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 1-10 of 18 results. Next

A007070 a(n) = 4*a(n-1) - 2*a(n-2) with a(0) = 1, a(1) = 4.

Original entry on oeis.org

1, 4, 14, 48, 164, 560, 1912, 6528, 22288, 76096, 259808, 887040, 3028544, 10340096, 35303296, 120532992, 411525376, 1405035520, 4797091328, 16378294272, 55918994432, 190919389184, 651839567872, 2225519493120, 7598398836736, 25942556360704, 88573427769344, 302408598355968
Offset: 0

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Author

Keywords

Comments

Joe Keane (jgk(AT)jgk.org) observes that this sequence (beginning at 4) is "size of raises in pot-limit poker, one blind, maximum raising."
It appears that this sequence is the BinomialMean transform of A002315 - see A075271. - John W. Layman, Oct 02 2002
Number of (s(0), s(1), ..., s(2n+3)) such that 0 < s(i) < 8 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+3, s(0) = 1, s(2n+3) = 4. - Herbert Kociemba, Jun 11 2004
a(n) = number of distinct matrix products in (A+B+C+D)^n where commutators [A,B]=[C,D]=0 but neither A nor B commutes with C or D. - Paul D. Hanna and Joshua Zucker, Feb 01 2006
The n-th term of the sequence is the entry (1,2) in the n-th power of the matrix M=[1,-1;-1,3]. - Simone Severini, Feb 15 2006
Hankel transform of this sequence is [1,-2,0,0,0,0,0,0,0,0,0,...]. - Philippe Deléham, Nov 21 2007
A204089 convolved with A000225, e.g., a(4) = 164 = (1*31 + 1*15 + 4*7 + 14*3 + 48*1) = (31 + 15 + 28 + 42 + 48). - Gary W. Adamson, Dec 23 2008
Equals INVERT transform of A000225: (1, 3, 7, 15, 31, ...). - Gary W. Adamson, May 03 2009
For n>=1, a(n-1) is the number of generalized compositions of n when there are 2^i-1 different types of the part i, (i=1,2,...). - Milan Janjic, Sep 24 2010
Binomial transform of A078057. - R. J. Mathar, Mar 28 2011
Pisano period lengths: 1, 1, 8, 1, 24, 8, 6, 1, 24, 24, 120, 8, 168, 6, 24, 1, 8, 24, 360, 24, ... . - R. J. Mathar, Aug 10 2012
a(n) is the diagonal of array A228405. - Richard R. Forberg, Sep 02 2013
From Wolfdieter Lang, Oct 01 2013: (Start)
a(n) appears together with A106731, both interspersed with zeros, in the representation of nonnegative powers of the algebraic number rho(8) = 2*cos(Pi/8) = A179260 of degree 4, which is the length ratio of the smallest diagonal and the side in the regular octagon.
The minimal polynomial for rho(8) is C(8,x) = x^4 - 4*x^2 + 2, hence rho(8)^n = A(n+1)*1 + A(n)*rho(8) + B(n+1)*rho(8)^2 + B(n)*rho(8)^3, n >= 0, with A(2*k) = 0, k >= 0, A(1) = 1, A(2*k+1) = A106731(k-1), k >= 1, and B(2*k) = 0, k >= 0, B(1) = 0, B(2*k+1) = a(k-1), k >= 1. See also the P. Steinbach reference given under A049310. (End)
The ratio a(n)/A006012(n) converges to 1+sqrt(2). - Karl V. Keller, Jr., May 16 2015
From Tom Copeland, Dec 04 2015: (Start)
An aerated version of this sequence is given by the o.g.f. = 1 / (1 - 4 x^2 + 2 x^4) = 1 / [x^4 a_4(1/x)] = 1 / determinant(I - x M) = exp[-log(1 -4 x + 2 x^4)], where M is the adjacency matrix for the simple Lie algebra B_4 given in A265185 with the characteristic polynomial a_4(x) = x^4 - 4 x^2 + 2 = 2 T_4(x/2) = A127672(4,x), where T denotes a Chebyshev polynomial of the first kind.
A133314 relates a(n) to the reciprocal of the e.g.f. 1 - 4 x + 4 x^2/2!. (End)
a(n) is the number of vertices of the Minkowski sum of n simplices with vertices e_(2*i+1), e_(2*i+2), e_(2*i+3), e_(2*i+4) for i=0,...,n-1, where e_i is a standard basis vector. - Alejandro H. Morales, Oct 03 2022

Examples

			a(3) = 48 = 3 * 4 + 4 + 1 + 1 = 3*a(2) + a(1) + a(0) + 1.
Example for the octagon rho(8) powers: rho(8)^4  = 2 + sqrt(2) = -2*1 + 4*rho(8)^2  = A(5)*1 + A(4)*rho(8) + B(5)*rho(8)^2 + B(4)*rho(8)^3, with a(5) = A106731(1) = -2, B(5) = a(1) = 4, A(4) = 0, B(4) = 0. - _Wolfdieter Lang_, Oct 01 2013
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Row sums of A059474. - David W. Wilson, Aug 14 2006
Equals 2 * A003480, n>0.
Row sums of A140071.

Programs

  • Haskell
    a007070 n = a007070_list !! n
    a007070_list = 1 : 4 : (map (* 2) $ zipWith (-)
       (tail $ map (* 2) a007070_list) a007070_list)
    -- Reinhard Zumkeller, Jan 16 2012
  • Magma
    Z:=PolynomialRing(Integers()); N:=NumberField(x^2-8); S:=[ ((4+r)^(1+n)-(4-r)^(1+n))/((2^(1+n))*r): n in [0..20] ]; [ Integers()!S[j]: j in [1..#S] ]; // Vincenzo Librandi, Mar 27 2011
    
  • Magma
    [n le 2 select 3*n-2 else 4*Self(n-1)-2*Self(n-2): n in [1..23]];  // Bruno Berselli, Mar 28 2011
    
  • Maple
    A007070 :=proc(n) option remember; if n=0 then 1 elif n=1 then 4 else 4*procname(n-1)-2*procname(n-2); fi; end:
    seq(A007070(n), n=0..30); # Wesley Ivan Hurt, Dec 06 2015
  • Mathematica
    LinearRecurrence[{4,-2}, {1,4}, 30] (* Harvey P. Dale, Sep 16 2014 *)
  • PARI
    a(n)=polcoeff(1/(1-4*x+2*x^2)+x*O(x^n),n)
    
  • PARI
    a(n)=if(n<1,1,ceil((2+sqrt(2))*a(n-1)))
    
  • Sage
    [lucas_number1(n,4,2) for n in range(1, 24)]# Zerinvary Lajos, Apr 22 2009
    

Formula

G.f.: 1/(1 - 4*x + 2*x^2).
Preceded by 0, this is the binomial transform of the Pell numbers A000129. Its e.g.f. is then exp(2*x)*sinh(sqrt(2)*x)/sqrt(2). - Paul Barry, May 09 2003
a(n) = ((2+sqrt(2))^(n+1) - (2-sqrt(2))^(n+1))/sqrt(8). - Al Hakanson (hawkuu(AT)gmail.com), Dec 27 2008, corrected Mar 28 2011
a(n) = (2 - sqrt(2))^n*(1/2 - sqrt(2)/2) + (2 + sqrt(2))^n*(1/2 + sqrt(2)/2). - Paul Barry, May 09 2003
a(n) = ceiling((2 + sqrt(2))*a(n-1)). - Benoit Cloitre, Aug 15 2003
a(n) = U(n, sqrt(2))*sqrt(2)^n. - Paul Barry, Nov 19 2003
a(n) = (1/4)*Sum_{r=1..7} sin(r*Pi/8)*sin(r*Pi/2)*(2*cos(r*Pi/8))^(2*n+3). - Herbert Kociemba, Jun 11 2004
a(n) = center term in M^n * [1 1 1], where M = the 3 X 3 matrix [1 1 1 / 1 2 1 / 1 1 1]. M^n * [1 1 1] = [A007052(n) a(n) A007052(n)]. E.g., a(3) = 48 since M^3 * [1 1 1] = [34 48 34], where 34 = A007052(3). - Gary W. Adamson, Dec 18 2004
This is the binomial mean transform of A002307. See Spivey and Steil (2006). - Michael Z. Spivey (mspivey(AT)ups.edu), Feb 26 2006
a(2n) = Sum_{r=0..n} 2^(2n-1-r)*(4*binomial(2n-1,2r) + 3*binomial(2n-1,2r+1)) a(2n-1) = Sum_{r=0..n} 2^(2n-2-r)*(4*binomial(2n-2,2r) + 3*binomial(2n-2,2r+1)). - Jeffrey Liese, Oct 12 2006
a(n) = 3*a(n - 1) + a(n - 2) + a(n - 3) + ... + a(0) + 1. - Gary W. Adamson, Feb 18 2011
G.f.: 1/(1 - 4*x + 2*x^2) = 1/( x*(1 + U(0)) ) - 1/x where U(k)= 1 - 2^k/(1 - x/(x - 2^k/U(k+1) )); (continued fraction 3rd kind, 3-step). - Sergei N. Gladkovskii, Dec 05 2012
G.f.: A(x) = G(0)/(1-2*x) where G(k) = 1 + 2*x/(1 - 2*x - x*(1-2*x)/(x + (1-2*x)/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Jan 04 2013
G.f.: G(0)/(2*x) - 1/x, where G(k) = 1 + 1/(1 - x*(2*k-1)/(x*(2*k+1) - (1-x)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 26 2013
a(n-1) = Sum_{k=0..n} binomial(2*n, n+k)*(k|8) where (k|8) is the Kronecker symbol. - Greg Dresden, Oct 11 2022
E.g.f.: exp(2*x)*(cosh(sqrt(2)*x) + sqrt(2)*sinh(sqrt(2)*x)). - Stefano Spezia, May 20 2024

A014176 Decimal expansion of the silver mean, 1+sqrt(2).

Original entry on oeis.org

2, 4, 1, 4, 2, 1, 3, 5, 6, 2, 3, 7, 3, 0, 9, 5, 0, 4, 8, 8, 0, 1, 6, 8, 8, 7, 2, 4, 2, 0, 9, 6, 9, 8, 0, 7, 8, 5, 6, 9, 6, 7, 1, 8, 7, 5, 3, 7, 6, 9, 4, 8, 0, 7, 3, 1, 7, 6, 6, 7, 9, 7, 3, 7, 9, 9, 0, 7, 3, 2, 4, 7, 8, 4, 6, 2, 1, 0, 7, 0, 3, 8, 8, 5, 0, 3, 8, 7, 5, 3, 4, 3, 2, 7, 6, 4, 1, 5, 7
Offset: 1

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Comments

From Hieronymus Fischer, Jan 02 2009: (Start)
Set c:=1+sqrt(2). Then the fractional part of c^n equals 1/c^n, if n odd. For even n, the fractional part of c^n is equal to 1-(1/c^n).
c:=1+sqrt(2) satisfies c-c^(-1)=floor(c)=2, hence c^n + (-c)^(-n) = round(c^n) for n>0, which follows from the general formula of A001622.
1/c = sqrt(2)-1.
See A001622 for a general formula concerning the fractional parts of powers of numbers x>1, which satisfy x-x^(-1)=floor(x).
Other examples of constants x satisfying the relation x-x^(-1)=floor(x) include A001622 (the golden ratio: where floor(x)=1) and A098316 (the "bronze" ratio: where floor(x)=3). (End)
In terms of continued fractions the constant c can be described by c=[2;2,2,2,...]. - Hieronymus Fischer, Oct 20 2010
Side length of smallest square containing five circles of diameter 1. - Charles R Greathouse IV, Apr 05 2011
Largest radius of four circles tangent to a circle of radius 1. - Charles R Greathouse IV, Jan 14 2013
An analog of Fermat theorem: for prime p, round(c^p) == 2 (mod p). - Vladimir Shevelev, Mar 02 2013
n*(1+sqrt(2)) is the perimeter of a 45-45-90 triangle with hypotenuse n. - Wesley Ivan Hurt, Apr 09 2016
This algebraic integer of degree 2, with minimal polynomial x^2 - 2*x - 1, is also the length ratio diagonal/side of the second largest diagonal in the regular octagon (not counting the side). The other two diagonal/side ratios are A179260 and A121601. - Wolfdieter Lang, Oct 28 2020
c^n = A001333(n) + A000129(n) * sqrt(2). - Gary W. Adamson, Apr 26 2023
c^n = c * A000129(n) + A000129(n-1), where c = 1 + sqrt(2). - Gary W. Adamson, Aug 30 2023

Examples

			2.414213562373095...
		

References

  • B. C. Berndt, Ramanujan's Notebooks Part II, Springer-Verlag, p. 140, Entry 25.

Crossrefs

Apart from initial digit the same as A002193.
See A098316 for [3;3,3,...]; A098317 for [4;4,4,...]; A098318 for [5;5,5,...]. - Hieronymus Fischer, Oct 20 2010

Programs

  • Maple
    Digits:=100: evalf(1+sqrt(2)); # Wesley Ivan Hurt, Apr 09 2016
  • Mathematica
    RealDigits[1 + Sqrt@ 2, 10, 111] (* Or *)
    RealDigits[Exp@ ArcSinh@ 1, 10, 111][[1]] (* Robert G. Wilson v, Aug 17 2011 *)
    Circs[n_] := With[{r = Sin[Pi/n]/(1 - Sin[Pi/n])}, Graphics[Append[
      Table[Circle[(r + 1) {Sin[2 Pi k/n], Cos[2 Pi k/n]}, r], {k, n}],   {Blue, Circle[{0, 0}, 1]}]]] Circs[4] (* Charles R Greathouse IV, Jan 14 2013 *)
  • PARI
    1+sqrt(2) \\ Charles R Greathouse IV, Jan 14 2013

Formula

Conjecture: 1+sqrt(2) = lim_{n->oo} A179807(n+1)/A179807(n).
Equals cot(Pi/8) = tan(Pi*3/8). - Bruno Berselli, Dec 13 2012, and M. F. Hasler, Jul 08 2016
Silver mean = 2 + Sum_{n>=0} (-1)^n/(P(n-1)*P(n)), where P(n) is the n-th Pell number (A000129). - Vladimir Shevelev, Feb 22 2013
Equals exp(arcsinh(1)) which is exp(A091648). - Stanislav Sykora, Nov 01 2013
Limit_{n->oo} exp(asinh(cos(Pi/n))) = sqrt(2) + 1. - Geoffrey Caveney, Apr 23 2014
exp(asinh(cos(Pi/2 - log(sqrt(2)+1)*i))) = exp(asinh(sin(log(sqrt(2)+1)*i))) = i. - Geoffrey Caveney, Apr 23 2014
Equals Product_{k>=1} A047621(k) / A047522(k) = (3/1) * (5/7) * (11/9) * (13/15) * (19/17) * (21/23) * ... . - Dimitris Valianatos, Mar 27 2019
From Wolfdieter Lang, Nov 10 2023:(Start)
Equals lim_{n->oo} A000129(n+1)/A000129(n) (see A000129, Pell).
Equals lim_{n->oo} S(n+1, 2*sqrt(2))/S(n, 2*sqrt(2)), with the Chebyshev S(n,x) polynomial (see A049310). (End)
From Peter Bala, Mar 24 2024: (Start)
An infinite family of continued fraction expansions for this constant can be obtained from Berndt, Entry 25, by setting n = 1/2 and x = 8*k + 6 for k >= 0.
For example, taking k = 0 and k = 1 yields
sqrt(2) + 1 = 15/(6 + (1*3)/(12 + (5*7)/(12 + (9*11)/(12 + (13*15)/(12 + ... + (4*n + 1)*(4*n + 3)/(12 + ... )))))) and
sqrt(2) + 1 = (715/21) * 1/(14 + (1*3)/(28 + (5*7)/(28 + (9*11)/(28 + (13*15)/(28 + ... + (4*n + 1)*(4*n + 3)/(28 + ... )))))). (End)

A047522 Numbers that are congruent to {1, 7} mod 8.

Original entry on oeis.org

1, 7, 9, 15, 17, 23, 25, 31, 33, 39, 41, 47, 49, 55, 57, 63, 65, 71, 73, 79, 81, 87, 89, 95, 97, 103, 105, 111, 113, 119, 121, 127, 129, 135, 137, 143, 145, 151, 153, 159, 161, 167, 169, 175, 177, 183, 185, 191, 193, 199, 201, 207, 209, 215, 217, 223, 225, 231, 233
Offset: 1

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Also n such that Kronecker(2,n) = mu(gcd(2,n)). - Jon Perry and T. D. Noe, Jun 13 2003
Also n such that x^2 == 2 (mod n) has a solution. The primes are given in sequence A001132. - T. D. Noe, Jun 13 2003
As indicated in the formula, a(n) is related to the even triangular numbers. - Frederick Magata (frederick.magata(AT)uni-muenster.de), Jun 17 2004
Cf. property described by Gary Detlefs in A113801: more generally, these a(n) are of the form (2*h*n + (h-4)*(-1)^n-h)/4 (h,n natural numbers). Therefore a(n)^2 - 1 == 0 (mod h); in this case, a(n)^2 - 1 == 0 (mod 8). Also a(n)^2 - 1 == 0 (mod 16). - Bruno Berselli, Nov 17 2010
A089911(3*a(n)) = 2. - Reinhard Zumkeller, Jul 05 2013
S(a(n+1)/2, 0) = (1/2)*(S(a(n+1), sqrt(2)) - S(a(n+1) - 2, sqrt(2))) = T(a(n+1), sqrt(2)/2) = cos(a(n+1)*Pi/4) = sqrt(2)/2 = A010503, identically for n >= 0, where S is the Chebyshev polynomial (A049310) here extended to fractional n, evaluated at x = 0. (For T see A053120.) - Wolfdieter Lang, Jun 04 2023

References

  • L. B. W. Jolley, Summation of Series, Dover Publications, 1961, p. 16.

Crossrefs

Programs

  • Haskell
    a047522 n = a047522_list !! (n-1)
    a047522_list = 1 : 7 : map (+ 8) a047522_list
    -- Reinhard Zumkeller, Jan 07 2012
    
  • Mathematica
    Select[Range[1, 191, 2], JacobiSymbol[2, # ]==1&]
  • PARI
    a(n)=4*n-2+(-1)^n \\ Charles R Greathouse IV, Sep 24 2015

Formula

a(n) = sqrt(8*A014494(n)+1) = sqrt(16*ceiling(n/2)*(2*n+1)+1) = sqrt(8*A056575(n)-8*(2n+1)*(-1)^n+1). - Frederick Magata (frederick.magata(AT)uni-muenster.de), Jun 17 2004
1 - 1/7 + 1/9 - 1/15 + 1/17 - ... = (Pi/8)*(1 + sqrt(2)). [Jolley] - Gary W. Adamson, Dec 16 2006
From R. J. Mathar, Feb 19 2009: (Start)
a(n) = 4n - 2 + (-1)^n = a(n-2) + 8.
G.f.: x(1+6x+x^2)/((1+x)(1-x)^2). (End)
a(n) = 8*n - a(n-1) - 8. - Vincenzo Librandi, Aug 06 2010
From Bruno Berselli, Nov 17 2010: (Start)
a(n) = -a(-n+1) = a(n-1) + a(n-2) - a(n-3).
a(n) = 8*A000217(n-1)+1 - 2*Sum_{i=1..n-1} a(i) for n > 1. (End)
E.g.f.: 1 + (4*x - 1)*cosh(x) + (4*x - 3)*sinh(x). - Stefano Spezia, May 13 2021
E.g.f.: 1 + (4*x - 3)*exp(x) + 2*cosh(x). - David Lovler, Jul 16 2022
From Amiram Eldar, Nov 22 2024: (Start)
Product_{n>=1} (1 - (-1)^n/a(n)) = sqrt(2+sqrt(2)) (A179260).
Product_{n>=2} (1 + (-1)^n/a(n)) = (Pi/8)*cosec(Pi/8) (A352125). (End)

A121601 Decimal expansion of cosecant of 22.5 degrees = csc(Pi/8).

Original entry on oeis.org

2, 6, 1, 3, 1, 2, 5, 9, 2, 9, 7, 5, 2, 7, 5, 3, 0, 5, 5, 7, 1, 3, 2, 8, 6, 3, 4, 6, 8, 5, 4, 3, 7, 4, 3, 0, 7, 1, 6, 7, 5, 2, 2, 3, 7, 6, 6, 9, 8, 5, 3, 9, 0, 5, 5, 0, 9, 7, 7, 9, 6, 7, 3, 3, 8, 1, 6, 1, 6, 2, 0, 8, 2, 9, 2, 2, 3, 8, 4, 1, 0, 1, 9, 0, 3, 7, 0, 7, 4, 4, 0, 3, 8, 5, 2, 5, 6, 2, 8, 6, 4, 9, 2, 7, 7
Offset: 1

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Author

Rick L. Shepherd, Aug 09 2006

Keywords

Comments

1 + csc(Pi/8) is the radius of the smallest circle into which 9 unit circles can be packed ("r=3.613+ Proved by Pirl in 1969", according to the Friedman link, which has a diagram).
csc(Pi/8) is the distance between the center of the larger circle and the center of each unit circle that touches the larger circle.
A rectangle of length L and width W is a called a silver rectangle if L=rW, where r is the silver ratio; i.e., r = 1+sqrt(2). The diagonal has length D = sqrt(W^2+L^2), so that D/W = sqrt(4+2*sqrt(2)) = csc(Pi/8). - Clark Kimberling, Apr 04 2011
This algebraic integer of degree 4 also gives the length ratio diagonal/side of the longest diagonal in the regular octagon. The minimal polynomial is x^4 - 8*x + 8. In the power basis of Gal(Q(rho(8))/Q), with rho(8) = sqrt(2 + sqrt(2)) = A179260 it is -2*rho(8) + 1*rho(8)^3 which equals sqrt(2)*rho(8). - Wolfdieter Lang, Oct 28 2020

Examples

			2.6131259297527530557132863468543743071675223766985390550977...
		

References

  • D. Mumford et al., Indra's Pearls, Cambridge 2002; see p. 362. - N. J. A. Sloane, Nov 22 2009

Crossrefs

Programs

  • Magma
    SetDefaultRealField(RealField(100)); R:=RealField(); 1/Sin(Pi(R)/8); // G. C. Greubel, Nov 02 2018
  • Maple
    evalf(1/sin(Pi/8),120); # Muniru A Asiru, Nov 02 2018
  • Mathematica
    RealDigits[Csc[Pi/8],10,130][[1]] (* corrected by Harvey P. Dale, Jul 28 2012 *)
  • PARI
    1/sin(Pi/8)
    

Formula

Equals 2*sqrt(2)*cos(Pi/8).
Equals Product_{k >= 0} (8*k + 4)^2/((8*k + 1)*(8*k + 7)). - Antonio Graciá Llorente, Mar 11 2024

A050983 de Bruijn's S(4,n).

Original entry on oeis.org

1, 14, 786, 61340, 5562130, 549676764, 57440496036, 6242164112184, 698300344311570, 79881547652046140, 9301427008157320036, 1098786921802152516024, 131361675994216221116836, 15863471168011822803270200, 1932252897656224864335299400
Offset: 0

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Comments

a(n) is divisible by (n+1). Prime p divides a(p-1). Prime p>2 divides all a(n) from a((p+1)/2) to a(p-1). - Alexander Adamchuk, Jul 05 2006

References

  • N. G. de Bruijn, Asymptotic Methods in Analysis, North-Holland Publishing Co., 1958. See chapters 4 and 6.

Crossrefs

Programs

  • Mathematica
    Sum[ (-1)^(k+n)Binomial[ 2n, k ]^4, {k, 0, 2n} ]
    RecurrenceTable[{a[0] == 1, a[1] == 14, 4 (n + 1) (2 n + 1)^3 (48 n^2 + 162 n + 137) a[n] + (n + 2)^3 (2 n + 3) (48 n^2 + 66 n + 23) a[n + 2] == 2 (4 (n + 1)^2 (2 n + 3)^2 (408 n^2 + 969 n + 431) - (n + 1) (2 n + 3) (69 n + 31) + 57 n + 92) a[n + 1]}, a[n], {n, 0, 20}] (* Vladimir Reshetnikov, Sep 26 2016 *)
  • PARI
    a(n)=sum(k=0,2*n,(-1)^(k+n)*binomial(2*n,k)^4) \\ Charles R Greathouse IV, Dec 28 2011

Formula

a(n) = Sum_{k=-n..+n} (-1)^k*C(2*n,n+k)^4. - Benoit Cloitre, Mar 02 2005
a(n) = (-1)^n * HypergeometricPFQ[ {-2n, -2n, -2n, -2n}, {1, 1, 1}, -1]. - Alexander Adamchuk, Jul 05 2006
E.g.f.: Sum(n>=0,I^n*x^n/n!^4) * Sum(n>=0,(-I)^n*x^n/n!^4) = Sum(n>=0,a(n)*x^(2*n)/n!^4) where I^2=-1. - Paul D. Hanna, Dec 21 2011
a(n) ~ 0.125 k^(8n+3)/(Pi*n)^(3/2) where k = 2 cos(Pi/8) = A179260. This formula is due to de Bruijn 1958. - Charles R Greathouse IV, Dec 28 2011
Recurrence: a(0) = 1, a(1) = 14, 4 * (n + 1) * (2*n + 1)^3 * (48*n^2 + 162*n + 137) * a(n) + (n + 2)^3 * (2*n + 3) * (48*n^2 + 66*n + 23) * a(n+2) = 2 * (4 * (n + 1)^2 * (2*n + 3)^2 * (408*n^2 + 969*n + 431) - (n + 1) * (2*n + 3) * (69*n + 31) + 57*n + 92) * a(n+1). - Vladimir Reshetnikov, Sep 26 2016
From Peter Bala, Nov 02 2024; (Start)
a(n) = 1/n * Sum_{k = 0..2*n} (-1)^(n+k) * k * binomial(2*n, k)^4 for n >= 1.
a(n) = binomial(2*n, n) * Sum_{k = 0..n} binomial(2*n, n+k)^2 * binomial(2*n+k,k) = binomial(2*n, n) * Sum_{k = 0..n} (-1)^(n+k) * binomial(2*n, n+k) * binomial(2*n+k, k)^2. (End)

A101464 Decimal expansion of sqrt(2-sqrt(2)), edge length of a regular octagon with circumradius 1.

Original entry on oeis.org

7, 6, 5, 3, 6, 6, 8, 6, 4, 7, 3, 0, 1, 7, 9, 5, 4, 3, 4, 5, 6, 9, 1, 9, 9, 6, 8, 0, 6, 0, 7, 9, 7, 7, 3, 3, 5, 2, 2, 6, 8, 9, 1, 2, 4, 9, 7, 1, 2, 5, 4, 0, 8, 2, 8, 6, 7, 6, 0, 1, 2, 7, 1, 2, 5, 5, 0, 9, 2, 0, 6, 7, 9, 2, 0, 1, 7, 9, 3, 8, 4, 4, 7, 4, 0, 2, 7, 5, 7, 0, 6, 8, 4, 5, 6, 7, 0, 9, 4, 2, 9, 6, 8, 4, 8
Offset: 0

Views

Author

Jun Mizuki (suzuki32(AT)sanken.osaka-u.ac.jp), Jan 20 2005

Keywords

Examples

			0.765366864730179543456919968060797733522689124971254082867601271255092067920...
		

Crossrefs

Cf. A047621, A101465, A179260 (sqrt(2+sqrt(2))), A182168, A285871, A329246.

Programs

Formula

Equals i^(3/4) + i^(-3/4). - Gary W. Adamson, Jul 07 2022
Equals 2*sin(Pi/8) = 2*A182168. - Amiram Eldar, Apr 06 2023
Equals Product_ {k >= 0} ((8*k - 2)*(8*k + 10))/((8*k - 5)*(8*k + 13)). - Antonio Graciá Llorente, Mar 11 2024
Equals Product_{k>=1} (1 + (-1)^k/A047621(k)). - Amiram Eldar, Nov 22 2024
Equals sqrt(A101465) = 1/A285871 = exp(-A329246). - Hugo Pfoertner, Nov 22 2024

A154739 Decimal expansion of sqrt(1 - 1/sqrt(2)), the abscissa of the point of bisection of the arc of the unit lemniscate (x^2 + y^2)^2 = x^2 - y^2 in the first quadrant.

Original entry on oeis.org

5, 4, 1, 1, 9, 6, 1, 0, 0, 1, 4, 6, 1, 9, 6, 9, 8, 4, 3, 9, 9, 7, 2, 3, 2, 0, 5, 3, 6, 6, 3, 8, 9, 4, 2, 0, 0, 6, 1, 0, 7, 2, 0, 6, 3, 3, 7, 8, 0, 1, 5, 4, 4, 4, 6, 8, 1, 2, 9, 7, 0, 9, 5, 6, 5, 2, 9, 8, 8, 9, 7, 3, 5, 4, 1, 0, 1, 2, 6, 6, 6, 4, 7, 7, 8, 2, 6, 1, 4, 9, 5
Offset: 0

Views

Author

Stuart Clary, Jan 14 2009

Keywords

Comments

A root of 2*x^4 - 4*x^2 + 1 = 0.

Examples

			0.541196100146196984399723205366...
		

References

  • C. L. Siegel, Topics in Complex Function Theory, Volume I: Elliptic Functions and Uniformization Theory, Wiley-Interscience, 1969, page 5.

Crossrefs

Cf. A154743 for the ordinate and A154747 for the radius vector.
Cf. A154740, A154741 and A154742 for the continued fraction and the numerators and denominators of the convergents.
Cf. A085565 for 1.311028777..., the first-quadrant arc length of the unit lemniscate.

Programs

  • Mathematica
    nmax = 1000; First[ RealDigits[ Sqrt[ 1 - 1/Sqrt[2] ], 10, nmax] ]
  • PARI
    sqrt(1 - 1/sqrt(2)) \\ G. C. Greubel, Sep 23 2017
    
  • PARI
    polrootsreal(2*x^4-4*x^2+1)[3] \\ Charles R Greathouse IV, Feb 04 2025

Formula

From Amiram Eldar, Nov 22 2024: (Start)
Equals sqrt(2) * sin(Pi/8) = A002193 * A182168.
Equals Product_{k>=0} (1 - (-1)^k/(4*k+2)) = Product_{k>=1} (1 + (-1)^k/A016825(k)). (End)
Equals 1/A179260 = sqrt(A268682). - Hugo Pfoertner, Nov 22 2024

Extensions

Offset corrected by R. J. Mathar, Feb 05 2009

A302711 Decimal expansion of 2*sin(15*Pi/32).

Original entry on oeis.org

1, 9, 9, 0, 3, 6, 9, 4, 5, 3, 3, 4, 4, 3, 9, 3, 7, 7, 2, 4, 8, 9, 6, 7, 3, 9, 0, 6, 2, 1, 8, 9, 5, 9, 8, 4, 3, 1, 5, 0, 9, 4, 9, 7, 3, 7, 4, 5, 9, 7, 1, 4, 1, 2, 3, 6, 6, 7, 2, 2, 5, 9, 3, 1, 5, 6, 9, 7, 8, 0, 3, 3, 3, 7, 8, 9, 1, 7, 3, 0, 7, 5, 9, 4, 5, 0, 5, 8, 1, 6, 8, 5, 3, 9, 2, 9, 6, 7, 8, 0
Offset: 1

Views

Author

Wolfdieter Lang, Apr 28 2018

Keywords

Comments

This constant appears in a historic problem posed by Adriaan van Roomen (Adrianus Romanus) in his Ideae mathematicae from 1593, solved by Viète (see the Vieta link) using trigonometry. See the Havil reference, problem 1 (for a correction see below), pp. 69-74, and the Maor reference for Viète's approach, pp. 58-60.
The problem involves the monic Chebyshev polynomial of the first kind R(45, x) (R coefficients are given in A127672). The present problem was stated as R(45, x) = sqrt(2 + sqrt(2 + sqrt(2 + sqrt(2)))) for x = sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 + sqrt(3))))) (see A302712). This is equivalent to R(45, 2*sin(Pi/96)) = 2*sin(15*Pi/32). It is a special case of the well known identity R(2*k+1, x) = x*(-1)^k*S(2*k, sqrt(4-x^2)), with the Chebyshev S polynomials (see A049310 for the coefficients). Take k = 22, x = 2*sin(Pi/96), and see the Havil reference, p. 71, for the proof of 2*sin(15*Pi/32) = sqrt(2 + sqrt(2 + sqrt(2 + sqrt(2)))). [In the Havil reference on p. 69, the second to last exponent is 43 (not 41), and in the first problem, for the argument x a further +sqrt(2... is missing. In the general identity given on p. 71 a sign factor is missing. It should read, with n = 2*k+1: P_{2*k+1}(2*sin(theta)) = 2*(-1)^k*sin((2*k+1)*theta).]
For the argument x = sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2 + sqrt(3))))) = 2*sin(Pi/96) = 0.65438165643552284... see A302712.
R(45, x) factorizes into minimal polynomials of 2*cos(Pi/k), named C(k, x), for short, C[k], with coefficients given in A187360 as follows. R(45, x) = C[90]*C[30]*C[18]*C[10]*C[6]*C[2]. See a comment in A127672.
All 45 zeros of R(45, x), which are real, are 2*cos((2*k+1)*Pi/90), for k = 0..44. See a comment in A127672.
Viète used the iteration, written in terms of R polynomials as R(45, x) = -R(3, -R(3, R(5, x))) (from the semigroup property of Chebyshev T polynomials). See the Maor reference, pp. 58-60. - Wolfdieter Lang, May 05 2018
An algebraic integer of degree 16. - Charles R Greathouse IV, Jan 29 2022

Examples

			2*sin(15*Pi/32) = 1.990369453344393772489673906218959843150949737459714123...
		

References

  • Julian Havil, The Irrationals, A Story of the Numbers You Can't Count On, Princeton University Press, Princeton and Oxford, 2012, pp. 69-74.
  • Eli Maor, Trigonometric Delights, Princeton University Press, NJ, 1998, pp. 56-62.

Crossrefs

Programs

Formula

This constant is 2*sin(15*Pi/32) = sqrt(2 + sqrt(2 + sqrt(2 + sqrt(2)))). (for a proof see Havil. p.71).

A001668 Number of self-avoiding n-step walks on honeycomb lattice.

Original entry on oeis.org

1, 3, 6, 12, 24, 48, 90, 174, 336, 648, 1218, 2328, 4416, 8388, 15780, 29892, 56268, 106200, 199350, 375504, 704304, 1323996, 2479692, 4654464, 8710212, 16328220, 30526374, 57161568, 106794084, 199788408, 372996450, 697217994, 1300954248
Offset: 0

Views

Author

Keywords

References

  • 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).

Crossrefs

Cf. A006851.

Programs

  • Maple
    a:= proc(n) local v, b;
          if n<2 then return 1 +2*n fi;
          v:= proc() false end: v(0, 0), v(1, 0):= true$2;
          b:= proc(n, x, y) local c;
                if v(x, y) then 0
              elif n=0 then 1
              else v(x, y):= true;
                   c:= b(n-1, x+1, y) + b(n-1, x-1, y) +
                       b(n-1, x, y-1+2*((x+y) mod 2));
                   v(x, y):= false; c
                fi
              end;
          6*b(n-2, 1, 1)
        end:
    seq(a(n), n=0..20);  # Alois P. Heinz, Jul 07 2011
  • Mathematica
    a[n_] := a[n] = Module[{v, b}, If[n < 2 , Return[1+2*n]]; v[0, 0] = v[1, 0] = True; v[, ] = False; b[m_, x_, y_] := Module[{c}, If[v[x, y], 0 , If[ m == 0 , 1, v[x, y] = True; c = b[m-1, x+1, y] + b[m-1, x-1, y] + b[m-1, x, y-1 + 2*Mod[x+y, 2]]; v[x, y] = False; c]]]; 6*b[n-2, 1, 1]]; Table[Print["a(", n, ") = ", a[n]]; a[n], {n, 0, 32}] (* Jean-François Alcover, Nov 25 2013, translated from Alois P. Heinz's Maple program *)

Formula

mu^n <= a(n) <= mu^n alpha^sqrt(n) for mu = A179260 and some alpha. It has been conjectured that a(n) ~ mu^n * n^(11/32). - Charles R Greathouse IV, Nov 08 2013

Extensions

More terms from Pab Ter (pabrlos(AT)yahoo.com), May 06 2004

A285871 Decimal expansion of 1/sqrt(2 - sqrt(2)) (reciprocal of A101464).

Original entry on oeis.org

1, 3, 0, 6, 5, 6, 2, 9, 6, 4, 8, 7, 6, 3, 7, 6, 5, 2, 7, 8, 5, 6, 6, 4, 3, 1, 7, 3, 4, 2, 7, 1, 8, 7, 1, 5, 3, 5, 8, 3, 7, 6, 1, 1, 8, 8, 3, 4, 9, 2, 6, 9, 5, 2, 7, 5, 4, 8, 8, 9, 8, 3, 6, 6, 9, 0, 8, 0, 8, 1, 0, 4, 1, 4, 6, 1, 1, 9, 2, 0, 5, 0, 9, 5, 1, 8, 5, 3, 7, 2, 0, 1, 9, 2, 6, 2, 8, 1, 4
Offset: 1

Views

Author

Wolfdieter Lang, May 11 2017

Keywords

Comments

This number is the length ratio of the radius of a circle and the side of the inscribed octagon.
In the Corbalán reference, pp. 61-62, this number is called Cordoba number or Cordoba proportion, attributed to the architect Rafael de la Hoz (1924-2000), who used the rectangle with this proportion to explain the structure of the Mihrab of Cordoba.

Examples

			1.30656296487637652785664317342718715358376118834926952754889836690808104146...
		

References

  • Fernando Corbalán, Der goldene Schnitt, Librero, 2017. Original: La proportión áurea, RBA Contenidos Editoriales y Audiovisuales S. A. U., 2010. English: The golden Ratio, 2012, RBA Coleccionables.

Crossrefs

Programs

  • Magma
    SetDefaultRealField(RealField(100)); 1/Sqrt(2 - Sqrt(2)); // G. C. Greubel, Oct 10 2018
  • Maple
    evalf((sqrt(2-sqrt(2)))^(-1),100); # Muniru A Asiru, Oct 11 2018
  • Mathematica
    RealDigits[1/Sqrt[2 - Sqrt[2]], 10, 100][[1]] (* Indranil Ghosh, May 11 2017 *)
  • PARI
    default(realprecision, 100); 1/sqrt(2 - sqrt(2)) \\ G. C. Greubel, Oct 10 2018
    
  • Python
    from sympy import N, sqrt
    print(N(1/sqrt(2 - sqrt(2)), 100)) # Indranil Ghosh, May 11 2017
    

Formula

Equals 1/(2*sin(Pi/8)) = 1/A101464.
Equals Product_{k>=0} (1 + (-1)^k/(4*k+2)). - Amiram Eldar, Aug 07 2020
The minimal polynomial is 2*x^4 - 4*x^2 + 1. - Joerg Arndt, May 10 2021
Equals Sum_{n>=0} binomial(2*n - 1/2, -1/2)/2^n. - Antonio Graciá Llorente, Nov 13 2024

Extensions

Offset and example corrected by Amiram Eldar, Aug 07 2020
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