%I #19 Apr 04 2024 09:50:16

%S 1,2,5,7,13,25,34,52,89,93,100,200,233,338,610,850,915,1028,1352,1508,

%T 1918,2105,3918,4181,5540,6396,6728,7250,9282,10100,10132,10946,15507,

%U 16609,17125,32708,32776,37107,42568,47770,58218,61230,72125,74948,75025,78608

%N Numbers k such that sigma_2(k^2 + 1) == 0 (mod k).

%C Conjecture: the sequence contains infinitely many Fibonacci numbers (see A360107).

%C For k < 10^7, we observe only 6 prime numbers in the sequence: {2, 5, 7, 13, 89, 233} including the Fibonacci numbers {2, 5, 13, 89, 233} and the Lucas number {7}.

%H Robert Israel, <a href="/A360105/b360105.txt">Table of n, a(n) for n = 1..222</a>

%e 7 is in the sequence because the divisors of 7^2+1 = 50 are {1, 2, 5, 10, 25, 50}, and 1^2 + 2^2 + 5^2 + 10^2 + 25^2 + 50^2 = 3255 = 7*465 == 0 (mod 7).

%p filter:= k -> NumberTheory:-SumOfDivisors(k^2+1,2) mod k = 0:

%p select(filter, [$1..10^5]); # _Robert Israel_, Feb 19 2024

%t Select[Range[50000], Divisible[DivisorSigma[2, #^2+1], #]&]

%o (PARI) isok(k) = sigma(k^2 + 1, 2) % k == 0; \\ _Michel Marcus_, Jan 26 2023

%Y Cf. A000032, A000045, A001157, A067719, A360107.

%K nonn

%O 1,2

%A _Michel Lagneau_, Jan 26 2023