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 1-2 of 2 results.

A066642 a(n) = floor(n^(n/2)).

Original entry on oeis.org

1, 1, 2, 5, 16, 55, 216, 907, 4096, 19683, 100000, 534145, 2985984, 17403307, 105413504, 661735513, 4294967296, 28761784747, 198359290368, 1406563064942, 10240000000000, 76436817165460, 584318301411328, 4569515072723572, 36520347436056576, 298023223876953125
Offset: 0

Views

Author

Amarnath Murthy, Dec 29 2001

Keywords

Examples

			a(5) = 55 as {5^(1/2)}^5 = 55.9016994374947424102293417182819...
		

Crossrefs

Bisection gives A062971 (even part).

Programs

  • Magma
    [Floor(n^(n/2)): n in [1..25]]; // G. C. Greubel, Dec 30 2017
    
  • Maple
    a:= n-> floor(n^(n/2)):
    seq(a(n), n=0..25);  # Alois P. Heinz, Jun 08 2025
  • Mathematica
    Table[ Floor[Sqrt[n]^n], {n, 1, 25} ]
  • PARI
    a(n) = sqrtint(n^n); \\ Michel Marcus, Nov 01 2022
    
  • Python
    from math import isqrt
    def A066642(n): return isqrt(n**n) # Chai Wah Wu, Jun 08 2025

Extensions

More terms from Robert G. Wilson v, Jan 03 2002
a(0)=1 prepended by Alois P. Heinz, Jun 08 2025

A271390 a(n) = (2*n + 1)^(2*floor((n-1)/2) + 1).

Original entry on oeis.org

1, 3, 5, 343, 729, 161051, 371293, 170859375, 410338673, 322687697779, 794280046581, 952809757913927, 2384185791015625, 4052555153018976267, 10260628712958602189, 23465261991844685929951, 59938945498865420543457, 177482997121587371826171875, 456487940826035155404146917
Offset: 0

Views

Author

Ilya Gutkovskiy, Apr 06 2016

Keywords

Comments

All members are odd, therefore:
........................
| k | a(n) mod k |
|.......|..............|
| n+1 | A001477(n) |
| 2*n+2 | A005408(n) |
| 2 | A000012(n) |
| 3 | A080425(n+2)|
| 4 | A010684(n) |
| 6 | A130793(n) |
........................
Final digit of (2*n + 1)^(2*floor((n-1)/2) + 1) gives periodic sequence -> period 20: repeat [1,3,5,3,9,1,3,5,3,9,1,7,5,7,9,1,7,5,7,9], defined by the recurrence relation b(n) = b(n-2) - b(n-4) + b(n+5) + b(n+6) - b(n-7) - b(n-8) + b(n-9) - b(n-11) + b(n-13).

Examples

			a(0) =  1;
a(1) =  3^1 = 3;
a(2) =  5^1 = 5;
a(3) =  7^3 = 343;
a(4) =  9^3 = 729;
a(5) = 11^5 = 161051;
a(6) = 13^5 = 371293;
a(7) = 15^7 = 170859375;
a(8) = 17^7 = 410338673;
...
a(10000) = 1.644...*10^43006;
...
a(100000) = 8.235...*10^530097, etc.
This sequence can be represented as a binary tree:
                                    1
                 ................../ \..................
                3^1                                   5^1
     7^3......../ \......9^3                11^5....../ \.......13^5
     / \                 / \                 / \                 / \
    /   \               /   \               /   \               /   \
   /     \             /     \             /     \             /     \
15^7    17^7        19^9    21^9        23^11   25^11       27^13   29^13
		

Crossrefs

Programs

  • Maple
    A271390:=n->(2*n + 1)^(n - 1/2 - (-1)^n/2): seq(A271390(n), n=0..30); # Wesley Ivan Hurt, Apr 10 2016
  • Mathematica
    Table[(2 n + 1)^(2 Floor[(n - 1)/2] + 1), {n, 0, 18}]
    Table[(2 n + 1)^(n - 1 + (1 + (-1)^(n - 1))/2), {n, 0, 18}]
  • PARI
    a(n) = (2*n + 1)^(2*((n-1)\2) + 1); \\ Altug Alkan, Apr 06 2016
    
  • Python
    for n in range(0,10**3):print((int)((2*n+1)**(2*floor((n-1)/2)+1)))
    # Soumil Mandal, Apr 10 2016

Formula

a(n) = (2*n + 1)^(n - 1 + (1 + (-1)^(n-1))/2).
a(n) = A005408(n)^A109613(n-1).
a(n) = (2*n + 1)^(n - 1/2 - (-1)^n/2). - Wesley Ivan Hurt, Apr 10 2016
Showing 1-2 of 2 results.