Grox's Sequence I: A Polynomial Coefficient Recurrence¶
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
Let \(d=2(b+c)\) and define
Writing the hyperbolic functions as exponentials gives the answer identity
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 \(d<p\) .
A Four-Term Recurrence¶
Polynomial convolution can be avoided. Differentiate \(F\) and clear denominators to get \(QF'=RF\) , where
Compare coefficients of \(z^k\) , treating negative-index coefficients as zero:
Start with \(f_0=(-1)^c\) , precompute inverses from one through \(d\) , and generate the coefficients in order while accumulating the answer. Only the last four coefficients need to be retained.
Correctness¶
The generating-function identity proves that the weighted coefficient sum counts valid sequences. The constant term of \(Q\) is negative one, and every \(k+1\) is invertible, so the differential equation and the initial coefficient uniquely determine each subsequent coefficient. Exact coefficient comparison gives the stated recurrence; hence it produces the coefficients of \(F\) . Handling zero bases and actual zero exponents separately then evaluates the identity for every input.
Complexity and Algorithm Choice¶
Let \(D\) be the total number of decimal input characters in \(a,N\) . Time is \(O(D+(d+1)\log p)\) and auxiliary space is \(O(D+d)\) , including the linear inverse table. Modular exponentiation accounts for most of the computation.
The problem setter's solution uses the same exponential-generating-function identity, expands two polynomials and convolves them by a number-theoretic transform. Its complete derivation, proof and implementation are in the NTT solution and alternative.cpp. Its time is \(O(D+b+c^2+L\log L+(d+1)\log p)\) and space is \(O(D+L+b)\) , where \(L\) is the smallest power of two strictly greater than \(d\) . Both programs support the full constraints.
The preferred std.cpp removes the quadratic expansion and convolution through the four-term recurrence and uses less storage. This changes coefficient computation; the generating-function model remains the problem setter's contribution. Explicit expansion and NTT also extend naturally to different polynomial products and are retained as a complete alternative. See the measurements.
Attribution and Validation¶
Problem setter's solution: the exponential generating function, exponential-sum identity and NTT convolution. GPT-6: the differential-identity recurrence, proof, both standard C++17 implementations and boundary repairs. Small cases use a combinatorial DP that chooses positions for each symbol. Formal cases use explicit trinomial expansion and an independent NTT implementation, including zero type counts, empty sequences, modulus-related exponents and power-of-two polynomial degrees.