# Can you find a(n) as explicitly function of n?

For natural number n=(1, 2, 3, ...) and sequence definition:

a(n) = a(2n)+a(2n+1) =

= a(3n)+a(3n+1)+a(3n+2) =

= a(4n)+a(4n+1)+a(4n+2)+a(4n+3) =

= a(5n)+a(5n+1)+a(5n+2)+a(5n+3)+a(5n+4) =

...

Trivial a(n)=0 is out of question.

a(n) = a(2n)+a(2n+1) =

= a(3n)+a(3n+1)+a(3n+2) =

= a(4n)+a(4n+1)+a(4n+2)+a(4n+3) =

= a(5n)+a(5n+1)+a(5n+2)+a(5n+3)+a(5n+4) =

...

Trivial a(n)=0 is out of question.

Update:
@mattmiller, there must be a(3) = a(9)+a(10)+a(11), and you have 1/2 = 3/8?

Update 2:
@Torquestomp, this is a real question, if there are enough independent equations?
I have sequence, rounded on 3'rd decimals, so some error might be ocure, but all conditions are pretty obey:
1,000; 0,585; 0,415; 0,322; 0,263; 0,222; 0,193; 0,170; 0,152; 0,137; 0,125; 0,116; 0,107; 0,099; 0,093; 0,087;...
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@Torquestomp, this is a real question, if there are enough independent equations?

I have sequence, rounded on 3'rd decimals, so some error might be ocure, but all conditions are pretty obey:

1,000; 0,585; 0,415; 0,322; 0,263; 0,222; 0,193; 0,170; 0,152; 0,137; 0,125; 0,116; 0,107; 0,099; 0,093; 0,087; 0,083; 0,078; 0,074; 0,070; 0,067; 0,064; 0,062; 0,059; 0,057; 0,054; 0,052; 0,050; 0,049; 0,047; 0,046; .....

I have sequence, rounded on 3'rd decimals, so some error might be ocure, but all conditions are pretty obey:

1,000; 0,585; 0,415; 0,322; 0,263; 0,222; 0,193; 0,170; 0,152; 0,137; 0,125; 0,116; 0,107; 0,099; 0,093; 0,087; 0,083; 0,078; 0,074; 0,070; 0,067; 0,064; 0,062; 0,059; 0,057; 0,054; 0,052; 0,050; 0,049; 0,047; 0,046; .....

Update 3:
Actually, I got this series when I trying to find distribution of the first digit in the sequence of 2^n, for large n. First, I'm notice (by calculation in Excel) that same distribution ocures for 3^n, 4^n, 5^n, etc., then I find that it's happend in the same way (or very similar) in sistems with base...
show more
Actually, I got this series when I trying to find distribution of the first digit in the sequence of 2^n, for large n. First, I'm notice (by calculation in Excel) that same distribution ocures for 3^n, 4^n, 5^n, etc., then I find that it's happend in the same way (or very similar) in sistems with base diferent than 10. So, my example above is from sistem 100-bases, and for first digit in 2^n, for n= 1 to 10000. Than I thinking on this way:

After number with first digit "1", (when multiplying by 2) there must be next number with first digit "2" or "3", so p(1) = p(2) + p(3), where p is probability. After "2" there must be "4" or "5", so p(2) = p(4) + p(5), etc. But, if distribution is realy the same for 3^n, then p(1) = p(3)+p(4)+p(5), because after "1" as first digit, and multiplying by 3, there could be only 3, 4 or 5 as a first digit in next number, then p(3) = p(9)+p(10)+p(11), and so on... I'm hope you will understand. :)

After number with first digit "1", (when multiplying by 2) there must be next number with first digit "2" or "3", so p(1) = p(2) + p(3), where p is probability. After "2" there must be "4" or "5", so p(2) = p(4) + p(5), etc. But, if distribution is realy the same for 3^n, then p(1) = p(3)+p(4)+p(5), because after "1" as first digit, and multiplying by 3, there could be only 3, 4 or 5 as a first digit in next number, then p(3) = p(9)+p(10)+p(11), and so on... I'm hope you will understand. :)

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