

A031435


Reversal point for powers of consecutive natural numbers.


3



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
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OFFSET

1,2


COMMENTS

a(n+1) is 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. AdamsWatters, Dec 16 2005


EXAMPLE

2 from: 3^2>2^3 but 3^1<2^2; 4 from: 4^4>3^5 but 4^3<3^4; 7 from: 5^7>4^8 but 5^6<4^7; 9 from: 6^9>5^10 but 6^8<5^9; 12 from: 7^12>6^13 but 7^11<6^12; etc.


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
(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



