A001065
Sum of proper divisors (or aliquot parts) of n: sum of divisors of n that are less than n.
Original entry on oeis.org
0, 1, 1, 3, 1, 6, 1, 7, 4, 8, 1, 16, 1, 10, 9, 15, 1, 21, 1, 22, 11, 14, 1, 36, 6, 16, 13, 28, 1, 42, 1, 31, 15, 20, 13, 55, 1, 22, 17, 50, 1, 54, 1, 40, 33, 26, 1, 76, 8, 43, 21, 46, 1, 66, 17, 64, 23, 32, 1, 108, 1, 34, 41, 63, 19, 78, 1, 58, 27, 74, 1, 123, 1, 40, 49, 64, 19, 90, 1, 106
Offset: 1
x^2 + x^3 + 3*x^4 + x^5 + 6*x^6 + x^7 + 7*x^8 + 4*x^9 + 8*x^10 + x^11 + ...
For n = 44, sum of divisors of n = sigma(n) = 84; so a(44) = 84-44 = 40.
Related concepts: (Start)
From 1 to 17, all n are deficient, except 6 and 12 seen below. See A005100.
Abundant numbers: a(12) = 16, a(18) = 21. See A005101.
Perfect numbers: a(6) = 6, a(28) = 28. See A000396.
Amicable numbers: a(220) = 284, a(284) = 220. See A259180.
Sociable numbers: 12496 -> 14288 -> 15472 -> 14536 -> 14264 -> 12496. See A122726. (End)
For n = 10 the sum of the divisors of 10 that are less than 10 is 1 + 2 + 5 = 8. On the other hand, the partitions of 10 into equal parts that do not contain 1 as a part are [10], [5,5], [2,2,2,2,2], there are 8 parts, so a(10) = 8. - _Omar E. Pol_, Nov 24 2019
- M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 840.
- George E. Andrews, Number Theory. New York: Dover, 1994; Pages 1, 75-92; p. 92 #15: Sigma(n) / d(n) >= n^(1/2).
- Carl Pomerance, The first function and its iterates, pp. 125-138 in Connections in Discrete Mathematics, ed. S. Butler et al., Cambridge, 2018.
- H. J. J. te Riele, Perfect numbers and aliquot sequences, pp. 77-94 in J. van de Lune, ed., Studieweek "Getaltheorie en Computers", published by Math. Centrum, Amsterdam, Sept. 1980.
- 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).
- James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, page 91.
- T. D. Noe, Table of n, a(n) for n = 1..10000
- M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972. [alternative scanned copy].
- Joerg Arndt, On computing the generalized Lambert series, arXiv:1202.6525v3 [math.CA], (2012).
- Henry Bottomley, Illustration of initial terms
- K. Chum, R. K. Guy, M. J. Jacobson, Jr., and A. S. Mosunov, Numerical and statistical analysis of aliquot sequences, Exper. Math. 29 (2020), no. 4, 414-425; arXiv:2110.14136, Oct. 2021 [math.NT].
- Don Coppersmith, An answer to the problem of Don Saari, 1987.
- Paul Erdős, Andrew Granville, Carl Pomerance and Claudia Spiro, On the normal behavior of the iterates of some arithmetic functions, Analytic number theory, Birkhäuser Boston, 1990, pp. 165-204.
- Paul Erdős, Andrew Granville, Carl Pomerance and Claudia Spiro, On the normal behavior of the iterates of some arithmetic functions, Analytic number theory, Birkhäuser Boston, 1990, pp. 165-204. [Annotated copy with A-numbers]
- Passawan Noppakaew and Prapanpong Pongsriiam, Product of Some Polynomials and Arithmetic Functions, J. Int. Seq. (2023) Vol. 26, Art. 23.9.1.
- P. Pollack and C. Pomerance, Some problems of Erdős on the sum-of-divisors function, For Richard Guy on his 99th birthday: May his sequence be unbounded, Trans. Amer. Math. Soc. Ser. B 3 (2016), 1-26; errata.
- Carl Pomerance and Hee-Sung Yang, Variant of a theorem of Erdős on the sum-of-proper-divisors function, Math. Comp., to appear (2014).
- Primefan, Sums of Restricted Divisors for n=1 to 1000
- F. Richman, Aliquot series: Abundant, deficient, perfect
- Eric Weisstein's World of Mathematics, Restricted Divisor Function
- Eric Weisstein's World of Mathematics, Divisor Function
- Index entries for "core" sequences
Cf.
A007956 (products of proper divisors).
-
a001065 n = a000203 n - n -- Reinhard Zumkeller, Sep 15 2011
-
[SumOfDivisors(n)-n: n in [1..100]]; // Vincenzo Librandi, May 06 2015
-
A001065 := proc(n)
numtheory[sigma](n)-n ;
end proc:
seq( A001065(n),n=1..100) ;
-
Table[ Plus @@ Select[ Divisors[ n ], #Zak Seidov, Sep 10 2009 *)
Table[DivisorSigma[1, n] - n, {n, 1, 80}] (* Jean-François Alcover, Apr 25 2013 *)
Array[Plus @@ Most@ Divisors@# &, 80] (* Robert G. Wilson v, Dec 24 2017 *)
-
numlib::sigma(n)-n$ n=1..81 // Zerinvary Lajos, May 13 2008
-
{a(n) = if( n==0, 0, sigma(n) - n)} /* Michael Somos, Sep 20 2011 */
-
from sympy import divisor_sigma
def A001065(n): return divisor_sigma(n)-n # Chai Wah Wu, Nov 04 2022
-
[sigma(n, 1)-n for n in range(1, 81)] # Stefano Spezia, Jul 14 2025
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
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
A236106
Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the twice odd numbers (A016825) interleaved with k-1 zeros, and the first element of column k is in row k(k+1)/2.
Original entry on oeis.org
2, 6, 10, 2, 14, 0, 18, 6, 22, 0, 2, 26, 10, 0, 30, 0, 0, 34, 14, 6, 38, 0, 0, 2, 42, 18, 0, 0, 46, 0, 10, 0, 50, 22, 0, 0, 54, 0, 0, 6, 58, 26, 14, 0, 2, 62, 0, 0, 0, 0, 66, 30, 0, 0, 0, 70, 0, 18, 10, 0, 74, 34, 0, 0, 0, 78, 0, 0, 0, 6, 82, 38, 22, 0, 0, 2
Offset: 1
Triangle begins:
2;
6;
10, 2;
14, 0;
18, 6;
22, 0, 2;
26, 10, 0;
30, 0, 0;
34, 14, 6;
38, 0, 0, 2;
42, 18, 0, 0;
46, 0, 10, 0;
50, 22, 0, 0;
54, 0, 0, 6;
58, 26, 14, 0, 2;
62, 0, 0, 0, 0;
66, 30, 0, 0, 0;
70, 0, 18, 10, 0;
74, 34, 0, 0, 0;
78, 0, 0, 0, 6;
82, 38, 22, 0, 0, 2;
86, 0, 0, 14, 0, 0;
90, 42, 0, 0, 0, 0;
94, 0, 26, 0, 0, 0;
...
For n = 9 the divisors of 2*9 = 18 are 1, 2, 3, 6, 9, 18, therefore the sum of the even divisors of 18 is 2 + 6 + 18 = 26. On the other hand the 9th row of triangle is 34, 14, 6, therefore the alternating row sum is 34 - 14 + 6 = 26, equaling the sum of the even divisors of 18.
If n is even then the alternating sum of the n-th row of triangle is simpler than the sum of the even divisors of 2n. Example: for n = 12 the sum of the even divisors of 2*12 = 24 is 2 + 4 + 6 + 8 + 12 + 24 = 56, and the alternating sum of the 12th row of triangle is 46 - 0 + 10 - 0 = 56.
Cf.
A000203,
A000217,
A001227,
A003056,
A016825,
A074400,
A196020,
A211343,
A228813,
A231345,
A231347,
A235791,
A235794,
A236104,
A236112.
A236112
Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists k+1 copies of the squares in nondecreasing order, and the first element of column k is in row k(k+1)/2.
Original entry on oeis.org
0, 0, 1, 0, 1, 0, 4, 0, 4, 1, 0, 9, 1, 0, 9, 1, 0, 16, 4, 0, 16, 4, 1, 0, 25, 4, 1, 0, 25, 9, 1, 0, 36, 9, 1, 0, 36, 9, 4, 0, 49, 16, 4, 1, 0, 49, 16, 4, 1, 0, 64, 16, 4, 1, 0, 64, 25, 9, 1, 0, 81, 25, 9, 1, 0, 81, 25, 9, 4, 0, 100, 36, 9, 4, 1, 0, 100, 36, 16, 4, 1, 0, 121, 36, 16, 4, 1, 0, 121, 49, 16, 4, 1, 0
Offset: 1
Triangle begins:
0;
0;
1, 0;
1, 0;
4, 0;
4, 1, 0;
9, 1, 0;
9, 1, 0;
16, 4, 0;
16, 4, 1, 0;
25, 4, 1, 0;
25, 9, 1, 0;
36, 9, 1, 0;
36, 9, 4, 0;
49, 16, 4, 1, 0;
49, 16, 4, 1, 0;
64, 16, 4, 1, 0;
64, 25, 9, 1, 0;
81, 25, 9, 1, 0;
81, 25, 9, 4, 0;
100, 36, 9, 4, 1, 0;
100, 36, 16, 4, 1, 0;
121, 36, 16, 4, 1, 0;
121, 49, 16, 4, 1, 0;
...
For n = 24 the 24th row of triangle is 121, 49, 16, 4, 1, 0 therefore the alternating row sum is 121 - 49 + 16 - 4 + 1 - 0 = 85 equaling A004125(24).
Cf.
A000203,
A000217,
A000290,
A003056,
A004125,
A120444,
A196020,
A211343,
A228813,
A231345,
A231347,
A235791,
A235794,
A236104,
A236106,
A237048,
A237591,
A237593,
A261699.
A231345
Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the odd numbers interleaved with k-1 zeros but T(n,1) = -1 and the first element of column k is in row k(k+1)/2.
Original entry on oeis.org
-1, -1, -1, 1, -1, 0, -1, 3, -1, 0, 1, -1, 5, 0, -1, 0, 0, -1, 7, 3, -1, 0, 0, 1, -1, 9, 0, 0, -1, 0, 5, 0, -1, 11, 0, 0, -1, 0, 0, 3, -1, 13, 7, 0, 1, -1, 0, 0, 0, 0, -1, 15, 0, 0, 0, -1, 0, 9, 5, 0, -1, 17, 0, 0, 0, -1, 0, 0, 0, 3, -1, 19, 11, 0, 0, 1
Offset: 1
Triangle begins:
-1;
-1;
-1, 1;
-1, 0;
-1, 3;
-1, 0, 1;
-1, 5, 0;
-1, 0, 0;
-1, 7, 3;
-1, 0, 0, 1;
-1, 9, 0, 0;
-1, 0, 5, 0;
-1, 11, 0, 0;
-1, 0, 0, 3;
-1, 13, 7, 0, 1;
-1, 0, 0, 0, 0;
-1, 15, 0, 0, 0;
-1, 0, 9, 5, 0;
-1, 17, 0, 0, 0;
-1, 0, 0, 0, 3;
-1, 19, 11, 0, 0, 1;
-1, 0, 0, 7, 0, 0;
-1, 21, 0, 0, 0, 0;
-1, 0, 13, 0, 0, 0;
...
For n = 15 the divisors of 15 are 1, 3, 5, 15 hence the abundance of 15 is 1 + 3 + 5 + 15 - 2*15 = 1 + 3 + 5 - 15 = -6. On the other hand the 15th row of triangle is -1, 13, 7, 0, 1, hence the alternating row sum is -1 - 13 + 7 - 0 + 1 = -6, equalling the abundance of 15.
If n is even then the alternating sum of the n-th row of triangle is simpler than the sum of divisors of n minus 2*n. Example: the sum of divisors of 24 minus 2*24 is 1 + 2 + 3 + 4 + 6 + 8 + 12 + 24 - 2*24 = 60 - 48 = 12, and the alternating sum of the 24th row of triangle is -1 - 0 + 13 - 0 + 0 - 0 = 12.
Cf.
A000203,
A000217,
A000396,
A003056,
A005100,
A005843,
A033879,
A033880,
A069283,
A196020,
A212119,
A228813,
A231347,
A235791,
A235794,
A236104,
A236106,
A236112,
A237593.
A235794
Triangle read by rows: T(n,k), n>=1, k>=1, in which column k starts with k zeros and then lists the odd numbers interleaved with k zeros, and the first element of column k is in row k(k+1)/2.
Original entry on oeis.org
0, 1, 0, 0, 3, 0, 0, 1, 5, 0, 0, 0, 0, 0, 7, 3, 0, 0, 0, 1, 9, 0, 0, 0, 0, 5, 0, 0, 11, 0, 0, 0, 0, 0, 3, 0, 13, 7, 0, 1, 0, 0, 0, 0, 0, 15, 0, 0, 0, 0, 0, 9, 5, 0, 0, 17, 0, 0, 0, 0, 0, 0, 0, 3, 0, 19, 11, 0, 0, 1, 0, 0, 7, 0, 0, 0, 21, 0, 0, 0, 0, 0, 0
Offset: 1
Triangle begins:
0;
1;
0, 0;
3, 0;
0, 1;
5, 0, 0;
0, 0, 0;
7, 3, 0;
0, 0, 1;
9, 0, 0, 0;
0, 5, 0, 0;
11, 0, 0, 0;
0, 0, 3, 0;
13, 7, 0, 1;
0, 0, 0, 0, 0;
15, 0, 0, 0, 0;
0, 9, 5, 0, 0;
17, 0, 0, 0, 0;
0, 0, 0, 3, 0;
19, 11, 0, 0, 1;
0, 0, 7, 0, 0, 0;
21, 0, 0, 0, 0, 0;
0, 13, 0, 0, 0, 0;
23, 0, 0, 5, 0, 0;
...
For n = 14 the 14th row of triangle is 13, 7, 0, 1, and the alternating sum is 13 - 7 + 0 - 1 = 5, the same as A120444(14) = 5.
Cf.
A000203,
A000217,
A003056,
A004125,
A120444,
A196020,
A211343,
A228813,
A231345,
A231347,
A235791,
A236104,
A236106,
A236112.
A237273
Triangle read by rows: T(n,k) = k+m, if k < m and k*m = n, or T(n,k) = k, if k^2 = n. Otherwise T(n,k) = 0. With n>=1 and 1<=k<=A000196(n).
Original entry on oeis.org
1, 3, 4, 5, 2, 6, 0, 7, 5, 8, 0, 9, 6, 10, 0, 3, 11, 7, 0, 12, 0, 0, 13, 8, 7, 14, 0, 0, 15, 9, 0, 16, 0, 8, 17, 10, 0, 4, 18, 0, 0, 0, 19, 11, 9, 0, 20, 0, 0, 0, 21, 12, 0, 9, 22, 0, 10, 0, 23, 13, 0, 0, 24, 0, 0, 0, 25, 14, 11, 10, 26, 0, 0, 0, 5
Offset: 1
Triangle begins:
1;
3;
4;
5, 2;
6, 0;
7, 5;
8, 0;
9, 6;
10, 0, 3;
11, 7, 0;
12, 0, 0;
13, 8, 7;
14, 0, 0;
15, 9, 0;
16, 0, 8;
17, 10, 0, 4;
18, 0, 0, 0;
19, 11, 9, 0;
20, 0, 0, 0;
21, 12, 0, 9;
22, 0, 10, 0;
23, 13, 0, 0;
24, 0, 0, 0;
25, 14, 11, 10;
26, 0, 0, 0, 5;
27, 15, 0, 0, 0;
28, 0, 12, 0, 0;
29, 16, 0, 11, 0;
30, 0, 0, 0, 0;
31, 17, 13, 0, 11;
...
For n = 9 the divisors of n are 1, 3, 9, so row 9 is 10, 0, 3, because 1*9 = 9 and 3^2 = 9. The sum of row 9 is A000203(9) = 13.
For n = 12 the divisors of 12 are 1, 2, 3, 4, 6, 12, so row 12 is 13, 8, 7, because 1*12 = 12, 2*6 = 12 and 3*4 = 12. The sum of row 12 is A000203(12) = 28.
Cf.
A000290,
A008578,
A018253,
A027750,
A196020,
A210959,
A212119,
A212120,
A228812-
A228814,
A231347,
A236104,
A236631,
A237519,
A237593.
-
T(n, k) = if (n % k, 0, if (k^2==n, k, k + n/k));
tabf(nn) = {for (n = 1, nn, v = vector(sqrtint(n), k, T(n, k)); print(v););} \\ Michel Marcus, Jun 19 2019
Showing 1-10 of 14 results.
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