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Values of m: where m^2 begins a run of consecutive squares satisfying r=p+4*m^2 with a sequence of primes
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%I #3 Jul 03 2016 00:17:33

%S 3,6,2,1,8,4,7,1,2,1,1,1,19,1,2,1,3,1,1,1,1,1,1,1,3,3,1,1,2,1,1,2,1,4,

%T 1,1,1,1,1,2,2,1,7,3,4,1,1,2,7,1,1,2,1,1,2,1,2,1,1,2,1,1,2,1,1,2

%N Values of m: where m^2 begins a run of consecutive squares satisfying r=p+4*m^2 with a sequence of primes

%C Suggested by Farideh Firoozbakht's Puzzle 464 in Carlos Rivera's The Prime Puzzles & Problems Connection

%e a(1)=3 because when m is 3 a sequence of three values of r end with prime 37; then 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 m, 4, does not produce a prime because r=1+4*4^2=65). For this one value 1 is assumed prime.

%o (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

%Y A145897 A145898 A145741

%K easy,nonn

%O 1,1

%A _Enoch Haga_, Oct 25 2008