A275651 -id:A275651 - cp's OEIS frontend

A049470 Decimal expansion of cos(1).

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

Offset: 0

Albert du Toit (dutwa(AT)intekom.co.za), N. J. A. Sloane

Also, decimal expansion of the real part of e^i. - Bruno Berselli, Feb 08 2013

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

			0.5403023058681397...

Muniru A Asiru, Table of n, a(n) for n = 0..2000
Mohammad K. Azarian, Forty-Five Nested Equilateral Triangles and cosecant of 1 degree, Problem 813, College Mathematics Journal, Vol. 36, No. 5, November 2005, p. 413-414.
Mohammad K. Azarian, Solution of Forty-Five Nested Equilateral Triangles and cosecant of 1 degree, Problem 813, College Mathematics Journal, Vol. 37, No. 5, November 2006, pp. 394-395.
I. S. Gradsteyn and I. M. Ryzhik, Table of integrals, series and products, (1980), page 10 (formula 0.245.7).
Simon Plouffe, cos(1)
Eric Weisstein's World of Mathematics, Factorial Sums
Index entries for transcendental numbers

Cf. A049469 (imaginary part of e^i), A211883 (real part of -(i^e)), A211884 (imaginary part of -(i^e)). - Bruno Berselli, Feb 08 2013
Cf. A073743 ( cosh(1) ), A073448, A275651.
Cf. A068985, A346441, A346440, A346439, A346438, A346437, A346436, A346435, A196498.

evalf(cos(1)); # Altug Alkan, Sep 22 2018

Mathematica
```
RealDigits[Cos[1], 10, 110] [[1]]
```

cos(1) \\ Charles R Greathouse IV, Jan 04 2016

Continued fraction representation: cos(1) = 1/(1 + 1/(1 + 2/(11 + 12/(29 + ... + (2*n - 2)*(2*n - 3)/((4*n^2 - 2*n - 1) + ... ))))). See A275651 for proof. Cf. A073743. - Peter Bala, Sep 02 2016
Equals Sum_{k >= 0} (-1)^k/A010050(k), where A010050(k) = (2k)! [See Gradshteyn and Ryzhik]. - A.H.M. Smeets, Sep 22 2018
Equals 1/A073448. - Alois P. Heinz, Jan 23 2023
From Gerry Martens, May 04 2024: (Start)
Equals (4*(cos(1/4)^4 + sin(1/4)^4) - 3).
Equals (16*(cos(1/4)^6 + sin(1/4)^6) - 10)/6. (End)

A051396 a(n) = (2n-2)(2n-3)a(n-1)+1.

Original entry on oeis.org

0, 1, 3, 37, 1111, 62217, 5599531, 739138093, 134523132927, 32285551902481, 9879378882159187, 3754163975220491061, 1734423756551866870183, 957401913616630512341017, 622311243850809833021661051, 470467300351212233764375754557, 409306551305554643375006906464591

Offset: 0

Aleksandar Petojevic

The sequence 1,0,3,0,37,... has e.g.f. cosh(x)/(1-x^2) with a(n) = Sum_{k=0..n} C(n,k)k!(1+(-1)^k)(1+(-1)^(n-k))/4. - Paul Barry, May 01 2005

Seiichi Manyama, Table of n, a(n) for n = 0..225
Romeo Mestrovic, Variations of Kurepa's left factorial hypothesis, arXiv preprint arXiv:1312.7037 [math.NT], 2013.
Romeo Mestrovic, The Kurepa-Vandermonde matrices arising from Kurepa's left factorial hypothesis, Filomat 29:10 (2015), 2207-2215; DOI 10.2298/FIL1510207M.
Aleksandar Petojevic, On Kurepa's Hypothesis for the Left Factorial, FILOMAT (Nis), 12:1 (1998), p. 29-37.

Bisection of abs(A009179(n)). Cf. A049470 (cos(1)), A073743 (cosh(1)), A275651.

A051396 := proc(n) option remember; if n <= 1 then n else (2*n-2)*(2*n-3)*A051396(n-1)+1; fi; end;

a[0] = 0; a[n_] := a[n] = (2*n-2)*(2*n-3)*a[n-1] + 1;
Table[a[n], {n, 0, 16}] (* Jean-François Alcover, Dec 11 2017 *)
nxt[{n_,a_}]:={n+1,a(4n^2-2n)+1}; NestList[nxt,{0,0},20][[;;,2]] (* Harvey P. Dale, Sep 26 2023 *)

a(n) = Sum_{k=0..n-1} (2*n-2)!/(2*k)! = floor((2*n-2)!*cosh(1)), n>=1. - Vladeta Jovovic, Aug 10 2002
a(n+1) = Sum_{k=0..2n}, C(2n, k)*k!*(1+(-1)^k)^2. - Paul Barry, May 01 2005
a(n) +(-4*n^2+10*n-7)*a(n-1) +2*(n-2)*(2*n-5)*a(n-2)=0. - R. J. Mathar, Nov 26 2012
From Peter Bala, Sep 05 2016: (Start)
The sequence b(n) := (2*n - 2)! also satisfies Mathar's recurrence with b(1) = 1, b(2) = 2. This leads to the continued fraction representation a(n) = (2*n - 2)!*(1 + 1/(2 - 2/(13 - 12/(31 - ... - (2*n - 4)*(2*n - 5)/(4*n^2 - 10*n + 7) )))) for n >= 3. Taking the limit gives the continued fraction representation cosh(1) = A073743 = 1 + 1/(2 - 2/(13 - 12/(31 - ... - (2*n - 4)*(2*n - 5)/((4*n^2 - 10*n + 7) - ... )))). (End)

A051397 a(n) = (2n-2)(2n-1)a(n-1)+1.

Original entry on oeis.org

0, 1, 7, 141, 5923, 426457, 46910271, 7318002277, 1536780478171, 418004290062513, 142957467201379447, 60042136224579367741, 30381320929637160076947, 18228792557782296046168201, 12796612375563171824410077103, 10390849248957295521420982607637

Offset: 0

Aleksandar Petojevic

Harvey P. Dale, Table of n, a(n) for n = 0..225
Romeo Mestrovic, Variations of Kurepa's left factorial hypothesis, arXiv preprint arXiv:1312.7037 [math.NT], 2013.
Romeo Mestrovic, The Kurepa-Vandermonde matrices arising from Kurepa's left factorial hypothesis, Filomat 29:10 (2015), 2207-2215; DOI 10.2298/FIL1510207M.
Aleksandar Petojevic, On Kurepa's Hypothesis for the Left Factorial, FILOMAT (Nis), 12:1 (1998), p. 29-37.

Bisection of abs(A009628). Also bisection of A087208 and of A186763. Cf. A073742, A074790, A275651.

nxt[{n_,a_}]:={n+1,(2(n+1)-2)(2(n+1)-1)a+1}; Transpose[NestList[nxt,{0,0},20]][[2]] (* Harvey P. Dale, Jun 13 2016 *)

a(n) = Sum_{k=0..n-1} (2*n-1)!/(2*k+1)!. a(n) = floor((2*n-1)!*sinh(1)). - Vladeta Jovovic, Aug 10 2002
Conjecture: a(n) +(-4*n^2+6*n-3)*a(n-1) +2*(2*n-3)*(n-2)*a(n-2)=0. - R. J. Mathar, Jan 31 2014
From Peter Bala, Sep 02 2016: (Start)
G.f. sinh(x)/(1 - x^2) = x + 7*x^3/3! + 141*x^5/5! + 5923*x^7/7! + ....
Mathar's conjectured recurrence a(n) = (4*n^2 - 6*n + 3)*a(n-1) - (2*n - 3)*(2*n - 4)*a(n-2) follows easily from the defining recurrence. The sequence b(n) := (2*n - 1)! also satisfies Mathar's recurrence but with b(1) = 1, b(2) = 6. This leads to the continued fraction representation a(n) = (2*n - 1)!*(1 + 1/(6 - 6/(21 - 20/(43 - ... - (2*n - 3)*(2*n - 4)/(4*n^2 - 6*n + 3) )))) for n >= 3. Taking the limit gives the continued fraction representation sinh(1) = A073742 = 1 + 1/(6 - 6/(21 - 20/(43 - ... - (2*n - 3)*(2*n - 4)/((4*n^2 - 6*n + 3) - ... )))). (End)

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A049470 Decimal expansion of cos(1).

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A051396 a(n) = (2n-2)(2n-3)a(n-1)+1.

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A051397 a(n) = (2n-2)(2n-1)a(n-1)+1.

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cp's OEIS Frontend

A049470 Decimal expansion of cos(1).

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Maple

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Formula

A051396 a(n) = (2*n-2)*(2*n-3)*a(n-1)+1.

Views

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Maple

Mathematica

Formula

A051397 a(n) = (2*n-2)*(2*n-1)*a(n-1)+1.

Views

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Links

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Programs

Mathematica

Formula

A051396 a(n) = (2n-2)(2n-3)a(n-1)+1.

A051397 a(n) = (2n-2)(2n-1)a(n-1)+1.