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 15 results. Next

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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Author

Keywords

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)

A098317 Decimal expansion of phi^3 = 2 + sqrt(5).

Original entry on oeis.org

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

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Author

Eric W. Weisstein, Sep 02 2004

Keywords

Comments

This sequence is also the decimal expansion of ((1+sqrt(5))/2)^3. - Mohammad K. Azarian, Apr 14 2008
This is the length/width ratio of a 4-extension rectangle; see A188640 for definitions. - Clark Kimberling, Apr 10 2011
Its continued fraction is [4, 4, ...] (see A010709). - Robert G. Wilson v, Apr 10 2011

Examples

			4.23606797749978969640917366873127623544061835961152572427...
		

References

  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 138-139.
  • Alexey Stakhov, The mathematics of harmony: from Euclid to contemporary mathematics and computer science, World Scientific, Singapore, 2009, p. 657.

Crossrefs

Programs

Formula

2 plus the constant in A002163. - R. J. Mathar, Sep 02 2008
Equals 3 + 4*sin(Pi/10) = 1 + 4*cos(Pi/5) = 1 + 4*sin(3*Pi/10) = 3 + 4*cos(2*Pi/5) = 1 + csc(Pi/10). - Arkadiusz Wesolowski, Mar 11 2012
Equals lim_{n -> infinity} F(n+3)/F(n) = lim_{n -> infinity} (1 + 2*F(n+1)/F(n)) = 2 + sqrt(5), with F(n) = A000045(n). - Arkadiusz Wesolowski, Mar 11 2012
Equals exp(arcsinh(2)), since arcsinh(x) = log(x+sqrt(x^2+1)). - Stanislav Sykora, Nov 01 2013
Equals Sum_{n>=1} n/phi^n = phi/(phi-1)^2 = phi^3. - Richard R. Forberg, Jun 29 2014
Equals 1 + 2*phi, with phi = A001622, an integer in the quadratic number field Q(sqrt(5)). - Wolfdieter Lang, Dec 10 2022
c^n = A001076(n-1) + c * A001076(n); where c = 2 + sqrt(5). - Gary W. Adamson, Oct 09 2023
Equals lim_{n -> infinity} = S(n, 2*(-1 + 2*phi))/S(n-1, 2*(-1 + 2*phi)), with the S-Chebyshev polynomials (see A049310). See also the above limit formula with Fibonacci numbers. - Wolfdieter Lang, Nov 15 2023

Extensions

Title expanded to include observation from Mohammad K. Azarian by Charles R Greathouse IV, Mar 11 2012

A087130 a(n) = 5*a(n-1)+a(n-2) for n>1, a(0)=2, a(1)=5.

Original entry on oeis.org

2, 5, 27, 140, 727, 3775, 19602, 101785, 528527, 2744420, 14250627, 73997555, 384238402, 1995189565, 10360186227, 53796120700, 279340789727, 1450500069335, 7531841136402, 39109705751345, 203080369893127
Offset: 0

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Author

Paul Barry, Aug 16 2003

Keywords

Comments

Sequence is related to the fifth metallic mean [5;5,5,5,5,...] (see A098318).
The solution to the general recurrence b(n) = (2*k+1)*b(n-1)+b(n-2) with b(0)=2, b(1) = 2*k+1 is b(n) = ((2*k+1)+sqrt(4*k^2+4*k+5))^n+(2*k+1)-sqrt(4*k^2+4*k+5))^n)/2; b(n) = 2^(1-n)*Sum_{j=0..n} C(n, 2*j)*(4*k^2+4*k+5)^j*(2*k+1)^(n-2*j); b(n) = 2*T(n, (2*k+1)*x/2)(-1)^i with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
Primes in this sequence include a(0) = 2; a(1) = 5; a(4) = 727; a(8) = 528527 (3) semiprimes in this sequence include a(7) = 101785; a(13) = 1995189565; a(16) = 279340789727; a(19) = 39109705751345; a(20) = 203080369893127 - Jonathan Vos Post, Feb 09 2005
a(n)^2 - 29*A052918(n-1)^2 = 4*(-1)^n, with n>0 - Gary W. Adamson, Oct 07 2008
For more information about this type of recurrence follow the Khovanova link and see A054413 and A086902. - Johannes W. Meijer, Jun 12 2010
Binomial transform of A072263. - Johannes W. Meijer, Aug 01 2010

Crossrefs

Programs

  • Magma
    I:=[2,5]; [n le 2 select I[n] else 5*Self(n-1)+Self(n-2): n in [1..30]]; // Vincenzo Librandi, Sep 19 2016
  • Mathematica
    RecurrenceTable[{a[0] == 2, a[1] == 5, a[n] == 5 a[n-1] + a[n-2]}, a, {n, 30}] (* Vincenzo Librandi, Sep 19 2016 *)
  • PARI
    {a(n) = if( n<0, (-1)^n * a(-n), polsym(x^2 - 5*x -1, n) [n + 1])} /* Michael Somos, Nov 04 2008 */
    
  • Sage
    [lucas_number2(n,5,-1) for n in range(0, 21)] # Zerinvary Lajos, May 14 2009
    

Formula

a(n) = ((5+sqrt(29))/2)^n+((5-sqrt(29))/2)^n.
a(n) = A100236(n) + 1.
E.g.f. : 2*exp(5*x/2)*cosh(sqrt(29)*x/2); a(n) = 2^(1-n)*Sum_{k=0..floor(n/2)} C(n, 2k)*29^k*5^(n-2*k). a(n) = 2T(n, 5i/2)(-i)^n with T(n, x) Chebyshev's polynomials of the first kind (see A053120) and i^2=-1. - Paul Barry, Nov 15 2003
O.g.f.: (-2+5*x)/(-1+5*x+x^2). - R. J. Mathar, Dec 02 2007
a(-n) = (-1)^n * a(n). - Michael Somos, Nov 01 2008
A090248(n) = a(2*n). 5 * A097834(n) = a(2*n + 1). - Michael Somos, Nov 01 2008
Limit(a(n+k)/a(k), k=infinity) = (A087130(n) + A052918(n-1)*sqrt(29))/2. Limit(A087130(n)/A052918(n-1), n= infinity) = sqrt(29). - Johannes W. Meijer, Jun 12 2010
a(3n+1) = A041046(5n), a(3n+2) = A041046(5n+3) and a(3n+3) = 2*A041046 (5n+4). - Johannes W. Meijer, Jun 12 2010
a(n) = 2*A052918(n) - 5*A052918(n-1). - R. J. Mathar, Oct 02 2020
From Peter Bala, Jul 09 2025 : (Start)
The following series telescope (Cf. A000032):
For k >= 1, Sum_{n >= 1} (-1)^((k+1)*(n+1)) * a(2*n*k)/(a((2*n-1)*k)*a((2*n+1)*k)) = 1/a(k)^2.
For positive even k, Sum_{n >= 1} 1/(a(k*n) - (a(k) + 2)/a(k*n)) = 1/(a(k) - 2) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) + (a(k) - 2)/a(k*n)) = 1/(a(k) + 2).
For positive odd k, Sum_{n >= 1} 1/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) + 2)/(2*(a(2*k) - 2)) and
Sum_{n >= 1} (-1)^(n+1)/(a(k*n) - (-1)^n*(a(2*k) + 2)/a(k*n)) = (a(k) - 2)/(2*(a(2*k) - 2)). (End)

A098316 Decimal expansion of [3, 3, ...] = (3 + sqrt(13))/2.

Original entry on oeis.org

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

Views

Author

Eric W. Weisstein, Sep 02 2004

Keywords

Comments

For reasons following from the formula section, this constant could be called "the bronze ratio". For this, compare with A001622 and A014176.
If c is this constant and n > 0, then for n even, c^n = [A100230(n), 1, A100230(n)-1, 1, A100230(n)-1, 1, A100230(n)-1, 1, ...], for n odd, c^n = [A100230(n)+1, A100230(n)+1, A100230(n)+1, ...]. - Gerald McGarvey, Dec 15 2007
This is the shape of a 3-extension rectangle; see A188640 for definitions. - Clark Kimberling, Apr 10 2011
From Vladimir Shevelev, Mar 02 2013: (Start)
An analog of Fermat theorem: for prime p, round(c^p) == 3 (mod p).
A generalization for "metallic" constants c_N = (N+sqrt(N^2+4))/2, N>=1: for prime p, round((c_N)^p) == N (mod p). (End)
This is the positive real algebraic number c of degree 2 with minimal polynomial x^3 - x - 1. The other negative root is 3 - c. - Wolfdieter Lang, Aug 29 2022
c^n = c*A006190(n) + A006190(n-1). - Gary W. Adamson, Apr 02 2024

Examples

			3.30277563...
		

Crossrefs

Programs

Formula

3 plus the constant in A085550. - R. J. Mathar, Sep 02 2008
From Hieronymus Fischer, Jan 02 2009: (Start)
Set c:=(3+sqrt(13))/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:=(3+sqrt(13))/2 satisfies c-c^(-1)=floor(c)=3, hence c^n + (-c)^(-n) = round(c^n) for n>0, which follows from the general formula of A001622.
1/c=(sqrt(13)-3)/2.
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 A014176 (the silver ratio: where floor(x)=2). (End)
c=3+sum{k>=1}(-1)^(k-1)/(A006190(k)*A006190(k+1)). - Vladimir Shevelev, Feb 23 2013
A generalization for "metallic" constants c_N = (N+sqrt(N^2+4))/2, N>=1. Let {A_N(n), n>=0} be the sequence 0, 1, N, N^2+1, N^3+2*N, N^4+3*N^2+1,..., a(N) = N*a(N-1) + a(N-2). Then c_N = N + sum_{n>=1} (-1)^(n-1)/(A_N(n)*A_N(n+1)) (cf. A001622, A014176, A098316, A098317, A098318). - Vladimir Shevelev, Feb 23 2013
Equals lim_{n->oo} S(n, sqrt(13))/S(n-1, sqrt(13)), with the S-Chebyshev polynomial (see A049310). - Wolfdieter Lang, Nov 15 2023

A261327 a(n) = (n^2 + 4) / 4^((n + 1) mod 2).

Original entry on oeis.org

1, 5, 2, 13, 5, 29, 10, 53, 17, 85, 26, 125, 37, 173, 50, 229, 65, 293, 82, 365, 101, 445, 122, 533, 145, 629, 170, 733, 197, 845, 226, 965, 257, 1093, 290, 1229, 325, 1373, 362, 1525, 401, 1685, 442, 1853, 485, 2029, 530, 2213, 577, 2405, 626, 2605, 677
Offset: 0

Views

Author

Paul Curtz, Aug 15 2015

Keywords

Comments

Using (n+sqrt(4+n^2))/2, after the integer 1 for n=0, the reduced metallic means are b(1) = (1+sqrt(5))/2, b(2) = 1+sqrt(2), b(3) = (3+sqrt(13))/2, b(4) = 2+sqrt(5), b(5) = (5+sqrt(29))/2, b(6) = 3+sqrt(10), b(7) = (7+sqrt(53))/2, b(8) = 4+sqrt(17), b(9) = (9+sqrt(85))/2, b(10) = 5+sqrt(26), b(11) = (11+sqrt(125))/2 = (11+5*sqrt(5))/2, ... . The last value yields the radicals in a(n) or A013946.
b(2) = 2.41, b(3) = 3.30, b(4) = 4.24, b(5) = 5.19 are "good" approximations of fractal dimensions corresponding to dimensions 3, 4, 5, 6: 2.48, 3.38, 4.33 and 5.45 based on models. See "Arbres DLA dans les espaces de dimension supérieure: la théorie des peaux entropiques" in Queiros-Condé et al. link. DLA: beginning of the title of the Witten et al. link.
Consider the symmetric array of the half extended Rydberg-Ritz spectrum of the hydrogen atom:
0, 1/0, 1/0, 1/0, 1/0, 1/0, 1/0, 1/0, ...
-1/0, 0, 3/4, 8/9, 15/16, 24/25, 35/36, 48/49, ...
-1/0, -3/4, 0, 5/36, 3/16, 21/100, 2/9, 45/196, ...
-1/0, -8/9, -5/36, 0, 7/144, 16/225, 1/12, 40/441, ...
-1/0, -15/16, -3/16, -7/144, 0, 9/400, 5/144, 33/784, ...
-1/0, -24/25, -21/100, -16/225, -9/400, 0, 11/900, 24/1225, ...
-1/0, -35/36, -2/9, -1/12, -5/144, -11/900, 0, 13/1764, ...
-1/0, -48/49, -45/196, -40/441, -33/784, -24/1225, -13/1764, 0, ... .
The numerators are almost A165795(n).
Successive rows: A000007(n)/A057427(n), A005563(n-1)/A000290(n), A061037(n)/A061038(n), A061039(n)/A061040(n), A061041(n)/A061042(n), A061043(n)/A061044(n), A061045(n)/A061046(n), A061047(n)/A061048(n), A061049(n)/A061050(n).
A144433(n) or A195161(n+1) are the numerators of the second upper diagonal (denominators: A171522(n)).
c(n+1) = a(n) + a(n+1) = 6, 7, 15, 18, 34, 39, 63, 70, 102, 111, ... .
c(n+3) - c(n+1) = 9, 11, 19, 21, 29, 31, ... = A090771(n+2).
The final digit of a(n) is neither 4 nor 8. - Paul Curtz, Jan 30 2019

Crossrefs

Programs

  • Magma
    [Numerator(1+n^2/4): n in [0..60]]; // Vincenzo Librandi, Aug 15 2015
    
  • Maple
    A261327:=n->numer((4 + n^2)/4); seq(A261327(n), n=0..60); # Wesley Ivan Hurt, Aug 15 2015
  • Mathematica
    LinearRecurrence[{0, 3, 0, -3, 0, 1}, {1, 5, 2, 13, 5, 29}, 60] (* Vincenzo Librandi, Aug 15 2015 *)
    a[n_] := (n^2 + 4) / 4^Mod[n + 1, 2]; Table[a[n], {n, 0, 52}] (* Peter Luschny, Mar 18 2022 *)
  • PARI
    vector(60, n, n--; numerator(1+n^2/4)) \\ Michel Marcus, Aug 15 2015
    
  • PARI
    Vec((1+5*x-x^2-2*x^3+2*x^4+5*x^5)/(1-x^2)^3 + O(x^60)) \\ Colin Barker, Aug 15 2015
    
  • PARI
    a(n)=if(n%2,n^2+4,(n/2)^2+1) \\ Charles R Greathouse IV, Oct 16 2015
    
  • Python
    [(n*n+4)//4**((n+1)%2) for n in range(60)] # Gennady Eremin, Mar 18 2022
  • Sage
    [numerator(1+n^2/4) for n in (0..60)] # G. C. Greubel, Feb 09 2019
    

Formula

a(n) = numerator(1 + n^2/4). (Previous name.) See A010685 (denominators).
a(2*k) = 1 + k^2.
a(2*k+1) = 5 + 4*k*(k+1).
a(2*k+1) = 4*a(2*k) + 4*k + 1.
a(4*k+2) = A069894(k). - Paul Curtz, Jan 30 2019
a(-n) = a(n).
a(n+2) = a(n) + A144433(n) (or A195161(n+1)).
a(n) = A168077(n) + period 2: repeat 1, 4.
a(n) = A171621(n) + period 2: repeat 2, 8.
From Colin Barker, Aug 15 2015: (Start)
a(n) = (5 - 3*(-1)^n)*(4 + n^2)/8.
a(n) = n^2/4 + 1 for n even;
a(n) = n^2 + 4 for n odd.
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n>5.
G.f.: (1 + 5*x - x^2 - 2*x^3 + 2*x^4 + 5*x^5)/ (1 - x^2)^3. (End)
E.g.f.: (5/8)*(x^2 + x + 4)*exp(x) - (3/8)*(x^2 - x + 4)*exp(-x). - Robert Israel, Aug 18 2015
Sum_{n>=0} 1/a(n) = (4*coth(Pi)+tanh(Pi))*Pi/8 + 1/2. - Amiram Eldar, Mar 22 2022

Extensions

New name by Peter Luschny, Mar 18 2022

A223140 Decimal expansion of (sqrt(29) + 1)/2.

Original entry on oeis.org

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

Views

Author

Jaroslav Krizek, Apr 02 2013

Keywords

Comments

Decimal expansion of sqrt(7 + sqrt(7 + sqrt(7 + sqrt(7 + ... )))).
Sequence with a(1) = 2 is decimal expansion of sqrt(7 - sqrt(7 - sqrt(7 - sqrt(7 - ... )))) - A223141.
From Wolfdieter Lang, Jan 05 2024: (Start)
This number phi29 = (1 + sqrt(29))/2 is the fundamental algebraic integer in the quadratic number field Q(sqrt(29)) with minimal polynomial x^2 - x - 7. The other root is -A223141.
phi29^n = 7*A(n-1) + A(n)*phi29, where A(n) = A015442(n) with A(-1) = 1/7, for n >= 0. For negative powers n see A367454 = 1/phi29. (End)

Examples

			3.1925824035672520156253552457701...
		

Crossrefs

Essentially the same as A098318 and A085551.

Programs

Formula

Closed form: (sqrt(29) + 1)/2 = A098318-2 = 10*A085551+3 = A223141+1.
sqrt(7 + sqrt(7 + sqrt(7 + sqrt(7 + ... )))) - 1 = sqrt(7 - sqrt(7 - sqrt(7 - sqrt(7 - ... )))). See A223141.

A244593 Decimal expansion of z_c = phi^5 (where phi is the golden ratio), a lattice statistics constant which is the exact value of the critical activity of the hard hexagon model.

Original entry on oeis.org

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

Views

Author

Jean-François Alcover, Jul 01 2014

Keywords

Comments

Essentially the same digit sequence as A239798, A019863 and A019827. - R. J. Mathar, Jul 03 2014
The minimal polynomial of this constant is x^2 - 11*x - 1. - Joerg Arndt, Jan 01 2017

Examples

			11.09016994374947424102293417182819058860154589902881431067724311352630...
		

References

  • Steven R. Finch, Mathematical Constants, Cambridge University Press, 2003, Section 5.12.1 Phase transitions in Lattice Gas Models, p. 347.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 138-139.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 83.

Crossrefs

Programs

  • Mathematica
    RealDigits[GoldenRatio^5, 10, 103] // First
  • PARI
    (5*sqrt(5)+11)/2 \\ Charles R Greathouse IV, Aug 10 2016

Formula

Equals ((1 + sqrt(5))/2)^5 = (11 + 5*sqrt(5))/2.
Equals phi^5 = 11 + 1/phi^5 = 3 + 5*phi, an integer in the quadratic number field Q(sqrt(5)). - Wolfdieter Lang, Nov 11 2023
Equals lim_{n->infinity} S(n, 5*(-1 + 2*phi))/ S(n-1, 5*(-1 + 2*phi)), with the S-Chebyshev polynomials (see A049310). - Wolfdieter Lang, Nov 15 2023

A261391 a(n) = n^5 + 5*n^3 + 5*n.

Original entry on oeis.org

0, 11, 82, 393, 1364, 3775, 8886, 18557, 35368, 62739, 105050, 167761, 257532, 382343, 551614, 776325, 1069136, 1444507, 1918818, 2510489, 3240100, 4130511, 5206982, 6497293, 8031864, 9843875, 11969386, 14447457, 17320268, 20633239, 24435150, 28778261, 33718432, 39315243, 45632114
Offset: 0

Views

Author

Raphael Ranna, Aug 17 2015

Keywords

Comments

Also numbers of the form (n-th metallic mean)^5 - 1/(n-th metallic mean)^5, see link to Wikipedia.

Crossrefs

Programs

  • Mathematica
    Array[#^5 + 5 #^3 + 5 # &, 34] (* Michael De Vlieger, Aug 18 2015 *)
    Table[n^5 + 5*n^3 + 5*n, {n,0, 50}] (* G. C. Greubel, Aug 21 2015 *)
    LinearRecurrence[{6,-15,20,-15,6,-1},{0,11,82,393,1364,3775},40] (* Harvey P. Dale, May 07 2018 *)
  • PARI
    concat(0, Vec(x*(11*x^4+16*x^3+66*x^2+16*x+11)/(x-1)^6 + O(x^100))) \\ Colin Barker, Aug 18 2015

Formula

a(n) = ( (n+sqrt(n^2+4))/2 )^5 - 1/( (n+sqrt(n^2+4))/2 )^5.
a(n) = -a(-n) = 6*a(n-1)-15*a(n-2)+20*a(n-3)-15*a(n-4)+6*a(n-5)-a(n-6). - Colin Barker, Aug 18 2015
G.f.: x*(11*x^4+16*x^3+66*x^2+16*x+11) / (x-1)^6. - Colin Barker, Aug 18 2015
E.g.f.: (x^5 + 15*x^4 + 70*x^3 + 120*x^2 + 71*x + 11)*e^x. - G. C. Greubel, Aug 21 2015

Extensions

Offset changed from 1 to 0, initial 0 added and b-file adapted from Bruno Berselli, Aug 25 2015

A261540 a(n) = n^7 + 7*n^5 + 14*n^3 + 7*n.

Original entry on oeis.org

0, 29, 478, 4287, 24476, 101785, 337434, 946043, 2333752, 5206581, 10714070, 20633239, 37597908, 65378417, 109216786, 176222355, 275832944, 420346573, 625528782, 911300591, 1302512140, 1829807049, 2530582538, 3450050347, 4642403496, 6172093925, 8115226054
Offset: 0

Views

Author

Raphael Ranna, Aug 24 2015

Keywords

Comments

Also numbers of the form (n-th metallic mean)^7 - 1/(n-th metallic mean)^7, see link to Wikipedia.

Crossrefs

Programs

  • Magma
    [n^7 + 7*n^5 + 14*n^3 + 7*n: n in [0..30]]; // Vincenzo Librandi, Aug 24 2015
    
  • Mathematica
    Table[n^7 + 7 n^5 + 14 n^3 + 7 n, {n, 0, 30}] (* Bruno Berselli, Aug 24 2015 *)
    LinearRecurrence[{8, -28, 56, -70, 56, -28, 8, -1}, {0, 29, 478, 4287, 24476, 101785, 337434, 946043}, 30] (* Vincenzo Librandi, Aug 24 2015 *)
  • PARI
    a(n)=n^7+7*n^5+14*n^3+7*n \\ Charles R Greathouse IV, Aug 24 2015
  • Sage
    [n^7+7*n^5+14*n^3+7*n for n in (0..30)] # Bruno Berselli, Aug 24 2015
    

Formula

a(n) = -a(-n) = ( (n+sqrt(n^2+4))/2 )^7 - 1/( (n+sqrt(n^2+4))/2 )^7.
G.f.: x*(29 + 246*x + 1275*x^2 + 1940*x^3 + 1275*x^4 + 246*x^5 + 29*x^6)/(1 - x)^8. - Bruno Berselli, Aug 24 2015

Extensions

Offset changed from 1 to 0 and initial 0 added by Bruno Berselli, Aug 25 2015

A305837 Triangle read by rows: T(0,0) = 1; T(n,k) = 5*T(n-1,k) + T(n-2,k-1) for k = 0..floor(n/2); T(n,k)=0 for n or k < 0.

Original entry on oeis.org

1, 5, 25, 1, 125, 10, 625, 75, 1, 3125, 500, 15, 15625, 3125, 150, 1, 78125, 18750, 1250, 20, 390625, 109375, 9375, 250, 1, 1953125, 625000, 65625, 2500, 25, 9765625, 3515625, 437500, 21875, 375, 1, 48828125, 19531250, 2812500, 175000, 4375, 30, 244140625, 107421875, 17578125, 1312500, 43750, 525, 1
Offset: 0

Views

Author

Shara Lalo, Jun 11 2018

Keywords

Comments

The numbers in rows of the triangle are along skew diagonals pointing top-left in center-justified triangle given in A013612 ((1+5*x)^n).
The coefficients in the expansion of 1/(1-5x-x^2) are given by the sequence generated by the row sums.
If s(n) is the row sum at n, then the ratio s(n)/s(n-1) is approximately 5.1925824035..., a metallic mean (see A098318), when n approaches infinity.

Examples

			Triangle begins:
            1;
            5;
           25,           1;
          125,          10;
          625,          75,          1;
         3125,         500,         15;
        15625,        3125,        150,         1;
        78125,       18750,       1250,        20;
       390625,      109375,       9375,       250,        1;
      1953125,      625000,      65625,      2500,       25;
      9765625,     3515625,     437500,     21875,      375,      1;
     48828125,    19531250,    2812500,    175000,     4375,     30;
    244140625,   107421875,   17578125,   1312500,    43750,    525,     1;
   1220703125,   585937500,  107421875,   9375000,   393750,   7000,    35;
   6103515625,  3173828125,  644531250,  64453125,  3281250,  78750,   700,  1;
  30517578125, 17089843750, 3808593750, 429687500, 25781250, 787500, 10500, 40;
		

References

  • Shara Lalo and Zagros Lalo, Polynomial Expansion Theorems and Number Triangles, Zana Publishing, 2018, ISBN: 978-1-9995914-0-3, pp. 70, 72, 92, 380, 382.

Crossrefs

Row sums give A052918.
Cf. A000351 (column 0), A053464 (column 1), A081135 (column 2), A081143 (column 3), A036071 (column 4).
Cf. A013612.
Cf. A098318.

Programs

  • Mathematica
    t[0, 0] = 1; t[n_, k_] := If[n < 0 || k < 0, 0, 5 t[n - 1, k] + t[n - 2, k - 1]]; Table[t[n, k], {n, 0, 12}, {k, 0, Floor[n/2]}] // Flatten

Formula

G.f.: 1/(1 - 5*t*x - t^2).
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