A027862 Primes of the form j^2 + (j+1)^2.

5, 13, 41, 61, 113, 181, 313, 421, 613, 761, 1013, 1201, 1301, 1741, 1861, 2113, 2381, 2521, 3121, 3613, 4513, 5101, 7321, 8581, 9661, 9941, 10513, 12641, 13613, 14281, 14621, 15313, 16381, 19013, 19801, 20201, 21013, 21841, 23981, 24421, 26681

Offset: 1

Author

Patrick De Geest

Keywords

• nonn
• easy
• nice

Comments

Also, primes of the form 4*k+1 which are the hypotenuse of one and only one right triangle with integral legs. - Cino Hilliard, Mar 16 2003

Centered square primes (i.e., prime terms of centered squares A001844). - Lekraj Beedassy, Jan 21 2005

Primes of the form 2*k*(k-1)+1. - Juri-Stepan Gerasimov, Apr 27 2010

Equivalently, primes of the form (m^2+1)/2 (take m=2*j+1). These primes a(n) have nontrivial solutions of x^2 == 1 (Modd a(n)) given by x=x(n)=A002731(n). For Modd n see a comment on A203571. See also A206549 for such solutions for primes of the form 4*k+1, given in A002144.

E.g., a(3)=41, A002731(3)=9, 9^2=81, floor(81/41)=1 (odd),

-81 = -2*41 + 1 == 1 (mod 2*41), hence 9^2 == 1 (Modd 41). - Wolfdieter Lang, Feb 24 2012

Also primes of the form 4*k+1 that are the smallest side length of one and only one integer Soddyian triangle (see A230812). - Frank M Jackson, Mar 13 2014

Also, primes of the form (m^2+1)/2. - Zak Seidov, May 01 2014

Note that ((2n+1)^2 + 1)/2 = n^2 + (n+1)^2. - Thomas Ordowski, May 25 2015

Primes p such that 2p-1 is a square. - Thomas Ordowski, Aug 27 2016

Primes in the main diagonal of A000027 when represented as an array read by antidiagonals. - Clark Kimberling, Mar 12 2023

The diophantine equation x^2 + ... + (x + r)^2 = p may be rewritten to A*x^2 + B*x + C = p, where A = (r + 1), B = r*(r + 1), C = r*(r + 1)*(2*r + 1)/6. If gcd(A, B, C) > 1 no solution for a prime p exists. The gcd(A, B, C) = 1 holds only for r = 1, 2, 5 (gcd is the greatest common divisor). For r = 1 we have x^2 + (x + 1)^2 = p, thus for x from A027861 we calculate primes p from A027862. For r = 2 we have x^2 + (x + 1)^2 + (x + 2)^2 = p, thus for x from A027863 we calculate primes p from A027864. For r = 5 we have x^2 + ... + (x + 5)^2 = p, thus for x from A027866 we calculate primes p from A027867. - Ctibor O. Zizka, Oct 04 2023

Examples

			13 is in the sequence because it is prime and 13 = 2^2 + 3^2. - _Michael B. Porter_, Aug 27 2016
		

References

• D. M. Burton, Elementary Number Theory, Allyn and Bacon Inc. Boston, MA, 1976, p. 271.
• Morris Kline, Mathematical Thought from Ancient to Modern Times, 1972. pp. 275.

Links

• T. D. Noe and Zak Seidov, Table of n, a(n) for n = 1..10000
• Patrick De Geest, World!Of Numbers
• Daniel Shanks, An analytic criterion for the existence of infinitely many primes of the form 1/2 * (n^2 + 1), Illinois Journal of Mathematics 8:3 (1964), p. 377-379.
• Wacław Sierpiński, Sur les nombres triangulaires qui sont sommes de deux nombres triangulaires, Elem. Math., 17 (1962), pp. 63-65.
• Panayiotis G. Tsangaris, A sieve for all primes of the form x^2 + (x+1)^2, Acta Academiae Paedagogicae Agriensis, Sectio Mathematicae, 25 (1998), pp. 39-53.

Crossrefs

Primes p such that A079887(p) = 1.

Cf. A002731 (m values), A027861 (j values), A091277 (prime indices).

Subsequence of A002144 (p=4k+1).

Cf. A001844 (centered squares), A027863, A027864, A027866, A027867, A203571, A206549, A230812.

Programs

Magma

[ a: n in [0..150] | IsPrime(a) where a is n^2+(n+1)^2 ]; // Vincenzo Librandi, Dec 18 2010

Mathematica

Select[Table[n^2+(n+1)^2,{n,200}],PrimeQ] (* Harvey P. Dale, Aug 22 2012 *)
Select[Total/@Partition[Range[200]^2,2,1],PrimeQ] (* Harvey P. Dale, Apr 20 2016 *)

PARI

je=[]; for(n=1,500, if(isprime(n^2+(n+1)^2),je=concat(je,n^2+(n+1)^2))); je

PARI

fermat(n) = { for(x=1,n, y=2*x*(x+1)+1; if(isprime(y),print1(y" ")) ) }

Formula

a(n) = ((A002731(n)^2 - 1)/2) + 1. - Torlach Rush, Mar 14 2014

a(n) = (A002731(n)^2 + 1)/2. - Zak Seidov, May 01 2014

Extensions

More terms from Cino Hilliard, Mar 16 2003