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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
From Nathan 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)]) # Nathan 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: A019862 A330865 A300444 * A371135 A021930 A200103
KEYWORD
nonn,base,easy
AUTHOR
Julia Zimmerman, May 29 2023
STATUS
approved

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Last modified April 19 02:45 EDT 2024. Contains 371782 sequences. (Running on oeis4.)