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 1-3 of 3 results.

A176594 a(n) = 5^(2^n).

Original entry on oeis.org

5, 25, 625, 390625, 152587890625, 23283064365386962890625, 542101086242752217003726400434970855712890625, 293873587705571876992184134305561419454666389193021880377187926569604314863681793212890625
Offset: 0

Views

Author

Vincenzo Librandi, Apr 21 2010

Keywords

Comments

Also the hypotenuse of primitive Pythagorean triangles obtained by repeated application of basic formula c(n)=p(n)^2+q(n)^2 starting p(0)=2, q(0)=1, see A100686, A098122. Example: a(2)=25 since starting (2,1) gives Pythagorean triple (3,4,5) using (3,4) as new generators gives triple (7,24,25) hypotenuse 25=a(2). - Carmine Suriano, Feb 04 2011

Links

Crossrefs

Cf. A001146, A011764, A078886, A098122, A100686, A165423.
Cf. A185457, A120905, A139011. - Carmine Suriano, Feb 04 2011

Programs

Formula

a(n) = A165423(n+3).
a(n+1) = a(n)^2 with a(0)=5.
a(n-1) = (Im((2+i)^(2^n))^2 + Re((2+i)^(2^n))^2)^(1/2). - Carmine Suriano, Feb 04 2011
Sum_{n>=0} 1/a(n) = A078886. - Amiram Eldar, Nov 09 2020
Product_{n>=0} (1 + 1/a(n)) = 5/4. - Amiram Eldar, Jan 29 2021

Extensions

Offset corrected by R. J. Mathar, Jun 18 2010

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

Views

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

Cf. A100441, A165423, A176594, A225164.

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.

A225164 Denominators of the sequence s(n) of the sum resp. product 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, 84, 45444, 15686405364, 2147492192737717340004, 45388476229808808857318702720533556450342484
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 A165423(n+2), hence s(n) = sum(A165423(i+1)/A225157(i),i=1..n) = product(A165423(i+1)/A225157(i),i=1..n) = A165423(n+2)/a(n) = A176594(n-1)/a(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; ...
s(n) = 1/b(n) = 5, 25/4, 625/84, ...

Crossrefs

Cf. A076628, A165423, A176594, A225157.

Programs

  • Maple
    b:=proc(n) option remember; b(n-1)-b(n-1)^2; end:
b(1):=1/5;
a:=n->5^(2^(n-1))*b(n);
seq(a(i),i=1..8);

Formula

a(n) = 5^(2^(n-1))*b(n) where b(n)=b(n-1)-b(n-1)^2 with b(1)=1/5.
Showing 1-3 of 3 results.

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