,
A237593,
A239660,
A239931-
A239934,
A240542,
A244580,
A245092,
A249351,
A259176,
A259177,
A261350,
A261699,
A262626,
A285356,
A286000,
A286001.
-
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 *)
-
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
-
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
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
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 + ...
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- Eric Weisstein's World of Mathematics, Irredundant Set
- Eric Weisstein's World of Mathematics, Path Complement Graph
- G. G. Wojnar, D. Sz. Wojnar, and L. Q. Brin, Universal Peculiar Linear Mean Relationships in All Polynomials, arXiv:1706.08381 [math.GM], 2017.
- Index entries for two-way infinite sequences
- Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
- Index entries for sequences related to Chebyshev polynomials.
Cf. numbers of the form n*(n*k-k+4)/2 listed in
A226488.
Similar sequences are listed in
A316466.
-
a := List([0..1000], n -> n*(n+3)/2); # Muniru A Asiru, Jan 25 2018
-
a000096 n = n * (n + 3) `div` 2
a000096_list = [x | x <- [0..], a023531 x == 1]
-- Reinhard Zumkeller, Feb 14 2015, Dec 04 2012
-
[n*(n+3)/2: n in [0..60]]; // Juri-Stepan Gerasimov, Apr 05 2016
-
[n: n in [0..2300] | IsSquare(8*n+9)]; // Juri-Stepan Gerasimov, Apr 05 2016
-
A000096 := n->n*(n+3)/2; seq(A000096(n), n=0..50);
A000096 :=z*(-2+z)/(z-1)**3; # Simon Plouffe in his 1992 dissertation
-
Table[n*(n+3)/2, {n, 0, 100}] (* Vladimir Joseph Stephan Orlovsky, Oct 25 2008 *)
LinearRecurrence[{3,-3,1}, {0,2,5}, 60] (* Harvey P. Dale, Apr 30 2013 *)
-
{a(n) = n * (n+3)/2}; \\ Michael Somos, May 26 2004
-
first(n) = Vec(x*(2-x)/(1-x)^3 + O(x^n), -n) \\ Iain Fox, Dec 12 2017
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
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
Cf.
A000203,
A000217,
A000290,
A001227,
A003056,
A008794,
A024916,
A004125,
A196020,
A211343,
A228813,
A231345,
A231347,
A235791,
A235794,
A235799,
A236106,
A236112,
A236540,
A237270,
A237591,
A237593,
A239660,
A244050,
A245092,
A262626,
A286000.
-
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 *)
-
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
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
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)
Cf.
A000203,
A000217,
A001227,
A196020,
A211343,
A228813,
A231345,
A231347,
A235794,
A236106,
A236112,
A237270,
A237271,
A237593,
A239660,
A245092,
A261699,
A262626,
A286000,
A286001,
A280850,
A280851,
A296508,
A335616.
-
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 *)
-
row(n) = vector((sqrtint(8*n+1)-1)\2, i, 1+(n-(i*(i+1)/2))\i); \\ Michel Marcus, Mar 27 2014
-
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
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
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)
Cf.
A000203,
A000217,
A001227,
A001318,
A003056,
A211343,
A212119,
A228813,
A231345,
A231347,
A235791,
A235794,
A236104,
A236106,
A236112,
A237048,
A237271,
A237591,
A237593,
A238005,
A239660,
A244050,
A245092,
A261699,
A262626,
A286000,
A286001,
A280850,
A335616,
A338721.
-
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
-
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 *)
-
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
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
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)
Cf.
A000203,
A000265,
A001065,
A001227,
A005279,
A024916,
A060831,
A061345,
A067742,
A071561,
A071562,
A175254,
A196020,
A221529,
A235791,
A236104,
A237048,
A237591,
A237593,
A239657,
A244050,
A244971,
A245092,
A249223,
A250068,
A261699,
A262045,
A262612,
A262626,
A274824,
A279387,
A279693,
A319073,
A340583,
A340846,
A342344,
A347186,
A379288.
Cf.
A027750,
A174973 (2-dense numbers),
A239663,
A240062,
A243982,
A379379,
A380580,
A384149,
A384222,
A384225,
A384226,
A384230,
A384930.
-
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 *)
-
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
-
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
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
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)
Cf.
A000203,
A004125,
A024916,
A005843,
A175254,
A196020,
A224880,
A235791,
A236104,
A237270,
A237271,
A237591,
A237593,
A239050,
A239660,
A239931-
A239934,
A243980,
A244050,
A244360-
A244363,
A244370,
A244371,
A244970,
A244971,
A245093,
A261350,
A262626,
A277437,
A279387,
A280223,
A280295.
-
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 *)
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
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)
- Hardy and Wright, "An introduction to the theory of numbers", Oxford University Press, fifth edition, p. 266.
- Daniel Mondot, Table of n, a(n) for n = 1..10000 (first 1000 terms from T. D. Noe)
- Vaclav Kotesovec, Plot of (a(n) - Pi^2*n^2/12) / (n*log(n)^(2/3)) for n = 2..100000.
- P. L. Patodia (pannalal(AT)usa.net), PARI program for A072692 and A024916.
- Peter Polm, C# program for A024916.
- A. Walfisz, Weylsche Exponentialsummen in der neueren Zahlentheorie, ZAMM - Journal of Applied Mathematics and Mechanics / Zeitschrift für Angewandte Mathematik und Mechanik, Volume 44, Issue 12, page 607, 1964.
Cf.
A056550,
A104471(2*n-1, n),
A123229,
A128489,
A000217,
A134867,
A072692,
A158905,
A237593,
A245092,
A006218,
A222548,
A092406,
A160664.
-
a024916 n = sum $ map (\k -> k * div n k) [1..n]
-- Reinhard Zumkeller, Apr 20 2015
-
[(&+[DivisorSigma(1, k): k in [1..n]]): n in [1..60]]; // G. C. Greubel, Mar 15 2019
-
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
-
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 *)
-
A024916(n)=sum(k=1,n,n\k*k) \\ M. F. Hasler, Nov 22 2007
-
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
-
A024916(n)={my(s=0,d=1,q=n);while(dPeter Polm, Aug 18 2014
-
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.
-
def A024916(n): return sum(k*(n//k) for k in range(1,n+1)) # Chai Wah Wu, Dec 17 2021
-
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
-
[sum(sigma(k) for k in (1..n)) for n in (1..60)] # G. C. Greubel, Mar 15 2019
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