Calculating 1^X + 2^X + ... + N^X mod 1000000007

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Is there any algorithm to calculate (1^x + 2^x + 3^x + ... + n^x) mod 1000000007? Note: a^b is the b-th power of a.

The constraints are 1 <= n <= 10^16, 1 <= x <= 1000. So the value of N is very large.

I can only solve for O(m log m) if m = 1000000007. It is very slow because the time limit is 2 secs.

Do you have any efficient algorithm?

There was a comment that it might be duplicate of this question, but it is definitely different.

You can sum up the series

1**X + 2**X + ... + N**X

with the help of Faulhaber's formula and you'll get a polynomial with an X + 1 power to compute for arbitrary N.

If you don't want to compute Bernoulli Numbers, you can find the the polynomial by solving X + 2 linear equations (for N = 1, N = 2, N = 3, ..., N = X + 2) which is a slower method but easier to implement.

Let's have an example for X = 2. In this case we have an X + 1 = 3 order polynomial:

    A*N**3 + B*N**2 + C*N + D

The linear equations are

      A +    B +   C + D = 1              =  1 
    A*8 +  B*4 + C*2 + D = 1 + 4          =  5
   A*27 +  B*9 + C*3 + D = 1 + 4 + 9      = 14
   A*64 + B*16 + C*4 + D = 1 + 4 + 9 + 16 = 30 

Having solved the equations we'll get

  A = 1/3
  B = 1/2
  C = 1/6
  D =   0 

The final formula is

  1**2 + 2**2 + ... + N**2 == N**3 / 3 + N**2 / 2 + N / 6

Now, all you have to do is to put an arbitrary large N into the formula. So far the algorithm has O(X**2) complexity (since it doesn't depend on N).

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There are a few ways of speeding up modular exponentiation. From here on, I will use ** to denote "exponentiate" and % to denote "modulus".

First a few observations. It is always the case that (a * b) % m is ((a % m) * (b % m)) % m. It is also always the case that a ** n is the same as (a ** floor(n / 2)) * (a ** (n - floor(n/2)). This means that for an exponent <= 1000, we can always complete the exponentiation in at most 20 multiplications (and 21 mods).

We can also skip quite a few calculations, since (a ** b) % m is the same as ((a % m) ** b) % m and if m is significantly lower than n, we simply have multiple repeating sums, with a "tail" of a partial repeat.

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I think Vatine’s answer is probably the way to go, but I already typed this up and it may be useful, for this or for someone else’s similar problem.

I don’t have time this morning for a detailed answer, but consider this. 1^2 + 2^2 + 3^2 + ... + n^2 would take O(n) steps to compute directly. However, it’s equivalent to (n(n+1)(2n+1)/6), which can be computed in O(1) time. A similar equivalence exists for any higher integral power x.

There may be a general solution to such problems; I don’t know of one, and Wolfram Alpha doesn’t seem to know of one either. However, in general the equivalent expression is of degree x+1, and can be worked out by computing some sample values and solving a set of linear equations for the coefficients.

However, this would be difficult for large x, such as 1000 as in your problem, and probably could not be done within 2 seconds.

Perhaps someone who knows more math than I do can turn this into a workable solution?

Edit: Whoops, I see Fabian Pijcke had already posted a useful link about Faulhaber's formula before I posted.

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Comments
  • @TobySpeight No, it is definitely different. 1^x, 2^x..., n^x is mod exponential, and I want to find FAST ALGORITHM because n <= 10^16 and mod = 10^9+7.
  • n^x mod m is the same as ((n^(x-1) mod m) * (n mod m)) mod m; (a + b) mod m is the same as ((a mod m) + (b mod m)) mod m.
  • @CiaPan Please note that 10^9+7 is a prime.
  • Give a look to en.wikipedia.org/wiki/Faulhaber's_formula. If your problem comes from a challenge site like HackerRank, I bet the inner product always simplifies to an integer and that there is a way to get rid of the global factor as well.
  • ((Woud like to make it an answer, alas the question got closed, so I can only post this as a comment.)) For small n you can compute the sum directly in reasonable time. For large n I think you can find such pairs of numbers, which add up to 1000000007 (for example 1 and 1000000006, 7 and 1000000000...). Those numbers equal 1 and (–1) mod 1000000007, respectively, so they cancel in summation – and so do their respective powers. You can skip them then, so you'll have to directly calculate no more than 1000000007 terms of the sum...
  • But unfortunately x might be 1000... Solving linear equation can be O(X^3), but I think it is slow. I am checking about Faulhaber's Formula right now.
  • @square1001: if you have to be very fast, I suggest precomputing Bernoulli Numbers (up to 1000); it's not an easy task (like linear equations solving) but in this case you'll get the polynom almost ready to use.
  • I checked about Bernoulli number, and finally understood that B[1], B[2],..., B[x] for O(x^2). So this problem can solve for O(N^2) !
  • @square1001: In O(X**2), to be true, it doesn't depend N (number of items to sum), but of power X they should be raised
  • Implementated. I got Accepted.
  • But your algorithm has to calculate 1^x, 2^x, ..., 1000000006^x. It has over 10^9 terms, so you have to iterate over 10^10 times. So, I think it can't pass because the time limit is 2 secs.
  • @square1001 Just tested a NOP loop on my laptop, takes ~0.1 seconds to iterate 10 ** 10 times and that is likely dominated by "fork" and "exec".
  • This becomes much easier if x is odd, because, for example, 1000000006 is congruent to -1, so 1^x and 1000000006^x cancel out, and so do many other terms. I don’t see a shortcut if x is even, though.
  • And if x is known in advance, you could build a closed formula for the (non-mod) sum of exponentials. But, yes, you can probably collapse this to "calculate 500000003 ** x % m; multiply by floor(n / m); loop for i from 1 to n % m, exponentiate and sum"
  • @Vatine 500000003 is cancelled by 500000004 for odd x. Therefore for odd x and n=m the sum equals zero.