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 1-4 of 4 results.

A225156 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 3/1; 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, 2, 7, 67, 5623, 37772347, 1653794703916063, 3104205768420613437667191487267, 10767416908549848056705041797805600349527548164015760674541223
Offset: 1

Views

Author

Martin Renner, Apr 30 2013

Keywords

Comments

Numerators of the sequence of fractions f(n) is A165421(n+1), hence sum(A165421(i+1)/a(i),i=1..n) = product(A165421(i+1)/a(i),i=1..n) = A165421(n+2)/A225163(n) = A011764(n-1)/A225163(n).

Examples

			f(n) = 3, 3/2, 9/7, 81/67, ...
3 + 3/2 = 3 * 3/2 = 9/2; 3 + 3/2 + 9/7 = 3 * 3/2 * 9/7 = 81/14; ...

Links

  • Paul Yiu, Recreational Mathematics, Department of Mathematics, Florida Atlantic University, 2003, Chapter 5.4, p. 207 (Project).

Crossrefs

Cf. A011764, A100441, A165421, A225163.

Programs

  • Maple
    b:=n->3^(2^(n-2)); # n > 1
b(1):=3;
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) = 3^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.
a(n) = 3^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.

A225161 Denominators of the sequence of fractions f(n) defined recursively by f(1) = 9/1; 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, 8, 73, 5977, 39556153, 1714946746986937, 3196895220321005409761642330233, 11033196234263169646028268239301916905952651329069957632398777
Offset: 1

Views

Author

Martin Renner, Apr 30 2013

Keywords

Comments

Numerators of the sequence of fractions f(n) is A165427(n+1), hence sum(A165427(i+1)/a(i),i=1..n) = product(A165427(i+1)/a(i),i=1..n) = A165427(n+2)/A225168(n) = A165421(n+3)/A225168(n) = A011764(n)/A225168(n).

Examples

			f(n) = 9, 9/8, 81/73, 6561/5977, ...
9 + 9/8 = 9 * 9/8 = 81/8; 9 + 9/8 + 81/73 = 9 * 9/8 * 81/73 = 6561/584; ...

Crossrefs

Cf. A011764, A100441, A165421, A165427, A225168.

Programs

  • Maple
    b:=n->9^(2^(n-2)); # n > 1
b(1):=9;
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..8);

Formula

a(n) = 9^(2^(n-2)) - product(a(i),i=1..n-1), n > 1 and a(1) = 1.
a(n) = 9^(2^(n-2)) - p(n) with a(1) = 1 and p(n) = p(n-1)*a(n-1) with p(1) = 1.

A225163 Denominators of the sequence s(n) of the sum resp. product of fractions f(n) defined recursively by f(1) = 3/1; 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, 2, 14, 938, 5274374, 199225484935778, 329478051871899046990657602014, 1022767669188735114815831063606918316150663428260080434555738
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 A165421(n+2), hence s(n) = sum(A165421(i+1)/A225156(i),i=1..n) = product(A165421(i+1)/A225156(i),i=1..n) = A165421(n+2)/a(n) = A011764(n-1)/a(n).

Examples

			f(n) = 3, 3/2, 9/7, 81/67, ...
3 + 3/2 = 3 * 3/2 = 9/2; 3 + 3/2 + 9/7 = 3 * 3/2 * 9/7 = 81/14; ...
s(n) = 1/b(n) = 3, 9/2, 81/14, ...

Links

  • Paul Yiu, Recreational Mathematics, Department of Mathematics, Florida Atlantic University, 2003, Chapter 5.4, p. 207 (Project).

Crossrefs

Cf. A011764, A076628, A165421, A225156.

Programs

  • Maple
    b:=proc(n) option remember; b(n-1)-b(n-1)^2; end:
b(1):=1/3;
a:=n->3^(2^(n-1))*b(n);
seq(a(i),i=1..9);

Formula

a(n) = 3^(2^(n-1))*b(n) where b(n)=b(n-1)-b(n-1)^2 with b(1)=1/3.

A225168 Denominators of the sequence s(n) of the sum resp. product of fractions f(n) defined recursively by f(1) = 9/1; 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, 8, 584, 3490568, 138073441864904, 236788599971507074896206759048, 756988343475413525492604622110601759725560263205883476698184
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 A165427(n+2), hence sum(A165427(i+1)/A225161(i),i=1..n) = product(A165427(i+1)/A225161(i),i=1..n) = A165427(n+2)/a(n) = A165421(n+3)/a(n) = A011764(n)/a(n).

Examples

			f(n) = 9, 9/8, 81/73, 6561/5977, ...
9 + 9/8 = 9 * 9/8 = 81/8; 9 + 9/8 + 81/73 = 9 * 9/8 * 81/73 = 6561/584; ...
s(n) = 1/b(n) = 9, 81/8, 6561/584, ...

Crossrefs

Cf. A011764, A076628, A165421, A165427, A225161.

Programs

  • Maple
    b:=proc(n) option remember; b(n-1)-b(n-1)^2; end:
b(1):=1/9;
a:=n->9^(2^(n-1))*b(n);
seq(a(i),i=1..8);

Formula

a(n) = 9^(2^(n-1))*b(n) where b(n)=b(n-1)-b(n-1)^2 with b(1)=1/9.
Showing 1-4 of 4 results.

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