Recently **Stan Wagon** posed a puzzle (POTW
1018) that I will switch a little bit to prime questions.

Let's define a balanced number n as these numbers such that writing the
series of numbers from 1 to n implies to use a quantity of ones (1's) q(n)
such that q(n)=n.

For example:

a) "1" is the earliest (trivial) balanced number because q(1)=1

b) 11 is not a balanced number because the q(11)=4.

b) 199981 is the earliest non-trivial balanced number because
q(199981)=199981

I have found that **
35200001, 500199989
and
535199981**
are the earliest three balanced primes.

**Questions:**

**1. Find the next three primes in this sequence.**

**2. Find the largest balanced number**

**3. Find the largest balanced prime.**

** **

Contributions came from **J. K. Andersen, Luke Pebody **and**
Giovanni Resta**.

**J. K. Andersen**'s main discovery was that this puzzle is the same
than my own Puzzle 200 (sorry folks!).
Consequently The three asked answers are:

1. Find the next three primes in this sequence: There are only those
three primes.

2. Find the largest balanced number: 1111111110

3. Find the largest balanced prime: The largest given in the puzzle:
535199981.

**Luke Pebody** wrote:

535199981 is the largest balanced prime, so there are no more in the
sequence. 1,111,111,110 is the largest balanced number.

First of all let us define a function, say U(n) which
counts the number of 1's in the numbers 1,2,..,n.

After a little effort it is easy to see that U(n) can be calculated as
follows.

Let us write n as d_k*10^k + d_(k-1)*10^(k-1) + + d_1*10+d_0, that is n is
written d_k .... d_2 d_1 d_0.

U(n) is obtained adding two values

1) if the digit d_p, with p>0, is equal to 1 then we add the number
obtained taking the digits from d_(p-1) to d_0.

For example,

if n = 43432 this contribute is 0 since there are no 1 in n.

If n = 33186 this contribute is 86 (which follows the 1) if n= 214189 then
the contribute is 4189 + 89

2) for each digit d * 10^q

we add the contribution:

if d=0 : 0

if d=1 : 1+q*10^(q-1)

else : 10^q + q*d*10^(q-1)

for example, for the first non trivial balanced number we have n=199981
and the first contribution is 99981 while for the other digits we have
1*10^0 -> 1+0*0 = 1

8*10^1 -> 10^1+8*1*10^0 = 18

9*10^2 -> 10^2+9*2*10^1 = 280

9*10^3 -> 10^3+9*3*10^2 = 3700

9*10^4 -> 10^4+9*4*10^3 = 46000

1*10^5 -> 1+5*10^4 = 50001

Summing up we have 1+18+280+3700+46000+50001+99981=n=199981.

Using this definition of U(n) it is possible to plot it easily against n.

For example, in plot1 I have plotted

Log(n/U(n)) for values of n between 10^4 and 10^(60) (Hence, a positive
value means n>U(n) and a negative value means

n<U(n) )

While it is probably not too difficult to prove that the U(n)=n cannot
have solutions larger than 10^15, from the graphics we can assume that the
solutions must lie in a smaller range.

Indeed in the plot2 we see a plot
between 10^4 and 10^12, where I think all solutions lies.

Since the interval was quite small I have searched them more or less by
brute force. There are 86 solutions:

1 1599989 500000001 501599989

199981 1599990 500199981 501599990

199982 2600000 500199982 502600000

199983 2600001 500199983 502600001

199984 13199998 500199984 513199998

199985 35000000 500199985 535000000

199986 35000001 500199986 535000001

199987 35199981 500199987 535199981

199988 35199982 500199988 535199982

199989 35199983 500199989 535199983

199990 35199984 500199990 535199984

200000 35199985 500200000 535199985

200001 35199986 500200001 535199986

1599981 35199987 501599981 535199987

1599982 35199988 501599982 535199988

1599983 35199989 501599983 535199989

1599984 35199990 501599984 535199990

1599985 35200000 501599985 535200000

1599986 35200001 501599986 535200001

1599987 117463825 501599987 1111111110

1599988 500000000 501599988

and the largest one is 1111111110. There are no other prime solutions
apart those you find.