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 A031435 Reversal point for powers of consecutive natural numbers. 4
 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 (list; graph; refs; listen; history; text; internal format)
 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. Adams-Watters, 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

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Last modified November 18 22:39 EST 2019. Contains 329305 sequences. (Running on oeis4.)