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
KEYWORD
nonn
AUTHOR
Donald Mintz (djmintz(AT)home.com)
EXTENSIONS
More terms from Benoit Cloitre, May 23 2002
STATUS
approved