A Certain Unscientific Repeated Geometric Sum: Iterate over the Smaller Dimension¶

Idea¶

Let \(p=2147483647\) and let \(S\) denote prefix summation. We choose the smaller of the position count \(n\) and the number of summations \(m\) .

Fewer positions: a binomial convolution kernel¶

After \(m\) prefix sums, the coefficient of the initial value at position \(x\) in the value at position \(n\) is

\[ \binom{n-x+m-1}{m-1}. \]

Starting with coefficient one for one summation, Pascal's identity proves this by induction. Equivalently, distribute \(n-x\) increments among \(m\) stages. Thus

\[ F_m(n)=\sum_{t=0}^{n-1}\binom{m+t-1}{t}F_0(n-t). \]

Start with \(w_0=1\) and update \(w_{t+1}=w_t(m+t)/(t+1)\) . For \(a\ne1\) ,

\[ F_0(x)=\frac{a^{2x+1}-a^x}{a-1}. \]

Iterate downwards from \(x=n\) , updating the powers with \(a^{-1}\) and \(a^{-2}\) . For \(a=1\) , use \(F_0(x)=x+1\) . This branch visits \(n\) terms.

Fewer summations: a recurrence in the number of rounds¶

Define \(H_t(n,q)=S^t(q^x)(n)\) . A geometric sum gives

\[ S(q^x)(n)=\frac{q}{q-1}(q^n-1). \]

Apply \(S^{t-1}\) and use \(S^{t-1}(1)(n)=\binom{n+t-2}{t-1}\) to obtain

\[ H_0(n,q)=q^n,\qquad H_t(n,q)=\frac{q}{q-1}\left(H_{t-1}(n,q)-\binom{n+t-2}{t-1}\right). \]

Maintain this recurrence for both \(q=a\) and \(q=a^2\) , then combine them:

\[ F_m(n)=\frac{aH_m(n,a^2)-H_m(n,a)}{a-1}. \]

Within the input range, \(a\ne1\) implies \(a^2\not\equiv1\pmod p\) : the only two roots are one and \(p-1\) , and the latter exceeds the input bound. All denominators are invertible. For \(a=1\) , use

\[ F_m(n)=\binom{n+m}{m+1}+\binom{n+m-1}{m}. \]

This is the sum of the repeated prefix sums of \(x\) and of the constant one. Compute its binomial coefficients by successive multiplication.

Correctness¶

The kernel branch counts the exact multiplicity of every initial entry. The round recurrence follows from linearity of prefix summation and the geometric-sum identity at every step. The unit-base formula covers the remaining case. Choosing the kernel for \(n\le m\) and the round recurrence otherwise therefore returns the required value for every input.

Implementation and Complexity¶

Work in the modular field. Precompute the needed small inverses using \(\operatorname{inv}(i)=-(p\mathbin{//}i)\operatorname{inv}(p\bmod i)\bmod p\) . Here \(n+m<p\) , and the table extends only to \(\min(n,m)+1\) , so no denominator is zero modulo \(p\) . Use signed 64-bit products and reduce after each multiplication: every product of two residues is below \(p^2<2^{63}\) .

Time is \(O(\min(n,m)+\log p)\) and auxiliary space is \(O(\min(n,m))\) . Both summation directions are implemented in std.cpp; short positions with many rounds use the kernel branch.

Attribution and Validation¶

Geometric sums, binomial convolution kernels and finite-difference identities are classical methods. GPT-6 supplied their combined application, derivation, implementation and independent checks; the available material contains no working problem setter's solution. Small cases use direct repeated prefix sums. Formal cases use a partial-fraction expansion or independently computed weights, including unit bases and the branch boundary. See the measurements.