A104457 -id:A104457 - cp's OEIS frontend

A139347 Decimal expansion of negated tangent of the golden ratio. That is, the decimal expansion of -tan((1+sqrt(5))/2).

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

Offset: 2

Mohammad K. Azarian, Apr 15 2008

By the Lindemann-Weierstrass theorem, this constant is transcendental. - Charles R Greathouse IV, May 13 2019

			-21.15380078249327464858628117032582559788124367464826...

Index entries for transcendental numbers.

Cf. A001622, A094214, A104457, A098317, A002390, A139339, A139340, A139341, A139342, A139345, A139346, A139348.

RealDigits[Tan[GoldenRatio], 10, 100][[1]] (* Amiram Eldar, Feb 07 2022 *)

-tan((sqrt(5)+1)/2) \\ Charles R Greathouse IV, May 13 2019

Equals tan(A001622).
From Amiram Eldar, Feb 07 2022: (Start)
Equals 1/A139348.
Equals A139345/A139346. (End)

Offset corrected by Mohammad K. Azarian, Dec 13 2008

Sign added to definition by R. J. Mathar, Feb 05 2009

A139348 Decimal expansion of negated cotangent of the golden ratio. That is, the decimal expansion of -cot((1+sqrt(5))/2).

Original entry on oeis.org

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

Offset: 0

Mohammad K. Azarian, Apr 15 2008

By the Lindemann-Weierstrass theorem, this constant is transcendental. - Charles R Greathouse IV, May 13 2019

			0.04727282866479448118935650960621633420056105722556...

Index entries for transcendental numbers.

Cf. A001622, A094214, A104457, A098317, A002390, A139339, A139340, A139341, A139342, A139345, A139346, A139347.

Join[{0},RealDigits[-Cot[GoldenRatio],10,120][[1]]] (* Harvey P. Dale, Sep 18 2013 *)

-cotan((sqrt(5)+1)/2) \\ Charles R Greathouse IV, May 13 2019

Equals cot(A001622).
From Amiram Eldar, Feb 07 2022: (Start)
Equals 1/A139347.
Equals A139346/A139345. (End)

Added sign in definition. Leading zero dropped by R. J. Mathar, Feb 05 2009

A219010 Numerators in a product expansion for sqrt(5).

Original entry on oeis.org

3, 123, 28143753123, 17656721319717734662791328845675730903632844218828123

Offset: 0

Peter Bala, Nov 09 2012

a(4) has 262 digits.
Iterating the algebraic identity sqrt(1 + 4/x) = (1 + 2*(x + 2)/(x^2 + 3*x + 1)) * sqrt(1 + 4/(x*(x^2 + 5*x + 5)^2)) produces the rapidly converging product expansion sqrt(1 + 4/x) = Product_{n >= 0} (1 + 2*a(n)/b(n)), where a(n) and b(n) are integer sequences when x is an integer: a(n) satisfies the recurrence a(n+1) = a(n)^5 - 5*a(n)^3 + 5*a(n) with the initial condition a(0) = x + 2 and b(n) satisfies the recurrence b(n+1) = 5/2*(b(n)^4 - b(n)^2)*sqrt(4*b(n) + 5) + b(n)^5 + 15/2*b(n)^4 - 25/2*b(n)^2 + 5 with the initial condition b(0) = x^2 + 3*x + 1.
The present case is when x = 1. The denominators b(n) are in A219011. See also A219012 (x = 2) and A219014 (x = 4).
For another product expansion for sqrt(5) see A002814.

			The first three terms of the product give 51 correct decimal places of sqrt(5): (1 + 2*3/5)*(1 + 2*123/15005)*(1 + 2*28143753123/792070839820228500005) = 2.23606 79774 99789 69640 91736 68731 27623 54406 18359 61152 5(4...).

Amiram Eldar, Table of n, a(n) for n = 0..4

Cf. A000032, A000045, A002814, A104457, A219011, A219012, A219014.

Table[(1 + 5*(Fibonacci[5^n] Fibonacci[5^n + 1] + Fibonacci[5^n - 1]^2))/2, {n, 0, 3}] (* or *)
Table[LucasL[2*5^n], {n, 0, 3}] (* Michael De Vlieger, Jul 30 2017 *)

A219010(n):=1/2*(1 + 5*(fib(5^n)*fib(5^n+1)+(fib(5^n - 1))^2))$
makelist(A219010(n),n,0,3); /* Martin Ettl, Nov 12 2012 */

Let tau = (3 + sqrt(5))/2. Then a(n) = tau^(5^n) + 1/tau^(5^n).
a(n) = (1/2)*(1 + 5*(Fibonacci(5^n)*Fibonacci(5^n + 1) + (Fibonacci(5^n - 1))^2)).
Recurrence equation: a(n+1) = a(n)^5 - 5*a(n)^3 + 5*a(n) with initial condition a(0) = 3.
a(n) = Lucas(2*5^n). - Ehren Metcalfe, Jul 29 2017

A283968 a(n) = a(n-1) + 1 + floor(n*(3 + sqrt(5))/2), a(0) = 1.

Original entry on oeis.org

1, 2, 3, 5, 7, 9, 12, 15, 19, 23, 27, 32, 37, 42, 48, 54, 61, 68, 75, 83, 91, 100, 109, 118, 128, 138, 148, 159, 170, 182, 194, 206, 219, 232, 245, 259, 273, 288, 303, 318, 334, 350, 367, 384, 401, 419, 437, 455, 474, 493, 513, 533, 553, 574, 595, 617, 639

Offset: 0

Clark Kimberling, Mar 18 2017

This is row 1 of the transposable interspersion A283938.

Clark Kimberling, Table of n, a(n) for n = 0..1000

Cf. A104457, A283938, A283961, A283969.

r = GoldenRatio^2; z = 120;
s[0] = 1; s[n_] := s[n] = s[n - 1] + 1 + Floor[n*r];
Table[n + 1 + Sum[Floor[(n - k)/r], {k, 0, n}], {n, 0, z}] (* A283968 *)
Table[s[n], {n, 0, z}] (* A283969 *)

r = (3 + sqrt(5))/2;
a(n) = n + 1 + sum(k=0, n, floor((n - k)/r));
for(n=0, 30, print1(a(n),", ")) \\ Indranil Ghosh, Mar 19 2017

from sympy import sqrt
import math
def a(n):
    return n + 1 + sum([int(math.floor((n - k)/r)) for k in range(n + 1)])
print([a(n) for n in range(61)]) # Indranil Ghosh, Mar 19 2017

a(n) = a(n-1) + 1 + floor(n*(3 + sqrt(5))/2), a(0) = 1.

A283969 a(n) = n + 1 + Sum_{k=0..n} floor((n-k)/r), where r = (3+sqrt(5))/2.

Original entry on oeis.org

1, 4, 10, 18, 29, 43, 59, 78, 99, 123, 150, 179, 211, 246, 283, 323, 365, 410, 458, 508, 561, 616, 674, 735, 798, 864, 933, 1004, 1078, 1154, 1233, 1315, 1399, 1486, 1576, 1668, 1763, 1860, 1960, 2063, 2168, 2276, 2386, 2499, 2615, 2733, 2854, 2978, 3104

Offset: 0

Clark Kimberling, Mar 18 2017

This is column 1 of the transposable interspersion A283938.

Clark Kimberling, Table of n, a(n) for n = 0..1000

Cf. A104457, A283968, A283938, A283961.

Partial sums of A026352.

r = GoldenRatio^2; z = 120;
s[0] = 1; s[n_] := s[n] = s[n - 1] + 1 + Floor[n*r];
Table[n + 1 + Sum[Floor[(n - k)/r], {k, 0, n}], {n, 0, z}] (* A283968 *)
Table[s[n], {n, 0, z}] (* A283969 *)

a(n) = if(n<1, 1, a(n - 1) + 1 + floor(n*(3 + sqrt(5))/2));
for(n = 0, 50, print1(a(n),", ")) \\ Indranil Ghosh, Mar 19 2017

import math
from sympy import sqrt
def a(n):
    return 1 if n<1 else a(n - 1) + 1 + int(math.floor(n*(3 + sqrt(5))/2))
print([a(n) for n in range(51)]) # Indranil Ghosh, Mar 19 2017

a(n) = n + 1 + Sum_{k=0..n} floor((n-k)/r), where r = (3+sqrt(5))/2.

A139349 Decimal expansion of negated secant of the golden ratio. That is, the decimal expansion of -sec((1+sqrt(5))/2).

Original entry on oeis.org

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

Offset: 2

Mohammad K. Azarian, Apr 15 2008

By the Lindemann-Weierstrass theorem, this constant is transcendental. - Charles R Greathouse IV, May 13 2019

			21.17742400636614440872804040937130213307185355364174...

Index entries for transcendental numbers.

Cf. A001622, A094214, A104457, A098317, A002390, A139339, A139340, A139341, A139342, A139345, A139346, A139347, A139348.

RealDigits[-Sec[GoldenRatio],10,120][[1]] (* Harvey P. Dale, Dec 08 2011 *)

-1/cos((sqrt(5)+1)/2) \\ Charles R Greathouse IV, May 13 2019

Equals sec(A001622).

Equals 1/A139346. - Amiram Eldar, Feb 07 2022

Offset corrected by Mohammad K. Azarian, Dec 13 2008

Sign in definition added by R. J. Mathar, Feb 05 2009

A139350 Decimal expansion of csc((1+sqrt(5))/2), where (1+sqrt(5))/2 is the golden ratio.

Original entry on oeis.org

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

Offset: 1

Mohammad K. Azarian, Apr 15 2008

By the Lindemann-Weierstrass theorem, this constant is transcendental. - Charles R Greathouse IV, May 13 2019

			1.00111673661465225489616711351705587794461531806624...

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for transcendental numbers.

Cf. A001622, A094214, A104457, A098317, A002390, A139339, A139340, A139341, A139342, A139345, A139346, A139347, A139348, A139349.

RealDigits[Csc[GoldenRatio], 10, 100][[1]] (* Vincenzo Librandi, Feb 19 2013 *)

1/sin((sqrt(5)+1)/2) \\ Charles R Greathouse IV, May 13 2019

Equals 1/A139345. - Amiram Eldar, Feb 07 2022

Edited by Bruno Berselli, Feb 19 2013

A140232 a(n) = ceiling(n*exp((1+sqrt(5))/2)).

Original entry on oeis.org

6, 11, 16, 21, 26, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, 81, 86, 91, 96, 101, 106, 111, 116, 122, 127, 132, 137, 142, 147, 152, 157, 162, 167, 172, 177, 182, 187, 192, 197, 202, 207, 212, 217, 222, 227, 232, 238, 243, 248, 253, 258, 263, 268, 273, 278, 283

Offset: 1

Mohammad K. Azarian, May 13 2008

nonn

G. C. Greubel, Table of n, a(n) for n = 1..10000

Cf. A001622, A094214, A104457, A098317, A002390, A139339, A139340, A139341, A139342, A139345, A139346, A139347, A139348, A139349, A139350, A140231, A016861.

phi:=(1+Sqrt(5))/2; [Ceiling(n*Exp(phi)): n in [1..60]]; // G. C. Greubel, Jun 30 2019

Ceiling[Exp[GoldenRatio]*Range[60]] (* G. C. Greubel, Jun 30 2019 *)

phi=(1+sqrt(5))/2; vector(60, n, ceil(n*exp(phi)) ) \\ G. C. Greubel, Jun 30 2019

[ceil(n*exp(golden_ratio)) for n in (1..60)] # G. C. Greubel, Jun 30 2019

a(n) = ceiling(n*A139341). - R. J. Mathar, Feb 06 2009

A184795 Numbers m such that prime(m) is of the form floor(k*s), where s=(3+sqrt(5))/2; complement of A184793.

Original entry on oeis.org

1, 3, 4, 6, 9, 11, 13, 15, 21, 23, 24, 28, 29, 35, 36, 37, 39, 43, 44, 49, 51, 54, 57, 59, 62, 64, 68, 71, 75, 78, 82, 83, 85, 90, 92, 99, 100, 101, 102, 109, 111, 113, 116, 120, 124, 125, 126, 127, 128, 129, 130, 132, 133, 135, 136, 141, 144, 145, 147, 150, 151, 154, 156, 158, 159, 161, 164, 168, 170, 172, 173, 175

Offset: 1

Clark Kimberling, Jan 22 2011

nonn

Indices of primes that appear in the Beatty sequence A001950, the Beatty sequence of A104457. - R. J. Mathar, Jan 22 2011

			Prime(1)=2 with k=1. Prime(3)=5 with k=2. Prime(4)=7 with k=3.

Cf. A184774, A184792, A184793.

Mathematica
```
(See A184792.)
```

A374883 Decimal expansion of phi(2phi + 1) (i.e., (7 + 3*sqrt(5))/2), where phi is the golden ratio.

Original entry on oeis.org

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

Offset: 1

Marco Ripà, Jul 22 2024

The author conjectures that this is the minimum volume of an axis-aligned bounding box which includes the shortest minimum-link circuit joining all the vertices of the cube {0,1}^3 (i.e., the closed polygonal chains consisting of exactly 6 edges visiting all the points of the set {(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1),(1,1,0),(1,1,1)}).
In detail, such a circuit of 6 links is given by (1/2,1+phi,1/2)-((1-phi)/2,0,(1+phi)/2)-((phi+1)/2,0, (1-phi)/2)-(1/2,1+phi,1/2)-((phi+1)/2,0,(phi+1)/2)-((1-phi)/2,0,(1-phi)/2(1/2,1+phi,1/2), where phi := (1+sqrt(5))/2 (see A001622).
Then, phi*(2*phi + 1) = phi^2*(phi + 1) since phi - 1 = 1/phi.

			6.8541019662496845446137605030969...

Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 138-139.

Roberto Rinaldi and Marco Ripà, Optimal cycles enclosing all the nodes of a k-dimensional hypercube, arXiv:2212.11216 [math.CO], 2022.
Marco Ripà, General uncrossing covering paths inside the Axis-Aligned Bounding Box, Journal of Fundamental Mathematics and Applications, Volume 4, 2021, Number 2, Pages 154-166.

Cf. A001622, A090550, A104457, A134973, A205769, A225227, A261547, A363755, A373537, A374149, A374260.

RealDigits[3*GoldenRatio + 2, 10, 120][[1]] (* Amiram Eldar, Jul 23 2024 *)

Equals (7 + 3*sqrt(5))/2.
Equals phi^2*(phi + 1), where phi = (1 + sqrt(5))/2.
Equals A104457^2 = 2*A205769. - Hugo Pfoertner, Jul 22 2024
Equals A090550 + 1 = A134973 + 5. - Amiram Eldar, Jul 23 2024
Equals phi^4. - Stefano Spezia, May 28 2025

cp's OEIS Frontend

A139347 Decimal expansion of negated tangent of the golden ratio. That is, the decimal expansion of -tan((1+sqrt(5))/2).

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A139348 Decimal expansion of negated cotangent of the golden ratio. That is, the decimal expansion of -cot((1+sqrt(5))/2).

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A219010 Numerators in a product expansion for sqrt(5).

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A283968 a(n) = a(n-1) + 1 + floor(n*(3 + sqrt(5))/2), a(0) = 1.

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A283969 a(n) = n + 1 + Sum_{k=0..n} floor((n-k)/r), where r = (3+sqrt(5))/2.

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A139349 Decimal expansion of negated secant of the golden ratio. That is, the decimal expansion of -sec((1+sqrt(5))/2).

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A139350 Decimal expansion of csc((1+sqrt(5))/2), where (1+sqrt(5))/2 is the golden ratio.

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A374883 Decimal expansion of phi(2phi + 1) (i.e., (7 + 3*sqrt(5))/2), where phi is the golden ratio.