Summation of series 1 + (1+2+1) + (1+2+1+3+1+2+1)

Question

The function is: F(n-1) n F(n-1)

Its a type of palindrome function called Zimmer Series.

The values would be: 1, 121, 1213121, ...

I want to figure the summation of the individual digits.

1 + (1+2+1) + (1+2+1+3+1+2+1) +...

Any help is welcome.

Answer 1

Breaking this down into steps, we first find out a formula for the summation of a single value of the series and then we can find out the summation of said formula.

Expanding the definition you gave and manipulating it:

F(n) = n + 2F(n-1)
F(n) = n + 2(n-1) + 22(n-2) + 23(n-3) + ... + 2n-1
2F(n) =    2n     + 22(n-1) + 23(n-2) + ... + 2n-1(2) + 2n

F(n) - 2F(n) = -F(n) = n - 2 - 22 - 23 - ... - 2n

From this and using the formula for Geometric Progression we can then get an expression for a single term of the series.

F(n) = (2n + 2n-1 + ... + 2) - n
     = (2n+1 - 2) - n

Now we just have to work out the summation of this expression.

G(n) = Σ F(n) = Σ (2n+1 - 2 - n)
G(n) = (2n+2 - 22) - (2n) - (n(n+1)/2)

Simplifying this should hopefully give you the answer you seek!

G(n) = (2n+2  - (n(n+5)/2) - 22)

Trying this out on a few of the terms just to double check.

G(1) = (21+2 - (1(1+5)/2) - 22)
G(1) = 1

G(2) = (22+2  - (2(2+5)/2) - 22)
G(2) = 5 = 1 + (1 + 2 + 1)

G(3) = (23+2  - (3(3+5)/2) - 22)
G(3) = 16 = 1 + (1 + 2 + 1) + (1 + 2 + 1 + 3 + 1 + 2 + 1)

Answer 2

EDIT: Mark Dickinson is right, I misinterpreted the question, this solution is incorrect.

I think after the second term the sequence is of the form difference in Arithmetic Progression.

Let me show you how

Second Term = 1+2+1
Third Term  = 1+2+1+3 + 1+2+1
Difference  = 1+2+1+3 = 7 

Third Term  = 1+2+1+3+1+2+1
Fourth Term = 1+2+1+3+  1+4+1+3 +1+2+1
Difference  = 1+4+1+3 = 9

Fourth Term = 1+2+1+3+1+4+1+3+1+2+1
Fifth Term  = 1+2+1+3+1+4+  1+5+1+4  +1+3+1+2+1
Difference  = 1+5+1+4 = 11

So as you can see the difference is in the arithmetic progression and you find the sum of the terms using the formula for the sum of numbers whose different are in Arithmetic Progression