A203617 -id:A203617 - cp's OEIS frontend

A342922 Numbers k such that A342925(k) = k + 2*A003415(k).

Original entry on oeis.org

6, 28, 29, 496, 857, 1721, 8128, 164284, 6511664, 33550336, 400902412, 8589869056

Offset: 1

Antti Karttunen, Apr 07 2021

Question: Are all odd terms in A001359?

Certainly any prime p such that A003415(p+1) = p + 2 satisfies the equation. See comments in A007850.

Index entries for sequences where odd perfect numbers must occur, if they exist at all

Cf. A000396 (subsequence), A001359, A003415, A007850.

Cf. A203617, A342923, A342924, A342925.

Select[Range[2*10^5], #3 == #1 + 2 #2 & @@ Prepend[Map[If[# < 2, 0, # Total[#2/#1 & @@@ FactorInteger[#]]] &, {#, DivisorSigma[1, #]}], #] &] (* Michael De Vlieger, Feb 25 2022 *)

A003415(n) = if(n<=1, 0, my(f=factor(n)); n*sum(i=1, #f~, f[i, 2]/f[i, 1]));
A342925(n) = A003415(sigma(n));
isA342922(n) = (A342925(n)==(n+(2*A003415(n))));

Terms a(11) and a(12) from Antti Karttunen, Feb 25 2022

A203618 Numbers m such that (m'+1)' = m-1, where m' is the arithmetic derivative of m.

Original entry on oeis.org

1, 2, 6, 42, 104, 120, 165, 245, 272, 561, 1806, 47058, 765625, 1137501, 3874128, 9131793, 2214502422, 52495396602

Offset: 1

Paolo P. Lava, Jan 20 2012

The differential equation whose solutions are the primary pseudoperfect numbers is m' = k*m-1, with k a positive integer. Let us rewrite the equation as m'+1 = k*m and then take the derivative: (m'+1)' = (k*m)' = k'*m + k*m' = k'*m + k*(k*m-1) = (k'+k^2)*m-k. Let k=1: (m'+1)' = m-1. The solutions of this equation are the primary pseudoperfect numbers plus pairs of numbers (x,y) for which x' = y-1 and y' = x-1.
A054377 is a subsequence of this sequence.
a(17) > 10^9. - Michel Marcus, Nov 05 2014
a(19) > 10^11. - Giovanni Resta, Jun 04 2016

			765625' = 1137500; (1137500 + 1)' = 1137501' = 765624 = 765625 - 1, so 765625 is a term.
1137501' = 765624; (765624 + 1)' = 765625' = 1137500 = 1137501 - 1, so 1137501 is a term.

Cf. A054377, A203617.

with(numtheory);
P:=proc(i)
local a,n,p,pfs;
for n from 1 to i do
  pfs:=ifactors(n)[2]; a:=n*add(op(2,p)/op(1,p),p=pfs);
  pfs:=ifactors(a+1)[2]; a:=(a+1)*add(op(2,p)/op(1,p),p=pfs);
  if a=n-1 then print(n); fi;
od;
end:
P(10000000);

A003415[n_]:=If[Abs[n]<2,0,n*Total[#2/#1&@@@FactorInteger[Abs[n]]]];
Select[Range[1,100000],A003415[A003415[#]+1]==#-1&] (* Julien Kluge, Jul 08 2016 *)

ad(n) = sum(i=1, #f=factor(n)~, n/f[1, i]*f[2, i]);
isok(n) = ad(ad(n)+1) == n-1; \\ Michel Marcus, Nov 05 2014

a(17)-a(18) from Giovanni Resta, Jun 04 2016

cp's OEIS Frontend

A342922 Numbers k such that A342925(k) = k + 2*A003415(k).

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