cp's OEIS Frontend

When compared to original Scrabble version, Super Scrabble is played with 200 tiles (4 of which are blanks), which is double original Scrabble version's distribution, and 21 X 21 square board, as compared to 15 X 15 square board in the original Scrabble version, with quadruple letter scoring and quadruple word scoring places available. Super scrabble has same face value of letter tiles as in original Scrabble (see A080993), and blanks are worth zero points each.

This is similar to A232551, except that this includes non-primitive quadratic forms like 2x^2+2xy+4y^2 and 2x^2+4y^2 because the prime 2 can be represented by both of them. But unlike A067752, we do not include quadratic forms like 4x^2+2xy+4y^2 and 4x^2+4xy+4y^2 by which no prime can be represented.

So, when n == 3 (mod 4), this includes the additional non-primitive quadratic form 2x^2+2xy+((n+1)/2)y^2 and when p^2 divides n, where p is prime, this includes the additional non-primitive quadratic form px^2+(n/p)y^2.

If p is a prime and if p^2 does not divide n, then there exist a unique non-primitive quadratic form of discriminant = -4n by which p can be represented if and only if -n is a quadratic residue (mod p) and there exists a multiple of p which can be written in the form x^2+ny^2 in which p appears raised to an odd power, except when p = 2 and n == 3 (mod 8).

Please look into A234001 for a more detailed description.

If n is squarefree and n == 1 (mod 4) or n == 2 (mod 4), then a(n) = 1.

If p^2 divides n for some prime p, a(n) is a multiple of p.

If n == 3 (mod 8), then a(n) is a multiple of 4 because numbers of the form x^2+n*y^2 cannot have any prime factors that are congruent to 2+n (mod 2n) raised to an odd power.

If n == 7 (mod 8), then a(n) is a multiple of 2 because numbers of the form x^2+n*y^2 can have prime factors that are congruent to 2+n (mod 2n) raised to an odd power, but they cannot be congruent to 2 (mod 4). So, we need to characterize the prime factor of 2 from the remaining prime factors that are congruent to 2+n (mod 2n) separately.

If n is a convenient number (A000926), the set of residue classes (mod 4n) that a prime p represented by x^2+n*y^2 belong to are those for which p is a quadratic residue (mod 4n) or p-n is a quadratic residue (mod 4n), assuming that p^2 does not divide n. For non-convenient numbers n, some of the primes in these set of residue classes (mod 4n) can be represented by x^2+n*y^2, but not all.

A prime p such that p^2 does not divide n, can be represented by some primitive quadratic form of discriminant = -4n, if and only if -n is a quadratic residue (mod p).

A prime p can be represented by some quadratic form of discriminant = -4n if and only if there is a multiple of p that can be written in the x^2+n*y^2 form, in which prime factor of p appears raised to an odd power or if p = 2 and n == 3 (mod 4).

a(n) is always a divisor of 4n.

If n is squarefree and n == 1 (mod 4) or n == 2 (mod 4), then a(n) = 4n.

If p^2 divides n for some prime p, a(n) is a divisor of (4n)/p.

If n == 3 (mod 8), then a(n) is a divisor of n because numbers of the form x^2+n*y^2 cannot have any prime factors that are congruent to 2+n (mod 2n) raised to an odd power.

If n == 7 (mod 8), then a(n) is a divisor of 2n because numbers of the form x^2+n*y^2 can have prime factors that are congruent to 2+n (mod 2n) raised to an odd power, but they cannot be congruent to 2 (mod 4). So, we need to characterize the prime factor of 2 from the remaining prime factors that are congruent to 2+n (mod 2n) separately.

Equivalently, numbers of the form 49^n*(7m+1), 49^n*(7m+2), or 49^n*(7m+4). [Corrected by Charles R Greathouse IV, Jan 12 2017]

From Peter Munn, Feb 08 2024: (Start)

Numbers whose squarefree part is congruent to a (nonzero) quadratic residue modulo 7.

The integers in a subgroup of the positive rationals under multiplication. As such the sequence is closed under multiplication and - where the result is an integer - under division. The subgroup has index 4 and is generated by the primes congruent to a quadratic residue (1, 2 or 4) modulo 7, the square of 7, and 3 times the other primes; that is a generator corresponding to each prime: 2, 3*3, 3*5, 7^2, 11, 3*13, 3*17, 3*19, 23, 29, 3*31, ... .

(End)

Equivalently, this sequence is the union of numbers of the form 25^n*(5*n+2) and numbers of the form 25^n*(5*n+3).

Squares modulo all powers of 2. - Robert Israel, Aug 26 2014

From Peter Munn, Dec 11 2019: (Start)

Closed under multiplication.

Contains all even powers of primes.

A subgroup of the positive integers under the binary operation A059897(.,.). For all n, a(n) has no Fermi-Dirac factor of 2 and if m_k denotes the number of Fermi-Dirac factors of a(n) that are congruent to k modulo 8, m_3, m_5 and m_7 have the same parity. It can further be shown (1) all numbers that meet these requirements are in the sequence and (2) this implies closure under A059897(.,.).

(End)

This is closely related to the class number problem.

A quadratic form is primitive if the GCD of the coefficients is 1. For example, the quadratic form 2*x^2+4*y^2 is not primitive.

Two quadratic forms f(x,y) = a*x^2+b*x*y+c*y^2 and g(x,y) = p*x^2+q*x*y+r*y^2 are distinct (or inequivalent) if and only if one cannot be obtained by a linear transformation (of the variables x, y) from the other. For example, the three quadratic forms u(x,y) = 3*x^2+2*x*y+3*y^2, v(x,y) = 3*x^2+4*x*y+4*y^2 and w(x,y) = 3*x^2+10*x*y+11*y^2 are equivalent because v(x,y) = u(x+y,-y) and w(x,y) = v(x+y,y). Also, w(x,y) = u(x+2*y,-y). Similarly, the two quadratic forms s(x,y) = 8*x^2+9*y^2 and t(x,y) = 17*x^2+50*x*y+41*y^2 are equivalent because t(x,y) = s(x+2*y,x+y).

The quadratic form x^2+n*y^2 is one such form and the only such form if n = 1, 2, 3, 4, 7.

a(n) = 1 if and only if n = 1, 2, 3, 4, 7.

If n is a squarefree convenient number (A000926), a(n) represents the class number of the ring Z[sqrt(n)] if n == 1 (mod 4) or if n == 2 (mod 4) and the class number of the ring Z[(1+sqrt(n))/2] if n == 3 (mod 4) and this class number is a power of 2.

Any prime p such that -n is a quadratic residue mod p can be represented by exactly one of the a(n) distinct primitive quadratic forms of discriminant = -4n in at most four different ways (if n >= 2) or in at most eight different ways (if n = 1).

If n is a prime congruent to 3 (mod 4), then a(n) = A232550(n).

If p is a prime, p^2 does not divide n, and p > 2 if n == 3 (mod 8), then there is a multiple of p in which p is raised to an odd power which can be written in the form x^2+n*y^2 if and only if -n is a quadratic residue mod p.

The product of two numbers (prime or composite, same or different) which can be represented by the same quadratic form of discriminant = -4n can be written in the form x^2+n*y^2, as the following identity shows:

(X*a^2+Y*a*b+Z*b^2)*(X*c^2+Y*c*d+Z*d^2) = (a*c*X+b*d*Z+a*d*(Y/2)+b*c*(Y/2))^2 + ((X*Z)-(Y^2/4))*(a*d-b*c)^2.

(X*a^2+Y*a*b+Z*b^2)*(X*c^2+Y*c*d+Z*d^2) = (a*c*X+b*d*((Y^2/(2*X))-Z)+a*d*(Y/2)+b*c*(Y/2))^2 + ((X*Z)-(Y^2/4))*(b*d*(Y/X)+a*d+b*c)^2.

Note that for the latter equation, (a*c*X+b*d*((Y^2/(2*X))-Z)+a*d*(Y/2)+b*c*(Y/2)) and (b*d*(Y/X)+a*d+b*c) need not always be integers. If they are both integers, then it will be a second representation of the product of (X*a^2+Y*a*b+Z*b^2) and (X*c^2+Y*c*d+Z*d^2) in the form x^2+((X*Z)-(Y^2/4))*y^2.

This sequence is the same as taking every fourth number in A107628. - T. D. Noe, Jan 02 2014

A quadratic form is primitive if the GCD of the coefficients is 1. For example, the quadratic form 2*x^2+4*y^2 is not primitive.

Two quadratic forms f(x,y) = a*x^2+b*x*y+c*y^2 and g(x,y) = p*x^2+q*x*y+r*y^2 are distinct (or inequivalent) if and only if one cannot be obtained by a linear transformation (of the variables x, y) from the other. For example, the three quadratic forms u(x,y) = 3*x^2+2*x*y+3*y^2, v(x,y) = 3*x^2+4*x*y+4*y^2 and w(x,y) = 3*x^2+10*x*y+11*y^2 are equivalent because v(x,y) = u(x+y,-y) and w(x,y) = v(x+y,y). Also, w(x,y) = u(x+2*y,-y). Similarly, the two quadratic forms s(x,y) = 8*x^2+9*y^2 and t(x,y) = 17*x^2+50*x*y+41*y^2 are equivalent because t(x,y) = s(x+2*y,x+y).

The quadratic form x^2+n*y^2 is one such form and the only such form if n is a convenient number (A000926).

a(n) = 1 if and only if n is a convenient number (A000926).

Any prime p such that p is a quadratic residue (mod 4n) or p-n is a quadratic residue (mod 4n) can be represented by exactly one of the a(n) distinct primitive quadratic forms of discriminant = -4n in at most four different ways (if n >= 2) or in at most eight different ways (if n = 1).

If a prime p can be written in the form x^2+n*y^2, then either p is a quadratic residue (mod 4n) or p-n is a quadratic residue (mod 4n), assuming that p^2 does not divide n.

For primes p such that p is a quadratic residue (mod 4n) or p-n is a quadratic residue (mod 4n), there is a lowest square m^2 such that m^2*p can be written in form x^2+n*y^2, where x and y are nonnegative integers (see A232529 and A232530).

If n is a prime congruent to 3 (mod 4), then a(n) = A232551(n).

The product of two numbers (prime or composite, same or different) which can be represented by the same quadratic form of discriminant = -4n can be written in the form x^2+n*y^2, as the following identity shows.

(X*a^2+Y*a*b+Z*b^2)*(X*c^2+Y*c*d+Z*d^2) = (a*c*X+b*d*Z+a*d*(Y/2)+b*c*(Y/2))^2 + ((X*Z)-(Y^2/4))*(a*d-b*c)^2.

(X*a^2+Y*a*b+Z*b^2)*(X*c^2+Y*c*d+Z*d^2) = (a*c*X+b*d*((Y^2/(2*X))-Z)+a*d*(Y/2)+b*c*(Y/2))^2 + ((X*Z)-(Y^2/4))*(b*d*(Y/X)+a*d+b*c)^2.

Note that for the latter equation, (a*c*X+b*d*((Y^2/(2*X))-Z)+a*d*(Y/2)+b*c*(Y/2)) and (b*d*(Y/X)+a*d+b*c) need not always be integers. If they are both integers, then it will be a second representation of the product of (X*a^2+Y*a*b+Z*b^2) and (X*c^2+Y*c*d+Z*d^2) in the form x^2+((X*Z)-(Y^2/4))*y^2.

If n is a convenient number (A000926) then a(n) = 1.

m^2 is also the smallest positive square that can be generated by using all the inequivalent primitive quadratic forms of discriminant = -4n.