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.

Showing 1-4 of 4 results.

A001079 a(n) = 10*a(n-1) - a(n-2); a(0) = 1, a(1) = 5.

Original entry on oeis.org

1, 5, 49, 485, 4801, 47525, 470449, 4656965, 46099201, 456335045, 4517251249, 44716177445, 442644523201, 4381729054565, 43374646022449, 429364731169925, 4250272665676801, 42073361925598085, 416483346590304049
Offset: 0

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Keywords

Comments

Also gives solutions to the equation x^2-1=floor(x*r*floor(x/r)) where r=sqrt(6). - Benoit Cloitre, Feb 14 2004
Appears to give all solutions >1 to the equation x^2=ceiling(x*r*floor(x/r)) where r=sqrt(6). - Benoit Cloitre, Feb 24 2004
a(n) and b(n) (A004189) are the nonnegative proper solutions to the Pell equation a(n)^2 - 6*(2*b(n))^2 = +1, n >= 0. The formula given below by Gregory V. Richardson follows. - Wolfdieter Lang, Jun 26 2013
a(n) are the integer square roots of (A032528 + 1). They are also the values of m where (A032528(m) - 1) has integer square roots. See A122653 for the integer square roots of (A032528 - 1), and see A122652 for the values of m where (A032528(m) + 1) has integer square roots. - Richard R. Forberg, Aug 05 2013
a(n) are also the values of m where floor(2m^2/3) has integer square roots, excluding m = 0. The corresponding integer square roots are given by A122652(n). - Richard R. Forberg, Nov 21 2013
Except for the first term, positive values of x (or y) satisfying x^2 - 10xy + y^2 + 24 = 0. - Colin Barker, Feb 09 2014
Dickson on page 384 gives the Diophantine equation "24x^2 + 1 = y^2" and later states "y_{n+1} = 10y_n - y_{n-1}" where y_n is this sequence. - Michael Somos, Jun 19 2023

Examples

			Pell equation: n = 0: 1^2 - 24*0^2 = +1, n = 1: 5^2 - 6*(1*2)^2 = 1, n = 2: 49^2 - 6*(2*10)^2 = +1. - _Wolfdieter Lang_, Jun 26 2013
G.f. = 1 + 5*x + 49*x^2 + 485*x^3 + 4801*x^4 + 47525*x^5 + 470449*x^6 + ...
		

References

  • Bastida, Julio R. Quadratic properties of a linearly recurrent sequence. Proceedings of the Tenth Southeastern Conference on Combinatorics, Graph Theory and Computing (Florida Atlantic Univ., Boca Raton, Fla., 1979), pp. 163-166, Congress. Numer., XXIII-XXIV, Utilitas Math., Winnipeg, Man., 1979. MR0561042 (81e:10009) - From N. J. A. Sloane, May 30 2012
  • L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, p. 384.
  • L. Euler, (E388) Vollstaendige Anleitung zur Algebra, Zweiter Theil, reprinted in: Opera Omnia. Teubner, Leipzig, 1911, Series (1), Vol. 1, p. 374.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • V. Thébault, Les Récréations Mathématiques. Gauthier-Villars, Paris, 1952, p. 281.

Crossrefs

Programs

  • Magma
    I:=[1,5]; [n le 2 select I[n] else 10*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 10 2016
    
  • Maple
    A001079 := proc(n)
        option remember;
        if n <= 1 then
            op(n+1,[1,5]) ;
        else
            10*procname(n-1)-procname(n-2) ;
        end if;
    end proc:
    seq(A001079(n),n=0..20) ; # R. J. Mathar, Apr 30 2017
  • Mathematica
    Table[(-1)^n Round[N[Cos[2 n ArcSin[Sqrt[3]]], 50]], {n, 0, 20}] (* Artur Jasinski, Oct 29 2008 *)
    a[ n_] := ChebyshevT[n, 5]; (* Michael Somos, Aug 24 2014 *)
    CoefficientList[Series[(1-5*x)/(1-10*x+x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 20 2017 *)
    a[n_] := 3^n*Sum[(2/3)^k*Binomial[2*n, 2*k], {k,0,n}]; Flatten[Table[a[n], {n,0,18}]] (* Detlef Meya, May 21 2024 *)
  • PARI
    {a(n) = subst(poltchebi(n), 'x, 5)}; /* Michael Somos, Sep 05 2006 */
    
  • PARI
    {a(n) = real((5 + 2*quadgen(24))^n)}; /* Michael Somos, Sep 05 2006 */
    
  • PARI
    {a(n) = n = abs(n); polsym(1 - 10*x + x^2, n)[n+1] / 2}; /* Michael Somos, Sep 05 2006 */
    
  • PARI
    x='x+O('x^30); Vec((1-5*x)/(1-10*x+x^2)) \\ G. C. Greubel, Dec 20 2017

Formula

For all members x of the sequence, 6*x^2 -6 is a square. Limit_{n->infinity} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 13 2002
a(n) = T(n, 5) = (S(n, 10)-S(n-2, 10))/2 with S(n, x) := U(n, x/2) and T(n), resp. U(n, x), are Chebyshev's polynomials of the first, resp. second, kind. See A053120 and A049310. S(n, 10) = A004189(n+1).
a(n) = sqrt(1+24*A004189(n)^2) (cf. Richardson comment).
a(n)*a(n+3) - a(n+1)*a(n+2) = 240. - Ralf Stephan, Jun 06 2005
Chebyshev's polynomials T(n,x) evaluated at x=5.
G.f.: (1-5*x)/(1-10*x+x^2). - Simon Plouffe in his 1992 dissertation
a(n)= ((5+2*sqrt(6))^n + (5-2*sqrt(6))^n)/2.
a(-n) = a(n).
a(n+1) = 5*a(n) + 2*(6*a(n)^2-6)^(1/2) - Richard Choulet, Sep 19 2007
(sqrt(2)+sqrt(3))^(2*n)=a(n)+A001078(n)*sqrt(6). - Reinhard Zumkeller, Mar 12 2008
a(n+1) = 2*A054320(n) + 3*A138288(n). - Reinhard Zumkeller, Mar 12 2008
a(n) = cosh(2*n* arcsinh(sqrt(2))). - Herbert Kociemba, Apr 24 2008
a(n) = (-1)^n * cos(2*n* arcsin(sqrt(3))). - Artur Jasinski, Oct 29 2008
a(n) = cos(2*n* arccos(sqrt(3))). - Artur Jasinski, Sep 10 2016
a(n) = A142238(2n-1) = A041006(2n-1) = A041038(2n-1), for all n > 0. - M. F. Hasler, Feb 14 2009
2*a(n)^2 = 3*A122652(n)^2 + 2. - Charlie Marion, Feb 01 2013
E.g.f.: cosh(2*sqrt(6)*x)*exp(5*x). - Ilya Gutkovskiy, Sep 10 2016
From Peter Bala, Aug 17 2022: (Start)
a(n) = (1/2)^n * [x^n] ( 10*x + sqrt(1 + 96*x^2) )^n.
The g.f. A(x) satisfies A(2*x) = 1 + x*B'(x)/B(x), where B(x) = 1/sqrt(1 - 20*x + 4*x^2) is the g.f. of A098270.
The Gauss congruences a(n*p^k) == a(n*p^(k-1)) (mod p^k) hold for all primes p >= 3 and positive integers n and k.
Sum_{n >= 1} 1/(a(n) - 3/a(n)) = 1/4.
Sum_{n >= 1} (-1)^(n+1)/(a(n) + 2/a(n)) = 1/6.
Sum_{n >= 1} 1/(a(n)^2 - 3) = 1/4 - 1/sqrt(24). (End)
a(n) = 3^n*Sum_{k=0..n} (2/3)^k*binomial(2*n, 2*k). - Detlef Meya, May 21 2024

Extensions

Chebyshev comments from Wolfdieter Lang, Nov 08 2002

A054320 Expansion of g.f.: (1 + x)/(1 - 10*x + x^2).

Original entry on oeis.org

1, 11, 109, 1079, 10681, 105731, 1046629, 10360559, 102558961, 1015229051, 10049731549, 99482086439, 984771132841, 9748229241971, 96497521286869, 955226983626719, 9455772314980321, 93602496166176491, 926569189346784589, 9172089397301669399, 90794324783669909401
Offset: 0

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Comments

Chebyshev's even-indexed U-polynomials evaluated at sqrt(3).
a(n)^2 is a star number (A003154).
Any k in the sequence has the successor 5*k + 2*sqrt(3(2*k^2 + 1)). - Lekraj Beedassy, Jul 08 2002
{a(n)} give the values of x solving: 3*y^2 - 2*x^2 = 1. Corresponding values of y are given by A072256(n+1). x + y = A001078(n+1). - Richard R. Forberg, Nov 21 2013
The aerated sequence (b(n))n>=1 = [1, 0, 11, 0, 109, 0, 1079, 0, ...] is a fourth-order linear divisibility sequence; that is, if n | m then b(n) | b(m). It is the case P1 = 0, P2 = -8, Q = -1 of the 3-parameter family of divisibility sequences found by Williams and Guy. See A100047. - Peter Bala, Mar 22 2015

Examples

			a(1)^2 = 121 is the 5th star number (A003154).
		

Crossrefs

A member of the family A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, which are the expansions of (1+x) / (1-kx+x^2) with k = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10. - Philippe Deléham, May 04 2004
Cf. A138281. Cf. A100047.
Cf. A142238.

Programs

  • GAP
    a:=[1,11];; for n in [3..30] do a[n]:=10*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Jul 22 2019
  • Magma
    I:=[1,11]; [n le 2 select I[n] else 10*Self(n-1)-Self(n-2): n in [1..30]]; // Vincenzo Librandi, Mar 22 2015
    
  • Mathematica
    CoefficientList[Series[(1+x)/(1-10x+x^2), {x,0,30}], x] (* Vincenzo Librandi, Mar 22 2015 *)
    a[c_, n_] := Module[{},
       p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
       d := Numerator[Convergents[Sqrt[c], n p]];
       t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
       Return[t];
    ] (* Complement of A142238 *)
    a[3/2, 20] (* Gerry Martens, Jun 07 2015 *)
  • PARI
    a(n)=subst(poltchebi(n+1)-poltchebi(n),x,5)/4;
    

Formula

(a(n)-1)^2 + a(n)^2 + (a(n)+1)^2 = b(n)^2 + (b(n)+1)^2 = c(n), where b(n) is A031138 and c(n) is A007667.
a(n) = 10*a(n-1) - a(n-2).
a(n) = (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1).
a(n) = U(2*(n-1), sqrt(3)) = S(n-1, 10) + S(n-2, 10) with Chebyshev's U(n, x) and S(n, x) := U(n, x/2) polynomials and S(-1, x) := 0. S(n, 10) = A004189(n+1), n >= 0.
6*a(n)^2 + 3 is a square. Limit_{n->oo} a(n)/a(n-1) = 5 + 2*sqrt(6). - Gregory V. Richardson, Oct 13 2002
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i), then (-1)^n*q(n, -12) = a(n). - Benoit Cloitre, Nov 10 2002
a(n) = L(n,-10)*(-1)^n, where L is defined as in A108299; see also A072256 for L(n,+10). - Reinhard Zumkeller, Jun 01 2005
From Reinhard Zumkeller, Mar 12 2008: (Start)
(sqrt(2) + sqrt(3))^(2*n+1) = a(n)*sqrt(2) + A138288(n)*sqrt(3);
a(n) = A138288(n) + A001078(n).
a(n) = A001079(n) + 3*A001078(n). (End)
a(n) = A142238(2n) = A041006(2n)/2 = A041038(2n)/4. - M. F. Hasler, Feb 14 2009
a(n) = sqrt(A006061(n)). - Zak Seidov, Oct 22 2012
a(n) = sqrt((3*A072256(n)^2 - 1)/2). - T. D. Noe, Oct 23 2012
(sqrt(3) + sqrt(2))^(2*n+1) - (sqrt(3) - sqrt(2))^(2*n+1) = a(n)*sqrt(8). - Bruno Berselli, Oct 29 2019
a(n) = A004189(n)+A004189(n+1). - R. J. Mathar, Oct 01 2021
E.g.f.: exp(5*x)*(2*cosh(2*sqrt(6)*x) + sqrt(6)*sinh(2*sqrt(6)*x))/2. - Stefano Spezia, May 16 2023
From Peter Bala, May 09 2025: (Start)
a(n) = Dir(n, 5), where Dir(n, x) denotes the n-th row polynomial of the triangle A244419.
a(n)^2 - 10*a(n)*a(n+1) + a(n+1)^2 = 12.
More generally, for arbitrary x, a(n+x)^2 - 10*a(n+x)*a(n+x+1) + a(n+x+1)^2 = 12 with a(n) := (sqrt(6) - 2)/4*(5 + 2*sqrt(6))^(n+1) - (sqrt(6) + 2)/4*(5 - 2*sqrt(6))^(n+1) as given above.
a(n+1/2) = sqrt(3) * A001078(n+1).
a(n+3/4) + a(n+1/4) = sqrt(6)*sqrt(sqrt(3) + 1) * A001078(n+1).
a(n+3/4) - a(n+1/4) = sqrt(sqrt(3) - 1) * A001079(n+1).
Sum_{n >= 1} (-1)^(n+1)/(a(n) - 1/a(n)) = 1/12 (telescoping series: for n >= 1, 1/(a(n) - 1/a(n)) = 1/A004291(n) + 1/A004291(n+1)).
Product_{n >= 1} (a(n) + 1)/(a(n) - 1) = sqrt(3/2) (telescoping product: Product_{n = 1..k} ((a(n) + 1)/(a(n) - 1))^2 = 3/2 * (1 - 1/A171640(k+2))). (End)

Extensions

Chebyshev comments from Wolfdieter Lang, Oct 31 2002

A001078 a(n) = 10*a(n-1) - a(n-2) with a(0) = 0, a(1) = 2.

Original entry on oeis.org

0, 2, 20, 198, 1960, 19402, 192060, 1901198, 18819920, 186298002, 1844160100, 18255302998, 180708869880, 1788833395802, 17707625088140, 175287417485598, 1735166549767840, 17176378080192802, 170028614252160180, 1683109764441408998
Offset: 0

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Keywords

Comments

Also 6*x^2+1 is a square. - Cino Hilliard, Mar 08 2003
This sequence has the following property: For each n, if A = a(n), B = 2*a(n+1), C = 3*a(n+1) then A*B+1, A*C+1, B*C+1 are perfect squares. - Deshpande M.N. (dpratap_ngp(AT)sancharnet.in), Sep 22 2004
n such that 6*n^2 = floor(sqrt(6)*n*ceiling(sqrt(6)*n)). - Benoit Cloitre, May 10 2003
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 19 2005
(sqrt(2) + sqrt(3))^(2*n) = A001079(n) + a(n)*sqrt(6); a(n) = A054320(n) + A138288(n). - Reinhard Zumkeller, Mar 12 2008
Numbers m such that A000217(m) plus A000326(m) equals an octagonal number (A000567). For a(3)=198, A000217(198)=19701, A000326(198)=58707, therefore 19701 + 58707 = 78408 = A000567(162). - Bruno Berselli, Apr 15 2013

References

  • O. Bottema: Verscheidenheden XXVI. Het vraagstuk van Malfatti, Euclides 25 (1949-50), pp. 144-149 [in Dutch].
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (pp. 283, 302, P_{16}).
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • V. Thébault, Les Récréations Mathématiques. Gauthier-Villars, Paris, 1952, p. 281.

Crossrefs

Programs

  • Haskell
    a001078 n = a001078_list !! n
    a001078_list =
       0 : 2 : zipWith (-) (map (10*) $ tail a001078_list) a001078_list
    -- Reinhard Zumkeller, Mar 18 2011
    
  • Magma
    I:=[0, 2]; [n le 2 select I[n] else 10*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Dec 19 2017
  • Maple
    A001078 := proc(n) option remember; if n=0 then 0 elif n=1 then 2 else 10*A001078(n-1)-A001078(n-2); fi; end;
    A001078:=2*z/(1-10*z+z**2); # conjectured by Simon Plouffe in his 1992 dissertation
  • Mathematica
    a[0]=0; a[1]=2; a[n_] := a[n] = 10*a[n-1] - a[n-2]; Table[a[n],{n,0,19}] (* Jean-François Alcover, Mar 18 2011 *)
    LinearRecurrence[{10,-1},{0,2},20] (* Harvey P. Dale, Jun 23 2011 *)
    CoefficientList[Series[2*x/(1 - 10*x + x^2), {x, 0, 50}], x] (* G. C. Greubel, Dec 19 2017 *)
  • PARI
    nxsqp1(m,n) = { for(x=1,m, y = n*x*x+1; if(issquare(y),print1(x" ")) ) }
    
  • PARI
    a(n)=imag((5+2*quadgen(24))^n) /* Michael Somos, Jul 05 2005 */
    
  • PARI
    a(n)=subst(poltchebi(n+1)-5*poltchebi(n),x,5)/12 /* Michael Somos, Jul 05 2005 */
    
  • PARI
    x='x+O('x^30); concat([0], Vec(2*x/(1 - 10*x + x^2))) \\ G. C. Greubel, Dec 19 2017
    

Formula

From Emeric Deutsch, Jun 19 2005: (Start)
a(n) = ((5 + 2*sqrt(6))^n - (5 - 2*sqrt(6))^n)/(2*sqrt(6)).
G.f.: 2*z/(1 - 10*z + z^2). (End)
a(-n) = -a(n).
From Mohamed Bouhamida, Sep 20 2006: (Start)
a(n) = 9*(a(n-1) + a(n-2)) - a(n-3).
a(n) = 11*(a(n-1) - a(n-2)) + a(n-3). (End)
a(n+1) = A054320(n) + A138288(n). - Reinhard Zumkeller, Mar 12 2008
a(n) = sinh(2n*arcsinh(sqrt(2)))/sqrt(6). - Herbert Kociemba, Apr 24 2008
a(n) = 2*A004189(n). - R. J. Mathar, Oct 26 2009
E.g.f.: 2*exp(5*x)*sinh(2*sqrt(6)*x)/(2*sqrt(6)). - Stefano Spezia, May 16 2023

Extensions

Thanks to Antreas P. Hatzipolakis (xpolakis(AT)otenet.gr) and Floor van Lamoen for the Bottema references.

A138288 a(n) = A054320(n) - A001078(n).

Original entry on oeis.org

1, 9, 89, 881, 8721, 86329, 854569, 8459361, 83739041, 828931049, 8205571449, 81226783441, 804062262961, 7959395846169, 78789896198729, 779939566141121, 7720605765212481, 76426118085983689, 756540575094624409, 7488979632860260401, 74133255753507979601, 733843577902219535609
Offset: 0

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Author

Reinhard Zumkeller, Mar 12 2008

Keywords

Comments

Numbers k such that 6*k^2 - 2 is a square. - Bruno Berselli, Feb 10 2014

Examples

			1 + 9*x + 89*x^2 + 881*x^3 + 8721*x^4 + 86329*x^5 + ...
		

References

  • H. Brocard, Note #2049, L'Intermédiaire des Mathématiciens, 8 (1901), pp. 212-213. - N. J. A. Sloane, Mar 02 2022

Crossrefs

Cf. similar sequences listed in A238379.

Programs

  • Mathematica
    CoefficientList[Series[(1 - x)/(1 - 10 x + x^2), {x, 0, 40}], x] (* Vincenzo Librandi, Feb 12 2014 *)
    a[c_, n_] := Module[{},
      p := Length[ContinuedFraction[ Sqrt[ c]][[2]]];
      d := Denominator[Convergents[Sqrt[c], n p]];
      t := Table[d[[1 + i]], {i, 0, Length[d] - 1, p}];
      Return[t];
      ] (* Complement of A041007, A041039 *)
    a[6, 20] (* Gerry Martens, Jun 07 2015 *)
  • PARI
    {a(n) = subst( poltchebi(n+1) + poltchebi(n), x, 5) / 6} /* Michael Somos, Jan 25 2013 */
  • Sage
    [lucas_number1(n,10,1)-lucas_number1(n-1,10,1) for n in range(1, 20)] # Zerinvary Lajos, Nov 10 2009
    

Formula

a(n) = A072256(n+1).
a(n) = A001079(n) + 2*A001078(n).
a(n) = 10*a(n-1) - a(n-2). a(-1) = a(0) = 1.
(sqrt(2)+sqrt(3))^(2*n+1) = A054320(n-1)*sqrt(2) + a(n)*sqrt(3).
From Michael Somos, Jan 25 2013: (Start)
G.f.: (1 - x) / (1 - 10*x + x^2).
a(-1-n) = a(n). (End)
a(n) = sqrt(2+(5-2*sqrt(6))^(1+2*n)+(5+2*sqrt(6))^(1+2*n))/(2*sqrt(3)). - Gerry Martens, Jun 04 2015
E.g.f.: exp(5*x)*(3*cosh(2*sqrt(6)*x) + sqrt(6)*sinh(2*sqrt(6)*x))/3. - Stefano Spezia, May 16 2023
Showing 1-4 of 4 results.