cp's OEIS Frontend

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

Let p(n) = A002144(n) be the n-th Pythagorean prime.

Pythagorean prime p can be divided into a pair of integers (a,b) such as p =a+b and a*b==1 mod p. And (p-2)!==1 mod p because of Wilson's Theorem (p-1)!==-1 mod p. It can be divided into two parts (a,b) such as {2*3*4*...*((p(n)-1)/2)==a(n) mod p(n)} and {((p(n)-1)/2+1)*...*(p(n)-4)*(p(n)-3)*(p(n)-2)==-a(n)==(p(n)-a(n)) mod p(n)}. The pair numbers make a(n)+(p(n)-a(n))=p(n) and a(n)*(p(n)-a(n))==1 mod p(n). The left integer of the pair numbers is a(n). The right integer (p(n)-a(n)) is A336884(n).

The set of selecting odd numbers from {a(n)} and A336884 is A206549. The set of selecting even numbers from {a(n)} and A336884 is A209874 except for the number 1. A256011 never appears in {a(n)} or A336884. It is related to nonexistence of numbers that the largest prime factor of n^2+1 is greater than n.

The odd number of the difference |a(n)-A336884(n)|=|a(n)-(p(n)-a(n))|=|2*a(n)-p(n)| is A186814(n). A282538 never appears in the set of the difference |a(n)-A336884(n)|.

If p(n) is unknown, p(n) can be derived from a(n) using following equation. From a*b==1 mod p, a*b=k*p+1. With p=a+b, it can transform to b(n)=(k*a(n)+1)/(a(n)-k), k is an odd integer parameter when the fraction makes an integer. If there are many k's, select the minimum k in those. Then a(n)+b(n)=p(n). b(n) is A336884(n).

For more information see A336883.

For Modd n (not to be confused with mod n) see a comment on A203571.

See A209388 for the number of elements of the reduced residue class Modd n, called delta(n).

a(prime(n)) = (prime(n)-2)!! Modd prime(n) = 1 if n=1 or (prime(n)-1)/2 is odd, and = r(prime(n)) if (prime(n)-1)/2 is even. Here r(prime(n)) is the smallest positive nontrivial solution of x^2==1 (Modd prime(n)), which exists only for primes of the form 4*k+1 given in A002144. For r(prime(n)) see A206549. This is the analog of Wilson's theorem for Modd prime(n).

For (prime(n)-2)!! see A207332. [Wolfdieter Lang, Mar 28 2012]

The companion sequence is A212354.

There are at most two incongruent solutions of this congruence due to the degree. The fact that there are precisely two such solutions for each prime of the form 4*k+1 (see A002144) is due to the reduction of this problem to one of quadratic residues, namely to X^2 == -1 (mod 2p), with p a prime (see the Nagell reference, given in A210848, pp. 132-3, especially theorem 77), adapted to the quadratic form f(x) = 2*x^2 + 2*x + 1, with discriminant D=-4. This congruence with composite modulus has exactly two incongruent solutions because X^2 == -1 (mod 2) has only the solution +1 modulo 2 (odd numbers), and X^2 == -1 (mod p) has (at least one) solution if the Legendre symbol (-1/p) = +1 (i.e., if -1 is a quadratic residue modulo p). Now (-1/p) = (-1)^(p-1)/2 (see, e.g., the Niven-Zuckerman-Montgomery reference given in A001844, Theorem 3.2 (1), p. 132). Hence there is a solution modulo p iff p == 1 (mod 4). Call the smallest positive one X0, with 0 < X0 < p-1. Then one also has the incongruent solution X1 := p-X0. This implies that there are precisely two incongruent solution of the original congruence modulo 2*p for each 1 (mod 4) prime (see, e.g., Nagell's book, pp. 83-4, Theorem 46). If u is a solution for p = A002144(n) (the existence of u has just been proved) then also the companion v := p-1-u satisfies this congruence, and v is incongruent to u modulo p.

Note that x^2 + (x+1)^2 = 4*T(x) + 1, with the triangular numbers A000217.

The primes with x^2 +(x+1)^2 = prime (necessarily from A002144) are found under A027862. The corresponding x values are found under A027861. These x values explain the positions n' where a(n') is smaller than a(n'-1) (for n'>=6): determine k with x=A027861(k), and then n' from A027862(k) = A002144(n'). Note that a(n') = x for such values n'. E.g., n'=6 with a(6)=4: x=4=A027861(3), p=41=A027862(3) = A002144(6). These values n' are n' = 1, 2, 6, 8, 14, 19, 30, ...

All positive solutions of this congruence are provided by the two sequences with entries u(n,k) = a(n) + k*A002144(n) and v(n,k) = A212354(n) + k*A002144(n), n >= 1, k >= 0. For the cases p = 5, 13 and 17 see A047219, A212160 and A212161, respectively, where the even-indexed numbers are the u(n,k) and the odd-indexed ones the v(n,k) (bisection).

2*a(n) + 1 = A206549(n), the smallest positive nontrivial solution of X^2 == +1 (Modd A002144(n)). For the next larger solution 2*A212354(n) + 1 >= p, hence it does not belong to the restricted residue system Modd A002144(n).

The companion sequence is A212353.

See the comments on A212353 for the proof of two incongruent solutions of this congruence for each prime A002144(n). One takes the smallest positive representatives in each case as A212353(n) and a(n), with A212353(n) < a(n).

All positive solutions of this congruence are provided by the two sequences with entries u(n,k) = A212353(n) + k*A002144(n) and v(n,k) = a(n) + k*A002144(n), n >= 1, k >= 0. For the cases p = 5, 13 and 17 see A047219, A212160 and A212161, respectively, where the even-indexed numbers are the u(n,k) and the odd-indexed ones the v(n,k) (bisection).

r2(n) := 2*a(n) + 1 >= A002144(n) iff r(n) := 2*A212353(n) + 1 <= A002144(n)- 2. r2(n)^2 == +1 (Modd A002144(n)) but only r(n) belongs to the relevant restricted residue class. See A206549. Note that floor(r2(n)^2/A002144(n)) is odd. The same holds for r2 replaced by r.