%I #12 Mar 26 2024 04:30:37

%S 1,7,13,19,37,43,67,79,97,103,109,127,163,193,223,229,277,307,313,349,

%T 379,397,439,457,463,487,499,613,643,673,739,757,769,823,853,859,877,

%U 883,907,937,967,1009,1087,1093,1213,1279,1297,1303,1423,1429,1447,1483

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

%C 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.

%C 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

%e 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).

%o (UBASIC)

%o 10 'p464

%o 20 N=1

%o 30 A=3:S=sqrt(N)

%o 40 B=N\A

%o 50 if B*A=N then 100

%o 60 A=A+2

%o 70 if A<=S then 40

%o 80 M=M+1:R=N+4*M^2:if R=prmdiv(R) and M<100 then print N;R;M:goto 80

%o 90 if M>=1 then stop

%o 100 M=0:N=N+2:goto 30

%Y Cf. A145896, A145898, A145741, A049492.

%K easy,nonn

%O 1,2

%A _Enoch Haga_, Oct 25 2008