Question from Shamik:
Question: A sequence of numbers is defined in the following way.
B(1) = 1
B(n) = B(n1) + {Sum of the digits of B(n1)} 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 nth 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
