cp's OEIS Frontend

Let R = Z[(1+sqrt{5})/2] denote the ring of integers in the real quadratic number field of discriminant 5. The main result of Maass (1941) is that every totally positive nu in R is a sum of 3 squares x^2+y^2+z^2 with x,y,z in R. The number N_{nu} of such representations is given by the formula in the theorem on page 191. The norms of the totally positive elements nu are rational integers m belonging to A031363, so we can order the terms of the sequence according to the values m = A031363(n). [Comment based on remarks from Gabriele Nebe.]

The terms were computed with the aid of Magma by David Durstoff, Nov 11 2015.

The attached file from David Durstoff gives list of pairs m=A031363(n), a(n), and also the initial terms of Maass's series theta(tau). David Durstoff says: "I expressed theta(tau) in terms of two variables q1 and q2. The coefficient of q1^k q2^m is a(nu) with k = trace(nu/delta) and m = trace(nu), where delta = (5+sqrt{5})/2 is a generator of the different ideal. I computed the terms for q1^0 to q1^10 and all possible powers of q2."

Note that there are examples of totally positive elements x and y in R which have the same norm, but for which the number of ways of writing x as a sum of three squares in R is different to the number of ways of writing y as a sum of three squares in R. E.g. there are 78 ways of writing (27 + 7*sqrt(5))/2 as a sum of three squares in R, and there are 192 ways of writing 11 as a sum of three squares in R, yet both elements have norm 121. The sequence of possible norms for which this can occur is 121, 209, 341, 361, 451, 484, 551, 589, 605, 649, ... - Robin Visser, Mar 28 2025

Also odd primes p such that 5 is a square mod p: (5/p) = +1 for p > 5.

Primes of the form x^2 + x*y - y^2 (as well as of the form x^2 + 3*x*y + y^2), both with discriminant = 5 and class number = 1. Binary quadratic forms a*x^2 + b*x*y + c*y^2 have discriminant d = b^2 - 4ac and gcd(a, b, c) = 1. [This was originally a separate entry, submitted by Laura Caballero Fernandez, Lourdes Calvo Moguer, Maria Josefa Cano Marquez, Oscar Jesus Falcon Ganfornina and Sergio Garrido Morales, Jun 06 2008. R. J. Mathar proved on Jul 22 2008 that this coincides with the present sequence.]

Also primes of the form 5x^2 - y^2 (cf. A031363). - N. J. A. Sloane, May 30 2014

Also primes of the form x^2 + 4*x*y - y^2. Every binary quadratic primitive form of discriminant 20 or 5 has proper solutions for positive integers N given in A089270, including the present primes. Proof from computing the corresponding representative parallel primitive forms, which leads to x^2 - 5 == 0 (mod N) or x^2 + x - 1 == 0 (mod N) which have solutions precisely for these positive N values, including these primes. - Wolfdieter Lang, Jun 19 2019

For a Pythagorean triple a, b, c, these primes (and 2) are the possible prime factors of 2a + b, |2a - b|, 2b + a, and 2b - a. - J. Lowell, Nov 05 2011

The prime factors of A028387(n^2+3n+1). - Richard R. Forberg, Dec 12 2014

Except for p = 5, these are primes p that divide Fibonacci(p-1). - Dmitry Kamenetsky, Jul 27 2015

Apart from the first term, these are rational primes that decompose in the field Q[sqrt(5)]. For example, 11 = ((7 + sqrt(5))/2)*((7 - sqrt(5))/2), 19 = ((9 + sqrt(5))/2)*((9 - sqrt(5))/2). - Jianing Song, Nov 23 2018

The possible prime factors of x^2 - x - 1. - Charles R Greathouse IV, Mar 18 2022

Let tau be the golden ratio (1+sqrt(5))/2; let zetaQ(tau)(s)=sum(1/(Z(tau):a)^s) the Dedekind zeta function where a runs through the nonzero ideals of Z(tau) and where (Z(tau):a) is the norm of a; then zetaQ(tau)(s)=sum(n>=1,a(n)/n^s). - Benoit Cloitre, Dec 29 2002

First occurrence of k beginning at zero, or 0 if not yet known: 2, 1, 11, 121, 209, 14641, 2299, 1771561, 6061, 43681, 278179, 0, 66671, 0, 33659659, 5285401, 187891, 0, 1266749, 0, 8067191, 639533521, 0, 0, 2066801, 0, 0, 36735721, 976130111, 0, 153276629, 0, 7703531, 0, 0, 0, 39269219, 0, 0, 0, 250082921, 0, 0, 0, 0, 0, 0, 0, 84738841, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 454508329, ..., .

If k is prime, the 0 above can be replaced by the smallest p^(k-1) with p a prime == {1,4} (mod 5), which is p=11. This follows from the multiplicative formula. - R. J. Mathar, Apr 02 2011

The terms often equal A001157(n) mod 5; the exceptions are at n = 2299, 3509, 3751, 3971, 4961, 6061, 6479, ... - R. J. Mathar, Apr 02 2011

Coefficients of Dedekind zeta function for the quadratic number field of discriminant 5. See A002324 for formula and Maple code. - N. J. A. Sloane, Mar 22 2022

Positive integers k such that x^2 - 4xy + y^2 + k = 0 has integer solutions. (See the CROSSREFS section for sequences relating to solutions for particular k.)

Comments on method used, from Colin Barker, Jun 06 2014: (Start)

In general, we want to find the values of f, from 1 to 400 say, for which x^2 + bxy + y^2 + f = 0 has integer solutions for a given b.

In order to solve x^2 + bxy + y^2 + f = 0 we can solve the Pellian equation x^2 - Dy^2 = N, where D = b*b - 4 and N = 4*(b*b - 4)*f.

But since sqrt(D) < N, the classical method of solving x^2 - Dy^2 = N does not work. So I implemented the method described in the 1998 sci.math reference, which says:

"There are several methods for solving the Pellian equation when |N| > sqrt(d). One is to use a brute-force search. If N < 0 then search on y = sqrt(abs(n/d)) to sqrt((abs(n)(x1 + 1))/(2d)) and if N > 0 search on y = 0 to sqrt((n(x1 - 1))/(2d)) where (x1, y1) is the minimum positive solution (x, y) to x^2 - dy^2 = 1. If N < 0, for each positive (x, y) found by the search, also take (-x, y). If N > 0, also take (x, -y). In either case, all positive solutions are generated from these using (x1, y1) in the standard way."

Incidentally all my Pell code is written in B-Prolog, and is somewhat voluminous. (End)

Also, positive integers of the form -x^2 + 2xy + 2y^2 of discriminant 12. - N. J. A. Sloane, May 31 2014 [Corrected by Klaus Purath, May 07 2023]

The equivalent sequence for x^2 - 3xy + y^2 + k = 0 is A031363.

The equivalent sequence for x^2 - 5xy + y^2 + k = 0 is A237351.

A positive k does not appear in this sequence if and only if there is no integer solution of x^2 - 3*y^2 = -k with (i) 0 < y^2 <= k/2 and (ii) 0 <= x^2 <= k/2. See the Nagell reference Theorems 108 a and 109, pp. 206-7, with D = 3, N = k and (x_1,y_1) = (2,1). - Wolfdieter Lang, Jan 09 2015

From Klaus Purath, May 07 2023: (Start)

There are no squares in this sequence. Products of an odd number of terms as well as products of an odd number of terms and any terms of A014209 belong to the sequence.

Products of an even number of terms are terms of A014209. The union of this sequence and A014209 is closed under multiplication.

A positive number belongs to this sequence if and only if it contains an odd number of prime factors congruent to {2, 3, 11} modulo 12. If it contains prime factors congruent to {5, 7} modulo 12, these occur only with even exponents. (End)

From Klaus Purath, Jul 09 2023: (Start)

Any term of the sequence raised to an odd power also belongs to the sequence. Proof: t^(2n+1) = t*t^2n = (3*x^2 - y^2)*t^2n = 3*(x*t^n)^2 - (y*t^n)^2. It seems that t^(2n+1) occurs only if t also is in the sequence.

Joerg Arndt has proved that there are no squares in this sequence: Assume s^2 = 3*y^2 - x^2, then s^2 + x^2 = 3 * y^2, but the sum of two squares cannot be 3 * y^2, qed. (End)

That products of any 3 terms belong to the sequence can be proved by the following identity: (na^2 - b^2) (nc^2 - d^2) (ne^2 - f^2) = n[a(nce + df) + b(cf + de)]^2 - [na(cf + de) + b(nce + df)]^2. This can be verified by expanding both sides of the equation. - Klaus Purath, Jul 14 2023

From Klaus Purath, Feb 17 2024: (Start)

Positive numbers of the form 15x^2 - y^2. The reduced form is -x^2 + 6xy + 6y^2.

Even powers of terms as well as products of an even number of terms belong to A243188. This can be proved with respect to the forms [a,0,-c] and [a, 0, +c] by the following identities: (au^2 - cv^2)(ax^2 - cy^2) = (aux + cvy)^2 - ac(uy + vx)^2 and (au^2 + cv^2)(ax^2 + cy^2) = (aux - cvy)^2 + ac(uy + vx)^2 for all a, c, u, v, x, y in R. This can be verified by expanding both sides of the equations. Generalization (conjecture): This multiplication rule applies to all sequences represented by any binary quadratic form [a, b, c].

Odd powers of terms as well as products of an odd number of terms belong to the sequence. This can be proved with respect to the forms [a,0,-c] and [a, 0, +c] by the following identities: (as^2 - ct^2)(au^2 - cv^2)(ax^2 - cy^2) = a[s(aux + cvy) + ct(uy + vx)]^2 - c[as(uy + vx) + t(aux + cvy)]^2 and (as^2 + ct^2)(au^2 + cv^2)(ax^2 + cy^2) = a[s(aux - cvy) - ct(uy + vx)]^2 + c[as(uy + vx) + t(aux - cvy)]^2 for all a, c, s, t, u, v, x, y in R. This can be verified by expanding both sides of the equations. Generalization (conjecture): This multiplication rule applies to all sequences represented by any binary quadratic form [a, b, c].

If we denote any term of this sequence by B and correspondingly of A243189 by C and of A243190 by D, then B*C = D, C*D = B and B*D = C. This can be proved by the following identities, where the sequence (B) is represented by [kn, 0, -1], (C) by [n, 0, -k] and (D) by [k, 0, -n].

Proof of B*C = D: (knu^2 - v^2)(nx^2 - ky^2) = k(nux + vy)^2 - n(kuy + vx)^2 for k, n, u, v, x, y in R.

Proof of C*D = B: (nu^2 - kv^2)(kx^2 - ny^2) = kn(ux + vy)^2 - (nuy + kvx)^2 for k, n, u, v, x, y in R.

Proof of B*D = C: (knu^2 - v^2)(kx^2 - ny^2) = n(kux + vy)^2 - k(nuy + vx)^2 for k, n, u, v, x, y in R. This can be verified by expanding both sides of the equations.

Generalization (conjecture): If there are three sequences of a given positive discriminant that are represented by the forms [a1, b1, c1], [a2, b2, c2] and [a1*a2, b3, c3] for a1, a2 != 1, then the BCD rules apply to these sequences. (End)

See comments on method used in A084917.

The equivalent sequence for x^2 - 3xy + y^2 + k = 0 is A031363.

The equivalent sequence for x^2 - 4xy + y^2 + k = 0 is A084917.

Positive numbers of the form 3x^2 - 7y^2. - Jon E. Schoenfield, Jun 03 2022

Nonnegative numbers of the form 8x^2 - y^2. - Jon E. Schoenfield, Jun 03 2022

Equivalently, positive numbers of the form k = x^2 + 2xy - 2y^2. These are equivalent forms, of discriminant 12.

Also numbers representable as x^2 + 4*x*y + y^2 with 0 <= x <= y. - Gheorghe Coserea, Jul 29 2018 [The restriction 0 <= x <= y is not necessary. - Klaus Purath, Feb 05 2023]

From Klaus Purath, Feb 05 2023: (Start)

Also positive numbers of the form x^2 + 2*m*x*y + (m^2 - 3)*y^2. This includes all forms given above so far.

All terms are congruent to {0, 1, 4, 6, 9, 10} modulo 12.

The product of any two terms belongs to the sequence - (empirically secured up to a(k)*a(m) for 2 <= k, m <= 85). Thus it appears that this sequence is closed under multiplication. Perhaps someone can find a proof? (End)