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A031435 Reversal point for powers of consecutive natural numbers.

%I A031435 #44 Jul 14 2023 12:37:48
%S A031435 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 A031435 83,87,91,95,100,104,109,113,118,122,127,131,136,141,145,150,155,159,
%U A031435 164,169,174,179,183,188,193,198,203,208,213,218,223,228,233,238,243,248
%N A031435 Reversal point for powers of consecutive natural numbers.
%C A031435 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 A031435 Peter Kagey, <a href="/A031435/b031435.txt">Table of n, a(n) for n = 1..10000</a>
%F A031435 If bases are N, N+1, the reversal point is floor( log(1+N)/log(1+1/N) ).
%F A031435 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 A031435 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 A031435 a(2) = 2: 3^2 > 2^3 but 3^1 < 2^2.
%e A031435 a(3) = 4: 4^4 > 3^5 but 4^3 < 3^4.
%e A031435 a(4) = 7: 5^7 > 4^8 but 5^6 < 4^7.
%e A031435 a(5) = 9: 6^9 > 5^10 but 6^8 < 5^9.
%e A031435 a(6) = 12: 7^12 > 6^13 but 7^11 < 6^12.
%t A031435 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 A031435 (PARI) for(n=1,100,print1(ceil((n+1/2)*log(n)),",")) \\ Valid for 1<n<77
%o A031435 (PARI) a(n) = my(k=1); while((1+1/n)^k < n, k++); k; \\ _Michel Marcus_, Mar 30 2019
%o A031435 (Ruby)
%o A031435 def a(n)
%o A031435   (1..Float::INFINITY).find { |i| (i * Math.log(n, n + 1)).to_i < i - 1 } - 1
%o A031435 end # _Peter Kagey_, Mar 29 2016
%Y A031435 Cf. A065560.
%K A031435 nonn
%O A031435 1,2
%A A031435 Donald Mintz (djmintz(AT)home.com)
%E A031435 More terms from _Benoit Cloitre_, May 23 2002