

A363377


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


1



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
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

0,1


COMMENTS

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


LINKS



FORMULA

a(n) = (89*(10^((n1)/2))8)/9 for odd n; a(n) = 8*(10^(n/2)1)/9 for even n >= 2.
a(n) = a(n1) + 10*a(n2)  10*a(n3), for n >= 4.
G.f.: (7+2*x71*x^2+70*x^3)/((1x)*(110*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


CROSSREFS



KEYWORD

nonn,base,easy


AUTHOR



STATUS

approved



