cp's OEIS Frontend

A261529 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.

%I A261529 #16 Aug 24 2015 04:11:58
%S A261529 17,37,91,235,683,1423,1675,2879,8101,9595,13711,18799,19601,21295,
%T A261529 25937,30059,32111,36251,39505,41071,49285,60719,79441,90575,93871,
%U A261529 94799,103429,112571,132085,136075,144965,180001,180251,188465,189679
%N A261529 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 A261529 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 A261529 17 is in the sequence because 17^2 + 1 = 2*5*29 and 2 + 5 + 29 = 6^2.
%p A261529 with(numtheory):
%p A261529 for n from 1 to 200000 do:
%p A261529   y:=factorset(n^2+1):n0:=nops(y):
%p A261529    if n0=3 and bigomega(n^2+1)=3 and
%p A261529    sqrt(y[1]+y[2]+y[3])=floor(sqrt(y[1]+y[2]+y[3]))
%p A261529    then
%p A261529    printf(`%d, `,n):
%p A261529    else
%p A261529    fi:
%p A261529 od:
%o A261529 (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 A261529 Cf. A002522, A180278.
%K A261529 nonn
%O A261529 1,1
%A A261529 _Michel Lagneau_, Aug 23 2015