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Question from Shamik: 
Question: A sequence of numbers is defined in the following way.  
  B(1) = 1 
  B(n) = B(n-1) + {Sum of the digits of B(n-1)} for n > 1.  
B(1) = 1  
  B(2) = 2  
  B(3) = 4  
  B(4) = 8  
  B(5) = 16  
  B(6) = 23  
  B(7) = 28  
  B(8) = 38  
  B(9) = 49  
  ::::::::: and so on ... 
Is the number 123456789 a term in the given sequence?  
Work Done: I could solve the problem as shown below. But, then I have a different 
  question at the end. 
Let B(n) be the n-th number in the given series and S(n) be the sum of its 
  digits. For example: B(5) = 16 ==> S(5) = 1 + 6 = 7. 
B(n) = S(n) (mod 9)  
  By definition, B(n+1) = B(n) + S(n) = {9*k + S(n)} + S(n) = 2*S(n) + 9*k 
  ==> B(n+1) = {2*S(n)} (mod 9) ... (1) 
As B(n+1) = r (mod 9) AND S(n+1) = r (mod 9, we can say  
  S(n+1) = {2*S(n)} (modulo 9) from (1). 
S(1) = B(1) = 1. 
  S(1+1) = S(2) = 2*1 (mod 9) = 2 (mod 9) 
  S(2+1) = S(3) = 2*2 (mod 9) = 4 (mod 9) 
  S(3+1) = S(4) = 2*4 (mod 9) = 8 (mod 9) 
  S(4+1) = S(5) = 2*8 (mod 9) = 16 (mod 9) = 7 (mod 9) 
  S(5+1) = S(6) = 2*16 (mod 9) = 32 (mod 9)  
  = 2*7 (mod 9) = 14 (mod 9) 
  = 5 (mod 9) 
  Similarly, S(7) = 1 (mod 9) and it continues in a similar fashion. 
B(1) = 1 modulo 9; B(2) = 2 modulo 9; B(3) = 4 modulo 9; B(4) = 8 modulo 9; 
  B(5) = 7 modulo 9; B(6) = 5 modulo 9; B(7) = 1 modulo 9; and so on ... 
  ==> B(n) = B(n - 6) modulo 9 for n > 6 
In simple words ... 
  S(1) = B(1) = 1 and S(n+1) = {2*S(n)} (modulo 9) implies  
  S(n+1) = (2^n) (mod 9) = 1, 2, 4, 8, 7, 5 (mod 9) 
The number is 123456789. Sum of its digits = (1+2+3+4+5+6+7+8+9) = 45 = 
  9*5. 
  123456789 = 0 (mod 9) 
  Sum of the digits of 1234565789 = 45 = 0 (mod 9) 
  Therefore, 123456789 does not appear in the sequence. To even hold a chance 
  its (modulo 9) should have been one of 1, 2, 4, 8, 7, or 5. 
This way I can be sure when a given number is not present in the sequence. 
  But, is there any way by which we can be sure that a given number is 
  present in the series when its 'modulo 9' is one of 1, 2, 4, 8, 7, or 5. 
For example:  
  (a) 123456737 = 2 (mod 9) 
  (b) 123456755 = 2 (mod 9) 
The above 2 numbers do not get disqualified in the first step. I used to 
  computer program to find out B(3641312) = 123456755 where as 123456737 is 
  not present in the series. 
I there any mathematical way to find this out. I am not able to proceed 
  properly in the right direction. Please, help me. 
Thanks and regards, 
  Shamik 
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