A158598 -id:A158598 - cp's OEIS frontend

A158597 a(n) = 400*n^2 - 20.

380, 1580, 3580, 6380, 9980, 14380, 19580, 25580, 32380, 39980, 48380, 57580, 67580, 78380, 89980, 102380, 115580, 129580, 144380, 159980, 176380, 193580, 211580, 230380, 249980, 270380, 291580, 313580, 336380, 359980, 384380, 409580, 435580, 462380, 489980, 518380

Offset: 1

Vincenzo Librandi, Mar 22 2009

The identity (40*n^2 - 1)^2 - (400*n^2 - 20)*(2*n)^2 = 1 can be written as A158598(n)^2 - a(n)*A005843(n)^2 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005843, A158491, A158598.

I:=[380, 1580, 3580]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 16 2012

LinearRecurrence[{3, -3, 1}, {380, 1580, 3580}, 50] (* Vincenzo Librandi, Feb 16 2012 *)
400*Range[40]^2-20 (* Harvey P. Dale, Nov 04 2015 *)

for(n=1, 40, print1(400*n^2- 20", ")); \\ Vincenzo Librandi, Feb 16 2012

G.f.: 20*x*(-19 - 22*x + x^2)/(x-1)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From Amiram Eldar, Mar 14 2023: (Start)
Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)))/40.
Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(2*sqrt(5)))*Pi/(2*sqrt(5)) - 1)/40. (End)
From Elmo R. Oliveira, Jan 16 2025: (Start)
E.g.f.: 20*(exp(x)*(20*x^2 + 20*x - 1) + 1).
a(n) = 20*A158491(n). (End)

Comment rewritten, formula replaced by R. J. Mathar, Oct 28 2009

A158773 a(n) = 1600*n^2 - 40.

Original entry on oeis.org

1560, 6360, 14360, 25560, 39960, 57560, 78360, 102360, 129560, 159960, 193560, 230360, 270360, 313560, 359960, 409560, 462360, 518360, 577560, 639960, 705560, 774360, 846360, 921560, 999960, 1081560, 1166360, 1254360, 1345560, 1439960, 1537560, 1638360, 1742360

Offset: 1

Vincenzo Librandi, Mar 26 2009

The identity (80*n^2 - 1)^2 - (1600*n^2 - 40)*(2*n)^2 = 1 can be written as A158774(n)^2 - a(n)*A005843(n)^2 = 1.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Vincenzo Librandi, X^2-AY^2=1, Math Forum, 2007. [Wayback Machine link]
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A005843, A158598, A158774.

I:=[1560, 6360, 14360]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Feb 21 2012

LinearRecurrence[{3, -3, 1}, {1560, 6360, 14360}, 50] (* Vincenzo Librandi, Feb 21 2012 *)
1600*Range[30]^2-40 (* Harvey P. Dale, Apr 30 2017 *)

for(n=1, 40, print1(1600*n^2 - 40", ")); \\ Vincenzo Librandi, Feb 21 2012

G.f.: 40*x*(-39 - 42*x + x^2)/(x-1)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
From Amiram Eldar, Mar 24 2023: (Start)
Sum_{n>=1} 1/a(n) = (1 - cot(Pi/(2*sqrt(10)))*Pi/(2*sqrt(10)))/80.
Sum_{n>=1} (-1)^(n+1)/a(n) = (cosec(Pi/(2*sqrt(10)))*Pi/(2*sqrt(10)) - 1)/80. (End)
From Elmo R. Oliveira, Jan 26 2025: (Start)
E.g.f.: 40*(exp(x)*(40*x^2 + 40*x - 1) + 1).
a(n) = 40*A158598(n). (End)

Comment rewritten and formula replaced by R. J. Mathar, Oct 22 2009

cp's OEIS Frontend

A158597 a(n) = 400*n^2 - 20.

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