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This sequence gives also the p-defects for the congruences y^2 == x^3 - x (mod p), y^2 == x^3 - 11*x - 14 (mod p) and y^2 == x^3 - 11*x + 14 (mod p). See the Cremona link, Table 1, N = 32. - Wolfdieter Lang, Dec 22 2016

This elliptic curve y^2 = x^3 + 4*x appears as strong Weil curve for the weight 2 newform (eta(4*tau)*eta(8*tau))^2 of level N=32, with Dedekind's eta function. See the Martin-Ono link, Theorem 2, p. 3173, the row with Conductor 32. See also A002171 for the expansion of this newform in powers of q^4 (but with different offset). The also Nr. 49 of the Martin Table 1.

From this L-series of this elliptic curve one has:

a(n) = 0 if prime(n) == 2 or 3 (mod 4). (see the conjecture by Robert Israel, Sep 28 2016 in A276730).

If prime(n) == 1 (mod 4) = A002144(m) (for a unique m = m(n)) then prime(n) = A(m)^2 + B(m)^2 with the odd A(m) = A002972(m) and the even B(m) = 2*A002973(m). It turns out that 4*A002144(m) - a(m^2) = (2*B(m))^2 for m=m(n), and the sign s(m) of a(m) is + if A(m) + B(m) == 1 (mod 4) and - if A(m) + B(m) == 3 (mod 4). For the primes == 1 (mod 4) leading to sign + or - see A279392 or A279393, respectively. One has thus s(m) = (-1)^((A(m)-1)/2 + B(m)/2). See the Martin-Ono formula for a_{32}(p) in Theorem 3, p. 3175. This leads to the a(n) formula given below.

The corresponding even legs are given in 4*A253803.

The legs of the other Pythagorean triangle with hypotenuse A080109(n) are given A253804(n) (odd) and A253805(n) (even).

Each fourth power of a prime of the form 1 (mod 4) (see A002144(n)^2 = A080175(n)) has exactly two representations as sum of two positive squares (Fermat). See the Dickson reference, (B) on p. 227.

This means that there are exactly two Pythagorean triangles (modulo leg exchange) for each hypotenuse A080109(n) = A002144(n)^2, n >= 1. See the Dickson reference, (A) on p. 227.

Note that the Pythagorean triangles are not always primitive. E.g., n = 2: (65, 4*39, 13^2) = 13*(5, 4*3, 13). For each prime congruent 1 (mod 4) (A002144) there is one and only one such non-primitive triangle with hypotenuse p^2 (just scale the unique primitive triangle with hypotenuse p with the factor p). Therefore, one of the two existing Pythagorean triangles with hypotenuse from A080109 is primitive and the other is imprimitive.

The corresponding even legs are given in 4*A253805.

The legs of the other Pythagorean triangle with hypotenuse A080109(n) are given A253802(n) (odd) and A253803(n) (even).

Each fourth power of a prime of the form 1 (mod 4) (see A002144(n)^= A080175(n)) has exactly two representations as sum of two positive squares (Fermat). See the Dickson reference, (B) on p. 227.

This means that there are exactly two Pythagorean triangles (modulo leg exchange) for each hypotenuse A080109(n) = A002144(n)^2, n >= 1. See the Dickson reference, (A) on p. 227.

Concerning the primitivity question of these triangles see a comment on A253802.

See the Jon Perry comment on A005098. The (even) promic number a(n)*(a(n)+1) (see A002378) when subtracted from A005098(n) (the n-th value k such that 4*k+1 is prime) leaves a square, namely A002973(n)^2. E.g., n=4: 7 - 2*3 = 1^1. n=5: 9 - 0 = 3^2.

2*a(n)+1 = A002972(n), n>=1. E.g., n=4: 2*2+1=5.

Conjecture: Let b(n) be the sum of all primes p <= n with p == 1 (mod 4). Then a(n)/b(n) = 1/Pi + O(1/sqrt(n)).

A classical theorem of Euler (conjectured by Fermat) states that any prime p == 1 (mod 4) can be written uniquely as x^2 + y^2 with 1 <= x <= y.

For any prime p == 1 (mod 4), we obviously have a(p) > a(p-1). Also, b(n) >= 2*a(n) for all n > 0 since x^2 + y^2 >= 2*x*y.

Via computation we find that a(10^10) = 353452066546904620, b(10^10) = 1110397615780409147, and 3.14157907 < b(10^10)/a(10^10) < 3.14157908.