A031435 Reversal point for powers of consecutive natural numbers.

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

1,2

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

Links

Peter Kagey, Table of n, a(n) for n = 1..10000

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

Example

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.

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 *)

Prog

(PARI) for(n=1, 100, print1(ceil((n+1/2)*log(n)), ", ")) \\ Valid for 1<n<77
(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

Crossrefs

Cf. A065560.

Sequence in context: A184583 A189379 A226595 * A065560 A134886 A024193

Adjacent sequences: A031432 A031433 A031434 * A031436 A031437 A031438

Keyword

nonn

Author

Donald Mintz (djmintz(AT)home.com)

Extensions

More terms from Benoit Cloitre, May 23 2002

Status

approved