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A299025
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a(n) = the fractional part of 1 / A003592(n) read backwards.
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1
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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
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OFFSET
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1,2
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COMMENTS
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Numbers in this sequence that also appear in A003592, sorted, include the product of numbers k | 10^e with integer e >= 0 and 10^m with m >= e. For instance, the proper divisors of 10 {1, 2, 5} appear and {10, 20, 40, 50} follow, finally {100, 200, 400, 500, 800} followed by any product k 10^m with k = {1, 2, 4, 5, 8} and m >= 3. - Michael De Vlieger, Feb 03 2018
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LINKS
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FORMULA
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a(A180953(n)) = 10^(n-1) for any n > 0.
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EXAMPLE
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The first terms, alongside A003592(n) and the fractional part of 1/A003592(n), are:
-- ---- ---------- ------------------
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
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn,base,easy
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AUTHOR
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STATUS
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approved
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