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

Author

Rémy Sigrist, Feb 01 2018

Keywords

• nonn
• base
• easy

Comments

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

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

Crossrefs

Cf. A003592, A004086, A180953.

Programs

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.