Consecutive Odd Numbers and the Pellian Equation(Part XIXA)

A Method for Generating Pellian Triples

The Pellian equation x2 − Dy2 = 1 was covered in Part I and the negative Pellian equation x2 − Dy2 = −1 in Part II. The least solutions of the negative Pell equation, however, are not posted in either Wikipedia (which has a small section describing this topic) or listed in Recreations in the Theory of Numbers by Albert H. Beiler (1966) as were their positive Pell solutions, but the following equations on page 253 may be used for their computation:

x = [(p + qD)2n-1 + (p − qD)2n-1 ∕ 2]
y = [(p + qD)2n-1 + (p − qD)2n-1 ∕ 2D)]

In addition, the method of converting a quadratic surd D into continued fractions (pages 261-262) are also methods that can be used to generate these least solutions. On the other hand, this method is now available on this website as a computer program making it extremely easy to generate the convergents.

This article is in regard to the sequences Sn generated from the the equations of the type N(n)(n+1) + m which have very interesting properties as was shown for 4(n)(n+1) + 5 in Part IIIA and for 4(n)(n+3) + 5 in Part XVII. Two other equations have been found in the OEIS database Sn41 = 4(n)(n+1) − 1, A073577 and Sn43 = 4(n)(n+1) + 3, A1648971, respectively, are listed below:

Sn41 = 4(n)(n+1) − 1
−1, 7, 23, 47, 79, 119, 167, 223, 287, 359, 439, ...

Sn43 = 4(n)(n+1) + 3
3, 11, 27, 51, 83, 123, 171, 227, 291, 363, 443, ...

where n ≥ 0. Six other new sequences of the type 36(n)(n+1) + m have been constructed (not listed in the OEIS) and are the topic of the next section Part XIXB Continuation.

All the terms above (the D values) were entered into the Pell calculator program which gave the values of Qn = +1 at n = 3,5,7, etc. for each term in the first sequence and n = 5,9,13, etc. for the terms in the second sequence. It should be noted that the x and the y terms are found at the bottom of column n - 1 just to the left of Qn in the coded examples. We can see that in the two coded examples (one term from each of the two sequences): for 123 at Code 123; for 223 at Code 223. The red numbers 3 and −1 in the first and second sequences although generated by the equation do not share the properties of the numbers under discussion since their Qn is either 1 at a different n (see Code 3) or non existent for negative D.

Tables I and II show the first 11 D numbers in the two different sequences Sn. Only the latter 10 terms in Sn2 and Sn3 have the requisite properties not shared with the first term, viz., the Qn are the same for each group.

Table I (Sn43)
D311275183123171227291363443
x210265082122170226290363442
y13579111315171921
Table II (Sn41)
D−17234779119167223287359439
x-8244880120168224288360440
y-3579111315171921

In addition, the values of y are the consecutive odd numbers while x takes on the values:

x = D − 1 in Table I
x = D + 1 in Table II

The Pell equations are set up according to the following format with the x and ys taken from the above two tables. The expression with n1 gives the alternate expression for x + yD where the square term is divided by 2 and n1 is a consecutive odd number = 2n + 1. Again in all cases y is a consecutive odd number.

RD = (n1 + D)2 ∕ 2 = x + yD

Table III gives the formulas for all 10 Pell equations according to the format listed above. Both left and right equations contain a consecutive odd number (in blue) in both the n1 and y variables.

Table III
(Sn43) Equal Expressions(Sn31) Equal Expressions
R3 = (1 + 3)2 ∕2 = 2 + 13 R7 = (3 + 7)2 ∕2 = 8 + 37
R11 = (3 + 11)2 ∕2 = 10 + 311 R23 = (5 + 23)2 ∕2 = 24 + 523
R27 = (5 + 27)2 ∕2 = 26 + 527 R47 = (7 + 47)2 ∕2 = 48 + 747
R51 = (7 + 51)2 ∕2 = 50 + 751 R79 = (9 + 79)2 ∕2 = 80 + 979
R83 = (9 + 83)2 ∕2 = 82 + 983 R119 = (11 + 119)2 ∕2 = 120 + 11119
R123 = (11 + 123)2 ∕2 = 122 + 11123 R167 = (13 + 167)2 ∕2 = 168 + 13167
R171 = (13 + 171)2 ∕2 = 170 + 13171 R223 = (15 + 223)2 ∕2 = 224 + 15223
R227 = (15 + 227)2 ∕2 = 226 + 15227 R287 = (17 + 287)2 ∕2 = 288 + 17287
R291 = (17 + 291)2 ∕2 = 290 + 17291 R359 = (19 + 359)2 ∕2 = 360 + 19359
R363 = (19 + 363)2 ∕2 = 362 + 19363 R439 = (21 + 439)2 ∕2 = 440 + 21439

This concludes Part XIX. Go to Part XIXB Continuation. To go back to Part XVII.

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