A168133 -id:A168133 - cp's OEIS frontend

A056550 Numbers k such that Sum_{j=1..k} sigma(j) is divisible by k, where sigma(j) = sum of divisors of j (A000203).

1, 2, 8, 11, 17, 63, 180, 259, 818, 2161, 4441, 8305, 11998, 694218, 3447076, 4393603, 57402883, 73459800, 121475393, 2068420025, 5577330586, 13320495021, 35297649260, 138630178659, 988671518737, 1424539472772, 3028785109162, 13702718147734, 21320824383487

Offset: 1

Asher Auel, Jun 06 2000

			a(3) = 8 is in the sequence because A024916(8) / 8 = 56 / 8 = 7 is an integer. [_Jaroslav Krizek_, Dec 07 2009]

Cf. A000203, A024916, A168133, A168132.

f := []: for i from 1 to 9000 do if add(sigma(n), n=1..i) mod i = 0 then f := [op(f),i] fi; od; f;

k=10^4;a[1]=1;a[n_]:=a[n]=DivisorSigma[1,n]+a[n-1]; s=a/@Range@k;Select[Range@k,Divisible[s[[#]],#]&] (* Ivan N. Ianakiev, Apr 30 2016 *)
Module[{nn=44*10^5,ds},ds=Accumulate[DivisorSigma[1,Range[nn]]];Select[ Thread[{ds,Range[nn]}],Divisible[#[[1]],#[[2]]]&]][[All,2]] (* The program generates the first 16 terms of the sequence. To generate more, increase the value of nn. *) (* Harvey P. Dale, Dec 04 2018 *)

is(n)=sum(k=1,n,n\k*k)%n==0 \\ Charles R Greathouse IV, Feb 14 2013

Values of k for which A024916(k)/k is an integer.

More terms from Jud McCranie, Jul 04 2000
a(19)-a(24) from Donovan Johnson, Dec 29 2008
a(25) from Donovan Johnson, Jun 16 2011
a(26) from Jud McCranie, Dec 17 2024
a(27) from Jud McCranie, Dec 22 2024
a(28) from Jud McCranie, Apr 03 2025
a(29) from Jud McCranie, May 04 2025

A168132 Numbers m = Sum_{k=1..n} sigma(k)/k such that Sum_{k=1..n} sigma(k)/k is an integer for any k.

Original entry on oeis.org

1, 2, 7, 9, 14, 52, 149, 213, 673, 1778, 3653, 6831, 9868, 570972, 2835107, 3613594, 47211979, 60418265, 99909506, 1701207282, 4587170542, 10955668024, 29031152874, 114018751785, 813149731047

Offset: 1

Jaroslav Krizek, Nov 18 2009, Dec 04 2009

Numbers m = A024916(k)/k such that A024916(k)/k is an integer for any k. If a(14) exist must be bigger than 82000.

Corresponding values of k, and Sum_{k=1..n} sigma(k) are given in A056550 and A168133. [Jaroslav Krizek, Nov 21 2009]

			Number a(3) = 7 is in sequence because A024916(8)/8 = 56/8 = 7 is an integer for k = 8.

a(14)-a(25) from Donovan Johnson, Jun 16 2011

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

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A056550 Numbers k such that Sum_{j=1..k} sigma(j) is divisible by k, where sigma(j) = sum of divisors of j (A000203).

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A168132 Numbers m = Sum_{k=1..n} sigma(k)/k such that Sum_{k=1..n} sigma(k)/k is an integer for any k.

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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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