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 A153671 Minimal exponents m such that the fractional part of (101/100)^m obtains a maximum (when starting with m=1). 14
 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 110, 180, 783, 859, 1803, 7591, 10763, 19105, 50172, 355146, 1101696, 1452050, 3047334, 3933030 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,2 COMMENTS Recursive definition: a(1)=1, a(n) = least number m>a(n-1) such that the fractional part of (101/100)^m is greater than the fractional part of (101/100)^k for all k, 1<=k 10^7. Robert Price, Mar 21 2019 LINKS FORMULA Recursion: a(1):=1, a(k):=min{ m>1 | fract((101/100)^m) > fract((101/100)^a(k-1))}, where fract(x) = x-floor(x). EXAMPLE a(5)=5, since fract((101/100)^5)=0.05101005, but fract((101/100)^k)=0.01, 0.0201, 0.030301, 0.04060401 for 1<=k<=4; thus fract((101/100)^5)>fract((101/100)^k) for 1<=k<5. MATHEMATICA p = 0; Select[Range[1, 20000], If[FractionalPart[(101/100)^#] > p, p = FractionalPart[(101/100)^#]; True] &] (* Robert Price, Mar 21 2019 *) PROG (Python) A153671_list, m, n, k, q = [], 1, 101, 100, 0 while m < 10**4:     r = n % k     if r > q:         q = r         A153671_list.append(m)     m += 1     n *= 101     k *= 100     q *= 100 # Chai Wah Wu, May 16 2020 CROSSREFS Cf. A153663, A154130, A153675, A153679, A153687, A153695, A091560, A153711, A153719. Sequence in context: A090108 A323048 A090109 * A090107 A130696 A146297 Adjacent sequences:  A153668 A153669 A153670 * A153672 A153673 A153674 KEYWORD nonn AUTHOR Hieronymus Fischer, Jan 06 2009 EXTENSIONS a(72)-a(83) from Robert Price, Mar 21 2019 STATUS approved

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Last modified April 18 16:51 EDT 2021. Contains 343089 sequences. (Running on oeis4.)