A004125 -id:A004125 - cp's OEIS frontend

A235799 a(n) = n^2 - sigma(n).

0, 1, 5, 9, 19, 24, 41, 49, 68, 82, 109, 116, 155, 172, 201, 225, 271, 285, 341, 358, 409, 448, 505, 516, 594, 634, 689, 728, 811, 828, 929, 961, 1041, 1102, 1177, 1205, 1331, 1384, 1465, 1510, 1639, 1668, 1805, 1852, 1947, 2044, 2161, 2180, 2344, 2407

Offset: 1

Omar E. Pol, Jan 24 2014

nonn

From Omar E. Pol, Apr 11 2021: (Start)
If n is prime (A000040) then a(n) = n^2 - n - 1.
If n is a power of 2 (A000079) then a(n) = (n-1)^2.
If n is a perfect number (A000396) then a(n) = (n-1)^2 - 1, assuming there are no odd perfect numbers.
In order to construct the diagram of the symmetric representation of a(n) we use the following rules:
At stage 1 in the first quadrant of the square grid we draw the symmetric representation of sigma(n) using the two Dyck paths described in the rows n and n-1 of A237593. The area of the region that is below the symmetric representation of sigma(n) equals A024916(n-1).
At stage 2 we draw a pair of orthogonal line segments (if it's necessary) such that in the drawing appears totally formed a square n X n. The area of the region that is above the symmetric representation of sigma(n) equals A004125(n).
At stage 3 we turn OFF the cells of the symmetric representation of sigma(n). Then we turn ON the rest of the cells that are in the square n X n. The result is that the ON cell form the diagram of the symmetric representation of a(n). See the Example section. (End)

			From _Omar E. Pol, Apr 04 2021: (Start)
Illustration of initial terms in the first quadrant for n = 1..6:
.
.                                                             y|        _ _
.                                              y|      _ _     |_ _ _  |_  |
.                                 y|      _     |_ _ _|   |    |     |   |_|
.                      y|    _     |_ _  |_|    |        _|    |     |_ _
.             y|        |_ _|_|    |   |_       |       |      |         |
.      y|      |_       |   |      |     |      |       |      |         |
.       |_ _   |_|_ _   |_ _|_ _   |_ _ _|_ _   |_ _ _ _|_ _   |_ _ _ _ _|_ _
.          x        x          x            x              x                x
.
n:        1       2         3           4             5               6
a(n):     0       1         5           9            19              24
.
Illustration of initial terms in the first quadrant for n = 7..9:
.                                                y|          _ _ _ _
.                          y|          _ _ _      |_ _ _ _ _|       |
.      y|        _ _ _      |_ _ _ _  |     |     |          _ _    |
.       |_ _ _ _|     |     |       | |_    |     |         |_  |   |
.       |             |     |       |_  |_ _|     |           |_|  _|
.       |            _|     |         |_ _        |               |
.       |           |       |             |       |               |
.       |           |       |             |       |               |
.       |           |       |             |       |               |
.       |_ _ _ _ _ _|_ _    |_ _ _ _ _ _ _|_ _    |_ _ _ _ _ _ _ _|_ _
.                      x                     x                       x
.
n:              7                    8                      9
a(n):          41                   49                     68
.
For n = 9 the figures 1, 2 and 3 below show respectively the three stages described in the Comments section as follows:
.
.   y|_ _ _ _ _ 5            y|_ _ _ _ _ _ _ _ _      y|          _ _ _ _
.    |_ _ _ _ _|              |_ _ _ _ _|       |      |_ _ _ _ _|       |
.    |         |_ _ 3         |         |_ _ R  |      |          _ _    |
.    |         |_  |          |         |_  |   |      |         |_  |   |
.    |           |_|_ _ 5     |           |_|_ _|      |           |_|  _|
.    |               | |      |               | |      |               |
.    |      Q        | |      |       Q       | |      |               |
.    |               | |      |               | |      |               |
.    |               | |      |               | |      |               |
.    |_ _ _ _ _ _ _ _|_|_     |_ _ _ _ _ _ _ _|_|_     |_ _ _ _ _ _ _ _|_ _
.                       x                        x                        x
.         Figure 1.                Figure 2.                Figure 3.
.         Symmetric                Symmetric                Symmetric
.       representation           representation           representation
.         of sigma(9)              of sigma(9)             of a(9) = 68
.       A000203(9) = 13          A000203(9) = 13
.           and of                   and of
.     Q = A024916(8) = 56      R = A004125(9) = 12
.                              Q = A024916(8) = 56
.
Note that the symmetric representation of a(9) contains a hole formed by three cells because these three cells were the central part of the symmetric representation of sigma(9). (End)

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A000040, A000079, A000203, A000217, A000290, A000396, A004125, A024816, A024916, A067436, A120444, A153485, A196020, A236104, A236112, A237593, A244048, A342344.

[n^2 - DivisorSigma(1,n): n in [1..50]]; // G. C. Greubel, Oct 31 2018

Table[n^2-DivisorSigma[1,n],{n,50}] (* Harvey P. Dale, Sep 02 2016 *)

vector(50, n, n^2 - sigma(n)) \\ G. C. Greubel, Oct 31 2018

a(n) = A000290(n) - A000203(n).
a(n) = A024916(n-1) + A004125(n), n > 1.
G.f.: x*(1 + x)/(1 - x)^3 - Sum_{k>=1} x^k/(1 - x^k)^2. - Ilya Gutkovskiy, Mar 17 2017
From Omar E. Pol, Apr 10 2021: (Start)
a(n) = A024816(n) + A000217(n-1).
a(n) = A067436(n) + A153485(n) + A244048(n). (End)

A342344 Number of parts in the symmetric representation of antisigma(n).

Original entry on oeis.org

0, 0, 2, 3, 1, 3, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2

Offset: 1

Omar E. Pol, Mar 08 2021

nonn

In order to construct this sequence and the diagram of the symmetric representation of antisigma(n) = A024816(n) we use the following rules:
At stage 1 in the first quadrant of the square grid we draw the symmetric representation of sigma(n) using the two Dyck paths described in the rows n and n-1 of A237593. The area of the region that is below the symmetric representation of sigma(n) equals A024916(n-1).
At stage 2 we draw a pair of orthogonal line segments (if it's necessary) such that in the drawing appears totally formed a square n X n. The area of the region that is above the symmetric representation of sigma(n) equals A004125(n). Then we draw a zig-zag path with line segments of length 1 from (0,n-1) to (n-1,0) such that appears a staircase with n-1 steps. The area of the region (or regions) that is below the symmetric representation of sigma(n) and above the staircase equals A244048(n) = A153485(n-1). The area of the region that is below the staircase equals A000217(n-1).
At stage 3 we turn OFF the cells of the symmetric representation of sigma(n) and also the cells that are below the staircase. Then we turn ON the rest of the cells that are in the square n X n. The result is that the ON cell form the diagram of the symmetric representation of antisigma(n) = A024816(n). See the Example section.
For n >= 7; if A237271(n) = 1 or n is a term of A262259 then a(n) = 2 otherwise a(n) = 1.

			Illustration of the symmetric representation of antisigma(n) = AS(n) = A024816(n), for n = 1..6:
.                                                             y|        _ _
.                                              y|      _ _     |  _ _  |_  |
.                                 y|      _     |  _ _|   |    | |_  |   |_|
.                      y|    _     |  _  |_|    | |_     _|    |   |_|_ _
.             y|        |  _|_|    | |_|_       |   |_  |      |     |_  |
.      y|      |        | |_|      |   |_|      |     |_|      |       |_|
.       |_ _   |_ _ _   |_ _ _ _   |_ _ _ _ _   |_ _ _ _ _ _   |_ _ _ _ _ _ _
.          x        x          x            x              x                x
.
n:        1       2         3           4             5               6
a(n):     0       0         2           3             1               3
AS(n):    0       0         2           3             9               9
.
Illustration of the symmetric representation of antisigma(n) = AS(n) = A024816(n), for n = 7..9:
.                                                y|          _ _ _ _
.                          y|          _ _ _      |  _ _ _ _|       |
.      y|        _ _ _      |  _ _ _  |     |     | |_       _ _    |
.       |  _ _ _|     |     | |_    | |_    |     |   |_    |_  |   |
.       | |_          |     |   |_  |_  |_ _|     |     |_    |_|  _|
.       |   |_       _|     |     |_  |_ _        |       |_      |
.       |     |_    |       |       |_    |       |         |_    |
.       |       |_  |       |         |_  |       |           |_  |
.       |         |_|       |           |_|       |             |_|
.       |_ _ _ _ _ _ _ _    |_ _ _ _ _ _ _ _ _    |_ _ _ _ _ _ _ _ _ _
.                      x                     x                       x
.
n:              7                    8                      9
a(n):           1                    2                      1
AS(n):         20                   21                     32
.
For n = 9 the figures 1, 2 and 3 below show respectively the three stages described in the Comments section as follows:
.
.   y|_ _ _ _ _ 5            y|_ _ _ _ _ _ _ _ _      y|          _ _ _ _
.    |_ _ _ _ _|              |_ _ _ _ _|       |      |  _ _ _ _|       |
.    |         |_ _ 3         | |_      |_ _ R  |      | |_       _ _    |
.    |         |_  |          |   |_    |_  |   |      |   |_    |_  |   |
.    |           |_|_ _ 5     |     |_ T  |_|_ _|      |     |_    |_|  _|
.    |               | |      |       |_      | |      |       |_      |
.    |      Q        | |      |         |_    | |      |         |_    |
.    |               | |      |    W      |_  | |      |           |_  |
.    |               | |      |             |_| |      |             |_|
.    |_ _ _ _ _ _ _ _|_|_     |_ _ _ _ _ _ _ _|_|_     |_ _ _ _ _ _ _ _ _ _
.                       x                        x                        x
.         Figure 1.                Figure 2.                Figure 3.
.         Symmetric                Symmetric                Symmetric
.       representation           representation           representation
.         of sigma(9)              of sigma(9)            of antisigma(9)
.       A000203(9) = 13          A000203(9) = 13          A024816(9) = 32
.           and of                   and of
.     Q = A024916(8) = 56      R = A004125(9) = 12
.                              T = A244048(9) = 20
.                              T = A153485(8) = 20
.                              W = A000217(8) = 36
.
Note that the symmetric representation of antisigma(9) contains a hole formed by three cells because these three cells were the central part of the symmetric representation of sigma(9).

Cf. A000203, A000217, A000290, A004125, A024816, A024916, A153485, A174973, A236104, A237270, A237271, A237593, A238443, A239660, A239931, A239932, A239933, A239934, A244048, A262259.

A050482 Sum of remainders when n-th prime is divided by all preceding integers.

Original entry on oeis.org

0, 1, 4, 8, 22, 28, 51, 64, 98, 151, 167, 233, 297, 325, 403, 505, 635, 645, 790, 904, 923, 1113, 1244, 1422, 1654, 1800, 1888, 2056, 2098, 2256, 2849, 3066, 3326, 3450, 3969, 4045, 4329, 4696, 5014, 5325, 5767, 5759, 6499, 6565, 6898

Offset: 1

Brian Wallace (wallacebrianedward(AT)yahoo.co.uk), Dec 26 1999

nonn

a(n)/(n*log(n))^2 appears to approach a constant ~0.22... for large n. - Benedict W. J. Irwin, Dec 07 2016

Irwin's comment is incorrect. - Bill McEachen, Feb 04 2024. [Indeed, according to the first formula in A004125, a(n)/(n*log(n))^2 approaches a constant, which is not 0.22 but 1-Pi^2/12 = 0.1775... - Amiram Eldar, Feb 04 2024]

			a(4) = 8 because remainders when 7 is divided by 1..6 are 0,1,1,3,2,1, which add to 8.
a(2) = 3 mod (3-1) = 1.
a(3) = (5 mod (5-1)) + (5 mod (5-2)) + (5 mod (5-3)) = 2 + 1 + 1 = 4.

Charles R Greathouse IV, Table of n, a(n) for n = 1..10000

Cf. A004125, A034953.

A050482 := proc(n) local a,i; a := 0; for i from 1 to ithprime(n)-1 do a := a+(ithprime(n) mod i); od: end;

Table[Sum[Mod[Prime[n],k],{k,Prime[n]-1}],{n,45}] (* James C. McMahon, Feb 08 2024 *)

a(n)=my(p=prime(n));sum(k=2, p, p%k) \\ Charles R Greathouse IV, Jun 03 2013

from math import isqrt
from sympy import prime
def A050482(n): return (p:=prime(n))**2+((s:=isqrt(p))**2*(s+1)-sum((q:=p//k)*((k<<1)+q+1) for k in range(1,s+1))>>1) # Chai Wah Wu, Nov 01 2023

a(n) = A004125(A000040(n)). - R. J. Mathar, Jun 12 2009

A099726 Sum of remainders of the n-th prime mod k, for k = 1,2,3,...,n.

Original entry on oeis.org

0, 1, 3, 5, 7, 7, 14, 18, 28, 30, 31, 26, 38, 45, 63, 71, 93, 75, 96, 115, 101, 142, 161, 167, 152, 159, 203, 224, 219, 222, 216, 250, 263, 296, 341, 320, 319, 349, 433, 427, 496, 419, 487, 481, 538, 537, 495, 631, 635, 676, 697, 777, 665, 820, 784, 874, 929, 856

Offset: 1

Joseph Biberstine (jrbibers(AT)indiana.edu), Nov 07 2004

			a(7)=14 because the 7th prime is 17 and its remainders modulo 1,2,3,4,5,6,7 are 0,1,2,1,2,5,3 respectively and 0+1+2+1+2+5+3=14.

Daniel Suteu, Table of n, a(n) for n = 1..10000

Cf. A000040, A004125.

umpf:=n->add(modp(floor(ithprime(n)),m),m=1..n); seq(umpf(k),k=1..120);

Table[Total[Mod[Prime[p],Range[p]]],{p,Range[60]}] (* Harvey P. Dale, Feb 09 2025 *)

a(n) = my(p=prime(n)); sum(k=1, n, p%k); \\ Daniel Suteu, Feb 02 2021

T(n) = n*(n+1)/2;
S(n) = my(s=sqrtint(n)); sum(k=1, s, T(n\k) + k*(n\k)) - s*T(s); \\ A024916
g(a,b) = my(s=0); while(a <= b, my(t=b\a); my(u=b\t); s += t*(T(u) - T(a-1)); a = u+1); s;
a(n) = my(p=prime(n)); n*p - S(p) + g(n+1, p); \\ Daniel Suteu, Feb 02 2021

a(n) = n*p - A024916(p) + Sum_{k=n+1..p} k*floor(p/k), where p = prime(n). - Daniel Suteu, Feb 02 2021

Definition corrected by Daniel Suteu, Feb 02 2021

A180491 Product of remainders of n mod k, for k = 2,3,4,...,n-1.

Original entry on oeis.org

1, 1, 1, 0, 2, 0, 6, 0, 0, 0, 720, 0, 2160, 0, 0, 0, 2419200, 0, 65318400, 0, 0, 0, 754427520000, 0, 0, 0, 0, 0, 32953394073600000, 0, 311409573995520000, 0, 0, 0, 0, 0, 37269497815783833600000, 0, 0, 0, 7890485108998805913600000000, 0

Offset: 1

Carl R. White, Sep 08 2010

nonn

a(n) is zero where n is composite and is trivially less than or equal to n! when n is prime or 1.

a(n)=0 iff n is composite. See A180492. - Robert G. Wilson v, Sep 09 2010

			a(7) = (7 mod 2) * (7 mod 3) * (7 mod 4) * (7 mod 5) * (7 mod 6) = 1 * 1 * 3 * 2 * 1 = 6.

Cf. A034386, A000142, A004125, A180492, A180493.

Cf. A080339, A173392.

a:=proc(n) if n=1 then 1; elif isprime(n)=true then mul(n mod i, i=2..n-1); else 0; fi: end: seq(a(n), n=1..60); # Ridouane Oudra, Nov 01 2024

f[n_] := Times @@ Mod[n, Range[2, n - 1]]; Array[f, 42] (* Robert G. Wilson v, Sep 09 2010 *)

a(n) = A080339(n)*A173392(n). - Ridouane Oudra, Nov 01 2024

A236630 Irregular triangle T(n,k) of alternating sums of squares of entries in the rows in the triangle of A235791, read by rows.

Original entry on oeis.org

1, 4, 9, 8, 16, 15, 25, 21, 36, 32, 33, 49, 40, 41, 64, 55, 56, 81, 65, 69, 100, 84, 88, 87, 121, 96, 100, 99, 144, 119, 128, 127, 169, 133, 142, 141, 196, 160, 169, 165, 225, 176, 192, 188, 189, 256, 207, 223, 219, 220, 289, 225, 241, 237, 238

Offset: 1

Omar E. Pol, Jan 29 2014

The original name was: Number of "ON" cells at n-th stage in a cellular automaton (or pseudo cellular automaton) related to sigma (see Comments for precise definition).

			Triangle begins:
    1;
    4;
    9,   8;
   16,  15;
   25,  21;
   36,  32,  33;
   49,  40,  41;
   64,  55,  56;
   81,  65,  69;
  100,  84,  88,  87;
  121,  96, 100,  99;
  144, 119, 128, 127;
  169, 133, 142, 141;
  196, 160, 169, 165;
  225, 176, 192, 188, 189;
  256, 207, 223, 219, 220;
  289, 225, 241, 237, 238;
  ...
From _Omar E. Pol_, Apr 20 2024: (Start)
Illustration of the 6th row as the area of a polygon (or the number of cells) in the fourth quadrant:
.     _ _ _ _ _ _       _ _ _ _ _ _       _ _ _ _ _ _
.    |           |     |           |     |           |
.    |           |     |           |     |           |
.    |           |     |           |     |           |
.    |           |     |        _ _|     |          _|
.    |           |     |       |         |        _|
.    |_ _ _ _ _ _|     |_ _ _ _|         |_ _ _ _|
.
.          36           36 - 4 = 32     36 - 4 + 1 = 33
.
(End)

Michael De Vlieger, Table of n, a(n) for n = 1..10075 (rows n = 1..500, flattened)
Hartmut F. W. Hoft, Proof that right border of triangle is A024916

Row n has length A003056(n).
The first element of column k is in row A000217(k).
Column 1 gives the positive terms of A000290.
Right border gives A024916.
Row n is the alternating sum of entries in row n of A236104.
Cf. A003056, A000203, A004125, A026914, A071561, A196020, A235791, A236631, A237048, A237591, A237593, A237270, A249223, A264116, A280851.

Map[Accumulate, Table[(-2 Boole[EvenQ[k]] + 1)*Ceiling[(n + 1)/k - (k + 1)/2]^2, {n, 20}, {k, Floor[(Sqrt[8*n + 1] - 1)/2]}]] // Flatten (* Michael De Vlieger, Apr 30 2024, after Hartmut F. W. Hoft at A235791 *)

From Hartmut F. W. Hoft, Apr 30 2024: (Start)
T(n, k) = Sum_{j = 1 .. k} (-1)^(j + 1) * S(n, j)^2, n >= 0, 1 <= k <= A003056(n), where S(n, j) is the j-th entry in the n-th row of the triangle of A235791.
T(n, k) = Sum_{j = 1 .. k} (-1)^(j+1) * S(n, j), n >= 0, 1 <= k <= A003056(n), where S(n, j) is the j-th entry in the n-th row of the triangle of A236104. (End)

New name from Hartmut F. W. Hoft, Apr 27 2024

0 removed, offset changed and minor edits from Omar E. Pol, Apr 28 2024

A236631 Triangle read by rows: T(j,k), j>=1, k>=1, in which column k lists the positive squares repeated k-1 times, except the column 1 which is A123327. The elements of the even-indexed columns are multiplied by -1. The first element of column k is in row k(k+1)/2.

Original entry on oeis.org

1, 3, 5, -1, 8, -1, 10, -4, 15, -4, 1, 16, -9, 1, 23, -9, 1, 25, -16, 4, 31, -16, 4, -1, 34, -25, 4, -1, 45, -25, 9, -1, 42, -36, 9, -1, 55, -36, 9, -4, 60, -49, 16, -4, 1, 67, -49, 16, -4, 1, 69, -64, 16, -4, 1, 86, -64, 25, -9, 1, 84, -81, 25, -9, 1, 103

Offset: 1

Omar E. Pol, Jan 29 2014

T(j,k) which row j has length A003056(j) hence the first element of column k is in row A000217(j).
Row sums give A000203.
Interpreted as a sequence with index n this is also the first differences of A236630. If a(n) is positive then a(n) is the number of cells turned ON at n-th stage in the structure of A236630. If a(n) is negative then a(n) is the number of cells turned OFF at n-th stage in the structure of A236630.

			Written as an irregular triangle the sequence begins:
1;
3;
5,     -1;
8,     -1;
10,    -4;
15,    -4,    1;
16,    -9,    1;
23,    -9,    1;
25,   -16,    4;
31,   -16,    4,   -1;
34,   -25,    4,   -1;
45,   -25,    9,   -1;
42,   -36,    9,   -1;
55,   -36,    9,   -4;
60,   -49,   16,   -4,   1;
67,   -49,   16,   -4,   1;
69,   -64,   16,   -4,   1;
86,   -64,   25,   -9,   1;
84,   -81,   25,   -9,   1;
103,  -81,   25,   -9,   4;
102, -100,   36,   -9,   4,  -1;
113, -100,   36,  -16,   4,  -1;
122, -121,   36,  -16,   4,  -1;
145, -121,   49,  -16,   4,  -1;
...
For j = 15 the divisors of 15 are 1, 3, 5, 15, therefore the sum of divisors of 15 is 1 + 3 + 5 + 15 = 24. On the other hand the 15th row of triangle is 60, -49, 16, -4, 1, therefore the row sum is 60 - 49 + 16 - 4 + 1 = 24, equalling the sum of divisors of 15.

Cf. A000203, A000217, A000290, A003056, A004125, A024916, A008794, A123327, A196020, A211547, A236104, A236630, A237593.

T(n,1) = A000203(n) + A004125(n).

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.

A256532 Product of n and the sum of remainders of n mod k, for k = 1, 2, 3, ..., n.

Original entry on oeis.org

0, 0, 3, 4, 20, 18, 56, 64, 108, 130, 242, 204, 364, 434, 540, 576, 867, 846, 1216, 1220, 1470, 1694, 2254, 2040, 2575, 2912, 3375, 3472, 4379, 4140, 5177, 5344, 6072, 6698, 7630, 7128, 8621, 9424, 10491, 10320, 12177, 11928, 13975, 14432, 15255, 16468, 18941, 17952, 20286, 21000, 22899, 23608, 26765, 26568, 29095

Offset: 1

Omar E. Pol, May 03 2015

nonn

a(n) is also the volume (or the total number of unit cubes) of a polycube which is a right prism whose base is the symmetric representation of A004125(n).

Note that the union of this right prism and the irregular staircase after n-th stage described in A244580 and the irregular stepped pyramid after (n-1)-th stage described in A245092, form a hexahedron (or cube) of side length n. This comment is represented by the third formula.

			a(5) = 20 because 5 * (0 + 1 + 2 + 1) = 5 * 4 = 20.
a(6) = 18 because 6 * (0 + 0 + 0 + 2 + 1) = 6 * 3 = 18.
a(7) = 56 because 7 * (0 + 1 + 1 + 3 + 2 + 1) = 7 * 8 = 56.

Indranil Ghosh, Table of n, a(n) for n = 1..10000

Cf. A000203, A000578, A004125, A024916, A143128, A175254, A236104, A236112, A237270, A237271, A237593, A244580, A245092, A256533.

Table[n*Sum[Mod[n,i],{i,2,n-1}],{n,55}] (* Ivan N. Ianakiev, May 04 2015 *)

vector(50, n, n*sum(k=1, n, n % k)) \\ Michel Marcus, May 05 2015

def A256532(n):
    s=0
    for k in range(1,n+1):
        s+=n%k
    return s*n # Indranil Ghosh, Feb 13 2017

from math import isqrt
def A256532(n): return n**3+n*((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 22 2023

a(n) = n * A004125(n).
a(n) = n^3 - A256533(n).
a(n) = n^3 - A143128(n) - A175254(n-1), n > 1.

cp's OEIS Frontend

A235799 a(n) = n^2 - sigma(n).

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A342344 Number of parts in the symmetric representation of antisigma(n).

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A050482 Sum of remainders when n-th prime is divided by all preceding integers.

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A099726 Sum of remainders of the n-th prime mod k, for k = 1,2,3,...,n.

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A180491 Product of remainders of n mod k, for k = 2,3,4,...,n-1.

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A236630 Irregular triangle T(n,k) of alternating sums of squares of entries in the rows in the triangle of A235791, read by rows.

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A236631 Triangle read by rows: T(j,k), j>=1, k>=1, in which column k lists the positive squares repeated k-1 times, except the column 1 which is A123327. The elements of the even-indexed columns are multiplied by -1. The first element of column k is in row k(k+1)/2.

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