A235796 -id:A235796 - cp's OEIS frontend

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

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

Omar E. Pol, Feb 19 2014

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

			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 *)

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

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

A004125 Sum of remainders of n mod k, for k = 1, 2, 3, ..., n.

Original entry on oeis.org

0, 0, 1, 1, 4, 3, 8, 8, 12, 13, 22, 17, 28, 31, 36, 36, 51, 47, 64, 61, 70, 77, 98, 85, 103, 112, 125, 124, 151, 138, 167, 167, 184, 197, 218, 198, 233, 248, 269, 258, 297, 284, 325, 328, 339, 358, 403, 374, 414, 420, 449, 454, 505, 492, 529, 520, 553, 578, 635, 586, 645, 672

Offset: 1

N. J. A. Sloane, Jeffrey Shallit

Row sums of A051778, A048158. Antidiagonal sums of A051127. - L. Edson Jeffery, Mar 03 2012
Let u_m(n) = Sum_{k=1..n} (n^m mod k^m) with m integer. As n-->+oo, u_m(n) ~ (n^(m+1))*(1-(1/(m+1))*Zeta(1+1/m)). Proof: using Riemann sums, we have u_m(n) ~ (n^(m+1))*int(((1/x)[nonascii character here])*(1-floor(x^m)/(x^m)),x=1..+oo) and the result follows. - Yalcin Aktar, Jul 30 2008 [x is the real variable of integration. The nonascii character (which was illegible in the original message) is probably some form of multiplication sign. I suggest that we leave it the way it is for now. - N. J. A. Sloane, Dec 07 2014]
Also the alternating row sums of A236112. - Omar E. Pol, Jan 26 2014
If n is prime then a(n) = a(n-1) + n - 2. - Omar E. Pol, Mar 19 2014
If n is a power of 2 greater than 1, then a(n) = a(n-1). - David Morales Marciel, Oct 21 2015
It appears that if n is an even perfect number, then a(n) = a(n-1) - 1. - Omar E. Pol, Oct 21 2015
Partial sums of A235796. - Omar E. Pol, Jun 26 2016
Aside from a(n) = a(n-1) for n = 2^m, the only values appearing more than once among the first 6*10^8 terms are those at n = 38184 +- 1, 458010 +- 1, 776112 +- 1, 65675408 +- 1, and 113393280 +- 2. - Trevor Cappallo, Jun 07 2021
The off-by-1 terms in the comment above are the terms of A068077. Proof: If a(n-1) = a(n+1), then (n-1)^2 - Sum_{k=1..n-1} sigma(k) = (n+1)^2 - Sum_{k=1..n+1} sigma(k) via the formula; rearranging terms gives sigma(n)+sigma(n+1)=4n. - Lewis Chen, Sep 24 2021

			a(5) = 4. The remainder when 5 is divided by 2,3,4 respectively is 1,2,1 and their sum = 4.

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

N. J. A. Sloane, Table of n, a(n) for n = 1..2000 (first 1000 terms from T. D. Noe)
Jeffrey Shallit, Problem E2817, Amer. Math. Monthly, vol. 87, p 137, 1980.

Cf. A000290, A006218, A023196, A048158, A050482, A051778, A120444 (first differences).

Cf. A067439, A072514, A111490.

List([1..70],n->n^2-Sum([1..n],k->Sigma(k))); # Muniru A Asiru, Mar 28 2018

a004125 n = sum $ map (mod n) [1..n]
-- Reinhard Zumkeller, Jan 28 2011

[&+[n mod r: r in [1..n]]: n in [1..70]]; // Bruno Berselli, Jul 06 2014

A004125 := n -> add( modp(n,k), k=2..n); /* much faster and unambiguous; "a mod b" may be mods(a,b) */ # M. F. Hasler, Nov 22 2007

Table[Sum[Mod[n,k],{k,2,n-1}],{n,70}] (* Harvey P. Dale, Nov 23 2011 *)
Accumulate[Table[2n-1-DivisorSigma[1,n],{n,70}]] (* Harvey P. Dale, Jul 11 2014 *)

A004125(n)=sum(k=2,n,n%k) \\ M. F. Hasler, Nov 22 2007

def a(n): return sum(n%k for k in range(1, n))
print([a(n) for n in range(1, 63)]) # Michael S. Branicky, Jun 08 2021

from math import isqrt
def A004125(n): return n**2+((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

def a(n): return sum(n.mod(k) for k in (1..n))
print([a(n) for n in (1..62)])  # Peter Luschny, May 12 2025

a(n) = n^2 - Sum_{k=1..n} sigma(k) = A000290(n) - A024916(n), hence asymptotically a(n) = n^2*(1-Pi^2/12) + O(n*log(n)^(2/3)). - Benoit Cloitre, Apr 28 2002. Asymptotics corrected/improved by Charles R Greathouse IV, Feb 22 2015
a(n) = A008805(n-3) + A049798(n-1), for n > 2. - Carl Najafi, Jan 31 2013
a(n) = A000217(n-1) - A153485(n). - Omar E. Pol, Jan 28 2014
G.f.: x^2/(1-x)^3 - (1-x)^(-1) * Sum_{k>=1} k*x^(2*k)/(1-x^k). - Robert Israel, Aug 13 2015
a(n) = Sum_{i=1..n} (n mod i). - Wesley Ivan Hurt, Sep 15 2017
From Ridouane Oudra, May 12 2025: (Start)
a(n) = A067439(n) + A072514(n).
a(n) = Sum_{d|n} d*A067439(n/d).
a(p) = A067439(p), for p prime.
a(p^k) = A072514(p^(k+1))/p, for p prime and k >= 0. (End)
a(n) = A111490(n) - n. - Peter Luschny, May 12 2025

Edited by M. F. Hasler, Apr 18 2015

A294898 Deficiency minus binary weight: a(n) = A033879(n) - A000120(n) = A005187(n) - A000203(n).

Original entry on oeis.org

0, 0, 0, 0, 2, -2, 3, 0, 3, 0, 7, -6, 9, 1, 2, 0, 14, -5, 15, -4, 7, 5, 18, -14, 16, 7, 10, -3, 24, -16, 25, 0, 16, 12, 19, -21, 33, 13, 18, -12, 37, -15, 38, 1, 8, 16, 41, -30, 38, 4, 26, 3, 48, -16, 33, -11, 30, 22, 53, -52, 55, 23, 16, 0, 44, -14, 63, 8, 39, -7, 66, -53, 69, 31, 22, 9, 54, -16, 73, -28, 38, 35, 78, -59, 58

Offset: 1

Antti Karttunen, Nov 25 2017

sign

"Least deficient numbers" or "almost perfect numbers" are those k for which A033879(k) = 1, or equally, for which a(k) = -A048881(k-1). The only known solutions are powers of 2 (A000079), all present also in A295296. See also A235796 and A378988. - Antti Karttunen, Dec 16 2024

Antti Karttunen, Table of n, a(n) for n = 1..16384
Index entries for sequences related to sigma(n).

Cf. A000120, A000203, A001065, A005187, A011371, A013661, A033879, A048881, A235796, A294896, A294899, A297114 (Möbius transform), A317844 (difference from a(n)), A326133, A326138, A324348 (a(n) applied to Doudna sequence), A379008 (a(n) applied to prime shift array), A378988.

Cf. A295296 (positions of zeros), A295297 (parity of a(n)).

Array[IntegerExponent[(2 #)!, 2] - DivisorSigma[1, #] &, 85] (* Michael De Vlieger, Nov 26 2017 *)

A294898(n) = ((2*n-sigma(n))-hammingweight(n)); \\ Antti Karttunen, Dec 16 2024

(define (A294898 n) (- (A005187 n) (A000203 n)))

a(n) = A005187(n) - A000203(n).
a(n) = A011371(n) - A001065(n).
a(n) = A033879(n) - A000120(n).
Sum_{k=1..n} a(k) ~ c * n^2, where c = 1 - zeta(2)/2 = 0.177532... . - Amiram Eldar, Feb 22 2024

Name edited by Antti Karttunen, Dec 16 2024

A294015 Sum of the even divisors of 2n, minus the (n-1)st odd number.

Original entry on oeis.org

1, 3, 3, 7, 3, 13, 3, 15, 9, 17, 3, 33, 3, 21, 19, 31, 3, 43, 3, 45, 23, 29, 3, 73, 13, 33, 27, 57, 3, 85, 3, 63, 31, 41, 27, 111, 3, 45, 35, 101, 3, 109, 3, 81, 67, 53, 3, 153, 17, 87, 43, 93, 3, 133, 35, 129, 47, 65, 3, 217, 3, 69, 83, 127, 39, 157, 3, 117, 55, 149, 3, 247, 3, 81, 99, 129, 39, 181, 3, 213, 81, 89, 3, 281

Offset: 1

Omar E. Pol, Oct 28 2017

nonn

a(n) = 3 if and only if n is prime.

Antti Karttunen, Table of n, a(n) for n = 1..20000

Partial sums give A294016.
Cf. A000203, A013661, A005408, A074400, A235796, A294017.
Cf. A001065, A091818.

a[n_] := 2*(DivisorSigma[1, n] - n) + 1; Array[a, 100] (* Amiram Eldar, Mar 30 2024 *)

a(n) = 2*sigma(n) - 2*n + 1; \\ Michel Marcus, Oct 29 2017

a(n) = A074400(n) - A005408(n-1) = 2*A000203(n) - 2*n + 1 = A000203(n) - A235796(n).
Sum_{k=1..n} a(k) = (Pi^2/6 - 1) * n^2 + O(n*log(n)). - Amiram Eldar, Mar 30 2024
a(n) = 2*A001065(n) + 1 = A091818(n) + 1. - Omar E. Pol, Dec 01 2024

A378988 a(n) = 2*n XOR 1+sigma(n), where XOR is bitwise-xor, A003987.

Original entry on oeis.org

0, 0, 3, 0, 13, 1, 7, 0, 28, 7, 27, 5, 21, 5, 7, 0, 49, 12, 51, 3, 11, 9, 55, 13, 18, 31, 31, 1, 37, 117, 31, 0, 115, 115, 119, 20, 109, 113, 119, 11, 121, 53, 123, 13, 21, 21, 111, 29, 88, 58, 47, 11, 93, 21, 39, 9, 35, 47, 75, 209, 69, 29, 23, 0, 215, 21, 195, 247, 235, 29, 199, 84, 217, 231, 235, 21, 251, 53, 207

Offset: 1

Antti Karttunen, Dec 16 2024

For any hypothetical quasiperfect number q (for which sigma(q) = 2*q+1, see A336701), a(q) would be equal to 2*q XOR 2*q+2 = 2*(q XOR q+1) = 2*A038712(1+q) = A100892(1+q).

See also A000079 and A235796 concerning the "almost perfect" or "least deficient" numbers that give positions of 0's here.

Cf. A000079 (conjectured to give positions of all 0's), A000396 (positions of 1's), A000203, A003987, A028982 (positions of even terms), A028983 (of odd terms), A038712, A100892, A318467, A336701, A378998, A379009 [= a(n^2)].

Cf. also A235796, A294898, A294899, A318457.

Array[BitXor[2*#, DivisorSigma[1, #] + 1] &, 100] (* Paolo Xausa, Dec 16 2024 *)

PARI
```
A378988(n) = bitxor(n+n,1+sigma(n));
```

For all n in A028983, a(n) = 2n+1 XOR sigma(n) = 1+A318467(n).

A237046 Numbers n such that sigma(n) < 2n-1.

Original entry on oeis.org

3, 5, 7, 9, 10, 11, 13, 14, 15, 17, 19, 21, 22, 23, 25, 26, 27, 29, 31, 33, 34, 35, 37, 38, 39, 41, 43, 44, 45, 46, 47, 49, 50, 51, 52, 53, 55, 57, 58, 59, 61, 62, 63, 65, 67, 68, 69, 71, 73, 74, 75, 76, 77, 79, 81, 82, 83, 85, 86, 87, 89, 91, 92, 93, 94, 95, 97, 98, 99, 101

Offset: 1

Omar E. Pol, Feb 19 2014

Numbers n such that A235796(n) > 0.
Complement of A103288.
If the only least-deficient numbers are the powers of 2 (which is an open problem) then the union of A000079 and A023196 and this sequence gives A000027 (see also A103288).

			10 is in the sequence because the sum of divisors of 10 is less than (2*10 - 1). The sum of divisors of 10 is 1 + 2 + 5 + 10 = 18 and 2*10 - 1 = 19.

Cf. A000203, A005100, A023196, A033879, A103288, A196020, A235796, A236104.

Select[Range[120],DivisorSigma[1,#]<2#-1&] (* Harvey P. Dale, Sep 12 2018 *)

s=[]; for(n=1, 200, if(sigma(n)<2*n-1, s=concat(s,n))); s \\ Colin Barker, Feb 20 2014

A237588 Sigma(n) - 2n + 1.

Original entry on oeis.org

0, 0, -1, 0, -3, 1, -5, 0, -4, -1, -9, 5, -11, -3, -5, 0, -15, 4, -17, 3, -9, -7, -21, 13, -18, -9, -13, 1, -27, 13, -29, 0, -17, -13, -21, 20, -35, -15, -21, 11, -39, 13, -41, -3, -11, -19, -45, 29, -40, -6, -29, -5, -51, 13, -37, 9, -33, -25, -57, 49, -59, -27, -21, 0

Offset: 1

Omar E. Pol, Feb 20 2014

sign

Also we can write Sigma(n) - (2n - 1).
a(n) = 2 - n iff n is prime.
a(n) = 1 iff n is a perfect number.
Conjecture: a(n) = 0 iff n is a power of 2.
The problem is not new. In fact, the following comments appeared on page 74 of Guy's book: "If Sigma(n) = 2*n - 1, n has been called almost perfect. Powers of 2 are almost perfect; it is not known if any other numbers are.". - Zhi-Wei Sun, Feb 23 2014

			-----------------------------------------------
.     The sum of       The positive
n    divisors of n     odd numbers        a(n)
-----------------------------------------------
1          1                1               0
2          3                3               0
3          4                5              -1
4          7                7               0
5          6                9              -3
6         12               11               1
7          8               13              -5
8         15               15               0
9         13               17              -4
10        18               19              -1
...

R. K. Guy, Unsolved Problems in Number Theory, 3rd Edition, Springer, New York, 2004.

Cf. A000079, A000203, A000396, A001065, A004125, A005408, A033879, A033880, A039653, A120444, A196020, A235796, A236104.

[1-2*n+SumOfDivisors(n): n in [1..100]]; // Vincenzo Librandi, Feb 25 2014

Table[DivisorSigma[1,n]-2n+1,{n,70}] (* Harvey P. Dale, Nov 15 2014 *)

vector(100, n, sigma(n)-2*n+1) \\ Colin Barker, Feb 21 2014

a(n) = A000203(n) - A005408(n-1) = 1 - n + A001065(n) = 1 - A033879(n) = 1 + A033880(n) = (-1)*A235796(n).

a(n) = A088580(n) - 2*n. - Omar E. Pol, Mar 23 2014

A239446 Triangle read by rows: T(n,k), n>=1, k>=1, in which column k lists the elements of A004273 interleaved with k zeros, and the first element of column k is in row k*(k+1)/2.

Original entry on oeis.org

0, 0, 1, 0, 0, 0, 3, 0, 0, 1, 0, 5, 0, 0, 0, 0, 0, 7, 3, 0, 0, 0, 1, 0, 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, 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, 0, 7, 0, 0, 0, 21, 0

Offset: 1

Omar E. Pol, Mar 20 2014

Alternating sum of row n equals A235796(n), i.e., sum_{k=1..A003056(n)} (-1)^(k-1)*T(n,k) = A235796(n).
Row n has length A003056(n) hence column k starts in row A000217(k).
Column k starts with k+1 zeros and then lists the odd numbers interleaved with k zeros.
It appears that row n lists all zeros iff n is a power of 2.

			Triangle begins:
0;
0;
1,  0;
0,  0;
3,  0;
0,  1,  0;
5,  0,  0;
0,  0,  0;
7,  3,  0;
0,  0,  1,  0;
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,  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,  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 = 15 the 15th row of triangle is 13, 7, 0, 1, and the alternating sum is 13 - 7 + 0 - 1 = A235796(15) = 5.

Cf. A000203, A000217, A003056, A004125, A004273, A196020, A231345, A231347, A235791, A235794, A235796, A236106, A236104, A236112, A237048, A237588, A237591, A239313.

A294628 a(n) = 8*(sigma(n) - n + (1/2)).

Original entry on oeis.org

4, 12, 12, 28, 12, 52, 12, 60, 36, 68, 12, 132, 12, 84, 76, 124, 12, 172, 12, 180, 92, 116, 12, 292, 52, 132, 108, 228, 12, 340, 12, 252, 124, 164, 108, 444, 12, 180, 140, 404, 12, 436, 12, 324, 268, 212, 12, 612, 68, 348, 172, 372, 12, 532, 140, 516, 188, 260, 12, 868, 12, 276

Offset: 1

Omar E. Pol, Nov 05 2017

nonn

Muniru A Asiru, Table of n, a(n) for n = 1..2000

Partial sums give A294629.

Cf. A001065, A017113, A235796, A237593, A239050, A294015, A294016, A294017, A294630.

List([1..10^5],n->8*(Sigma(n)-n+(1/2))); # Muniru A Asiru, Mar 04 2018

with(numtheory): seq(sigma(8*n-1)/8, n=1..10^3); # Muniru A Asiru, Mar 04 2018

a[n_] := 8 (DivisorSigma[1, n] - n) + 4; Array[a, 62] (* Robert G. Wilson v, Dec 12 2017 *)

a(n) = 4*A294015(n).
a(n) = 8*(A001065(n) + (1/2)).
a(n) = 8*(A000203(n) - n + (1/2)).
a(n) = A239050(n) - 4*A235796(n).
a(n) = A017113(n-1) - 8*A235796(n).

A362081 Numbers k achieving record abundance (sigma(k) > 2*k) via a residue-based measure M(k) (see Comments), analogous to superabundant numbers A004394.

Original entry on oeis.org

1, 2, 4, 6, 12, 24, 30, 36, 72, 120, 360, 420, 840, 1680, 2520, 4032, 5040, 10080, 25200, 32760, 65520, 98280, 194040, 196560, 388080, 942480, 1801800, 3160080, 3603600, 6320160, 12640320, 24504480, 53721360, 61981920, 73513440, 115315200, 122522400, 189909720, 192099600, 214885440

Offset: 1

Richard Joseph Boland, Apr 08 2023

nonn

The residue-based quantifier function, M(k) = (k+1)*(1 - zeta(2)/2) - 1 - ( Sum_{j=1..k} k mod j )/k, measures either abundance (sigma(k) > 2*k), or deficiency (sigma(k) < 2*k), of a positive integer k. It follows from the known facts that Sum_{j=1..k} (sigma(j) + k mod j) = k^2 and that the average order of sigma(k)/k is Pi^2/6 = zeta(2) (see derivation below).

M(k) ~ 0 when sigma(k) ~ 2*k and for sufficiently large k, M(k) is positive when k is an abundant number (A005101) and negative when k is a deficient number (A005100). The terms of this sequence are the abundant k for which M(k) > M(m) for all m < k, analogous to the superabundant numbers A004394, which utilize sigma(k)/k as the measure. However, sigma(k)/k does not give a meaningful measure of deficiency, whereas M(k) does, thus a sensible notion of superdeficient (see A362082).

			The abundance measure is initially negative, becoming positive for k > 30. Initial measures with factorizations from the Mathematica program:
   1  -0.64493406684822643647   {{1,1}}
   2  -0.46740110027233965471   {{2,1}}
   4  -0.36233516712056609118   {{2,2}}
   6  -0.25726923396879252765   {{2,1},{3,1}}
  12  -0.10873810118013850374   {{2,2},{3,1}}
  24  -0.10334250226949712257   {{2,3},{3,1}}
  30  -0.096478036147509765322  {{2,1},{3,1},{5,1}}
  36   0.068719763307810925260  {{2,2},{3,2}}
  72   0.12657322670640173542   {{2,3},{3,2}}

Jeffrey C. Lagarias, An Elementary Problem Equivalent to the Riemmann Hypothesis, arXiv:math/0008177 [math.NT], 2000-2001; Amer. Math. Monthly, 109 (2002), 534-543.

Cf. A362082 (superdeficient), A362083 (analogous to A335067, A326393).

Cf. A004394, A004490, A002201, A005100, A005101, A004125, A024916, A000290, A120444, A235796, A000396, A000079.

Clear[max, Rp, R, seqtable, M];
max = -1; Rp = 0; seqtable = {};
Do[R = Rp + 2 k - 1 - DivisorSigma[1, k];
  M = N[(k + 1)*(1 - Zeta[2]/2) - 1 - R/k, 20];
  If[M > max, max = M; Print[k, "   ", max, "   ", FactorInteger[k]];
   AppendTo[seqtable, k]];
  Rp = R, {k, 1, 1000000000}];
Print[seqtable]

M(n) = (n+1)*(1 - zeta(2)/2) - 1 - sum(k=2, n, n%k)/n;
lista(nn) = my(m=-oo, list=List()); for (n=1, nn, my(mm = M(n)); if (mm > m, listput(list, n); m = mm);); Vec(list); \\ Michel Marcus, Apr 21 2023

Derived starting with lemmas 1-3:
1) Sum_{j=1..k} (sigma(j) + k mod j) = k^2.
2) The average order of sigma(k)/k is Pi^2/6 = zeta(2).
3) R(k) = Sum_{j=1..k} k mod j, so R(k)/k is the average order of (k mod j).
Then:
Sum_{j=1..k} sigma(j) ~ zeta(2)*Sum_{j=1..k} j = zeta(2)*(k^2+k)/2.
R(k)/k ~ k - k*zeta(2)/2 - zeta(2)/2.
0 ~ (k+1)*(1 - zeta(2)/2) - 1 - R(k)/k.
Thus M(k) = (k+1)*(1 - zeta(2)/2) - 1 - R(k)/k is a measure of variance about sigma(k) ~ 2*k corresponding to M(k) ~ 0.

cp's OEIS Frontend

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

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A004125 Sum of remainders of n mod k, for k = 1, 2, 3, ..., n.

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A294898 Deficiency minus binary weight: a(n) = A033879(n) - A000120(n) = A005187(n) - A000203(n).

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A294015 Sum of the even divisors of 2n, minus the (n-1)st odd number.

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A378988 a(n) = 2*n XOR 1+sigma(n), where XOR is bitwise-xor, A003987.

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A237046 Numbers n such that sigma(n) < 2n-1.

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A237588 Sigma(n) - 2n + 1.