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As my code times out, I wonder how can we be sure that the sum s(d) is finite?

For example, if I test the case b=10 and d=1, I get:

f(n,1)=n for n=0

f(n,1)=n for n=1

f(n,1)=n for n=199981

f(n,1)=n for n=199982

f(n,1)=n for n=199983

f(n,1)=n for n=199984

f(n,1)=n for n=199985

f(n,1)=n for n=199986

f(n,1)=n for n=199987

f(n,1)=n for n=199988

f(n,1)=n for n=199989

f(n,1)=n for n=199990

f(n,1)=n for n=200000

f(n,1)=n for n=200001

f(n,1)=n for n=1599981

...

f(n,1)=n for n=35199990

f(n,1)=n for n=35200000

f(n,1)=n for n=35200001

....

## Project Euler #156: Counting Digits

You are viewing a single comment's thread. Return to all comments →

As my code times out, I wonder how can we be sure that the sum s(d) is finite?

For example, if I test the case b=10 and d=1, I get:

f(n,1)=n for n=0

f(n,1)=n for n=1

f(n,1)=n for n=199981

f(n,1)=n for n=199982

f(n,1)=n for n=199983

f(n,1)=n for n=199984

f(n,1)=n for n=199985

f(n,1)=n for n=199986

f(n,1)=n for n=199987

f(n,1)=n for n=199988

f(n,1)=n for n=199989

f(n,1)=n for n=199990

f(n,1)=n for n=200000

f(n,1)=n for n=200001

f(n,1)=n for n=1599981

...

f(n,1)=n for n=35199990

f(n,1)=n for n=35200000

f(n,1)=n for n=35200001

....