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 1-6 of 6 results.

A002530 a(n) = 4*a(n-2) - a(n-4) for n > 1, a(n) = n for n = 0, 1.

Original entry on oeis.org

0, 1, 1, 3, 4, 11, 15, 41, 56, 153, 209, 571, 780, 2131, 2911, 7953, 10864, 29681, 40545, 110771, 151316, 413403, 564719, 1542841, 2107560, 5757961, 7865521, 21489003, 29354524, 80198051, 109552575, 299303201, 408855776, 1117014753, 1525870529, 4168755811
Offset: 0

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Author

Keywords

Comments

Denominators of continued fraction convergents to sqrt(3), for n >= 1.
Also denominators of continued fraction convergents to sqrt(3) - 1. See A048788 for numerators. - N. J. A. Sloane, Dec 17 2007. Convergents are 1, 2/3, 3/4, 8/11, 11/15, 30/41, 41/56, 112/153, ...
Consider the mapping f(a/b) = (a + 3*b)/(a + b). Taking a = b = 1 to start with and carrying out this mapping repeatedly on each new (reduced) rational number gives the following sequence 1/1, 2/1, 5/3, 7/4, 19/11, ... converging to 3^(1/2). Sequence contains the denominators. The same mapping for N, i.e., f(a/b) = (a + Nb)/(a + b) gives fractions converging to N^(1/2). - Amarnath Murthy, Mar 22 2003
Sqrt(3) = 2/2 + 2/3 + 2/(3*11) + 2/(11*41) + 2/(41*153) + 2/(153*571), ...; the sum of the first 6 terms of this series = 1.7320490367..., while sqrt(3) = 1.7320508075... - Gary W. Adamson, Dec 15 2007
From Clark Kimberling, Aug 27 2008: (Start)
Related convergents (numerator/denominator):
lower principal convergents: A001834/A001835
upper principal convergents: A001075/A001353
intermediate convergents: A005320/A001075
principal and intermediate convergents: A143642/A140827
lower principal and intermediate convergents: A143643/A005246. (End)
Row sums of triangle A152063 = (1, 3, 4, 11, ...). - Gary W. Adamson, Nov 26 2008
From Alois P. Heinz, Apr 13 2011: (Start)
Also number of domino tilings of the 3 X (n-1) rectangle with upper left corner removed iff n is even. For n=4 the 4 domino tilings of the 3 X 3 rectangle with upper left corner removed are:
. ._. . ._. . ._. . ._.
.|__| .|__| .| | | .|___|
| |_| | | | | | ||| |_| |
||__| |||_| ||__| |_|_| (End)
This is the sequence of Lehmer numbers u_n(sqrt(R),Q) with the parameters R = 2 and Q = -1. It is a strong divisibility sequence, that is, gcd(a(n),a(m)) = a(gcd(n,m)) for all natural numbers n and m. - Peter Bala, Apr 18 2014
2^(-floor(n/2))*(1 + sqrt(3))^n = A002531(n) + a(n)*sqrt(3); integers in the real quadratic number field Q(sqrt(3)). - Wolfdieter Lang, Feb 11 2018
Let T(n) = 2^(n mod 2), U(n) = a(n), V(n) = A002531(n), x(n) = V(n)/U(n). Then T(n*m) * U(n+m) = U(n)*V(m) + U(m)*V(n), T(n*m) * V(n+m) = 3*U(n)*U(m) + V(m)*V(n), x(n+m) = (3 + x(n)*x(m))/(x(n) + x(m)). - Michael Somos, Nov 29 2022

Examples

			Convergents to sqrt(3) are: 1, 2, 5/3, 7/4, 19/11, 26/15, 71/41, 97/56, 265/153, 362/209, 989/571, 1351/780, 3691/2131, ... = A002531/A002530 for n >= 1.
1 + 1/(1 + 1/(2 + 1/(1 + 1/2))) = 19/11 so a(5) = 11.
G.f. = x + x^2 + 3*x^3 + 4*x^4 + 11*x^5 + 15*x^6 + 41*x^7 + ... - _Michael Somos_, Mar 18 2022
		

References

  • Serge Lang, Introduction to Diophantine Approximations, Addison-Wesley, New York, 1966.
  • Russell Lyons, A bird's-eye view of uniform spanning trees and forests, in Microsurveys in Discrete Probability, AMS, 1998.
  • I. Niven and H. S. Zuckerman, An Introduction to the Theory of Numbers. 2nd ed., Wiley, NY, 1966, p. 181.
  • Murat Sahin and Elif Tan, Conditional (strong) divisibility sequences, Fib. Q., 56 (No. 1, 2018), 18-31.
  • 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).
  • A. Tarn, Approximations to certain square roots and the series of numbers connected therewith, Mathematical Questions and Solutions from the Educational Times, 1 (1916), 8-12.

Crossrefs

Cf. A002531 (numerators of convergents to sqrt(3)), A048788, A003297.
Bisections: A001353 and A001835.
Cf. A152063.
Analog for sqrt(m): A000129 (m=2), A001076 (m=5), A041007 (m=6), A041009 (m=7), A041011 (m=8), A005668 (m=10), A041015 (m=11), A041017 (m=12), ..., A042935 (m=999), A042937 (m=1000).

Programs

  • Magma
    I:=[0,1,1,3]; [n le 4 select I[n] else 4*Self(n-2) - Self(n-4): n in [1..50]]; // G. C. Greubel, Feb 25 2019
    
  • Maple
    a := proc(n) option remember; if n=0 then 0 elif n=1 then 1 elif n=2 then 1 elif n=3 then 3 else 4*a(n-2)-a(n-4) fi end; [ seq(a(i),i=0..50) ];
    A002530:=-(-1-z+z**2)/(1-4*z**2+z**4); # conjectured (correctly) by Simon Plouffe in his 1992 dissertation
  • Mathematica
    Join[{0},Table[Denominator[FromContinuedFraction[ContinuedFraction[Sqrt[3],n]]], {n,1,50}]] (* Stefan Steinerberger, Apr 01 2006 *)
    Join[{0},Denominator[Convergents[Sqrt[3],50]]] (* or *) LinearRecurrence[ {0,4,0,-1},{0,1,1,3},50] (* Harvey P. Dale, Jan 29 2013 *)
    a[ n_] := If[n<0, -(-1)^n, 1] SeriesCoefficient[ x*(1+x-x^2)/(1-4*x^2+x^4), {x, 0, Abs@n}]; (* Michael Somos, Apr 18 2019 *)
    a[ n_] := ChebyshevU[n-1, Sqrt[-1/2]]*Sqrt[2]^(Mod[n, 2]-1)/I^(n-1) //Simplify; (* Michael Somos, Nov 29 2022 *)
  • PARI
    {a(n) = if( n<0, -(-1)^n * a(-n), contfracpnqn(vector(n, i, 1 + (i>1) * (i%2)))[2, 1])}; /* Michael Somos, Jun 05 2003 */
    
  • PARI
    { for (n=0, 50, a=contfracpnqn(vector(n, i, 1+(i>1)*(i%2)))[2, 1]; write("b002530.txt", n, " ", a); ); } \\ Harry J. Smith, Jun 01 2009
    
  • PARI
    my(w=quadgen(12)); A002530(n)=real((2+w)^(n\/2)*if(bittest(n,0),1-w/3,w/3));
    apply(A002530, [0..30]) \\ M. F. Hasler, Nov 04 2019
    
  • Python
    from functools import cache
    @cache
    def a(n): return [0, 1, 1, 3][n] if n < 4 else 4*a(n-2) - a(n-4)
    print([a(n) for n in range(36)]) # Michael S. Branicky, Nov 13 2022
  • Sage
    (x*(1+x-x^2)/(1-4*x^2+x^4)).series(x, 50).coefficients(x, sparse=False) # G. C. Greubel, Feb 25 2019
    

Formula

G.f.: x*(1 + x - x^2)/(1 - 4*x^2 + x^4).
a(n) = 4*a(n-2) - a(n-4). [Corrected by László Szalay, Feb 21 2014]
a(n) = -(-1)^n * a(-n) for all n in Z, would satisfy the same recurrence relation. - Michael Somos, Jun 05 2003
a(2*n) = a(2*n-1) + a(2*n-2), a(2*n+1) = 2*a(2*n) + a(2*n-1).
From Benoit Cloitre, Dec 15 2002: (Start)
a(2*n) = ((2 + sqrt(3))^n - (2 - sqrt(3))^n)/(2*sqrt(3)).
a(2*n) = A001353(n).
a(2*n-1) = ceiling((1 + 1/sqrt(3))/2*(2 + sqrt(3))^n) = ((3 + sqrt(3))^(2*n - 1) + (3 - sqrt(3))^(2*n - 1))/6^n.
a(2*n-1) = A001835(n). (End)
a(n+1) = Sum_{k=0..floor(n/2)} binomial(n - k, k) * 2^floor((n - 2*k)/2). - Paul Barry, Jul 13 2004
a(n) = Sum_{k=0..floor(n/2)} binomial(floor(n/2) + k, floor((n - 1)/2 - k))*2^k. - Paul Barry, Jun 22 2005
G.f.: (sqrt(6) + sqrt(3))/12*Q(0), where Q(k) = 1 - a/(1 + 1/(b^(2*k) - 1 - b^(2*k)/(c + 2*a*x/(2*x - g*m^(2*k)/(1 + a/(1 - 1/(b^(2*k + 1) + 1 - b^(2*k + 1)/(h - 2*a*x/(2*x + g*m^(2*k + 1)/Q(k + 1)))))))))). - Sergei N. Gladkovskii, Jun 21 2012
a(n) = (alpha^n - beta^n)/(alpha - beta) for n odd, and a(n) = (alpha^n - beta^n)/(alpha^2 - beta^2) for n even, where alpha = 1/2*(sqrt(2) + sqrt(6)) and beta = (1/2)*(sqrt(2) - sqrt(6)). Cf. A108412. - Peter Bala, Apr 18 2014
a(n) = (-sqrt(2)*i)^n*S(n, sqrt(2)*i)*2^(-floor(n/2)) = A002605(n)*2^(-floor(n/2)), n >= 0, with i = sqrt(-1) and S the Chebyshev polynomials (A049310). - Wolfdieter Lang, Feb 10 2018
a(n+1)*a(n+2) - a(n+3)*a(n) = (-1)^n, n >= 0. - Kai Wang, Feb 06 2020
E.g.f.: sinh(sqrt(3/2)*x)*(sinh(x/sqrt(2)) + sqrt(2)*cosh(x/sqrt(2)))/sqrt(3). - Stefano Spezia, Feb 07 2020
a(n) = ((1 + sqrt(3))^n - (1 - sqrt(3))^n)/(2*2^floor(n/2))/sqrt(3) = A002605(n)/2^floor(n/2). - Robert FERREOL, Apr 13 2023

Extensions

Definition edited by M. F. Hasler, Nov 04 2019

A075839 Numbers k such that 11*k^2 - 2 is a square.

Original entry on oeis.org

1, 19, 379, 7561, 150841, 3009259, 60034339, 1197677521, 23893516081, 476672644099, 9509559365899, 189714514673881, 3784780734111721, 75505900167560539, 1506333222617099059, 30051158552174420641, 599516837820871313761, 11960285597865251854579
Offset: 1

Views

Author

Gregory V. Richardson, Oct 14 2002

Keywords

Comments

Lim_{n -> infinity} a(n)/a(n-1) = 10 + 3*sqrt(11).
Positive values of x (or y) satisfying x^2 - 20xy + y^2 + 18 = 0. - Colin Barker, Feb 18 2014

References

  • A. H. Beiler, "The Pellian", ch. 22 in Recreations in the Theory of Numbers: The Queen of Mathematics Entertains. Dover, New York, New York, pp. 248-268, 1966.
  • L. E. Dickson, History of the Theory of Numbers, Vol. II, Diophantine Analysis. AMS Chelsea Publishing, Providence, Rhode Island, 1999, pp. 341-400.
  • Peter G. L. Dirichlet, Lectures on Number Theory (History of Mathematics Source Series, V. 16); American Mathematical Society, Providence, Rhode Island, 1999, pp. 139-147.

Crossrefs

Row 20 of array A094954.
Cf. similar sequences listed in A238379.

Programs

  • GAP
    a:=[1,19];; for n in [3..20] do a[n]:=20*a[n-1]-a[n-2]; od; a; # G. C. Greubel, Dec 06 2019
  • Magma
    I:=[1,19]; [n le 2 select I[n] else 20*Self(n-1)-Self(n-2): n in [1..20]]; // Vincenzo Librandi, Feb 20 2014
    
  • Maple
    seq(coeff(series( x*(1-x)/(1-20*x+x^2), x, n+1), x, n), n = 1..20); # G. C. Greubel, Dec 06 2019
  • Mathematica
    LinearRecurrence[{20,-1},{1,19},20] (* Harvey P. Dale, Apr 13 2012 *)
    Rest@CoefficientList[Series[x*(1-x)/(1-20x+x^2), {x, 0, 20}], x] (* Vincenzo Librandi, Feb 20 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 A041015 *)
    a[11, 20] (* Gerry Martens, Jun 07 2015 *)
  • PARI
    a(n)=subst(poltchebi(n+1)+poltchebi(n),x,10)/11
    
  • Sage
    def A075839_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( x*(1-x)/(1-20*x+x^2) ).list()
    a=A075839_list(20); a[1:] # G. C. Greubel, Dec 06 2019
    

Formula

11*a(n)^2 - 9*A083043(n)^2 = 2.
a(n) = ((3+sqrt(11))*(10+3*sqrt(11))^(n-1) - (3-sqrt(11))*(10-3*sqrt(11))^(n-1) )/(2*sqrt(11)). - Dean Hickerson, Dec 09 2002
From Michael Somos, Oct 29 2002: (Start)
G.f.: x*(1-x)/(1-20*x+x^2).
a(n) = 20*a(n-1) - a(n-2), n>1. (End)
Let q(n, x) = Sum_{i=0..n} x^(n-i)*binomial(2*n-i, i) then a(n) = q(n, 18). - Benoit Cloitre, Dec 06 2002
a(-n+1) = a(n). - Michael Somos, Apr 18 2003
E.g.f.: (1/11)*exp(10*x)*(11*cosh(3*sqrt(11)*x) - 3*sqrt(11)*sinh(3*sqrt(11)*x)) - 1. - Stefano Spezia, Dec 06 2019

Extensions

More terms from Colin Barker, Feb 18 2014
Offset changed to 1 by G. C. Greubel, Dec 06 2019

A041014 Numerators of continued fraction convergents to sqrt(11).

Original entry on oeis.org

3, 10, 63, 199, 1257, 3970, 25077, 79201, 500283, 1580050, 9980583, 31521799, 199111377, 628855930, 3972246957, 12545596801, 79245827763, 250283080090, 1580944308303, 4993116004999, 31539640338297
Offset: 0

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Keywords

Crossrefs

Cf. A010468, A041015 (denominators).
Analog for other sqrt(m): A001333 (m=2), A002531 (m=3), A001077 (m=5), A041006 (m=6), A041008 (m=7), A041010 (m=8), A005667 (m=10), A041016 (m=12), ..., A042936 (m=1000).

Programs

  • Mathematica
    Table[Numerator[FromContinuedFraction[ContinuedFraction[Sqrt[11],n]]],{n,1,50}] (* Vladimir Joseph Stephan Orlovsky, Mar 16 2011 *)
    Numerator[Convergents[Sqrt[11], 30]] (* Vincenzo Librandi, Oct 28 2013 *)
  • PARI
    A041014=contfracpnqn(c=contfrac(sqrt(11)), #c)[1,][^-1] \\ Discard last element which may be incorrect. Use e.g. \p999 to get more terms, or extend as follows:
    {A041014_upto(N,A=Vec(A041014,N))=for(n=#A041014+1,N, A[n]=20*A[n-2]-A[n-4]); A041014=A} \\ M. F. Hasler, Nov 01 2019

Formula

G.f.: (3 + 10*x + 3*x^2 - x^3)/(1 - 20*x^2 + x^4).

A123482 Coefficients of the series giving the best rational approximations to sqrt(11).

Original entry on oeis.org

60, 23940, 9528120, 3792167880, 1509273288180, 600686976527820, 239071907384784240, 95150018452167599760, 37869468272055319920300, 15071953222259565160679700, 5998599512991034878630600360, 2387427534217209622129818263640, 950190160018936438572789038328420
Offset: 1

Views

Author

Gene Ward Smith, Oct 02 2006

Keywords

Comments

The partial sums of the series 10/3 - 1/a(1) - 1/a(2) - 1/a(3) - ... give the best rational approximations to sqrt(11), which constitute every second convergent of the continued fraction. The corresponding continued fractions are [3;3,6,3], [3;3,6,3,6,3], [3;3,6,3,6,3,6,3] and so forth.

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[-60*x/((x - 1)*(x^2 - 398*x + 1)), {x, 0, 50}], x] (* G. C. Greubel, Oct 13 2017 *)
  • PARI
    Vec(-60*x/((x-1)*(x^2-398*x+1)) + O(x^100)) \\ Colin Barker, Jun 23 2014

Formula

a(n+3) = 399*a(n+2) - 399*a(n+1) + a(n).
a(n) = -5/33 + (5/66 + 1/44*11^(1/2))*(199 + 60*11^(1/2))^n + (5/66 - 1/44*11^(1/2))*(199 - 60*11^(1/2))^n.
G.f.: -60*x / ((x-1)*(x^2-398*x+1)). - Colin Barker, Jun 23 2014

Extensions

More terms from Colin Barker, Jun 23 2014

A042937 Denominators of continued fraction convergents to sqrt(1000).

Original entry on oeis.org

1, 1, 2, 3, 5, 8, 53, 114, 281, 4329, 8939, 22207, 142181, 164388, 306569, 470957, 777526, 1248483, 78183472, 79431955, 157615427, 237047382, 394662809, 631710191, 4184923955, 9001558101, 22188040157, 341822160456, 705832361069, 1753486882594
Offset: 0

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Keywords

Examples

			sqrt(1000) = 31.62... = 31 + 1/(1 + 1/(1 + ...)) with convergents 31/1, 32/1, 63/2, 95/3, 158/5, ... - _M. F. Hasler_, Nov 02 2019
		

Crossrefs

Cf. A042936 (numerators), A040968 (continued fraction), A010467 (decimals).
Analog for sqrt(m): A000129 (m=2), A002530 (m=3), A001076 (m=5), A041007 (m=6), A041009 (m=7), A041011 (m=8), A005663 (m=10), A041015 (m=11), A041017 (m=12), ..., A042933 (m=998), A042935 (m=999).

Programs

  • Mathematica
    Denominator[Convergents[Sqrt[1000], 30]] (* Vincenzo Librandi, Feb 01 2014 *)
  • PARI
    A42937=contfracpnqn(c=contfrac(sqrt(1000)),#c-1)[2,] \\ Possibly incorrect last term ignored. NB: a(n) = A42937[n+1]. For more terms use e.g. \p999, or compute any a(n) from this as in A042936. - M. F. Hasler, Nov 01 2019

Extensions

More terms from Vincenzo Librandi, Feb 01 2014

A180029 Eight white queens and one red queen on a 3 X 3 chessboard. G.f.: (1 + 2*x)/(1 - 6*x - 2*x^2).

Original entry on oeis.org

1, 8, 50, 316, 1996, 12608, 79640, 503056, 3177616, 20071808, 126786080, 800860096, 5058732736, 31954116608, 201842165120, 1274961223936, 8053451673856, 50870632491008, 321330698293760, 2029725454744576
Offset: 0

Views

Author

Johannes W. Meijer, Aug 09 2010

Keywords

Comments

The a(n) represent the number of n-move routes of a fairy chess piece starting in the central square (m = 5) on a 3 X 3 chessboard. This fairy chess piece behaves like a white queen on the eight side and corner squares but on the central square the queen explodes with fury and turns into a red queen, see A180028.
The sequence above corresponds to 8 red queen vectors, i.e., A[5] vector, with decimal values 255, 383, 447, 479, 503, 507, 509 and 510. The other squares lead for these vectors to A135030.

Programs

  • Magma
    I:=[1,8]; [n le 2 select I[n] else 6*Self(n-1)+2*Self(n-2): n in [1..20]]; // Vincenzo Librandi, Nov 15 2011
  • Maple
    with(LinearAlgebra): nmax:=19; m:=5; A[5]:= [0,1,1,1,1,1,1,1,1]: A:=Matrix([[0,1,1,1,1,0,1,0,1], [1,0,1,1,1,1,0,1,0], [1,1,0,0,1,1,1,0,1], [1,1,0,0,1,1,1,1,0], A[5], [0,1,1,1,1,0,0,1,1], [1,0,1,1,1,0,0,1,1], [0,1,0,1,1,1,1,0,1], [1,0,1,0,1,1,1,1,0]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax);
  • Mathematica
    LinearRecurrence[{6,2},{1,8},50 ] (* Vincenzo Librandi, Nov 15 2011 *)

Formula

G.f.: (1+2*x)/(1 - 6*x - 2*x^2).
a(n) = 6*a(n-1) + 2*a(n-2) with a(0) = 1 and a(1) = 8.
a(n) = ((5-4*A)*A^(-n-1) + (5-4*B)*B^(-n-1))/22 with A = (-3+sqrt(11))/2 and B = (-3-sqrt(11))/2.
Lim_{k->infinity} a(n+k)/a(k) = (-1)^(n-1)*A016116(n+1)/(A041015(n-1)*sqrt(11) - A041014(n-1)) for n >= 1.
Showing 1-6 of 6 results.