Problem 551

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Bordimor
Posts: 2
Joined: Wed Oct 04, 2023 7:38 pm

Problem 551

Post by Bordimor »

I think the statement of this problem is wrong

its says for n >= 1, an is the sum of the digits of all preceding terms.
so for n = 6 its should be 1+1+2+4+8+16 = 32 but the result example says 23 and going on the next value should be 64 but the problem says that is 38

i dont know if its an error in the statement or in the solutions value, also the result for n = 10^15 its a little big for get the result
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bruce_love
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Joined: Sat Sep 24, 2011 11:59 am

Re: Problem 551

Post by bruce_love »

Instead of adding $16$ you add the "digits of" $16$.
$1+1+2+4+8+(1+6) = 23$
Bordimor
Posts: 2
Joined: Wed Oct 04, 2023 7:38 pm

Re: Problem 551

Post by Bordimor »

OH i see it now, thanks.
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