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.

Showing 1-5 of 5 results.

A172489 a(0)=1, a(1)=6; for n>=2, a(n) = 6^A042950(n-2).

Original entry on oeis.org

1, 6, 36, 216, 46656, 2176782336, 4738381338321616896, 22452257707354557240087211123792674816, 504103876157462118901767181449118688686067677834070116931382690099920633856
Offset: 0

Views

Author

Giovanni Teofilatto, Feb 05 2010

Keywords

Crossrefs

Cf. A042950, A167182, A173260, A175129.

Extensions

Revised by N. J. A. Sloane, Jun 20 2021

A173260 a(0)=1, a(1)=5; for n>=2, a(n) = 5^A042950(n-2).

Original entry on oeis.org

1, 5, 25, 125, 15625, 244140625, 59604644775390625, 3552713678800500929355621337890625, 12621774483536188886587657044524579674771302961744368076324462890625
Offset: 0

Views

Author

Giovanni Teofilatto, Feb 24 2010

Keywords

Crossrefs

Cf. A042950, A167182, A172489, A175129.

Extensions

Revised by N. J. A. Sloane, Jun 20 2021

A175129 a(0)=1, a(1)=3; for n>=2, a(n) = 3^A042950(n-2).

Original entry on oeis.org

1, 3, 9, 27, 729, 531441, 282429536481, 79766443076872509863361, 6362685441135942358474828762538534230890216321
Offset: 0

Views

Author

Giovanni Teofilatto, Feb 17 2010

Keywords

Crossrefs

Cf. A042950, A167182, A172489, A173260.

Extensions

Revised by N. J. A. Sloane, Jun 20 2021

A225160 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 8/1; 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, 7, 57, 3697, 15302113, 258902783918017, 73384158961115901868286873473, 5848244449673109813614947741525727934929692392922517757697
Offset: 1

Views

Author

Martin Renner, Apr 30 2013

Keywords

Comments

Numerators of the sequence of fractions f(n) is A165426(n+1), hence sum(A165426(i+1)/a(i),i=1..n) = product(A165426(i+1)/a(i),i=1..n) = A165426(n+2)/A225167(n) = A167182(n+2)/A225167(n).

Examples

			f(n) = 8, 8/7, 64/57, 4096/3697, ...
8 + 8/7 = 8 * 8/7 = 64/7; 8 + 8/7 + 64/57 = 8 * 8/7 * 64/57 = 4096/399; ...

Crossrefs

Cf. A100441, A165426, A167182, A225167.

Programs

  • Maple
    b:=n->8^(2^(n-2)); # n > 1
b(1):=8;
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) = 8^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.
a(n) = 8^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.

A225167 Denominators of the sequence s(n) of the sum resp. product of fractions f(n) defined recursively by f(1) = 8/1; 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, 7, 399, 1475103, 22572192792639, 5844003553148435725257076863, 428857285713570950220841681681938481172663051541516755199
Offset: 1

Views

Author

Martin Renner, Apr 30 2013

Keywords

Comments

Numerators of the sequence s(n) of the sum resp. product of fractions f(n) is A165426(n+2), hence sum(A165426(i+1)/A225160(i),i=1..n) = product(A165426(i+1)/A225160(i),i=1..n) = A165426(n+2)/a(n) = A167182(n+2)/a(n).

Examples

			f(n) = 8, 8/7, 64/57, 4096/3697, ...
8 + 8/7 = 8 * 8/7 = 64/7; 8 + 8/7 + 64/57 = 8 * 8/7 * 64/57 = 4096/399; ...
s(n) = 1/b(n) = 8, 64/7, 4096/399, ...

Crossrefs

Cf. A076628, A165426, A167182, A225160.

Programs

  • Maple
    b:=proc(n) option remember; b(n-1)-b(n-1)^2; end:
b(1):=1/8;
a:=n->8^(2^(n-1))*b(n);
seq(a(i),i=1..8);

Formula

a(n) = 8^(2^(n-1))*b(n) where b(n)=b(n-1)-b(n-1)^2 with b(1)=1/8.
Showing 1-5 of 5 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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