A004125 -id:A004125 - cp's OEIS frontend

A309176 a(n) = n^2 * (n + 1)/2 - Sum_{k=1..n} sigma_2(k).

0, 0, 2, 3, 12, 13, 33, 40, 66, 81, 135, 135, 212, 249, 319, 354, 489, 511, 681, 725, 876, 981, 1233, 1235, 1509, 1660, 1920, 2032, 2437, 2472, 2936, 3091, 3488, 3755, 4275, 4290, 4955, 5292, 5854, 6024, 6843, 6968, 7870, 8190, 8839, 9340, 10420, 10442, 11568, 12038, 13014, 13474, 14851, 15098, 16436

Offset: 1

Ilya Gutkovskiy, Jul 15 2019

nonn

Amiram Eldar, Table of n, a(n) for n = 1..10000

Cf. A000326, A001157, A004125, A048158, A051126, A154585, A256532.

Cf. A002411, A064602.

Table[n^2 (n + 1)/2 - Sum[DivisorSigma[2, k], {k, 1, n}], {n, 1, 55}]
nmax = 55; CoefficientList[Series[x (1 + 2 x)/(1 - x)^4 - 1/(1 - x) Sum[k^2 x^k/(1 - x^k), {k, 1, nmax}], {x, 0, nmax}], x] // Rest
Table[Sum[Mod[n, k] k, {k, 1, n}], {n, 1, 55}]

a(n) = n^2*(n+1)/2 - sum(k=1, n, sigma(k, 2)); \\ Michel Marcus, Sep 18 2021

from math import isqrt
def A309176(n): return (n**2*(n+1)>>1)+((s:=isqrt(n))**2*(s+1)*(2*s+1)-sum((q:=n//k)*(6*k**2+q*(2*q+3)+1) for k in range(1,s+1)))//6 # Chai Wah Wu, Oct 21 2023

G.f.: x * (1 + 2*x)/(1 - x)^4 - (1/(1 - x)) * Sum_{k>=1} k^2 * x^k/(1 - x^k).
a(n) = Sum_{k=1..n} (n mod k) * k.
a(n) = A002411(n) - A064602(n).

A354801 n^2 minus the sum of all aliquot divisors of all positive integers <= n.

Original entry on oeis.org

1, 3, 7, 11, 19, 24, 36, 44, 57, 68, 88, 95, 119, 136, 156, 172, 204, 218, 254, 271, 301, 330, 374, 385, 428, 463, 503, 530, 586, 603, 663, 695, 745, 792, 848, 864, 936, 989, 1049, 1078, 1158, 1187, 1271, 1318, 1374, 1439, 1531, 1550, 1639, 1695, 1775, 1832, 1936, 1977, 2069, 2116

Offset: 1

Omar E. Pol, Jun 06 2022

After the Dyck paths described in A237593 we can see that a(n) has a symmetric representation as follows: a(n) is the sum of the areas of two polygons. In the fourth quadrant of the infinite square grid the first polygon has a vertex at (0,0) and its area is equal to A000217(n). The second polygon appears if n >= 3 and it has a vertex at (n,-n) and its area is equal to A004125(n). So the area of the arrowhead-shaped polygon is equal to A153485(n). See the illustration of initial terms in the Links section.

Omar E. Pol, Illustration of initial terms

Partial sums of A353190.

Cf. A000217, A000290, A001065, A004125, A024916, A153485, A236104, A237591, A237593.

Accumulate[Table[3*n - 1 - DivisorSigma[1, n], {n, 1, 60}]] (* Amiram Eldar, Jun 12 2022 *)

a(n) = n^2 - sum(k=1, n, sigma(k)-k); \\ Michel Marcus, Jun 13 2022

from math import isqrt
def A354801(n): return n*(3*n+1)+(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) = A000217(n) + A004125(n).
a(n) = A000290(n) - A153485(n).
a(n) = A024916(n) + A004125(n) - A153485(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.

A362082 Numbers k achieving record deficiency via a residue-based measure, M(k) = (k+1)*(1 - zeta(2)/2) - 1 - ( Sum_{j=1..k} k mod j )/k.

Original entry on oeis.org

1, 5, 11, 23, 47, 59, 167, 179, 359, 503, 719, 1439, 5039, 6719, 7559, 15119, 20159, 52919, 75599, 83159, 166319, 415799, 720719, 831599, 1081079, 2162159, 4324319, 5266799, 7900199, 10533599, 18345599, 28274399, 41081039, 136936799, 205405199, 410810399

Offset: 1

Richard Joseph Boland, Apr 17 2023

nonn

M(k) = (k+1)*(1 - zeta(2)/2) - 1 - ( Sum_{j=1..k} k mod j )/k is a measure of either abundance (sigma(k) > 2*k), or deficiency (sigma(k) < 2*k), of a positive integer k. The measure 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 deficient k for which M(k) < M(m) for all m < k and may be thought of as "superdeficient", contra-analogous to the superabundant numbers A004394 utilizing sigma(k)/k as the measure of abundance, which is otherwise not particularly meaningful as a deficiency measure.
15119=13*1163 is the first term that is composite and subsequently, up to 1000000000, roughly half of the terms are composite.

			First few terms with their M(k) measure and factorizations as generated by the Mathematica program:
    1   -0.64493406684822643647   {{1,1}}
    5   -0.73480220054467930942   {{5,1}}
   11   -0.86960440108935861883  {{11,1}}
   23   -1.0000783673961085420   {{23,1}}
   47   -1.0528856894638174541   {{47,1}}
   59   -1.1107338698535727552   {{59,1}}
  167   -1.1984137110594038972  {{167,1}}
  179   -1.2619431113124463216  {{179,1}}
  359   -1.3499704727921791778  {{359,1}}
  503   -1.3722914063892448936  {{503,1}}
  719   -1.4363475145965658088  {{719,1}}

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. A362081 (analogous to superabundant A004394).
Cf. A362083 (analogous to A335067, A326393).
Cf. A004490, A002201, A005100, A005101, A004125, A024916, A000290, A120444, A235796, A000396, A000079.

Clear[min, Rp, R, seqtable, M]; min = 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 < min, min = M; Print[k, "   ", min, "   ", 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.

A362083 Numbers k such that, via a residue based measure M(k) (see Comments), k is deficient, k+1 is abundant, and abs(M(k)) + abs(M(k+1)) reaches a new maximum.

Original entry on oeis.org

11, 17, 19, 47, 53, 103, 347, 349, 557, 1663, 1679, 2519, 5039, 10079, 15119, 25199, 27719, 55439, 110879, 166319, 277199, 332639, 554399, 665279, 720719, 1441439, 2162159, 3603599, 4324319, 7207199, 8648639, 10810799, 21621599, 36756719, 61261199, 73513439, 122522399, 147026879

Offset: 1

Richard Joseph Boland, Apr 17 2023

nonn

The residue-based quantifier function, M(k), measures either abundance (sigma(k) > 2*k), or deficiency (sigma(k) < 2*k), of a positive integer k. The measure is defined by M(k) = (k+1)*(1 - zeta(2)/2) - 1 - (Sum_{j=1..k} k mod j)/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 deficient k such that k+1 is abundant and abs(M(k)) + abs(M(k+1)) achieves a new maximum, somewhat analogous to A335067 and A326393.

			The first few terms with measure sums and factorizations generated by the Mathematica program:
0.90610439514731535319   35  {{5,1},{7,1}}   36   {{2,2},{3,2}}
1.1735781643159997761    59  {{59,1}}        60   {{2,2},{3,1},{5,1}}
1.3642976724582397229   119  {{7,1},{17,1}} 120   {{2,3},{3,1},{5,1}}
1.3954100615479538209   179  {{179,1}}      180   {{2,2},{3,2},{5,1}}
1.4600817810807682323   239  {{239,1}}      240   {{2,4},{3,1},{5,1}}
1.6088158511317518390   359  {{359,1}}      360   {{2,3},{3,2},{5,1}}
1.7153941935887132383   719  {{719,1}}      720   {{2,4},{3,2},{5,1}}
1.7851979872921589879   839  {{839,1}}      840   {{2,3},{3,1},{5,1},{7,1}}

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. A362081 (analogous to superabundant A004394), A362082 (superdeficient).

Cf. A335067, A326393, A004490, A002201, A326393, A005100, A005101, A004125, A024916, A000290, A120444, A235796, A000396, A000079.

Clear[max, Rp, R, seqtable, Mp, M];max = -1; Rp = 0; Mp = -0.644934066; seqtable = {};
Do[R = Rp + 2 k - 1 - DivisorSigma[1, k];
 M = N[(k)*(1 - Zeta[2]/2) - 1  - R/k, 20];
 If[DivisorSigma[1, k - 1] < 2 (k - 1) && DivisorSigma[1, k] > 2 k &&
   Abs[Mp] + Abs[M] > max, max = Abs[Mp] + Abs[M];
  Print[max, "   ", k - 1, "   ", FactorInteger[k - 1], "   ", k,
   "   ", FactorInteger[k]]; AppendTo[seqtable, {k - 1, k}]]; Rp = R;
 Mp = M, {k, 2, 1000000000}]; seq = Flatten[seqtable]; Table[seq[[2 j - 1]], {j, 1, Length[seq]/2}]

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.

A154586 Numbers n for which abs((-1)^k*Sum_{k=1..n} ((n-k+1) mod k)) = 0.

Original entry on oeis.org

1, 4, 8, 25, 27, 75, 209, 3507, 8466, 16179, 29285, 33987, 175904, 326764, 1161207

Offset: 1

Paolo P. Lava and Giorgio Balzarotti, Jan 12 2009

Subset of A154585.

a(16) > 10^7. - Donovan Johnson, Oct 03 2011

			n=8 -> abs(-(8 mod 1) + (7 mod 2) - (6 mod 3) + (5 mod 4) - (4 mod 5) + (3 mod 6) - (2 mod 7) + (1 mod 8)) = abs(-0 + 1 - 0 + 1 - 4 + 3 - 2 + 1) = abs(0) = 0.

Cf. A004125, A154585.

#include  int main(int argc, char * argv[]) { for(int n=1;;n++) { unsigned long long a = 0; for(int k=1;k <=n;k += 2) a -= (n-k+1) % k ; for(int k=2;k <=n;k += 2) a += (n-k+1) % k ; if ( a == 0) printf("%d,\n",n) ; } } /* R. J. Mathar, Jan 14 2009 */

P:=proc(i) local a,n; for n from 1 to i do a:=abs(add((-1)^k*((n-k+1) mod k),k=1..n)); if a=0 then print(n); fi; od; end: P(100);

abs{(-1)^k*A004125} = 0

{a(n): A154585(a(n))=0}. - R. J. Mathar, Jan 14 2009

8466 inserted, and sequence extended up to a(13), by R. J. Mathar, Jan 14 2009

a(14)-a(15) from Donovan Johnson, Oct 03 2011

A166248 a(n) is the absolute value of n minus sum of all the remainders modulo the numbers below n.

Original entry on oeis.org

1, 2, 2, 3, 1, 3, 1, 0, 3, 3, 11, 5, 15, 17, 21, 20, 34, 29, 45, 41, 49, 55, 75, 61, 78, 86, 98, 96, 122, 108, 136, 135, 151, 163, 183, 162, 196, 210, 230, 218, 256, 242, 282, 284, 294, 312, 356, 326, 365, 370, 398, 402, 452, 438, 474, 464, 496, 520, 576, 526, 584, 610

Offset: 1

Juri-Stepan Gerasimov, Oct 10 2009

nonn

			a(1) = abs(1-0) = 1;
a(2) = abs(2-0) = 2;
a(3) = abs(3-1) = 2;
a(4) = abs(4-1) = 3;
a(5) = abs(5-4) = 1;
a(6) = abs(6-3) = 3;
a(7) = abs(7-8) = 1.

Cf. A000027, A004125, A163180.

A004125 := proc(n) add( modp(n,k),k=1..n) ; end proc: A166248 := proc(n) abs(n-A004125(n)) ; end: seq(A166248(n),n=1..100) ; # R. J. Mathar, Oct 24 2009

from math import isqrt
def A166248(n): return abs(n*(n-1)+((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, Nov 01 2023

a(n) = abs(n - Sum_{k=1..n} (n mod k)).

a(n) = abs(n - A004125(n)). - Michel Marcus, May 08 2019

a(19), a(20), a(37) etc. corrected by R. J. Mathar, Oct 24 2009

A176314 Sum of remainders of mod(n, k), for 1 <= k <= sqrt(n).

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 3, 0, 2, 2, 1, 1, 4, 2, 5, 2, 2, 3, 6, 0, 3, 5, 6, 4, 8, 2, 6, 4, 5, 7, 6, 1, 6, 9, 11, 5, 10, 4, 9, 8, 5, 8, 13, 3, 8, 7, 10, 10, 16, 11, 12, 5, 8, 12, 18, 4, 10, 14, 10, 10, 12, 8, 15, 16, 20, 13, 20, 4, 11, 16, 15, 16, 16, 12, 19, 7, 11, 17, 25, 11, 14, 20, 25

Offset: 1

Franklin T. Adams-Watters, Apr 15 2010

nonn

It appears, as one would expect, that a(n) is asymptotically 1/4 n.

24 is the last n for which a(n) = 0.

Cf. A004125, A070039.

Total/@Table[Mod[n,k],{n,90},{k,Sqrt[n]}] (* Harvey P. Dale, Nov 20 2013 *)

PARI
```
a(n) = sum(k=2,sqrtint(n),n%k)
```

a(n+1) = a(n) + floor(sqrt(n)) - A070039(n+1).

A244409 Numbers x such that it is possible to find a value k for which Sum_{j=1..x} (x mod j) = Sum_{j=1..k} j.

Original entry on oeis.org

3, 4, 6, 13, 15, 16, 43, 112, 278, 346, 527, 845, 1214, 1612, 2189, 2863, 10278, 610410, 981350, 2054106, 3286515, 3764767, 4293562, 5543363, 5728393, 20142483, 66790186, 67652048, 72730730, 137252581, 198373964, 338557754, 406463074, 687452210, 911028356

Offset: 1

Paolo P. Lava, Jun 27 2014

nonn

a(38) > 2 * 10^9. - Hiroaki Yamanouchi, Sep 29 2014

			If x = 6 we have 6 mod 1 + 6 mod 2 + 6 mod 3 + 6 mod 4 + 6 mod 5 + 6 mod 6 = 0 + 0 + 0 + 2 + 1 + 0 = 3 and 1 + 2 = 3 (k = 2).
If x = 15 we have 15 mod 1 + 15 mod 2 + ... + 15 mod 14 + 15 mod 15 = 0 + 1 + 0 + 3 + 0 + 3 + 1 + 7 + 6 + 5 + 4 + 3 + 2 + 1 + 0 = 36 = 1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 (k = 8).

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..37

Cf. A000217, A004125.

with(numtheory); P:=proc(q)local a,b,c,k,n;
for n from 1 to q do a:=add(n mod k,k=1..n); b:=n; c:=0;
while c<=a do if c=a then lprint(n,b); break; else b:=b+1;
c:=c+(b mod n); fi; od: od; end: P(10^9);

A004125(x) = A000217(k).

a(18)-a(35) from Hiroaki Yamanouchi, Sep 29 2014

A244912 Sum of leading digits in representations of n in bases 2,3,...,n.

Original entry on oeis.org

1, 2, 3, 4, 6, 7, 9, 9, 11, 12, 15, 16, 18, 20, 20, 21, 25, 26, 29, 31, 33, 34, 38, 36, 38, 39, 42, 43, 47, 48, 52, 54, 56, 58, 58, 59, 61, 63, 67, 68, 72, 73, 76, 79, 81, 82, 88, 84, 88, 90, 93, 94, 99, 101, 105, 107, 109, 110, 116, 117, 119, 122, 117, 119, 123

Offset: 2

Alex Ratushnyak, Jul 08 2014

			8 in bases 2...8 is:
  1000 (base 2)
  22 (base 3)
  20 (base 4)
  13 (base 5)
  12 (base 6)
  11 (base 7)
  10 (base 8)
The sum of first digits is 1+2+2+1+1+1+1 = 9, so a(8)=9.

Jens Kruse Andersen, Table of n, a(n) for n = 2..1000

Cf. A004125 (sum of last digits), A043306 (sum of all digits).

f[n_] := Sum[ IntegerDigits[n, k][[1]], {k, 2, n}]; Array[f, 70, 2] (* Robert G. Wilson v, Aug 02 2014 *)

a(n) = sum(i=2, n, digits(n, i)[1]); \\ Michel Marcus, Jul 17 2014

import math
def modlg(a, b):
    return a // b**int(math.log(a, b))
for n in range(2,77):
    s=0
    for k in range(2,n+1):
        s += modlg(n,k)
    print(s, end=', ')

a(n) = Sum_{k=2..n} floor(n/f(n,k)), with f(n,k) = k^floor(log_k(n)). - Ridouane Oudra, Apr 26 2025

cp's OEIS Frontend

A309176 a(n) = n^2 * (n + 1)/2 - Sum_{k=1..n} sigma_2(k).

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A354801 n^2 minus the sum of all aliquot divisors of all positive integers <= n.

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

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Mathematica

PARI

Formula

A362082 Numbers k achieving record deficiency via a residue-based measure, M(k) = (k+1)*(1 - zeta(2)/2) - 1 - ( Sum_{j=1..k} k mod j )/k.

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Mathematica

PARI

Formula

A362083 Numbers k such that, via a residue based measure M(k) (see Comments), k is deficient, k+1 is abundant, and abs(M(k)) + abs(M(k+1)) reaches a new maximum.

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Mathematica

Formula

A154586 Numbers n for which abs((-1)^k*Sum_{k=1..n} ((n-k+1) mod k)) = 0.

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C

Maple

Formula

Extensions

A166248 a(n) is the absolute value of n minus sum of all the remainders modulo the numbers below n.

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Maple

Python