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.

A083377 a(n) = the largest integer whose square has n digits and first digit 1.

Original entry on oeis.org

1, 4, 14, 44, 141, 447, 1414, 4472, 14142, 44721, 141421, 447213, 1414213, 4472135, 14142135, 44721359, 141421356, 447213595, 1414213562, 4472135954, 14142135623, 44721359549, 141421356237, 447213595499, 1414213562373, 4472135954999
Offset: 1

Views

Author

Werner S. Hürlimann (whurlimann(AT)bluewin.ch), Jun 05 2003

Keywords

Crossrefs

Programs

Formula

a(n) = floor(sqrt(10^n/5)).

Extensions

Edited by Don Reble, Nov 05 2005
Reference fixed by Charles R Greathouse IV, Oct 30 2009
More terms from Vincenzo Librandi, Oct 01 2011

A083379 a(n) = the number of squares with at most n digits and first digit 1.

Original entry on oeis.org

1, 2, 7, 20, 62, 193, 608, 1918, 6061, 19160, 60582, 191568, 605782, 1915640, 6057776, 19156359, 60577716, 191563545, 605777108, 1915635402
Offset: 1

Views

Author

Werner S. Hürlimann (whurlimann(AT)bluewin.ch), Jun 05 2003

Keywords

Comments

Asymptotically, the probability that a square begins with 1 is (sqrt(2)-1)/(sqrt(10)-1).
A generalization to arbitrary powers is found in Hürlimann, 2004. As the power increases, the probability distribution approaches Benford's law.

Crossrefs

Programs

  • Maple
    ListTools:-PartialSums([seq(floor(sqrt(2*10^n))-ceil(sqrt(10^n))+1, n=0..20)]); # Robert Israel, Feb 15 2021

Extensions

Edited by Don Reble, Nov 05 2005

A083380 a(n) is the number of cubes with at most n digits and first digit 1.

Original entry on oeis.org

1, 1, 2, 5, 11, 23, 49, 105, 225, 485, 1045, 2252, 4852, 10452, 22517, 48510, 104508, 225153, 485075, 1045058, 2251505, 4850716, 10450546, 22515012, 48507117, 104505409, 225150073, 485071123, 1045054049, 2251500692
Offset: 1

Views

Author

Werner S. Hürlimann (whurlimann(AT)bluewin.ch), Jun 05 2003

Keywords

Comments

Asymptotically, the probability that a cube begins with 1 is (2^(1/3) - 1)/(10^(1/3) - 1).
A generalization to arbitrary powers is found in Hürlimann, 2004. As the power increases, the probability distribution approaches Benford's law.

Crossrefs

Extensions

Edited by Don Reble, Nov 05 2005
Showing 1-3 of 3 results.