A299025 a(n) = the fractional part of 1 / A003592(n) read backwards.
0, 5, 52, 2, 521, 1, 5260, 50, 40, 52130, 520, 20, 526510, 5210, 10, 800, 5218700, 52600, 500, 400, 52609300, 521300, 5200, 200, 521359100, 6100, 5265100, 52100, 100, 5265679000, 8000, 52187000, 526000, 5000, 52182884000, 4000, 526093000, 23000, 5213000, 52000
Offset: 1
Examples
The first terms, alongside A003592(n) and the fractional part of 1/A003592(n), are: n a(n) A003592(n) frac(1/A003592(n)) -- ---- ---------- ------------------ 1 0 1 0 2 5 2 0.5 3 52 4 0.25 4 2 5 0.2 5 521 8 0.125 6 1 10 0.1 7 5260 16 0.0625 8 50 20 0.05 9 40 25 0.04 10 52130 32 0.03125 11 520 40 0.025 12 20 50 0.02 13 526510 64 0.015625 14 5210 80 0.0125 15 10 100 0.01 16 800 125 0.008 17 5218700 128 0.0078125 18 52600 160 0.00625 19 500 200 0.005 20 400 250 0.004
Links
- Rémy Sigrist, Table of n, a(n) for n = 1..10000
Programs
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Mathematica
With[{e = 12}, Table[FromDigits@ Reverse@ PadLeft[#1, Length@ #1 + Abs@ #2] - 10 Boole[n == 1] & @@ RealDigits[1/n], {n, Sort@ Flatten@ Table[2^i*5^j, {i, 0, e}, {j, 0, Log[5, 2^(e - i)]}]}]] (* Michael De Vlieger, Feb 03 2018, after Robert G. Wilson v at A003592 *) -
PARI
mx = 4000; A003592 = vecsort(concat(vector(1+logint(mx,2), i, vector(1+logint(floor(mx/2^(i-1)), 5), j, 2^(i-1) * 5^(j-1))))) backward(n) = my (v=0, i=frac(1/n), r=1/10); while (i, v += r*floor(i); i=frac(i)*10; r*=10); v print (apply(backward, A003592))
Formula
a(A180953(n)) = 10^(n-1) for any n > 0.
Comments