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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A120355 a(n) = min{k>0: the n-th convergent to e equals m/k! for some m}.

Original entry on oeis.org

1, 1, 3, 4, 7, 8, 13, 71, 31, 67, 13, 89, 83, 18089, 5441, 17377, 36269, 26021, 4909, 10391023, 1097, 28879, 1846921, 519691, 1329313, 793279, 7553783, 3308341, 65676881, 662407, 677311, 2425388512913, 4403182913, 10832561
Offset: 0

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Author

Jonathan Sondow, Aug 16 2006

Keywords

Examples

			The 6th convergent to e is 87/32 and 32 divides 8! but not 7!, so a(6) = 8.
		

Crossrefs

Formula

a(n) = A002034(A007677(n)).

Extensions

Extended by Max Alekseyev, Jul 28 2009
Missing a(7)=71 inserted by Georg Fischer, Oct 15 2024

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

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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.

A339267 Level of the Calkin-Wilf tree in which the n-th convergent of the continued fraction for e appears.

Original entry on oeis.org

2, 3, 5, 6, 7, 11, 12, 13, 19, 20, 21, 29, 30, 31, 41, 42, 43, 55, 56, 57, 71, 72, 73, 89, 90, 91, 109, 110, 111, 131, 132, 133, 155, 156, 157, 181, 182, 183, 209, 210, 211, 239, 240, 241, 271, 272, 273, 305, 306, 307, 341, 342, 343, 379, 380, 381, 419, 420, 421, 461
Offset: 1

Views

Author

Gary E. Davis, Nov 29 2020

Keywords

Comments

The depth level of a rational in the Calkin-Wilf tree is the sum of its continued fraction terms, with the root (1/1) as level 1 for this purpose. So the present sequence is partial sums of the continued fraction terms of e (A003417). Depth levels are the same in the related trees Stern-Brocot, Bird, etc. - Kevin Ryde, Dec 26 2020

Examples

			a(1) = 2 since 1st convergent 2, to e, appears at level 2 of the Calkin-Wilf tree.
a(2) = 3 since 2nd convergent 3 appears at level 3, a(3) = 5 since 3rd convergent 8/3 appears at level 5.
		

Crossrefs

Cf. A002487, A003417 (continued fraction for e), A007676/A007677 (convergents).

Programs

  • Mathematica
    children[{a_,b_}]:={{a,a+b},{a+b,b}};
    frac[{a_,b_}]:=a/b;
    L[1]={{1,1}};
    L[n_]:=Flatten[Map[children,L[n-1]],1];
    CWLevel[n_]:=Map[frac,If[n==1,L[1],Complement[L[n],L[n-1]]]];
    WhereCW[{a0_,b0_}]:=Module[{a=a0,b=b0,steps},steps =1;While[a>1 || b>1,{a,b}=parent[{a,b}];steps++];steps];
    fracpair[k_]:={Numerator[FromContinuedFraction[ContinuedFraction[E,k]]],Denominator[FromContinuedFraction[ContinuedFraction[E,k]]]};
    Table[WhereCW[fracpair[k]],{k,1,60}]
  • PARI
    a(n) = sqr(n\3) + n + 1; \\ Kevin Ryde, Dec 26 2020

Formula

a(n) = a(n-1) + 2*a(n-3) - 2*a(n-4) - a(n-6) + a(n-7) for n > 7.
a(n) = floor(n/3)^2 + n + 1. - Kevin Ryde, Dec 26 2020
G.f.: x*(2 + x + 2*x^2 - 3*x^3 - x^4 + x^6)/((1 - x)^3*(1 + x + x^2)^2). - Stefano Spezia, Dec 27 2020
Previous Showing 31-33 of 33 results.