A077420 -id:A077420 - cp's OEIS frontend

A280761 Solutions y_n to the negative Pell equation y^2 = 72*x^2 - 8.

8, 280, 9512, 323128, 10976840, 372889432, 12667263848, 430314081400, 14618011503752, 496582077046168, 16869172608065960, 573055286597196472, 19467010571696614088, 661305304151087682520, 22464913330565284591592, 763145747935068588431608

Offset: 0

N. J. A. Sloane, Jan 16 2017

Although this is a list, it has offset zero because one of the references numbered the solutions starting at 0.

Seiichi Manyama, Table of n, a(n) for n = 0..652
S. Vidhyalakshmi, V. Krithika, K. Agalya, On The Negative Pell Equation y^2 = 72*x^2 - 8, International Journal of Emerging Technologies in Engineering Research (IJETER), Volume 4, Issue 2, February (2016).
Index entries for linear recurrences with constant coefficients, signature (34,-1).

For the x_n values see A077420.

I:=[8,280]; [n le 2 select I[n] else 34*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Jan 18 2017

LinearRecurrence[{34, -1}, {8, 280}, 20] (* Vincenzo Librandi, Jan 18 2017 *)

a(n)=([0,1;-1,34]^n*[-8;8])[1,1] \\ Charles R Greathouse IV, Jan 17 2017

G.f.: 8*(1 + x)/(1 - 34*x + x^2). - Ilya Gutkovskiy, Jan 17 2017
a(n) = 34*a(n-1) - a(n-2), a(0)=8, a(1)=280. - Seiichi Manyama, Jan 17 2017
a(n) = (17+12*sqrt(2))^(-n)*(-4-3*sqrt(2) + (-4+3*sqrt(2))*(17+12*sqrt(2))^(2*n)) for n>0. - Colin Barker, Jan 17 2017

More terms from Ilya Gutkovskiy, Jan 17 2017

A278476 a(n) = floor((1 + sqrt(2))^3*a(n-1)) for n>0, a(0) = 1.

Original entry on oeis.org

1, 14, 196, 2757, 38793, 545858, 7680804, 108077113, 1520760385, 21398722502, 301102875412, 4236838978269, 59616848571177, 838872718974746, 11803834914217620, 166092561518021425, 2337099696166517569, 32885488307849267390, 462733936006056261028

Offset: 0

Ilya Gutkovskiy, Nov 23 2016

In general, the ordinary generating function for the recurrence relation b(n) = floor((1 + sqrt(2))^k*b(n - 1)) with n>0 and b(0) = 1, is (1 - x)/(1 - round((1 + sqrt(2))^k)*x + x^2) if k is nonzero even, and (1 - x - x^2)/((1 - x)*(1 - round((1 + sqrt(2))^k)*x - x^2)) if k is odd or k = 0.

G. C. Greubel, Table of n, a(n) for n = 0..869
Index entries for linear recurrences with constant coefficients, signature (15,-13,-1).

Cf. A014176.

Cf. similar sequences with recurrence relation b(n) = floor((1 + sqrt(2))^k*b(n-1)) for n>0, b(0) = 1: A024537 (k = 1), A001653 (k = 2), this sequence (k = 3), A077420 (k = 4), A097733 (k = 6).

m:=25; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1-x-x^2)/((1-x)*(1-14*x-x^2)))); // G. C. Greubel, Oct 10 2018

seq(coeff(series((1-x-x^2)/((1-x)*(1-14*x-x^2)),x,n+1), x, n), n = 0 .. 20); # Muniru A Asiru, Oct 11 2018

RecurrenceTable[{a[0] == 1, a[n] == Floor[(1 + Sqrt[2])^3 a[n - 1]]}, a, {n, 18}]
LinearRecurrence[{15, -13, -1}, {1, 14, 196}, 19]
CoefficientList[Series[(1-x-x^2)/((1-x)*(1-14*x-x^2)), {x,0,50}], x] (* G. C. Greubel, Oct 10 2018 *)

Vec((1 - x - x^2)/((1 - x)*(1 - 14*x - x^2)) + O(x^50)) \\ G. C. Greubel, Nov 24 2016

G.f.: (1 - x - x^2)/((1 - x)*(1 - 14*x - x^2)).
a(n) = 15*a(n-1) - 13*a(n-2) - a(n-3).
a(n) = ((65 - 52*sqrt(2))*(7 - 5*sqrt(2))^n + 13*(5 + 4*sqrt(2))*(7 + 5*sqrt(2))^n + 10)/140.

A280181 Indices of centered 9-gonal numbers (A060544) that are also squares (A000290).

Original entry on oeis.org

1, 17, 561, 19041, 646817, 21972721, 746425681, 25356500417, 861374588481, 29261379507921, 994025528680817, 33767606595639841, 1147104598723073761, 38967788749988868017, 1323757712900898438801, 44968794449880558051201, 1527615253583038075302017

Offset: 1

Colin Barker, Dec 28 2016

Also positive integers y in the solutions to 2*x^2 - 9*y^2 + 9*y - 2 = 0, the corresponding values of x being A046176.

Consider all ordered triples of consecutive integers (k, k+1, k+2) such that k is a square and k+1 is twice a square; then the values of k are the squares of the NSW numbers (A002315), the values of k+1 are twice the squares of the odd Pell numbers (A001653), and the values of k+2 are thrice the terms of this sequence. (See the Example section.) - Jon E. Schoenfield, Sep 06 2019

			17 is in the sequence because the 17th centered 9-gonal number is 1225, which is also the 35th square.
From _Jon E. Schoenfield_, Sep 06 2019: (Start)
The following table illustrates the relationship between the NSW numbers (A002315), the odd Pell numbers (A001653), and the terms of this sequence:
.
  |  A002315(n-1)^2  |   2*A001653(n)^2  |
n |   = 3*a(n) - 2   |    = 3*a(n) - 1   |       3*a(n)
--+------------------+-------------------+-------------------
1 |    1^2 =       1 |   1^2*2 =       2 |      1*3 =       3
2 |    7^2 =      49 |   5^2*2 =      50 |     17*3 =      51
3 |   41^2 =    1681 |  29^2*2 =    1682 |    561*3 =    1683
4 |  239^2 =   57121 | 169^2*2 =   57122 |  19041*3 =   57123
5 | 1393^2 = 1940449 | 985^2*2 = 1940450 | 646817*3 = 1940451
(End)

Colin Barker, Table of n, a(n) for n = 1..650
Index entries for linear recurrences with constant coefficients, signature (35,-35,1).

Cf. A000290, A001653, A002315, A046176, A046177, A056771, A060544, A077420.

Cf. A156164.

LinearRecurrence[{35, -35, 1}, {1, 17, 561}, 50] (* G. C. Greubel, Dec 28 2016 *)

Vec(x*(1 - 18*x + x^2) / ((1 - x)*(1 - 34*x + x^2)) + O(x^20))

a(n) = (6 + (3-2*sqrt(2))*(17+12*sqrt(2))^(-n) + (3+2*sqrt(2))*(17+12*sqrt(2))^n) / 12.
a(n) = 35*a(n-1) - 35*a(n-2) + a(n-3) for n>3.
G.f.: x*(1 - 18*x + x^2) / ((1 - x)*(1 - 34*x + x^2)).
a(n) = (A002315(n-1)^2 + 2)/3 = (2*A001653(n)^2 + 1)/3. - Jon E. Schoenfield, Sep 06 2019
a(n) = A077420(floor((n-1)/2)) * A056771(floor(n/2)). - Jon E. Schoenfield, Sep 08 2019
E.g.f.: -1+(1/12)*(6*exp(x)+(3-2*sqrt(2))*exp((17-12*sqrt(2))*x)+(3+2*sqrt(2))*exp((17+12*sqrt(2))*x)). - Stefano Spezia, Sep 08 2019
Limit_{n->oo} a(n+1)/a(n) = 17 + 12*sqrt(2) = A156164. - Andrea Pinos, Oct 07 2022

A281234 Solutions y to the negative Pell equation y^2 = 72*x^2 - 288 with x,y >= 0.

Original entry on oeis.org

0, 48, 288, 1680, 9792, 57072, 332640, 1938768, 11299968, 65861040, 383866272, 2237336592, 13040153280, 76003583088, 442981345248, 2581884488400, 15048325585152, 87708069022512, 511200088549920, 2979492462277008, 17365754685112128, 101215035648395760

Offset: 1

Colin Barker, Jan 18 2017

The corresponding values of x are in A003499.

			48 is in the sequence because (x, y) = (6,48) is a solution to y^2 = 72*x^2 - 288.

Colin Barker, Table of n, a(n) for n = 1..1000
S. Vidhyalakshmi, V. Krithika, K. Agalya, On The Negative Pell Equation y^2 = 72*x^2 - 8, International Journal of Emerging Technologies in Engineering Research (IJETER), Volume 4, Issue 2, February (2016).
Index entries for linear recurrences with constant coefficients, signature (6,-1).

Cf. A001109, A003499, A077420, A280761.

concat(0, Vec(48*x^2 / (1 - 6*x + x^2) + O(x^25)))

G.f.: 48*x^2 / (1 - 6*x + x^2).
a(n) = 6*a(n-1) - a(n-2) for n>2.
a(n) = 48*A001109(n-1).
a(n) = 6*sqrt(2)*(-(3 - 2*sqrt(2))^n*(3+2*sqrt(2)) + (3 - 2*sqrt(2))*(3 + 2*sqrt(2))^n).

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A280761 Solutions y_n to the negative Pell equation y^2 = 72*x^2 - 8.

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A278476 a(n) = floor((1 + sqrt(2))^3*a(n-1)) for n>0, a(0) = 1.

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A280181 Indices of centered 9-gonal numbers (A060544) that are also squares (A000290).

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A281234 Solutions y to the negative Pell equation y^2 = 72*x^2 - 288 with x,y >= 0.

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