Question

I have this recurrence formula:

   P(n) = ( P(n-1) + 2^(n/2) ) % (X)
   s.t. P(1) = 2;

where n/2 is computer integer division i.e. floor of x/2

Since i am taking mod X, this relation should repeat at least with in X outputs.

but it can start repeating before that.

How to find this value?

Answer 1

It needn’t repeat within x terms, consider x = 3:

P(1)  = 2
P(2)  = (P(1)  + 2^(2/2))  % 3 = 4 % 3        = 1
P(3)  = (P(2)  + 2^(3/2))  % 3 = (1 + 2) % 3  = 0
P(4)  = (P(3)  + 2^(4/2))  % 3 = 4 % 3        = 1
P(5)  = (P(4)  + 2^(5/2))  % 3 = (1 + 4) % 3  = 2
P(6)  = (P(5)  + 2^(6/2))  % 3 = (2 + 8) % 3  = 1
P(7)  = (P(6)  + 2^(7/2))  % 3 = (1 + 8) % 3  = 0
P(8)  = (P(7)  + 2^(8/2))  % 3 = 16 % 3       = 1
P(9)  = (P(8)  + 2^(9/2))  % 3 = (1 + 16) % 3 = 2
P(10) = (P(9)  + 2^(10/2)) % 3 = (2 + 32) % 3 = 1
P(11) = (P(10) + 2^(11/2)) % 3 = (1 + 32) % 3 = 0
P(12) = (P(11) + 2^(12/2)) % 3 = (0 + 64) % 3 = 1

and you see that the period is 4.

Generally (suppose X is odd, it’s a bit more involved for even X), let k be the period of 2 modulo X, i.e. k > 0, 2^k % X = 1, and k is minimal with these properties (see below).

Consider all arithmetic modulo X. Then

           n
P(n) = 2 + ∑ 2^(j/2)
          j=2

It is easier to see when we separately consider odd and even n:

                   m           m
P(2*m+1) = 2 + 2 * ∑ 2^i = 2 * ∑ 2^i = 2*(2^(m+1) - 1) = 2^((n+2)/2) + 2^((n+1)/2) - 2
                  i=1         i=0

since each 2^j appears twice, for j = 2*i and j = 2*i+1. For even n = 2*m, there’s one summand 2^m missing, so

P(2*m) = 2^(m+1) + 2^m - 2 = 2^((n+2)/2) + 2^((n+1)/2) - 2

and we see that the length of the period is 2*k, since the changing parts 2^((n+1)/2) and 2^((n+2)/2) have that period. The period immediately begins, there is no pre-period part (there can be a pre-period for even X).

Now k <= φ(X) by Euler’s generalisation of Fermat’s theorem, so the period is at most 2 * φ(X).
(φ is Euler’s totient function, i.e. φ(n) is the number of integers 1 <= k <= n with gcd(n,k) = 1.)

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What makes it possible that the period is longer than X is that P(n+1) is not completely determined by P(n), the value of n also plays a role in determining P(n+1), in this case the dependence is simple, each power of 2 being used twice in succession doubles the period of the pure powers of 2.

Consider the sequence a[k] = (2^k) % X for odd X > 1. It has the simple recurrence

a[0] = 1
a[k+1] = (2 * a[k]) % X

so each value completely determines the next, thus the entire following part of the sequence. (Since X is assumed odd, it also determines the previous value [if k > 0] and thus the entire previous part of the sequence. With H = (X+1)/2, we have a[k-1] = (H * a[k]) % X.)

Hence if the sequence assumes one value twice (and since there are only X possible values, that must happen within the first X+1 values), at indices i and j = i+p > i, say, the sequence repeats and we have a[k+p] = a[k] for all k >= i. For odd X, we can go back in the sequence, therefore a[k+p] = a[k] also holds for 0 <= k < i. Thus the first value that occurs twice in the sequence is a[0] = 1.

Let p be the smallest positive integer with a[p] = 1. Then p is the length of the smallest period of the sequence a, and a[k] = 1 if and only if k is a multiple of p, thus the set of periods of a is the set of multiples of p. Euler’s theorem says that a[φ(X)] = 1, from that we can conclude that p is a divisor of φ(X), in particular p <= φ(X) < X.

Now back to the original sequence.

P(n) = 2 + a[1] + a[1] + a[2] + a[2] + ... + a[n/2]
     = a[0] + a[0] + a[1] + a[1] + a[2] + a[2] + ... + a[n/2]

Since each a[k] is used twice in succession, it is natural to examine the subsequences for even and odd indices separately,

E[m] = P(2*m)
O[m] = P(2*m+1)

then the transition from one value to the next is more regular. For the even indices we find

E[m+1] = E[m] + a[m] + a[m+1] = E[m] + 3*a[m]

and for the odd indices

O[m+1] = O[m] + a[m+1] + a[m+1] = O[m] + 2*a[m+1]

Now if we ignore the modulus for the moment, both E and O are geometric sums, so there’s an easy closed formula for the terms. They have been given above (in slightly different form),

E[m] = 3 * 2^m - 2     = 3 * a[m] - 2
O[m] = 2 * 2^(m+1) - 2 = 2 * a[m+1] - 2 = a[m+2] - 2

So we see that O has the same (minimal) period as a, namely p, and E also has that period. Unless maybe if X is divisible by 3, that is also the minimal (positive) period of E (if X is divisible by 3, the minimal positive period of E could be a proper divisor of p, for X = 3 e.g., E is constant).

Thus we see that 2*p is a period of the sequence P obtained by interlacing E and O.

It remains to be seen that 2*p is the minimal positive period of P. Let m be the minimal positive period. Then m is a divisor of 2*p.

Suppose m were odd, m = 2*j+1. Then

P(1) = P(m+1) = P(2*m+1)
P(2) = P(m+2) = P(2*m+2)

and consequently

P(2) - P(1) = P(m+2) - P(m+1) = P(2*m+2) - P(2*m+1)

But P(2) - P(1) = a[1] and

P(m+2)   - P(m+1)   = a[(m+2)/2]            = a[j+1]
P(2*m+2) - P(2*m+1) = a[(2*m+2)/2] = a[m+1] = a[2*j+2]

So we must have a[1] = a[j+1], hence j is a period of a, and a[j+1] = a[2*j+2], hence j+1 is a period of a too. But that means that 1 is a period of a, which implies X = 1, a contradiction.

Therefore m is even, m = 2*j. But then j is a period of O (and of E), thus a multiple of p. On the other hand, m <= 2*p implies j <= p, and the only (positive) multiple of p satisfying that inequality is p itself, hence j = p, m = 2*p.