A145896 -id:A145896 - cp's OEIS frontend

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

Enoch Haga, Oct 25 2008

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

			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).

Cf. A145896, A145898, A145741, A049492.

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

A145898 Ending prime: where number of consecutive squares m^2 satisfy r = p + 4*m^2, prime.

Original entry on oeis.org

37, 151, 29, 23, 293, 107, 263, 83, 113, 107, 113, 131, 1607, 197, 239, 233, 313, 311, 317, 353, 383, 401, 443, 461, 499, 523, 503, 617, 659, 677, 743, 773, 773, 887, 857, 863, 881, 887, 911, 953, 983, 1013, 1283, 1129, 1277, 1283, 1301, 1319, 1619, 1433

Offset: 1

Enoch Haga, Oct 25 2008

Farideh Firoozbakht noticed the unusually high number of 19 primes ending in a(13)=1607 in A145741 where she specified an m sequence of 1..10. My m sequence is open. 1,2,3,...,m+1, all squared.

			a(1)=37 because when m is 3, the first prime is 5 and the 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).

Cf. A145896, A145897, A145741.

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

cp's OEIS Frontend

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

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