A332445 Numbers k of the form 4m+1 for which A087808(sigma(k)) is equal to 2*A087808(k).

2009, 19377, 37809, 59373, 74673, 115677, 270041, 310329, 354609, 357309, 720425, 732321, 841437, 2071737, 2612269, 3131149, 3866461, 3930929, 5172093, 5593981, 7118753, 7903961, 8224173, 9327393, 9438129, 11452321, 12708025, 18857209, 18861889, 18875313, 19110321, 20278269, 20709225, 20950061, 23963597, 24895153

Offset: 1

Author

Antti Karttunen, Feb 14 2020

Keywords

• nonn

Comments

Numbers k such that A332224(k) is equal to A087808(2*k) and k == 1 mod 4.

Notably, the only square among the first 299 terms is a(248) = 5808421369 = 76213^2. sigma(5808421369) = 5808497583 == 3 (mod 4) == 7 (mod 8). Of the remaining 298 terms < 2^33, 92 are such that sigma(k) == 6 (mod 8) and 206 are such that sigma(k) == 2 (mod 8), that is, are terms of A332227.

Question: Why the terms come in clusters? Compare also the scatterplots of A087808 and A332224, and a similar sequence A332465.

Links

• Antti Karttunen, Table of n, a(n) for n = 1..299; all terms <= 2^33
• Index entries for sequences where any odd perfect numbers must occur
• Index entries for sequences related to sigma(n)

Crossrefs

Intersection of A016813 and A332446.

Cf. also A228058, A332227, A332465.

Cf. A000203, A087808, A286357, A332224, A332225, A332458.

Programs

PARI

A087808(n) = if(n<1, 0, if(n%2==0, 2*A087808(n/2), A087808((n-1)/2)+1));
isA332445(n) = ((1==(n%4))&&(2*A087808(n)==A087808(sigma(n))));