A153671 Minimal exponents m such that the fractional part of (101/100)^m obtains a maximum (when starting with m=1).
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
Offset: 1
Keywords
A153679 Minimal exponents m such that the fractional part of (1024/1000)^m obtains a maximum (when starting with m=1).
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, 82, 134, 1306, 2036, 6393, 34477, 145984, 2746739, 2792428, 8460321
Offset: 1
Comments
Recursive definition: a(1)=1, a(n) = least number m>a(n-1) such that the fractional part of (1024/1000)^m is greater than the fractional part of (1024/1000)^k for all k, 1<=k
The next such number must be greater than 5*10^5.
a(40) > 10^7. Robert Price, Mar 16 2019
Examples
a(30)=82, since fract((1024/1000)^82)= 0.99191990..., but fract((1024/1000)^k)<0.9893 for 1<=k<=81; thus fract((1024/1000)^82)>fract((1024/1000)^k) for 1<=k<82 and 82 is the minimal exponent >29 with this property.
Programs
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Mathematica
$MaxExtraPrecision = 10000; p = 0; Select[Range[1, 50000], If[FractionalPart[(1024/1000)^#] > p, p = FractionalPart[(1024/1000)^#]; True] &] (* Robert Price, Mar 16 2019 *)
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Python
A153679_list, m, n, k, q = [], 1, 1024, 1000, 0 while m < 10**4: r = n % k if r > q: q = r A153679_list.append(m) m += 1 n *= 1024 k *= 1000 q *= 1000 # Chai Wah Wu, May 16 2020
Formula
Recursion: a(1):=1, a(k):=min{ m>1 | fract((1024/1000)^m) > fract((1024/1000)^a(k-1))}, where fract(x) = x-floor(x).
Extensions
a(37)-a(39) from Robert Price, Mar 16 2019
A154136 Greatest number m such that the fractional part of (4/3)^A154132(m) >= 1-(1/m).
1, 4, 88, 1228, 2253, 4562, 8183, 167378
Offset: 1
Examples
a(3)=88, since 1-(1/89)=0.988764...>fract((4/3)^A154132(3))=fract((4/3)^8)=0.988721...>0.988636...=1-(1/88).
Formula
a(n):=floor(1/(1-fract((4/3)^A154132(n)))), where fract(x) = x-floor(x).
Comments
Examples
Crossrefs
Programs
Mathematica
Python
Formula
Extensions