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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A225157 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 5/1; f(n+1) is chosen so that the sum and the product of the first n terms of the sequence are equal.

Original entry on oeis.org

1, 4, 21, 541, 345181, 136901485261, 21135572172649245550621, 496712610012943408146407697714437299262548141, 271328559212953102170688304392824035451911661168940831351173011072850527195615099225368381
Offset: 1

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Author

Martin Renner, Apr 30 2013

Keywords

Comments

Numerators of the sequence of fractions f(n) is A165423(n+1), hence sum(A165423(i+1)/a(i),i=1..n) = product(A165423(i+1)/a(i),i=1..n) = A165423(n+2)/A225164(n) = A176594(n-1)/A225164(n).

Examples

			f(n) = 5, 5/4, 25/21, 625/541, ...
5 + 5/4 = 5 * 5/4 = 25/4; 5 + 5/4 + 25/21 = 5 * 5/4 * 25/21 = 625/84; ...
		

Crossrefs

Programs

  • Maple
    b:=n->5^(2^(n-2)); # n > 1
    b(1):=5;
    p:=proc(n) option remember; p(n-1)*a(n-1); end;
    p(1):=1;
    a:=proc(n) option remember; b(n)-p(n); end;
    a(1):=1;
    seq(a(i),i=1..9);

Formula

a(n) = 5^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.
a(n) = 5^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.
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