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.

A081725 A065560(n) + n - prime(n).

Original entry on oeis.org

1, 2, 4, 3, 5, 5, 7, 7, 6, 8, 15, 7, 10, 10, 9, 8, 11, 10, 11, 14, 13, 14, 13, 11, 12, 15, 16, 20, 21, 13, 14, 14, 17, 13, 16, 16, 16, 17, 17, 17, 20, 16, 20, 22, 26, 19, 13, 15, 19, 21, 21, 25, 21, 21, 21, 21, 25, 25, 27, 31, 27, 20, 22, 26, 28, 20, 21, 17, 21, 23, 24, 22, 22, 22
Offset: 1

Views

Author

Benoit Cloitre, Apr 06 2003

Keywords

Comments

Conjecture : a(n)>0

A031435 Reversal point for powers of consecutive natural numbers.

Original entry on oeis.org

1, 2, 4, 7, 9, 12, 15, 18, 21, 25, 28, 32, 35, 39, 42, 46, 50, 54, 58, 62, 66, 70, 74, 78, 83, 87, 91, 95, 100, 104, 109, 113, 118, 122, 127, 131, 136, 141, 145, 150, 155, 159, 164, 169, 174, 179, 183, 188, 193, 198, 203, 208, 213, 218, 223, 228, 233, 238, 243, 248
Offset: 1

Views

Author

Donald Mintz (djmintz(AT)home.com)

Keywords

Comments

a(n+1) is the smallest k such that floor((1+1/n)^k) == 0 (mod n). A065560(n) is not a strictly increasing sequence, but this one is. - Benoit Cloitre, May 23 2002

Examples

			a(2) = 2: 3^2 > 2^3 but 3^1 < 2^2.
a(3) = 4: 4^4 > 3^5 but 4^3 < 3^4.
a(4) = 7: 5^7 > 4^8 but 5^6 < 4^7.
a(5) = 9: 6^9 > 5^10 but 6^8 < 5^9.
a(6) = 12: 7^12 > 6^13 but 7^11 < 6^12.
		

Crossrefs

Cf. A065560.

Programs

  • Mathematica
    nn = 60; Table[SelectFirst[Range[5 nn], Mod[Floor[(1 + 1/n)^#], n] == 0 &], {n, nn}] (* Michael De Vlieger, Mar 30 2016, Version 10 *)
  • PARI
    for(n=1,100,print1(ceil((n+1/2)*log(n)),",")) \\ Valid for 1
    				
  • PARI
    a(n) = my(k=1); while((1+1/n)^k < n, k++); k; \\ Michel Marcus, Mar 30 2019
    
  • Ruby
    def a(n)
      (1..Float::INFINITY).find { |i| (i * Math.log(n, n + 1)).to_i < i - 1 } - 1
    end # Peter Kagey, Mar 29 2016

Formula

If bases are N, N+1, the reversal point is floor( log(1+N)/log(1+1/N) ).
For n>1, ceiling((n+1/2)*log(n)) is an approximation to a(n) which is valid for all n <= 1000 except n=77 and n=214. - Benoit Cloitre, May 23 2002; corrected by Franklin T. Adams-Watters, Dec 16 2005
a(n) = floor(1/(1-log(n)/log(n+1))), empirical observation verified for n = 1 to 10000. - Fred Patrick Doty, Jul 13 2023

Extensions

More terms from Benoit Cloitre, May 23 2002

A081726 Let f(x,y)=floor((1+1/y)^x); sequence gives least k such that f(k+2,n)/f(k,n)=(1+1/n)^2.

Original entry on oeis.org

1, 36, 20, 117, 37, 42, 70, 68, 57, 139, 186, 149, 119, 542, 389, 135, 328, 297, 327, 1055, 299, 273, 782, 328, 1117, 445, 1916, 1951, 493, 1148, 1238, 4156, 1616, 2328, 1003, 3106, 312, 536, 3320, 383, 1470, 528, 375, 3870, 555, 990, 1278, 1846, 851
Offset: 1

Views

Author

Benoit Cloitre, Apr 06 2003

Keywords

Crossrefs

Cf. A065560.
Showing 1-3 of 3 results.