A031435 - OEIS

%I #44 Jul 14 2023 12:37:48

%S 1,2,4,7,9,12,15,18,21,25,28,32,35,39,42,46,50,54,58,62,66,70,74,78,

%T 83,87,91,95,100,104,109,113,118,122,127,131,136,141,145,150,155,159,

%U 164,169,174,179,183,188,193,198,203,208,213,218,223,228,233,238,243,248

%N Reversal point for powers of consecutive natural numbers.

%C 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

%H Peter Kagey, <a href="/A031435/b031435.txt">Table of n, a(n) for n = 1..10000</a>

%F If bases are N, N+1, the reversal point is floor( log(1+N)/log(1+1/N) ).

%F 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

%F a(n) = floor(1/(1-log(n)/log(n+1))), empirical observation verified for n = 1 to 10000. - _Fred Patrick Doty_, Jul 13 2023

%e a(2) = 2: 3^2 > 2^3 but 3^1 < 2^2.

%e a(3) = 4: 4^4 > 3^5 but 4^3 < 3^4.

%e a(4) = 7: 5^7 > 4^8 but 5^6 < 4^7.

%e a(5) = 9: 6^9 > 5^10 but 6^8 < 5^9.

%e a(6) = 12: 7^12 > 6^13 but 7^11 < 6^12.

%t nn = 60; Table[SelectFirst[Range[5 nn], Mod[Floor[(1 + 1/n)^#], n] == 0 &], {n, nn}] (* _Michael De Vlieger_, Mar 30 2016, Version 10 *)

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

%o (PARI) a(n) = my(k=1); while((1+1/n)^k < n, k++); k; \\ _Michel Marcus_, Mar 30 2019

%o (Ruby)

%o def a(n)

%o   (1..Float::INFINITY).find { |i| (i * Math.log(n, n + 1)).to_i < i - 1 } - 1

%o end # _Peter Kagey_, Mar 29 2016

%Y Cf. A065560.

%K nonn

%O 1,2

%A Donald Mintz (djmintz(AT)home.com)

%E More terms from _Benoit Cloitre_, May 23 2002