cp's OEIS Frontend

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

Previous name was: The first summation of row 3 of Euler's triangle - a row that will recursively accumulate to the power of 3.

Decimal expansion of 47/30. - Elmo R. Oliveira, Aug 09 2024

The sum of signed coefficients for each k-th row is divisible by (2*k-2)!. Moreover, another variant (but an incomplete one, and sorted differently) of the above sequence is presented in A101752.

Incidentally, the sum of signed coefficients for each k-th row is divisible by (2*k-2)!.

In general, the last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: let last(0)=0, last(1)=k*(2k+3) and, for n > 1, last(n) = (4k+2)*last(n-1) - last(n-2)-2*k*(k-1); e.g., if k=6, then last(2) = 2274 = 26*90 - 6 - 60.

In general, the first and last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: let first(0)=0 and last(0)=k; for n > 0, let first(n) = (2k+1)*first(n-1) + 2k*last(n-1) + k and last(n) = (2k+2)*first(n-1) + (2k+1)*last(n-1) + 2k; e.g., if k=6 and n=2, then first(2) = 2100 = 13*78 + 12*90 + 6 and last(2) = 2274 = 14*78 + 13*90 + 12.

In general, the last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: if q=(k+1)/k, then last(n) = k^n*(k+1)*((1+sqrt(q))^(2*n+1) - (1-sqrt(q))^(2*n+1))/(4*sqrt(q)) + (k-1)/2; e.g., if k=6 and n=2, then last(2) = 2274 = 6^2*7*((1+sqrt(7/6))^5 - (1-sqrt(7/6))^5)/(4*sqrt(7/6)) + 5/2.

In general, if u(n) is the numerator and e(n) is the denominator of the n-th continued fraction convergent to sqrt((k+1)/k), then the last terms of Consecutive Integer Pythagorean 2k+1-tuples may be found as follows: last(2n+1) = (e(2n+1)^2 + k^2*e(2n)^2 + k*(k-1)*e(2n+1)*e(n))/k and, for n > 0, last(2n) = (k*(u(2n)^2 + u(2n-1)^2 + (k-1)*u(2n)*u(2n-1)))/(k+1); e.g., a(3) = 30365 = (220^2 + 5^2*21^2 + 5*4*220*21)/5 and a(4) = 666605 = (5*(505^2 + 241^2 + 4*505*241))/6.

In general, if b(0)=1, b(1)=4k+2 and, for n > 1, b(n) = (4k+2)*b(n-1) - b(n-2), and last(n) is the last term of the n-th Consecutive Integer Pythagorean 2k+1-tuple as defined above, then Sum_{i=0..n} (k*last(i) - k(k-1)/2) = k(k+1)/2*b(n); e.g., if n=3, then 1+2+3+4+5+61+62+63+64+65+1381+1382+1383+1384+1385 = 7245 = 15*483.

In general, if last(n) is the last term of the n-th Consecutive Integer Pythagorean 2k+1-tuple, then lim_{n->infinity} last(n+1)/last(n) = k*(1+sqrt((k+1)/k))^2 = 2k + 1 + 2*sqrt(k^2+k).

When the points are marked on drawn lines the concavity is apparent.

The lines are indicated with capital letters A through G (see Fig. 6 in Link)

- A

B 1 7 12 16 19 21

C 2 8 13 17 20

D 3 9 14 18

E 4 10 15

F 5 11

G 6

Reading diagonally across the bottom of the first of 4 diagonals:

6,11,15,18,20,21. The next 3 diagonals are formed by adding 1 to 21, e.g.,

22,27,31,34,36,37

38,43,47,50,52,53

54,59,63,66,68,69. This grid is numbered 1, and the next, 2, starts at 70.

Each numbered set of 4 grids fills the corners of a square delineating and surrounding a circle suggested by the 24 numbers above on its circumference.

The definition seems unnecessarily obscure. What is really going on here? - N. J. A. Sloane, Nov 13 2009

The complement is A172513 with first differences in A172515. - R. J. Mathar, Feb 27 2010

The original definition was: "(argument(exp(-(log(2)+W(n, -log(sqrt(2))))/log(2)))*log(2) + Im(W(n, -log(sqrt(2)))))/(2*Pi*log(2)) where W is the Lambert W function". The expression simplifies to that given in NAME. From the documents in LINKS, it appears that W(n,z) denotes the n-th branch of a complex LambertW function. It remains to understand the intended meaning of the distinction between arg(exp(z)) and Im(z). - M. F. Hasler, Apr 12 2019

From Travis Scott, Oct 09 2022: (Start)

One's first impression of this sequence and its complement (q.v.) is that of a Beatty duet. Indeed, a(n) never strays far from ceiling(n/log(2)), differing by 1 only at the 7, 16, 25, 34, 43, 50, 52, 59, ...-th terms.

By the identity arg(e^z) = Im(z)(mod 2*Pi)_(-Pi,Pi] -- where the subscripted range indicates an offset modulo rolling over at -Pi -> Pi rather than at 2*Pi -> 0 [this can be formalized as Im(z) - 2*Pi*ceiling((Im(z) - Pi)/(2*Pi))] -- we see that the argument component of our expression doesn't add any new information but rather acts on the imaginary component as part of a quotient device that reduces to floor(Im(w)/(2*Pi)+1/2) [or to round(Im(w)/(2*Pi)), with the caveat to always round up in the unlikely event that we encounter a half-integer].

Now consider v = W(n,-log(2)/2), taking the same product logarithm as for w but not dividing the result by log(2). Our expression then simply counts branch cuts and we get n. In very abusive but perhaps more visual language, if the sequence on v keeps track of the number of times the Im(W(n,-log(2)/2))-th power of the 2*Pi-th root of unity laps the negative real axis as we follow it counterclockwise around the unit circle, then the sequence on w keeps track of how many laps that would be on a circle of radius log(2) or by a log(4)*Pi-th root of unity.

It remains to guess if this tally has an intent or if it is a tally for tally's sake. (End)