A358439 - OEIS

%I #37 Mar 31 2024 17:11:04

%S 4,85,1300,17500,220000,2650000,31000000,355000000,4000000000,

%T 44500000000,490000000000,5350000000000,58000000000000,

%U 625000000000000,6700000000000000,71500000000000000,760000000000000000,8050000000000000000,85000000000000000000,895000000000000000000

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

%C If nonnegative n-digit integers were considered, then a(1) would be 5.

%C 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).

%C 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).

%C For a(1), if 0 were included then there would be 5 holes in the 1-digit numbers 0..9.

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (20,-100).

%F a(n) = 5*(9n-1)*10^(n-2).

%F Formulas coming from the name with even digits:

%F a(n) = A358854(10^n-1) - A358854(10^(n-1)-1).

%F a(n) = A113119(n) - A359271(n) for n >= 2.

%F Formula coming from the comment with holes:

%F a(n) = Sum_{k=10^(n-1)..10^n-1} A064532(k).

%e 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.

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

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

%Y Cf. A064532, A081045, A212704.

%Y Cf. A113119, A196563, A358854, A359271.

%K nonn,base,easy

%O 1,1

%A _Bernard Schott_, Nov 16 2022