A378481 Integers k such that A378414(k) == k (mod A066417(k)).

33, 77, 153, 372, 1540, 2300, 2692, 2736, 7812, 8721, 12593, 26025, 26481, 27972, 39321, 64009, 104409, 175441, 325180, 335616, 422593, 455625, 564376, 575040, 756460, 800073, 1104521, 2180545, 2304332, 3502665, 3691968, 5130909, 5515121, 9331441, 9546265

Offset: 1

Author

Paolo P. Lava, Nov 28 2024

Keywords

• nonn
• easy

Comments

Also integers k such that A000217(k) == k (mod A066417(k)).

So far, only 33 belongs both to A232538 and A378414.

Examples

			Antidivisors of 77 are 2, 3, 5, 9, 14, 17, 22, 31, 51 and their sum is 154.
Then 77*78/2 mod 154 = 3003 mod 154 = 77.
		

Crossrefs

Cf. A000217, A066417, A232538, A378414.

Programs

Maple

with(numtheory): P:=proc(q) local j,k,n,v; v:=[];
for n from 3 to q do k:=0; j:=n; while j mod 2<>1 do k:=k+1; j:=j/2; od;
if n*(n+1)/2 mod (sigma(2*n+1)+sigma(2*n-1)+sigma(n/2^k)*2^(k+1)-6*n-2)=n
then v:=[op(v),n]; fi; od; op(v); end: P(10^5);