cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A173175 a(n) = sinh^2( 2n*arcsinh(sqrt n)).

Original entry on oeis.org

0, 8, 2400, 1825200, 2687489280, 6503780163000, 23436548406180000, 117725514040791821024, 786292024016459316676608, 6739465778247681589030301160, 72110357818535214970387726284000, 942092946853627620313318842336862608, 14758709413836719039368938494112056160000
Offset: 0

Views

Author

Artur Jasinski, Feb 11 2010

Keywords

Crossrefs

Programs

  • Maple
    A173175 := proc(n) sinh(2*n*arcsinh(sqrt(n))) ; %^2 ; expand(%); simplify(%) ; end proc: # R. J. Mathar, Feb 26 2011
  • Mathematica
    Table[Round[N[Sinh[(2 n) ArcSinh[Sqrt[n]]]^2, 100]], {n, 0, 20}]
  • PARI
    {a(n) = (polchebyshev(2*n, 1, 2*n+1)-1)/2} \\ Seiichi Manyama, Jan 02 2019
    
  • PARI
    {a(n) = 1/2*(-1+sum(k=0, 2*n, binomial(4*n, 2*k)*(n+1)^(2*n-k)*n^k))} \\ Seiichi Manyama, Jan 02 2019

Formula

From Seiichi Manyama, Jan 02 2019: (Start)
a(n) = A322699(n,2*n).
a(n) = (T_{2*n}(2*n+1) - 1)/2 where T_{n}(x) is a Chebyshev polynomial of the first kind.
a(n) = 1/2 * (-1 + Sum_{k=0..2*n} binomial(4*n,2*k)*(n+1)^(2*n-k)*n^k). (End)
a(n) ~ exp(1) * 2^(4*n - 2) * n^(2*n). - Vaclav Kotesovec, Jan 02 2019

Extensions

a(11)-a(12) from Seiichi Manyama, Jan 02 2019

A382209 Numbers k such that 10+k and 10*k are perfect squares.

Original entry on oeis.org

90, 136890, 197402490, 284654260890, 410471246808090, 591899253243012090, 853518312705176632890, 1230772815021611461622490, 1774773545742851022483004890, 2559222222188376152809031436090, 3690396669622092669499600847844090, 5321549438372835441042271613559748890
Offset: 1

Views

Author

Emilio Martín, Mar 18 2025

Keywords

Comments

The limit of a(n+1)/a(n) is 1441.99930651839... = 721+228*sqrt(10) = (19+6*sqrt(10))^2.
If 10*A158490(n) is a perfect square, then A158490(n) is a term.

Examples

			90 is a term because 10+90=100 is a square and 10*90=900 is a square.
(3,1) is a solution to x^2 - 10*y^2 = -1, from which a(n) = 100*y^2-10 = 10*x^2 = 90.
		

Crossrefs

Subsequence of A158490.
Cf. A383734 = 2*A008843 (2+k and 2*k are squares).
Cf. 5*A075796^2 (5+k and 5*k are squares).
Cf. 5*A081071 (20+k and 20*k are squares).
Cf. A245226 (m such that k+m and k*m are squares).

Programs

  • Mathematica
    CoefficientList[Series[ 90*(1 + 78*x + x^2)/((1 - x)*(1 - 1442*x + x^2)),{x,0,11}],x] (* or *) LinearRecurrence[{1443,-1443,1},{90,136890,197402490},12] (* James C. McMahon, May 08 2025 *)
  • Python
    from itertools import islice
    def A382209_gen(): # generator of terms
        x, y = 30, 10
        while True:
            yield x**2//10
            x, y = x*19+y*60, x*6+y*19
    A382209_list = list(islice(A382209_gen(),30)) # Chai Wah Wu, Apr 24 2025

Formula

a(n) = 10 * ((1/2) * (3+sqrt(10))^(2*n-1) + (1/2) * (3-sqrt(10))^(2*n-1))^2.
a(n) = 10 * (sinh((2n-1) * arcsinh(3)))^2.
a(n) = 10 * A173127(n)^2 = 100 * A097315(n)^2 - 10 (negative Pell's equation solutions).
a(n+2) = 1442 * a(n+1) - a(n) + 7200.
G.f.: 90*(1 + 78*x + x^2)/((1 - x)*(1 - 1442*x + x^2)). - Stefano Spezia, Apr 24 2025

A173194 a(n) = -sin^2 (2*n*arccos n) = - sin^2 (2*n*arcsin n).

Original entry on oeis.org

0, 0, 9408, 384199200, 54471499791360, 20405558846592060000, 16793517249722147195701440, 26730228454204365035835498694848, 75019085697452515216001640927169855488, 346154755746154620929434271983392498083891520
Offset: 0

Views

Author

Artur Jasinski, Feb 12 2010

Keywords

Crossrefs

Programs

  • Maple
    A173194 := proc(n) ((n+sqrt(n^2-1))^(2*n)-(n-sqrt(n^2-1))^(2*n))^2 ; expand(%/4) ; simplify(%) ; end proc: # R. J. Mathar, Feb 26 2011
  • Mathematica
    Round[Table[ -N[Sin[2 n ArcSin[n]], 100]^2, {n, 0, 15}]] (* Artur Jasinski *)
    Table[FullSimplify[(-1/2 (x - Sqrt[ -1 + x^2])^(2 x) + 1/2 (x + Sqrt[ -1 + x^2])^(2 x))^2], {x, 0, 7}] (* Artur Jasinski, Feb 17 2010 *)
    Table[(n^2-1)*ChebyshevU[2*n-1, n]^2, {n, 0, 20}] (* Vaclav Kotesovec, Jan 05 2019 *)
  • PARI
    {a(n) = (n^2-1)*n^2*(sum(k=0, n-1, binomial(2*n, 2*k+1)*(n^2-1)^(n-1-k)*n^(2*k)))^2} \\ Seiichi Manyama, Jan 05 2019
    
  • PARI
    {a(n) = (n^2-1)*polchebyshev(2*n-1, 2, n)^2} \\ Seiichi Manyama, Jan 05 2019

Formula

4*a(n) = ( (n+sqrt(n^2-1))^(2*n) - (n-sqrt(n^2-1))^(2*n) )^2. - Artur Jasinski, Feb 17 2010
From Seiichi Manyama, Jan 05 2019: (Start)
a(n) = (n^2-1) * n^2 * (Sum_{k=0..n-1} binomial(2*n,2*k+1)*(n^2-1)^(n-1-k)*n^(2*k))^2.
For n > 0, a(n) = (n^2-1) * U_{2*n-1}(n)^2 where U_{n}(x) is a Chebyshev polynomial of the second kind. (End)
a(n) ~ 2^(4*n - 2) * n^(4*n). - Vaclav Kotesovec, Jan 05 2019

Extensions

a(9) from Seiichi Manyama, Jan 05 2019

A173150 a(n) = sinh^2 (2n*arccosh(sqrt n)).

Original entry on oeis.org

0, 0, 288, 235224, 354079488, 865363202000, 3134808545188320, 15796198853361763368, 105717380511014096025600, 907380314352243226001152800, 9718304978537581699085289156000
Offset: 0

Views

Author

Artur Jasinski, Feb 11 2010

Keywords

Comments

Also a(n) = -sin(2n*arccos(sqrt(n)))^2 = -sin(2n*arcsin(sqrt(n)))^2.

Crossrefs

Programs

  • Maple
    A173150 := proc(n) sinh(2*n*arccosh(sqrt(n))) ; %^2 ; expand(%) ; simplify(%) ;end proc: # R. J. Mathar, Feb 26 2011
  • Mathematica
    Table[Round[-Sin[2 n ArcCos[Sqrt[n]]]^2], {n, 0, 20}] (* Artur Jasinski, Feb 11 2010 *)

Extensions

Minor edits by Vaclav Kotesovec, Apr 05 2016
Previous Showing 11-14 of 14 results.