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.

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A129978 Numbers k such that A120265(k) = numerator(Sum_{j=1..k} 1/j!) is a prime.

Original entry on oeis.org

2, 3, 4, 5, 6, 7, 12, 16, 19, 21, 22, 25, 41, 114, 181, 236, 2003, 6138
Offset: 1

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Author

Alexander Adamchuk, Jun 13 2007

Keywords

Comments

Corresponding primes are A120265(a(n)) = {3, 5, 41, 103, 1237, 433, 164611949, 35951249665217, 52255141388393, 43894318766250120011, 386270005143001056097, 53952693026046706215979, 1249584099900912571604389306768231303904375213027, ...}.
a(17) > 1000; A120265(1000) ~ 2.9*10^2564 = (e-1)*A061355(1000). - M. F. Hasler, Jun 18 2007

Crossrefs

Programs

  • Mathematica
    Do[ f=Numerator[ Sum[ 1/k!, {k,1,n} ] ]; If[ PrimeQ[f], Print[{n,f}] ], {n,1,236} ]
    Flatten[Position[Numerator[Accumulate[1/Range[2150]!]],?PrimeQ]] (* _Harvey P. Dale, May 03 2013 *)
  • PARI
    my(t=0); for( n=1,1000, if( ispseudoprime( numerator( t+=1/n!)), print1( n", " ))) \\ M. F. Hasler, Jun 18 2007

Extensions

Edited by M. F. Hasler, Jun 18 2007
a(17) from Alexander Adamchuk, May 02 2010
a(18) from Michael S. Branicky, Sep 24 2024

A196080 Numerators of the sum of the n-th partial sums of the expansions of e and 1/e.

Original entry on oeis.org

2, 2, 3, 3, 37, 37, 1111, 1111, 6913, 6913, 799933, 799933, 739138093, 739138093, 44841044309, 44841044309, 32285551902481, 32285551902481, 9879378882159187, 9879378882159187, 1251387991740163687
Offset: 0

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Author

Paul Curtz, Sep 27 2011

Keywords

Comments

The n-th partial sums of the Taylor expansion of E are f(n) = A061354(n)/A061355(n) = 1, 2, 5/2, 8/3, 65/24, 163/60,.. .
The partial sums of an expansion of 1/e are essentially A000255(n-2)/A001048(n-1) preceded by 1 and 0, namely g(n)= 1, 0, 1/2, 1/3, 3/8, 11/30, 53/144, 103/280, 2119/5760,... (Jolley's partial sums of 1/E in A068985 is the bisection 0, 1/3, 11/30, 103/280, 16687/45360,... of g(n).)
The current sequence are the numerators of f(n)+g(n), converging to E+1/E, namely 2, 2, 3, 3, 37/12, 37/12, 1111/360, 1111/360, 6913/2240 = 62217/21060, 6913/2240 = 62217/21060, 799933/259200 = 5599531/1814400,... The unreduced fractions are apparently given by duplicated A051396(n+1)/A002674(n).

Examples

			a(0)=1+1, a(1)=2+0, a(2)=(5+1)/2, a(3)=(8+1)/3.
		

Crossrefs

Cf. A001113, A068985, A137204 (e+1/e).

Programs

  • Mathematica
    a[n_] := (E*Gamma[n+1, 1] + (1/E)*Gamma[n+1, -1])/n! // FullSimplify // Numerator; Table[a[n], {n, 0, 20}] (* Jean-François Alcover, Aug 02 2012 *)

Extensions

Redefined by reduced fractions. - R. J. Mathar, Jul 02 2012

A233044 Pairs p, q for those partial sums p/q of the series e = sum_{n>=0} 1/n! that are not convergents to e.

Original entry on oeis.org

1, 1, 5, 2, 65, 24, 163, 60, 1957, 720, 685, 252, 109601, 40320, 98641, 36288, 9864101, 3628800, 13563139, 4989600, 260412269, 95800320, 8463398743, 3113510400, 47395032961, 17435658240, 888656868019, 326918592000
Offset: 1

Views

Author

Jonathan Sondow, Dec 07 2013

Keywords

Comments

Sondow (2006) conjectured that 2/1 and 8/3 are the only partial sums of the Taylor series for e that are also convergents to the simple continued fraction for e. Sondow and Schalm (2008, 2010) proved partial results toward the conjecture. Berndt, Kim, and Zaharescu (2012) proved it in full.

Examples

			1/1, 5/2, 65/24, 163/60, 1957/720, 685/252, 109601/40320, 98641/36288, 9864101/3628800, 13563139/4989600, 260412269/95800320, 8463398743/3113510400, 47395032961/17435658240, 888656868019/326918592000
		

References

  • J. Sondow and K. Schalm, Which partial sums of the Taylor series for e are convergents to e?, (and a link to the primes 2, 5, 13, 37, 463), part I, in Tapas in Experimental Mathematics, T. Amdeberhan and V. H. Moll, eds., Contemp. Math., vol. 457, American Mathematical Society, Providence, RI, 2008, pp. 273-284.

Crossrefs

Formula

a(2n-1)/a(2n) = A061354(k)/A061355(k) for some k <> 1 and 3.
a(2n-1)/a(2n) <> A007676(k)/A007677(k) for all k.
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