Triangular and the Odd Numbers and the Pellian Equation(Part XIXC)

Continuation: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, a continuation of Part XIXB, 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 in Part XVII for 4(n)(n+3) + 5 and in Part IIIA for 4(n)(n+1) + 5. Six other sequences with similar equations are displayed in Part XIXB and these two new sequences in this article are part of this group, in numerical order:

Sn-3 = 36(n)(n+1) − 3
−3, 69, 213, 429, 717, 1077, 1509, 2013, 2589, 3237, 3957, ...

Sn21 = 36(n)(n+1) + 21
21, 93, 237, 453, 741, 1101, 1533, 2037, 2613, 3261, 3981, ...

where all the sequences were evaluated at n ≥ 0. The two sequences will be shown to behave like Sn5 and Sn13 of Part XIXB where one of their equivalents is based on the square root of a cubed expression.

All the terms above (the D values) were entered into the Pell calculator program to give values of Qn = +1 for different values of n. 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: for 717 at Code 717 and for 741 Code 741. The red number −3 although generated by the equation in sequence Sn-3 being negative is not usable. The two numbers 21 and 69 have their Qn = 1 at n = 7 and 9, respectively (see Code 21 and Code 69), which differ from Qn = 1 at n = 11 and 13, respectively for the rest of the Ds in Sn21 and Sn-3.

Tables of x, y and D values

Tables I and II show the first 11 D numbers in the two different sequences Sn. Although the values of D equal 69 and 21 in Tables I and II, respectively, they are tabulated even though they do not have the requisite Qn = 1 at n = 13 and 11, respectively, shared with the other D terms in the equations.

Table I (Sn-3)
D−36921342971710771509201325893237
x-7775194399152409569983992352239964393055152409599 323606015631605599
y-9361332073584261360716760165765633969606359904 11101320
yf-3(312)5(2664)7(10512)9(29040)11(65160)13(127512) 15(226464)17(374112)19(584280)
Table II (Sn21)
D219323745374111011533203726133261
x5512151228151165375173526952431301565935351155143351 328116151638642935
y121260148207770027010873273216840203437460 641886011183628
yf1(12)3(420)5(2964)7(11100)9(30012)11(66612) 13(129540)15(229164)17(377580)19(588612)

where the values of y for Sn-3 and Sn21 are the consecutive odd numbers and D and y are divisible by three.

Tables of Pell Equations

The Pell equations are set up according to the following format where RD is equal to the equivalent expressions and where n1 is a constant multiplied by a consecutive triangular number and (2n+1) is a consecutive odd number:

RD = [(n1 + (2n+1)D)2 ∕4]3∕2 = x + yD
where n ≥ 0
(Sn-3) n1 = 24×½(n2 + n) + 1
(Sn21) n1 = 24×½(n2 + n) + 5

where n ≥ 2 for all but 21 and 69.

Tables I and II give the formulas for all 10 Pell equations according to the format listed above with the triangular and odd numbers in blue. Note that the values of n1 are off for R21 and R69.

Table III
(Sn-3) Equal Expressions(Sn21) Equal Expressions
R21 = [(3 + 121)2 ∕4]3/2 = 55 + 1(12)21
R69 = [(25 + 369)2 ∕4]3/2 = 7775 + 3(312)69 R93 = [(29 + 393)2 ∕4]3/2 = 12151 + 3(420)93
R213 = [(73 + 5213)2 ∕4]3/2 = 19439 + 5(2664)213 R237 = [(77 + 5237)2 ∕4]3/2 = 228151 + 5(2964)237
R429 = [(145 + 7429)2 ∕4]3/2 = 1524095 + 7(10512)429 R453 = [(149 + 7453)2 ∕4]3/2 = 1653751 + 7(11100)453
R717 = [(241 + 9717)2 ∕4]3/2 = 6998399 + 9(29040)717 R741 = [(245 + 9741)2 ∕4]3/2 = 7352695 + 9(30012)741
R1077 = [(361 + 111077)2 ∕4]3/2 = 23522399 + 11(65160)1077 R1101 = [(365 + 111101)2 ∕4]3/2 = 24313015 + 11(66612)1101
R1509 = [(505 + 131509)2 ∕4]3/2 = 64393055 + 13(127512)1509 R1533 = [(509 + 131533)2 ∕4]3/2 = 65935351 + 13(129540)1533
R2013 = [(673 + 152013)2 ∕4]3/2 = 152409599 + 15(226464)2013 R2037 = [(677 + 152037)2 ∕4]3/2 = 155143351 + 15(229164)2037
R2589 = [(865 + 172589)2 ∕4]3/2 = 323606015 + 17(374112)2589 R2613 = [(869 + 172613)2 ∕4]3/2 = 328116151 + 17(377580)2613
R3237 = [(1081 + 193237)2 ∕4]3/2 = 631605599 + 19(584280)3237 R3261 = [(1085 + 193261)2 ∕4]3/2 = 638642935 + 19(588612)3261

This concludes Part XIXC. To go back to Part XIXB.

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