A261530 - OEIS

%I #19 Aug 28 2015 15:52:25

%S 173,187,477,565,965,1237,1277,1437,1525,1636,2452,2587,2608,2653,

%T 2827,2885,2971,3197,3388,3412,3435,3477,3848,3891,4188,4237,4492,

%U 4724,5333,5728,5899,6272,7088,7108,7421,8363,8541,9379,9652,10227,10872,11581,12237

%N Numbers k such that k^2 + 1 = p*q*r*s where p,q,r,s are distinct primes and the sum p+q+r+s is a perfect square.

%C The primes in the sequence are 173, 1237, 1277, 2971, 5333, 8363, 19387, 20773, ...

%C The corresponding squares p+q+r+s are 121, 289, 441, 289, 529, 9025, 841, 5625, 529, 196, 5476, 3025, ...

%e 173 is in the sequence because 173^2 + 1 = 2*5*41*73 and 2 + 5 + 41 + 73 = 11^2.

%p with(numtheory):

%p for n from 1 to 20000 do:

%p   y:=factorset(n^2+1):n0:=nops(y):

%p    if n0=4 and bigomega(n^2+1)=4 and

%p    sqrt(y[1]+y[2]+y[3]+y[4])=floor(sqrt(y[1]+y[2]+y[3]+y[4]))

%p    then

%p    printf(`%d, `, n):

%p    else

%p    fi:

%p od:

%o (PARI) isok(n) = my(f = factor(n^2+1)); (#f~== 4) && (vecmax(f[,2]) == 1) && issquare(vecsum(f[,1])) ; \\ _Michel Marcus_, Aug 24 2015

%Y Cf. A002522, A261529.

%K nonn

%O 1,1

%A _Michel Lagneau_, Aug 24 2015