A212704 -id:A212704 - cp's OEIS frontend

A358439 Number of even digits necessary to write all positive n-digit integers.

4, 85, 1300, 17500, 220000, 2650000, 31000000, 355000000, 4000000000, 44500000000, 490000000000, 5350000000000, 58000000000000, 625000000000000, 6700000000000000, 71500000000000000, 760000000000000000, 8050000000000000000, 85000000000000000000, 895000000000000000000

Offset: 1

Bernard Schott, Nov 16 2022

If nonnegative n-digit integers were considered, then a(1) would be 5.
Also, a(n) is the total number of holes in all positive n-digit integers, assuming 4 has no hole. Digits 0, 6 and 9 have 1 hole, digit 8 has 2 holes, and other digits have no holes or (circular) loops (as in A064532).
Proof of the first formula: For n>=2, to write all positive n-digit integers, digits 6, 8, 9 occur A081045(n-1) = (9n+1)*10^(n-2) times each, and digit 0 occurs A212704(n-1) = 9*(n-1)*10^(n-2) times; so a(n) = 4*A081045(n-1) + A212704(n-1).
For a(1), if 0 were included then there would be 5 holes in the 1-digit numbers 0..9.

			To write the integers from 10 up to 99, each of the digits 2, 4, 6 and 8 must be used 19 times, and digit 0 must be used 9 times hence a(2) = 4*19 + 9 = 85.

Index entries for linear recurrences with constant coefficients, signature (20,-100).

Cf. A064532, A081045, A212704.

Cf. A113119, A196563, A358854, A359271.

Maple
```
seq((5*(9*n-1))*10^(n-2), n = 1 .. 30);
```

a[n_] := 5*(9*n - 1)*10^(n - 2); Array[a, 22] (* Amiram Eldar, Nov 16 2022 *)

a(n) = 5*(9n-1)*10^(n-2).
Formulas coming from the name with even digits:
a(n) = A358854(10^n-1) - A358854(10^(n-1)-1).
a(n) = A113119(n) - A359271(n) for n >= 2.
Formula coming from the comment with holes:
a(n) = Sum_{k=10^(n-1)..10^n-1} A064532(k).

A358620 Number of nonzero digits needed to write all nonnegative n-digit integers.

Original entry on oeis.org

9, 171, 2520, 33300, 414000, 4950000, 57600000, 657000000, 7380000000, 81900000000, 900000000000, 9810000000000, 106200000000000, 1143000000000000, 12240000000000000, 130500000000000000, 1386000000000000000, 14670000000000000000, 154800000000000000000

Offset: 1

Bernard Schott, Nov 23 2022

			a(1) = 9 because there are 9 one-digit numbers that are > 0.
a(2) = 171 because there are 90 two-digit numbers, so 90*2 = 180 digits are needed to write these integers, nine of these integers end with 0, and 180-9 = 171.

Index entries for linear recurrences with constant coefficients, signature (20,-100).

Cf. A081045, A113119, A212704.

Maple
```
seq((9*(9*n+1))*10^(n-2), n = 1 .. 20);
```

a[n_] := 9*(9*n + 1)*10^(n - 2); Array[a, 20] (* Amiram Eldar, Nov 23 2022 *)

a(n)=(81*n+9)*10^(n-2) \\ Charles R Greathouse IV, Nov 29 2022

def A358620(n): return 9 if n == 1 else 9*(9*n+1)*10**(n-2) # Chai Wah Wu, Nov 29 2022

a(n) = 9 * (9*n+1) * 10^(n-2).
a(n) = 20*a(n-1) - 100*a(n-2); a(1)=9, a(2)=171.
a(n) = 9 * A081045(n-1).
a(n) = A113119(n) - A212704(n-1), for n >= 2.

cp's OEIS Frontend

A358439 Number of even digits necessary to write all positive n-digit integers.

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