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

Author

Donald Mintz (djmintz(AT)home.com)

Keywords

• nonn

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.
		

Links

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

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