A145897 Starting prime (and 1): where number of consecutive squares m^2 satisfy r=p+4*m^2, prime.

1, 7, 13, 19, 37, 43, 67, 79, 97, 103, 109, 127, 163, 193, 223, 229, 277, 307, 313, 349, 379, 397, 439, 457, 463, 487, 499, 613, 643, 673, 739, 757, 769, 823, 853, 859, 877, 883, 907, 937, 967, 1009, 1087, 1093, 1213, 1279, 1297, 1303, 1423, 1429, 1447, 1483

Offset: 1

Author

Enoch Haga, Oct 25 2008

Keywords

• easy
• nonn

Comments

Suggested by Farideh Firoozbakht's Puzzle 464 in Carlos Rivera's The Prime Puzzles & Problems Connection. In this sequence Haga accepts 1 as a prime because then m^2 begins the first run of consecutive primes.

This looks like (apparent from the ad-hoc introduced leading 1) an erroneous version of A023200, because the definition says that it registers prime chains p+4*m^2, m=1,2,3,.. but apparently does not consider whether m is actually larger than 1. So 3 should be in the sequence because 3+4*1^2 is prime. - R. J. Mathar, Mar 25 2024

Examples

			a(1)=1 because when there are 3 consecutive m^2, first prime is 5 and ending prime is 37: r=1+4*1^1=5, prime; and r=1+4*2^2=17, prime; and r=1+4*3^2=37, prime (and the next value of r does not produce a prime).
		

Crossrefs

Cf. A145896, A145898, A145741, A049492.

Programs

UBASIC

10 'p464
20 N=1
30 A=3:S=sqrt(N)
40 B=N\A
50 if B*A=N then 100
60 A=A+2
70 if A<=S then 40
80 M=M+1:R=N+4*M^2:if R=prmdiv(R) and M<100 then print N;R;M:goto 80
90 if M>=1 then stop
100 M=0:N=N+2:goto 30