A363377 Largest positive integer having n holes that can be made using the fewest possible digits.

7, 9, 8, 98, 88, 988, 888, 9888, 8888, 98888, 88888, 988888, 888888, 9888888, 8888888, 98888888, 88888888, 988888888, 888888888, 9888888888, 8888888888, 98888888888, 88888888888, 988888888888, 888888888888, 9888888888888, 8888888888888, 98888888888888, 88888888888888, 988888888888888

Offset

0,1

Comments

Each decimal digit has 0, 1 or 2 holes so that n holes requires A065033(n) digits.

Links

Table of n, a(n) for n=0..29.

Index entries for sequences related to holes in digits.

Index entries for linear recurrences with constant coefficients, signature (1,10,-10).

Formula

From Natalia L. Skirrow, Jun 26 2023: (Start)

a(n) = (89*(10^((n-1)/2))-8)/9 for odd n; a(n) = 8*(10^(n/2)-1)/9 for even n >= 2.

a(n) = a(n-1) + 10*a(n-2) - 10*a(n-3), for n >= 4.

G.f.: (7+2*x-71*x^2+70*x^3)/((1-x)*(1-10*x^2)).

E.g.f.: (80*cosh(sqrt(10)*x) + 89*sqrt(10)*sinh(sqrt(10)*x) - 80*e^x)/90 + 7. (End)

Example

For n=0, the largest integer with no holes in it that is as short as possible is 7 (9 is larger, but has 1 hole; 11 is larger and has no holes, but is longer at length 2 > length 1).
For n=1, the largest integer with 1 hole that is as short as possible is 9 (following the same kind of reasoning as with n=0).

Mathematica

CoefficientList[Series[(7 + 2 x - 71 x^2 + 70 x^3)/((1 - x) (1 - 10 x^2)), {x, 0, 30}], x] (* Michael De Vlieger, Jul 05 2023 *)

Prog

(Python)
A363377=lambda n: (8+n%2*81)*10**(n>>1)//9 if n else 7
print([A363377(n) for n in range(30)]) # Natalia L. Skirrow, Jun 26 2023

Crossrefs

Cf. A002281 and A002282 (number of holes), A065033 (digits required).

Cf. A249572 and A250256 (smallest number).

Cf. A337099 (largest 7-segment).

Sequence in context: A330865 A390244 A300444 * A371135 A021930 A200103

Adjacent sequences: A363374 A363375 A363376 * A363378 A363379 A363380

Keyword

nonn,base,easy

Author

Julia Zimmerman, May 29 2023

Status

approved