OFFSET
1,1
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) = (9*n+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.
LINKS
Index entries for linear recurrences with constant coefficients, signature (20,-100).
FORMULA
a(n) = 5*(9*n-1)*10^(n-2).
Formulas coming from the name with even digits:
Formula coming from the comment with holes:
a(n) = Sum_{k=10^(n-1)..10^n-1} A064532(k).
From Elmo R. Oliveira, Nov 18 2025: (Start)
G.f.: x*(5*x+4)/(10*x-1)^2.
E.g.f.: (exp(10*x)*(90*x-1) + 1)/20.
a(n) = 20*a(n-1) - 100*a(n-2) for n > 2. (End)
EXAMPLE
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.
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 *)
CROSSREFS
KEYWORD
nonn,base,easy
AUTHOR
Bernard Schott, Nov 16 2022
STATUS
approved
