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

Author

Bernard Schott, Nov 16 2022

Keywords

• nonn
• base
• easy

Comments

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.

Examples

			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.
		

Links

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

Crossrefs

Cf. A064532, A081045, A212704.

Cf. A113119, A196563, A358854, A359271.

Programs

Maple

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

Mathematica

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

Formula

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).