

A339007


Least k such that p = k^2 + 1 and q = (k+2n)^2 + 1 are prime numbers with q  p square.


2



24, 6, 312984, 16896, 120, 734994, 10640, 10, 1946016, 150, 171864, 180, 31200, 17136, 120, 84, 8976, 54, 137256, 300, 231504, 66, 184, 360126, 24, 5824, 2496, 224, 261696, 90, 4359344, 66, 50160, 68816, 280, 864, 1524696, 570, 219336, 11520, 8487984, 126, 22704
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OFFSET

1,1


COMMENTS

4*n*(k + n) is a square. If n is a square, then k + n is also a square.
If n is prime, then n divides k.
If we add the additional condition that p and q are two consecutive primes of the form m^2 + 1, then we obtain the sequence A339008, with A339008(n) = a(n) for n = 1, 2, 3, 4, 6, 7 and 9.


LINKS



EXAMPLE

a(1) = 24 because 24^2 + 1 = 577, (24 + 2)^2 + 1 = 677 and 677  577 = 10^2 is a square. The other values m such that p = m^2 + 1 and q = (m+2)^2 + 1 are primes with q  p square are 11024, 133224, 156024, 342224, 416024,...
a(2) = 6 because 6^2 + 1 = 37, (6 + 4)^2 + 1 = 101 and 101  37 = 8^2 is a square. The other values m such that p = m^2 + 1 and q = (m+4)^2 + 1 are primes with q  p square are 16, 126, 1350, 1456, 1566, 2310, 5200,...


MAPLE

for n from 1 to 50 do:
ii:=0:
for k from 2 by 2 to 10^9 while(ii=0) do:
p:=k^2+1:q:=(k+2*n)^2 +1:
if isprime(p) and isprime(q) and sqrt(qp)=floor(sqrt(qp))
then
ii:=1:printf(`%d %d \n`, n, k):
else
fi:
od:
od:


PROG

(PARI) a(n) = my(k=1); while (!(isprime(p=k^2+1) && isprime(q=(k+2*n)^2 + 1) && issquare(qp)), k++); k; \\ Michel Marcus, Nov 18 2020


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



