cp's OEIS Frontend

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

%I A145897 #12 Mar 26 2024 04:30:37
%S A145897 1,7,13,19,37,43,67,79,97,103,109,127,163,193,223,229,277,307,313,349,
%T A145897 379,397,439,457,463,487,499,613,643,673,739,757,769,823,853,859,877,
%U A145897 883,907,937,967,1009,1087,1093,1213,1279,1297,1303,1423,1429,1447,1483
%N A145897 Starting prime (and 1): where number of consecutive squares m^2 satisfy r=p+4*m^2, prime.
%C A145897 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 A145897 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 A145897 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 A145897 (UBASIC)
%o A145897 10 'p464
%o A145897 20 N=1
%o A145897 30 A=3:S=sqrt(N)
%o A145897 40 B=N\A
%o A145897 50 if B*A=N then 100
%o A145897 60 A=A+2
%o A145897 70 if A<=S then 40
%o A145897 80 M=M+1:R=N+4*M^2:if R=prmdiv(R) and M<100 then print N;R;M:goto 80
%o A145897 90 if M>=1 then stop
%o A145897 100 M=0:N=N+2:goto 30
%Y A145897 Cf. A145896, A145898, A145741, A049492.
%K A145897 easy,nonn
%O A145897 1,2
%A A145897 _Enoch Haga_, Oct 25 2008