A281543 -id:A281543 - cp's OEIS frontend

A036468 Number of ways to represent 2n+1 as a+b with 0 < a < b and a^2 + b^2 prime.

1, 2, 2, 2, 3, 3, 4, 4, 4, 3, 4, 8, 4, 6, 5, 4, 9, 8, 6, 9, 7, 7, 7, 5, 7, 9, 14, 8, 9, 11, 7, 17, 11, 10, 9, 11, 9, 8, 13, 9, 15, 20, 11, 14, 13, 8, 18, 14, 10, 18, 16, 10, 17, 16, 13, 20, 20, 13, 14, 17, 12, 23, 18, 14, 22, 15, 17, 18, 21, 12, 19, 29, 16, 23, 21, 14, 27, 24

Offset: 1

R. K. Guy

nonn

Zhang Ming-Zhi (zamiz(AT)mail.sc.cninfo.net) asks if a(m) is always > 0.
I have confirmed that a(n) > 0 for 0 < n < 10^7. - T. D. Noe, Oct 17 2004
This open problem is mentioned by Guy at the end of section C1. - T. D. Noe, Apr 22 2009
a(n) <= phi(2n+1)/2, where phi(m) = A000010(m), while a(n) = phi(2n+1)/2 only for n = 1, 2, and 7. - Thomas Ordowski, Jan 25 2014
Records in a(n) are for 2n+1 = 3, 5, 11, 15, 25, 35, 55, 65, 85, 125, 145, 185, 205, 215, 235, 265, 295, 325, 365, 415, ... cf. A001750. - Thomas Ordowski, Mar 02 2017
a(n) tends to be larger for n == 2 (mod 5): see plot in Links. - Robert Israel, Mar 02 2017
Number of primes p = ((2n+1)^2 + x^2)/2 for positive integers x < 2n+1. - Thomas Ordowski, Mar 06 2017

R. K. Guy, Unsolved Problems in Theory of Numbers, Section C1.

T. D. Noe, Table of n, a(n) for n = 1..10000
Gordon Hamilton, Unsolved K-12: Grade 7, 2014. (video)
Robert Israel, a(5k+j) for j=0,1,2,3,4

Cf. A082519, A099468, A281543.

a:= n-> add(`if`(isprime(i^2+(2*n+1-i)^2), 1, 0), i=1..n):
seq(a(n), n=1..80);  # Alois P. Heinz, Jul 09 2016

Table[cnt=0; m=2n+1; Do[If[PrimeQ[k^2+(m-k)^2], cnt++ ], {k, n}]; cnt, {n, 100}]

a(n)=sum(k=1,n,isprime(k^2+(2*n-k+1)^2)) \\ Charles R Greathouse IV, Jan 09 2014

a(n) = O(n/log(n)). - Thomas Ordowski, Feb 11 2013

More terms from David W. Wilson and Michael Kleber

A069004 Number of times n^2 + s^2 is prime for positive integers s < n.

Original entry on oeis.org

0, 1, 1, 1, 2, 2, 1, 3, 1, 4, 2, 1, 4, 3, 3, 3, 4, 3, 4, 6, 2, 4, 5, 3, 7, 6, 4, 4, 4, 4, 7, 6, 5, 6, 8, 5, 6, 7, 3, 9, 5, 5, 8, 8, 7, 9, 6, 7, 10, 8, 6, 9, 10, 5, 8, 8, 6, 10, 11, 8, 11, 10, 6, 9, 15, 5, 10, 11, 4, 11, 13, 6, 12, 10, 12, 11, 9, 8, 11, 19, 10, 15, 9, 8, 19, 11, 8, 11, 14, 15, 13

Offset: 1

T. D. Noe, Apr 02 2002

Conjecture: a(n)>0 for all n>1. - Entries checked by Franklin T. Adams-Watters, May 05 2006
The graph of this sequence inspires the following conjecture: A > a(n)/pi(n) > B, where A and B are constants and pi(n) is the prime counting function (A000720). - T. D. Noe, Feb 26 2007
Stronger conjecture: Let pi(n) be the prime counting function (A000720). Then pi(n) >= a(n) >= pi(n)/5 for n>1, with the following equalities: pi(2)=a(2), pi(10)=a(10) and a(12)=pi(12)/5. - T. D. Noe, Feb 26 2007
Records in a(n) are for n = 1, 2, 5, 8, 10, 20, 25, 35, 40, 49, 59, 65, 80, 115, 125, 130, 158, 200, 250, 265, 310, ... - Thomas Ordowski, Mar 05 2017
Number of primes p = (x^2 + y^2)/2 with 0 < x < y such that x + y = 2n. - Thomas Ordowski, Mar 06 2017

			a(5)=2 because there are 2 values of s (2 and 4) such that 5^2 + s^2 is a prime number.

T. D. Noe, Table of n, a(n) for n = 1..10000

Cf. A000010, A036468, A057368, A281543.

maxN=100; lst={}; For[n=1, n<=maxN, n++, cnt=0; For[d=1, d?PrimeQ],{n,100}] (* _Harvey P. Dale, Mar 01 2023 *)

a(n) = sum(s=1, n-1, isprime(n^2+s^2)); \\ Michel Marcus, Jan 15 2017

a(n) = O(n/log(n)). a(n) <= phi(n), a(n) = phi(n) for n = 2, 6, and 10. a(n) <= phi(2n)/2, a(n) = phi(2n)/2 for n = 2, 3, 5, 6, and 10. - Thomas Ordowski, Mar 01 2017

Entries checked by Franklin T. Adams-Watters, May 05 2006

cp's OEIS Frontend

A036468 Number of ways to represent 2n+1 as a+b with 0 < a < b and a^2 + b^2 prime.

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