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A342922
Numbers k such that A342925(k) = k + 2*A003415(k).
7
6, 28, 29, 496, 857, 1721, 8128, 164284, 6511664, 33550336, 400902412, 8589869056
OFFSET
1,1
COMMENTS
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.
MATHEMATICA
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 *)
PROG
(PARI)
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))));
CROSSREFS
Cf. A000396 (subsequence), A001359, A003415, A007850.
Sequence in context: A379500 A261868 A379499 * A357462 A105402 A362805
KEYWORD
nonn,more
AUTHOR
Antti Karttunen, Apr 07 2021
EXTENSIONS
Terms a(11) and a(12) from Antti Karttunen, Feb 25 2022
STATUS
approved