Grox's Sequence I: Problem Setter's Solution with EGF and NTT¶

Counting Identity¶

First work over the rationals with exponential generating functions. Arbitrary counts, even counts, and zero or positive odd counts correspond to \(e^x\) , \(\cosh x\) , and \(1+\sinh x\) . Multiplication accounts for the binomial choices of labelled positions for different symbols. The answer is therefore

\[ N![x^N]\ e^{ax}(\cosh x)^b(1+\sinh x)^c. \]

Let \(d=2(b+c)\) and define

\[ F(z)=(1+z^2)^b(z^2+2z-1)^c=\sum_{i=0}^{d}f_i z^i. \]

Writing the hyperbolic functions as exponentials gives the answer identity

\[ 2^{-(b+c)}\sum_{i=0}^{d}f_i(a-b-c+i)^N. \]

This is first an identity over the rationals. Clearing the power-of-two denominator gives an integer identity, so it can be reduced modulo \(p\) for arbitrarily large \(N\) . We never divide by \(N!\) in the modular field; only the inverse of two is needed. The zero-occurrence option of the third group creates the coefficient \(2z\) , which must be included.

Large Exponents and Implementation Boundaries¶

Read \(a\bmod p\) and \(N\bmod(p-1)\) digit by digit, while separately tracking whether \(N\) is actually zero. Fermat's theorem reduces exponents only for nonzero bases. A zero base raised to a positive exponent gives zero; an actual zero exponent gives one. In particular, a positive multiple of \(p-1\) is not an empty sequence length.

Normalize residues into \([0,p-1]\) . Use 64-bit products and reduce immediately; \(p^2<2^{63}\) ensures safety. Every small integer inverted below is at most \(d<p\) .

Computing the Polynomial¶

Use binomial coefficients for \(A(z)=(1+z^2)^b\) : its odd coefficients vanish and \([z^{2i}]A=\binom bi\) . Successive coefficients are obtained using \((b-i)/(i+1)\) in linear time.

Start with one and multiply by \(z^2+2z-1\) exactly \(c\) times to construct \(B(z)\) . Update coefficients from high degree to low degree to preserve the values still needed. The total cost is \(O(c^2)\) .

Let \(L\) be the smallest power of two strictly greater than \(d=2(b+c)\) . Pad both arrays to \(L\) , transform them, multiply pointwise, then apply the inverse transform and divide by \(L\) to obtain \(F=AB\) . Use \(L=1\) for \(d=0\) . A degree-\(d\) polynomial has \(d+1\) coefficients; choosing \(L=d\) would wrap the leading term into the constant coefficient.

The modulus is \(479\times2^{21}+1\) , with primitive root three. The maximum transform length here is \(2^{18}\) , so all needed roots exist. NTT changes convolution into pointwise multiplication, and inverse NTT recovers the coefficients in the field. Since the product degree is below \(L\) , cyclic convolution equals ordinary polynomial multiplication.

Correctness and Complexity¶

The binomial recurrence constructs \(A\) exactly, repeated quadratic multiplication constructs \(B\) exactly, and NTT obtains their complete product. Substitution into the proved exponential-sum identity gives the answer.

Let \(D\) be the number of decimal input characters in \(a,N\) . Time is \(O(D+b+c^2+L\log L+(d+1)\log p)\) and space is \(O(D+L+b)\) . alternative.cpp supports the full constraints. The preferred four-term recurrence removes the quadratic expansion and convolution.

The problem setter's solution contributed the generating-function and NTT approach. GPT-6 supplied the complete derivation and standard C++17 implementation, including the zero-occurrence coefficient, transform length and zero-exponent boundaries. Both programs were checked by a combinatorial DP, independent trinomial expansion and formal stress cases. See the measurements.