A074790 -id:A074790 - cp's OEIS frontend

A049469 Decimal expansion of sin(1).

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

Offset: 0

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

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

By the Lindemann-Weierstrass theorem, this constant is transcendental. - Charles R Greathouse IV, May 12 2019

			0.8414709848078965...

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, I. M. Ryzhik, Table of integrals, series and products, (1980), page 10 (formula 0.245.8).
Paul J. Nahin, Inside interesting integrals, Undergrad. Lecture Notes in Physics, Springer (2020), chapter 1.5
Simon Plouffe, sin(1)
Eric Weisstein's World of Mathematics, Factorial Sums
Index entries for transcendental numbers

Cf. A049470 (real part of e^i), A211883 (real part of -(i^e)), A211884 (imaginary part of -(i^e)). - Bruno Berselli, Feb 08 2013

Cf. A074790.

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

Mathematica
```
RealDigits[N[Sin[1], 110]] [[1]]
```

sin(1) \\ Charles R Greathouse IV, Aug 20 2012

sumalt(n=0, (-1)^(n%2)/(2*n+1)!) \\ Gheorghe Coserea, Sep 23 2018

Continued fraction representation: sin(1) = 1 - 1/(6 + 6/(19 + 20/(41 + ... + (2*n - 1)*(2*n - 2)/((4*n^2 + 2*n - 1) + ... )))). See A074790 for details. - Peter Bala, Jan 30 2015
Equals Sum_{k > 0} (-1)^(k-1)/((2k-1)!) = Sum_{k > 0} (-1)^(k-1)/A009445(k-1) [See Gradshteyn and Ryzhik]. - A.H.M. Smeets, Sep 22 2018
Equals Product{k>=1} cos(1/2^k). - Amiram Eldar, Aug 20 2020
Equals Integral_{x=-1..1} cos(x)/[exp(1/x)+1] dx. [Nahin]. - R. J. Mathar, May 16 2024

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)

A275651 a(n) = (2n)!Sum_{k = 0..n} (-1)^k/(2*k)!.

Original entry on oeis.org

1, 1, 13, 389, 21785, 1960649, 258805669, 47102631757, 11304631621681, 3459217276234385, 1314502564969066301, 607300185015708631061, 335229702128671164345673

Offset: 0

Peter Bala, Sep 02 2016

Compare with the derangement numbers A000166(n) := n!*sum_{k = 0..n} (-1)^k/k! and also A074790.

Seiichi Manyama, Table of n, a(n) for n = 0..225

Cf. A000166, A049470 (cos(1)), A051396, A073743, A074790.

A275651 := proc(n) option remember; if (n = 0) then 1 else 2*n*(2*n - 1)*A275651(n-1)+(-1)^n end if; end proc:
seq(A275651(n), n = 0..20);

Table[(2 n)!*Sum[(-1)^k/(2 k)!, {k, 0, n}], {n, 12}] (* Michael De Vlieger, Sep 04 2016 *)

a(n) ~ (2*n)!*cos(1).
E.g.f. for the aerated sequence: cos(x)/(1 - x^2) = 1 + x^2/2! + 13*x^4/4! + 389*x^6/6! + ....
Recurrence equations:
a(n) = 2*n*(2*n - 1)*a(n-1) + (-1)^n with a(0) = 1.
a(n) = (4*n^2 - 2*n - 1)*a(n - 1) + (2*n - 2)*(2*n - 3)*a(n - 2) with a(0) = 1, a(1) = 1.
The latter recurrence is also satisfied by the sequence b(n) := (2*n)! with b(0) = 1, b(1) = 2. This leads to the continued fraction representation a(n) = (2*n )!*( 1/(1 + 1/(1 + 2/(11 + 12/(29 + ... + (2*n - 2)*(2*n - 3)/(4*n^2 - 2*n - 1) )))) ) for n >= 3. Taking the limit gives the continued fraction representation cos(1) = A049470 = 1/(1 + 1/(1 + 2/(11 + 12/(29 + ... + (2*n - 2)*(2*n - 3)/((4*n^2 - 2*n - 1) + ... ))))). Cf. A073743.

cp's OEIS Frontend

A049469 Decimal expansion of sin(1).

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

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A275651 a(n) = (2n)!Sum_{k = 0..n} (-1)^k/(2*k)!.

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cp's OEIS Frontend

A049469 Decimal expansion of sin(1).

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Programs

Maple

Mathematica

PARI

PARI

Formula

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

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Mathematica

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A275651 a(n) = (2*n)!*Sum_{k = 0..n} (-1)^k/(2*k)!.

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Maple

Mathematica

Formula

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

A275651 a(n) = (2n)!Sum_{k = 0..n} (-1)^k/(2*k)!.