A050226 -id:A050226 - cp's OEIS frontend

A073129 Partial sums of cototients arising in A063986.

0, 4, 5, 120, 125, 12201, 70800, 74657, 29526432, 37132574, 89210424, 124777688, 120392102955, 48617257062792, 343289046464505, 7519663359716376, 1214022599940709056, 40644839476190305216, 87437200646372849005, 98628693371623948080, 375871306587181970568

Offset: 1

Labos Elemer, Jul 16 2002

nonn

			n=15, a(15)=343289046464505, sum of first 41846733 cototients A063986(15)*A073128(15)=a(15) To continue A063986, A073128 or A073129 without recomputing previous terms, corresponding entries from 2 of above sequences is required.

Cf. A051953, A000010, A002088, A048920, A000217, A050226, A073128, A063986.

a(n)=Sum[cototient[j], j=1..A063986(n)]

Changed A063896 to A063986 and a(16)-a(21) from Donovan Johnson, May 11 2010

A218464 Numbers m = (Sum_(j=1..k) tau(j)) with m divisible by k, where tau(j) is the number of divisors of j.

Original entry on oeis.org

1, 8, 10, 45, 168, 176, 188, 605, 2016, 2040, 2082, 6510, 20384, 62433, 62523, 564542, 4928261, 4928703, 4928729, 42018075, 351871865, 1012753620, 1012755546, 2905896480, 2905898228, 192057921660, 1542529159875, 12309661243665, 12309661255437, 34700429419432

Offset: 1

Paolo P. Lava, Mar 26 2013

nonn

See A050226 for the values of k. - T. D. Noe, Mar 27 2013

			10 is in sequence because k=5 divides the sum of tau(1) + tau(2) + tau(3) + tau(4) + tau(5) = 1+2+2+3+2 = 10.

Giovanni Resta, Table of n, a(n) for n = 1..39 (based on A050226 values)

Cf. A000005, A168133.

Cf. A050226 (has the values of k).

with(numtheory);
A218464:=proc(q)  local n;  a:=0;
for n from 1 to q do a:=a+tau(n) if type(a/n,integer) then print(a); fi; od; end:
A218464 (10^10); # Paolo P. Lava, Mar 26 2013

sm = 0; t = {}; Do[sm = sm + DivisorSigma[0, n]; If[Mod[sm, n] == 0, AppendTo[t, sm]], {n, 1000}]; t (* T. D. Noe, Mar 27 2013 *)

a(22)-a(30) from Giovanni Resta, Mar 28 2013

A355544 Numbers k such that the arithmetic mean of the first k squarefree numbers is an integer.

Original entry on oeis.org

1, 3, 6, 37, 75, 668, 1075, 37732, 742767, 1811865, 3140083, 8937770, 108268896, 282951249, 633932500, 1275584757, 60455590365

Offset: 1

Amiram Eldar, Jul 06 2022

Numbers k such that A173143(k) is divisible by k.

The corresponding quotients A173143(k)/k are 1, 2, 4, 29, ..., and the corresponding values of A005117(k) are 1, 3, 7, 59, ... (see the link for more values).

			3 is a term since the arithmetic mean of the first 3 squarefree numbers, (1+2+3)/3 = 2, is an integer.

Amiram Eldar, Table of n, a(n), A173143(a(n))/a(n), A005117(a(n)) for n = 1..16.

Cf. A005117, A173143.

Similar sequences: A045345, A050226, A056550, A064605, A064606, A064607, A064610, A064611, A064612, A048290.

s = Select[Range[10^6], SquareFreeQ]; r = Accumulate[s]/Range[Length[s]]; ind = Position[r, _?IntegerQ] // Flatten

upto(n) = my(s=0,k=0); forsquarefree(m=1, n, s+=m[1]; k+=1; if(s%k == 0, print1(k, ", "))); \\ Daniel Suteu, Jul 06 2022

a(17) from Daniel Suteu, Jul 06 2022

A359028 Integers m such that A006218(m+1)/(m+1) > A006218(m)/m.

Original entry on oeis.org

1, 2, 3, 5, 7, 8, 9, 11, 13, 14, 15, 17, 19, 20, 21, 23, 25, 26, 27, 29, 31, 32, 33, 34, 35, 37, 38, 39, 41, 43, 44, 47, 49, 51, 53, 55, 59, 62, 63, 65, 67, 69, 71, 74, 75, 77, 79, 80, 83, 87, 89, 91, 95, 97, 98, 99, 101, 103, 104, 107, 109, 111, 113, 115, 116, 119, 123, 125, 127, 129

Offset: 1

Bernard Schott, Dec 12 2022

nonn

Equivalently: Indices m such that f(m + 1) > f(m) where f(m) = Sum_{k=1..m} d(k) / m, where d(k) is the number of divisors of k (A000005).
This sequence comes from a problem proposed by South Africa during the 47th International Mathematical Olympiad, in 2006 at Ljubljana, Slovenia, but not used for the competition (see link).
In fact, the problem asked for a proof that, for the sequence {f(m)} defined by f(m) = (1/m) * ([m/1] + [m/2] + ... + [m/m]), where [x] denotes the integer part of x,
(a) f(m + 1) > f(m) occurs infinitely often (these are the terms m of this sequence),
(b) f(m + 1) < f(m) occurs infinitely often (see A359029).
Differs from A047255 when a(24) = 34 while A047255(24) = 35.
Some results:
1. For every m, f(m) = (1/m) * ([m/1] + [m/2] + ... + [m/m]) proposed in the problem is the arithmetic mean of d(1), d(2), ..., d(m) = A006218(m)/m.
2. f(m + 1) > f(m) is equivalent to d(m + 1) > f(m).
3. Each m = c - 1, where c is a highly composite number (A002182) is a term.
Proof: in this case, d(m+1) = d(c) > max{d(1), ..., d(m)}; as f(m) = (d(1)+...+d(m)) / m < m*d(c)/m = d(c), it follows that d(m+1) = d(c) > f(m).
4. As there are infinitely many highly composite numbers, that also proves that f(m + 1) > f(m) occurs infinitely often, answer to IMO problem (a).
5. There exist other terms not of the form A002182 - 1: 2, 7, 8, 9, 13, 14, 15, ...
Note that f(m) = f(m+1) is possible iff f(m) = tau(m+1), so f(m) must be an integer (A050226) but this is not sufficient. The only known term such that f(m) = f(m+1) is at m=4, with f(4) = 2 and f(5) = tau(5) = 2.

			f(7) = (d(1)+d(2)+d(3)+d(4)+d(5)+d(6)+d(7)) / 7 = (1+2+2+3+2+4+2) / 7 = 16/7 < f(6) = (d(1)+d(2)+d(3)+d(4)+d(5)+d(6)) / 6 = (1+2+2+3+2+4) / 6 = 14/6 = 7/3, so 6 is a term.

47th International Mathematical Olympiad, Slovenia, 2006, Problem N3, page 57, Shortlisted problems with solutions.
Index to sequences related to Olympiads.

Cf. A002182, A006218, A047255, A050226, A359029.

with(numtheory):
for n from 1 to 100 do
m := (1/(n+1))*sum(tau(k),k=1..n+1) - (1/n)*sum(tau(k),k=1..n);
if m>0 then print(n); else fi; od:

With[{m = 130}, Position[Differences[Accumulate[DivisorSigma[0, Range[m]]]/Range[m]], ?(# > 0 &)] // Flatten] (* _Amiram Eldar, Dec 12 2022 *)

A359029 Integers m such that A006218(m+1)/(m+1) < A006218(m)/m.

Original entry on oeis.org

6, 10, 12, 16, 18, 22, 24, 28, 30, 36, 40, 42, 45, 46, 48, 50, 52, 54, 56, 57, 58, 60, 61, 64, 66, 68, 70, 72, 73, 76, 78, 81, 82, 84, 85, 86, 88, 90, 92, 93, 94, 96, 100, 102, 105, 106, 108, 110, 112, 114, 117, 118, 120, 121, 122, 124, 126, 128, 130, 132, 133, 136

Offset: 1

Bernard Schott, Dec 18 2022

nonn

Equivalently: Indices m such that f(m + 1) < f(m) where f(m) = Sum_{k=1..m} d(k) / m, where d(k) is the number of divisors of k (A000005).
This sequence comes from a problem proposed by South Africa during the 47th International Mathematical Olympiad, in 2006 at Ljubljana, Slovenia, but not used for the competition (see link).
In fact, the problem asked for a proof that, for the sequence {f(m)} defined by f(m) = (1/m) * ([m/1] + [m/2] + ... + [m/m]), where [x] denotes the integer part of x,
(a) f(m + 1) > f(m) occurs infinitely often (see A359028),
(b) f(m + 1) < f(m) occurs infinitely often (this sequence).
Some results:
1. For every m, f(m) = (1/m) * ([m/1] + [m/2] + ... + [m/m]) proposed in the problem is the arithmetic mean of d(1), d(2), ..., d(m) = A006218(m)/m.
2. f(m + 1) < f(m) is equivalent to d(m + 1) < f(m).
3. Each m = p - 1, p prime >= 7 is a term, so A006093 \ {1,2,4} is a subsequence.
Proof: in this case, d(m+1) = 2 < f(6) = 7/3. Since f(6) > 2, it follows that f(m) > 2 holds for all m >= 6; so for m = p - 1, p prime >= 7, d(m+1) = 2 < f(m) because when m >= 6, f(m) is > 2.
4. As there are infinitely many primes, that also proves that f(m + 1) < f(m) occurs infinitely often, which answers IMO problem (b).
5. There exist other terms not belonging to A006093: 24, 45, 48, 50, 54, ...
Note that f(m) = f(m+1) is possible iff f(m) = tau(m+1), so f(m) must be an integer (A050226) but this is not sufficient. The only known term such that f(m) = f(m+1) is at m=4, with f(4) = 2 and f(5) = tau(5) = 2.

			f(7) = (d(1)+d(2)+d(3)+d(4)+d(5)+d(6)+d(7)) / 7 = (1+2+2+3+2+4+2) / 7 = 16/7 < f(6) = (d(1)+d(2)+d(3)+d(4)+d(5)+d(6)) / 6 = (1+2+2+3+2+4) / 6 = 14/6 = 7/3, so 6 is a term.

47th International Mathematical Olympiad, Slovenia, 2006, Problem N3, page 57, Shortlisted problems with solutions.
Index to sequences related to Olympiads.

Cf. A006093, A006218, A050226, A359028.

with(numtheory):
for n from 1 to 150 do
m:= (1/(n+1))*sum(tau(k),k=1..n+1) - (1/n)*sum(tau(k),k=1..n);
if m<0 then print(n); else fi; od:

With[{m = 140}, Position[Differences[Accumulate[DivisorSigma[0, Range[m]]]/Range[m]], ?(# < 0 &)] // Flatten] (* _Amiram Eldar, Dec 18 2022 *)

f(n) = sum(k=1, n, n\k); \\ A006218
isok(m) = f(m+1)/(m+1) < f(m)/m; \\ Michel Marcus, Dec 19 2022

A362864 Numbers k that divide Sum_{i=1..k} (i - d(i)), where d(n) is the number of divisors of n (A000005).

Original entry on oeis.org

1, 2, 5, 8, 15, 24, 26, 47, 121, 204, 347, 562, 4204, 6937, 6947, 31108, 379097, 379131, 379133, 2801205, 12554202, 20698345, 56264197, 13767391064, 37423648626, 37423648726, 61701166395, 276525443156, 276525443176, 455913379395, 455913379831, 751674084802

Offset: 1

Ctibor O. Zizka, May 06 2023

nonn

Numbers k such that the mean number of nondivisors in the range 1..k is an integer.
Numbers k such that A161664(k) is divisible by k.
Numbers k such that (A000217(k) - A006218(k)) is divisible by k.
The subsequence of odd terms k equals the intersection of A050226 and this sequence.

			k = 5: Sum_{i=1..5} (i - d(i))/k = 5/5 = 1, so k = 5 is a term.

Martin Ehrenstein, Table of n, a(n) for n = 1..38

Cf. A000005, A000217, A006218, A049820, A050226, A161664.

seq[kmax_] := Module[{sum = 0, s = {}}, Do[sum += k - DivisorSigma[0, k]; If[Divisible[sum, k], AppendTo[s, k]], {k, 1, kmax}]; s]; seq[10^6] (* Amiram Eldar, May 06 2023 *)

isok(k) = !(sum(i=1, k, i - numdiv(i)) % k); \\ Michel Marcus, May 06 2023

from itertools import count, islice
from sympy import divisor_count
def A362864_gen(): # generator of terms
    c = 0
    for k in count(1):
        if not (c:=c+k-divisor_count(k))%k:
            yield k
A362864_list = list(islice(A362864_gen(),15)) # Chai Wah Wu, May 20 2023

More terms from Amiram Eldar, May 06 2023

a(24)-a(32) from Martin Ehrenstein, May 22 2023

A309272 Numbers m such that m divides A173290(m) = Sum_{k=1..m} psi(k), where psi is the Dedekind psi function (A001615).

Original entry on oeis.org

1, 2, 5, 15, 31, 40, 66, 81, 315, 966, 1398, 1768, 30166, 32335, 98734, 388033, 591597, 1375056, 14966304, 15160528, 50793208, 51302236, 99253376, 110994356, 230465053, 402340268, 497982399, 2027319577, 2879855394, 18450762682, 29922126368, 31711273834, 40583934786

Offset: 1

Amiram Eldar, Oct 23 2019

nonn

The corresponding quotients are 1, 2, 4, 12, 24, 31, 51, 62, 240, 735, 1063, 1344, 22924, 24572, 75029, 294870, 449560, 1044918, 11373028, 11520620, 38598210, 38985025, 75423522, 84345597, 175132440, 305741942, 378421246, 1540578144, 2188427680, 14020898356, 22738089456, 24097678498, 30840092321, ...

			2 is in the sequence since psi(1) + psi(2) = 1 + 3 = 4 is divisible by 2.
5 is in the sequence since psi(1) + psi(2) + ... + psi(5) = 1 + 3 + 4 + 6 + 6 = 20 is divisible by 5.

Cf. A001615, A173290.

Cf. A045345, A048290, A050226, A056550, A063971, A064605, A064606, A064607, A064610, A064611, A064612, A306950, A307043, A307161, A326488.

psi[1] = 1; psi[n_] := n * Times @@ (1 + 1/Transpose[FactorInteger[n]][[1]]); seq = {}; s = 0; Do[s += psi[n]; If[Divisible[s, n], AppendTo[seq, n]], {n, 1, 10^4}]; seq

a(31)-a(33) from Giovanni Resta, Oct 24 2019

A339009 Numbers k such that the average number of odd divisors of {1..k} is an integer.

Original entry on oeis.org

1, 2, 165, 170, 1274, 9437, 69720, 69732, 69734, 69736, 515230, 515236, 515246, 28132043, 28132063, 28132079

Offset: 1

Ilya Gutkovskiy, Nov 18 2020

Numbers k that divide A060831(k) where A060831(k) = Sum_{j=1..k} A001227(j).

The sequence also includes: 83860580242, 4578632504347, 4578632504465, 4578632504515. - Daniel Suteu, Nov 24 2020

			165 is in the sequence because the average number of odd divisors of {1..165} is an integer: A060831(165) / 165 = 495 / 165 = 3.

Eric Weisstein's World of Mathematics, Odd Divisor Function.

Cf. A001227, A048290, A050226, A060831, A063971, A064610, A064612.

s[n_] := Module[{c = 0, k = 1, sum = 0, seq = {}}, While[c < n, sum += DivisorSigma[0, k/2^IntegerExponent[k, 2]]; If[Divisible[sum, k], c++; AppendTo[seq, k]]; k++]; seq]; s[13] (* Amiram Eldar, Nov 18 2020 *)

f(n) = my(n2=n\2); sum(k=1, sqrtint(n), n\k)*2-sqrtint(n)^2-sum(k=1, sqrtint(n2), n2\k)*2+sqrtint(n2)^2; \\ A060831
isok(k) = (f(k) % k) == 0; \\ Michel Marcus, Nov 25 2020

A355541 Numbers k such that A061201(k) is divisible by k.

Original entry on oeis.org

1, 2, 7, 31, 1393, 5012, 7649, 50235, 147296, 426606, 611769, 3491681, 9324642, 11815109, 53962364, 82680301, 96789197, 230882246, 378444764, 1489280093, 1489280606, 3651325650, 5891877914, 5891877947, 5891877966, 58604540872

Offset: 1

Amiram Eldar, Jul 06 2022

Numbers k such that the mean value of A007425 over the range 1..k is an integer.
The corresponding quotients are 1, 2, 4, 9, 32, 43, 47, 67, 80, 94, 99, 125, 141, 145, 172, 180, 183, 200, 210, 239, 239, 259, 270, 270, 270, 326, ... .
a(27) > 7.5*10^10, if it exists.

			7 is a term since A061201(7) = 28 = 4 * 7 is divisible by 7.

Cf. A007425, A061201.

Similar sequences: A045345, A050226, A056550, A064605, A064606, A064607, A064610, A064611, A064612, A048290.

f[p_, e_] := (e+1)*(e+2)/2;  d3[1] = 1; d3[n_] := Times @@ f @@@ FactorInteger[n]; sum = 0; seq = {}; Do[sum += d3[n]; If[Divisible[sum, n], AppendTo[seq, n]], {n, 1, 10^6}]; seq

A355542 Numbers k such that A272718(k) is divisible by k.

Original entry on oeis.org

1, 2, 3, 11, 13, 50, 81, 96, 395, 640, 59136, 65719, 632621, 1342813, 2137073, 2755370, 3446370, 10860093, 321939569, 1872591111, 8858043355

Offset: 1

Amiram Eldar, Jul 06 2022

Numbers k such that the mean value of A018804 over the range 1..k is an integer.
The corresponding quotients are 1, 2, 3, 13, 16, 80, 141, 172, 865, 1500, 219530, 246058, 2804048, 6259092, 10263121, 13445321, 17051542, 57521176, 2036840289, 12849666590, 64967828053, ... .
a(22) > 6.5*10^10, if it exists.

			11 is a term since A061201(11) = 143 = 11 * 13 is divisible by 11.

Cf. A018804, A272718.

Similar sequences: A045345, A050226, A056550, A064605, A064606, A064607, A064610, A064611, A064612, A048290.

f[p_,e_] := (e*(p-1)/p+1)*p^e; pillai[1] = 1; pillai[n_] := Times @@ f @@@ FactorInteger[n]; seq = {}; sum = 0; Do[sum += pillai[n]; If[Divisible[sum, n], AppendTo[seq, n]], {n, 1, 10^6}]; seq

cp's OEIS Frontend

A073129 Partial sums of cototients arising in A063986.

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A218464 Numbers m = (Sum_(j=1..k) tau(j)) with m divisible by k, where tau(j) is the number of divisors of j.

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A355544 Numbers k such that the arithmetic mean of the first k squarefree numbers is an integer.

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A359028 Integers m such that A006218(m+1)/(m+1) > A006218(m)/m.

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A359029 Integers m such that A006218(m+1)/(m+1) < A006218(m)/m.

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A362864 Numbers k that divide Sum_{i=1..k} (i - d(i)), where d(n) is the number of divisors of n (A000005).

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A309272 Numbers m such that m divides A173290(m) = Sum_{k=1..m} psi(k), where psi is the Dedekind psi function (A001615).

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A339009 Numbers k such that the average number of odd divisors of {1..k} is an integer.