%I #15 Feb 05 2018 03:00:18
%S 0,5,52,2,521,1,5260,50,40,52130,520,20,526510,5210,10,800,5218700,
%T 52600,500,400,52609300,521300,5200,200,521359100,6100,5265100,52100,
%U 100,5265679000,8000,52187000,526000,5000,52182884000,4000,526093000,23000,5213000,52000
%N a(n) = the fractional part of 1 / A003592(n) read backwards.
%C 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
%H Rémy Sigrist, <a href="/A299025/b299025.txt">Table of n, a(n) for n = 1..10000</a>
%F a(A180953(n)) = 10^(n-1) for any n > 0.
%e The first terms, alongside A003592(n) and the fractional part of 1/A003592(n), are:
%e n a(n) A003592(n) frac(1/A003592(n))
%e -- ---- ---------- ------------------
%e 1 0 1 0
%e 2 5 2 0.5
%e 3 52 4 0.25
%e 4 2 5 0.2
%e 5 521 8 0.125
%e 6 1 10 0.1
%e 7 5260 16 0.0625
%e 8 50 20 0.05
%e 9 40 25 0.04
%e 10 52130 32 0.03125
%e 11 520 40 0.025
%e 12 20 50 0.02
%e 13 526510 64 0.015625
%e 14 5210 80 0.0125
%e 15 10 100 0.01
%e 16 800 125 0.008
%e 17 5218700 128 0.0078125
%e 18 52600 160 0.00625
%e 19 500 200 0.005
%e 20 400 250 0.004
%t 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 *)
%o (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)))))
%o 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
%o print (apply(backward, A003592))
%Y Cf. A003592, A004086, A180953.
%K nonn,base,easy
%O 1,2
%A _Rémy Sigrist_, Feb 01 2018