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Number k such that k^2 + 1 = p*q*r where p,q,r are distinct primes and the sum p+q+r is a perfect square.
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%I #16 Aug 24 2015 04:11:58

%S 17,37,91,235,683,1423,1675,2879,8101,9595,13711,18799,19601,21295,

%T 25937,30059,32111,36251,39505,41071,49285,60719,79441,90575,93871,

%U 94799,103429,112571,132085,136075,144965,180001,180251,188465,189679

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

%C a(n) is odd. The prime numbers of the sequence are 17, 37, 683, 1423, 2879, 8101, 13711, 30059, 36251, 60719, 93871, 112571, 180001, ...

%e 17 is in the sequence because 17^2 + 1 = 2*5*29 and 2 + 5 + 29 = 6^2.

%p with(numtheory):

%p for n from 1 to 200000 do:

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

%p if n0=3 and bigomega(n^2+1)=3 and

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

%p then

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

%p else

%p fi:

%p od:

%o (PARI) isok(n) = my(f = factor(n^2+1)); (#f~ == 3) && (vecmax(f[,2]) == 1) && issquare(vecsum(f[,1])); \\ _Michel Marcus_, Aug 24 2015

%Y Cf. A002522, A180278.

%K nonn

%O 1,1

%A _Michel Lagneau_, Aug 23 2015