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A261530
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.
0
173, 187, 477, 565, 965, 1237, 1277, 1437, 1525, 1636, 2452, 2587, 2608, 2653, 2827, 2885, 2971, 3197, 3388, 3412, 3435, 3477, 3848, 3891, 4188, 4237, 4492, 4724, 5333, 5728, 5899, 6272, 7088, 7108, 7421, 8363, 8541, 9379, 9652, 10227, 10872, 11581, 12237
OFFSET
1,1
COMMENTS
The primes in the sequence are 173, 1237, 1277, 2971, 5333, 8363, 19387, 20773, ...
The corresponding squares p+q+r+s are 121, 289, 441, 289, 529, 9025, 841, 5625, 529, 196, 5476, 3025, ...
EXAMPLE
173 is in the sequence because 173^2 + 1 = 2*5*41*73 and 2 + 5 + 41 + 73 = 11^2.
MAPLE
with(numtheory):
for n from 1 to 20000 do:
y:=factorset(n^2+1):n0:=nops(y):
if n0=4 and bigomega(n^2+1)=4 and
sqrt(y[1]+y[2]+y[3]+y[4])=floor(sqrt(y[1]+y[2]+y[3]+y[4]))
then
printf(`%d, `, n):
else
fi:
od:
PROG
(PARI) isok(n) = my(f = factor(n^2+1)); (#f~== 4) && (vecmax(f[, 2]) == 1) && issquare(vecsum(f[, 1])) ; \\ Michel Marcus, Aug 24 2015
CROSSREFS
Sequence in context: A259017 A097845 A364937 * A246135 A140002 A178652
KEYWORD
nonn
AUTHOR
Michel Lagneau, Aug 24 2015
STATUS
approved