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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A057438 a(1) = 1; a(n+1) = (Product_{k = 1..n} [a(k)]) * (Sum_{j = 1..n} [1/a(j)]).

Original entry on oeis.org

1, 1, 2, 5, 27, 739, 546391, 298543324411, 89128116550480609893151, 7943821159836055611643954282977557048699079331, 63104294619459055797454850600852928915607093463575707111291209057699988334565551829102647591
Offset: 1

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Author

Leroy Quet, Sep 08 2000

Keywords

Examples

			a(5) = a(1)*a(2)*a(3)*a(4)*(1/a(1) + 1/a(2) + 1/a(3) + 1/a(4)) = 1*1*2*5*(1 + 1 + 1/2 + 1/5) = 27.
		

Crossrefs

Programs

  • Mathematica
    a[1] = 1; a[n_] := a[n] = Sum[1/a[n - k], {k, n - 1}] Product[a[n - k], {k, n - 1}]; Table[ a[n], {n, 11}] (* Robert G. Wilson v, Jun 14 2005 *)

Formula

a(n) = a(n-1)^2+a(n-1)a(n-2)-a(n-2)^3 (valid for all n>3). - Ivan Sadofschi, Feb 22 2011
a(n) = a(n-1)^2+A074056(n-2) where A074056 is partial product of A057438. Close to a(n-1)^2+a(n-1)*0.365177806085453... and 1.1087260396143829635274191...^(2^n). - Henry Bottomley, Aug 14 2002

Extensions

More terms from Leroy Quet, Sep 08 2000

A074047 a(n)=a(n-1)*a(n-2)*a(n-3)*(1/a(n-1)+1/a(n-2)+1/a(n-3)) starting with a(1)=a(2)=a(3)=1.

Original entry on oeis.org

1, 1, 1, 3, 7, 31, 331, 12795, 4642051, 60935796571, 283646808320375611, 17285560913056915909539455163, 4902995236325455290013100337511909917402705547
Offset: 1

Views

Author

Henry Bottomley, Aug 14 2002

Keywords

Comments

Using the simplified formula which extends the original one to terms that may be zero, one could prefix the values (1, 1, 0), cf. A121810. See also A203761 and references therein. - M. F. Hasler, Jan 01 2013

Examples

			a(7)=31*7*3*(1/31+1/7+1/3)=331.
		

Crossrefs

Cf. A074046.

Programs

  • Mathematica
    RecurrenceTable[{a[n]==a[n-1]*a[n-2]+a[n-3]*a[n-1]+a[n-2]*a[n-3],a[1]==1,a[2]==1,a[3]==1},a[n],{n,1,15}] (* Vaclav Kotesovec, Jan 20 2014 *)

Formula

a(n) tends towards a(n-1)^phi and 1.22376...^(phi^n) where phi=(1+sqrt(5))/2=1.6180339887...
a(n)=a(n-1)*a(n-2)+a(n-3)*a(n-1)+a(n-2)*a(n-3). - M. F. Hasler, Jan 01 2013
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