4 Threshold-free reciprocal transfer at prime-power section orders
On this page · Permalink · Source line 3
The completion of Nishibuchi's conjecture fixes the requested eventual congruence. The stronger question is whether the sufficient starting row can be removed, and how much precision the difference order and the step actually provide. This section answers the first question at every prime-power section order. Subsequent sections give a factorial improvement and exact results at section order four.
Throughout these sections, the normalized source and its reciprocal are
where \(N\geq1\) and \(0\leq j<N\). In particular, \(C_{N,j}(0)=1\). The coefficients in the original convention satisfy
Thus every valuation of a nonzero difference gains \(v_p(j!)\) on returning to the original coefficients. This factor must be retained when stating optimal periods or the final congruence.
Write \(S f(n)=f(n+1)\) and \(\Delta_H=S^H-1\). A mixed difference \(\Delta_{H_1}\cdots\Delta_{H_k}\) acts on the row variable \(n\), rather than on the full Euler subscript. We use \(v_p(0)=+\infty\). Orders and starting rows are nonnegative integers; steps are positive integers unless stated otherwise. An empty mixed difference is the original value, and a zero ordinary step makes every nonempty product containing it zero.
4.1 The Hurwitz lattice and translated differential flatness
On this page · Permalink · Source line 38
Definition 4.1: Hurwitz integral series
For a prime \(p\), let
For \(N=p^a\) and \(\lambda\in\mathbb Z_p\), put
A series \(A\) is \(\lambda\)-flat if \(L_\lambda^k A\in\mathcal H_p\) for every \(k\geq0\).
Permalink · Source line 40 · Claim record · Lean coverage · Code · Worked examples
Products preserve \(\mathcal H_p\) because the coefficient at factorial degree \(d\) is
The derivative \(D\) acts by the coefficient shift \((DA)_d=a_{d+1}\). Consequently every nonnegative power of \(D\) preserves the lattice. If \(A\in\mathcal H_p\) has unit constant term, its reciprocal belongs to the same lattice: with \(B=A^{-1}\),
If \(A\) is supported at degrees divisible by \(N\), this recurrence also proves that \(B\) has the same support.
Lemma 4.2: Translated product formula
For \(N=p^a\), \(a\geq1\), and \(\lambda,\mu\in\mathbb Z_p\),
Flatness types add under multiplication.
Read proof · Read statement and proof side by side · Permalink · Source line 72 · Claim record · Lean coverage · Code · Worked examples
Proof of Lemma 4.2
The identity \((1+X)^{p^a}=1+X^{p^a}\) over \(\mathbb F_p\) shows that every interior binomial coefficient \(\binom Ni\) is divisible by \(p\). Subtract \((\lambda+\mu)AB\) from the ordinary \(N\)th derivative product formula and divide by \(p\) to obtain (4.4).
For completeness, the iterated formula is an integral multinomial formula, rather than a product rule with unspecified remainders. On the tensor of the two factors use the mutually commuting operators
Applying multiplication after the \(k\)th power of their sum gives exactly \(L_{\lambda+\mu}^k(AB)\). Expanding it yields terms indexed by \(a,b,e_1,\ldots,e_{N-1}\geq0\), with \(a+b+\sum_i e_i=k\):
Every scalar is integral. If \(A\) and \(B\) have the stated flatness types, both factors in every term belong to \(\mathcal H_p\), as does their product.
Back to statement · Read statement and proof side by side · Permalink · Source line 85 · Claim record · Lean coverage · Code · Worked examples
Lemma 4.3: Reciprocal transfer
If \(A\) is \(\lambda\)-flat and has unit constant term, then \(A^{-1}\) is \((-\lambda)\)-flat.
Read proof · Read statement and proof side by side · Permalink · Source line 113 · Claim record · Lean coverage · Code · Worked examples
Proof of Lemma 4.3
Set \(B=A^{-1}\). Its order-zero integrality was established in (4.3). Assume by induction that \(L_{-\lambda}^bB\in\mathcal H_p\) for \(b<k\). Since \(AB=1\),
Use (4.5) with \(\mu=-\lambda\). Exactly one term has inverse order \(b=k\): it has \(a=0\) and all \(e_i=0\), and is \(A L_{-\lambda}^k B\). In every other term \(b<k\), so its second factor is integral by induction and derivative-shift integrality. Its first factor is integral by the flatness of \(A\), and its scalar is integral. Hence \(A L_{-\lambda}^k B\in\mathcal H_p\). Multiplying by the already integral \(B\) proves \(L_{-\lambda}^k B\in\mathcal H_p\). This induction takes place in the entire Hurwitz lattice. It places no restriction on the coefficient degree or the starting row.
Back to statement · Read statement and proof side by side · Permalink · Source line 119 · Claim record · Lean coverage · Code · Worked examples
4.2 The actual Euler source
On this page · Permalink · Source line 137
The packed factorial coefficient of \(G_{N,j}\) is
The empty product is one. If \(N=p^a\) and \(1\leq i<N\), then \(v_p(i)<a\), so \(N/i\in p\mathbb Z_p\). Every affine denominator in (4.6) is therefore a unit on \(\mathbb Z_p\).
For \(s\in p\mathbb Z_p\), an exact finite-difference identity is
It follows by induction, subtracting adjacent rational functions; each successive subtraction adds the factor \(-(k+1)s\) in the numerator. Its denominator is a unit, so it belongs to \(p^k\mathbb Z_p\). The discrete Leibniz formula
then proves \(\Delta_1^k g_{N,j}(n)\in p^k\mathbb Z_p\) for every \(k,n\). The coefficient at degree \(Nn\) of \(L_1^kG_{N,j}\) is precisely \(p^{-k}\Delta_1^k g_{N,j}(n)\), and the other residue classes vanish. Thus the actual source is \(1\)-flat.
Theorem 4.4: All-row prime-power transfer
Let \(p\) be prime, \(a\geq1\), \(N=p^a\), and \(0\leq j<N\). The coefficients \(C_{N,j}(n)\) belong to \(\mathbb Z_p\) for all \(n\geq0\). Set \(A_{N,j}(n)=(-1)^n C_{N,j}(n)\) and \(T_H=S^H-(-1)^H\). For \(k\geq1\) and positive \(H_1,\ldots,H_k\),
At \(p=2\), the ordinary differences of \(C_{N,j}\) and of the original coefficients satisfy the same bounds. These dyadic statements also hold for \(N=1\).
Read proof · Visual guide: Each step contributes its own precision · Visual guide: Odd primes require the sign twist · Read statement and proof side by side · Permalink · Source line 170 · Claim record · Lean coverage · Code · Worked examples
Proof of Theorem 4.4
Reciprocal transfer gives \(L_{-1}^kF_{N,j}\in\mathcal H_p\). Extraction of the packed coefficient gives \((S+1)^k C_{N,j}(n)\in p^k\mathbb Z_p\). Moreover,
so \(A_{N,j}\) has the ordinary \(p^k\) difference property. At \(p=2\), \(L_1=L_{-1}-1\); its integral binomial expansion on \(F_{N,j}\) gives \(\Delta_1^k C_{N,j}(n)\in2^k\mathbb Z_2\) directly. For \(N=1\) the interior sum in the product formula is empty, so the same argument applies without change; equivalently, \(F_{1,0}(t)=e^{-t}\).
To retain the valuation of a step, let \(B\) be any sequence satisfying \(\Delta_1^m B\in p^m\mathbb Z_p\) at every order and row. Use the finite operator identity
The relation \(i\binom Hi=H\binom{H-1}{i-1}\) implies
Multiplying these identities for the \(k\) steps, every term of total order \(M\) has its scalar multiplied by \(p^M\), and is divisible by \(p^{k+\sum_\nu v_p(H_\nu)}\). This proves the mixed-step bound. Finally,
Iteration contributes only a global sign. Multiplying by \(j!\) proves (4.10).
Back to statement · Read statement and proof side by side · Permalink · Source line 189 · Claim record · Lean coverage · Code · Worked examples
Corollary 4.5: Threshold-free dyadic congruence
For \(N=2^a\), \(a\geq0\), all permitted \(j\), and \(n\geq0\),
In particular, \(\Delta_H^k\) gains \(k(1+v_2(H))\) bits before normalization. For \(r>v_2(j!)\), the row displacement \(H=2^{\max(0,r-v_2(j!)-1)}\) is a sufficient period modulo \(2^r\) from row zero. If \(r\leq v_2(j!)\), all original coefficients vanish modulo \(2^r\).
Visual guide: Removing the starting-row threshold · Permalink · Source line 224 · Claim record · Lean coverage · Code · Worked examples
Remark 4.6: Necessary boundaries
At an odd prime the ordinary step-one difference cannot replace the sign-twisted operator in the general theorem. For \(j=0\), \(C_{N,0}(0)=1\) and \(C_{N,0}(1)=-1\), so their ordinary difference is \(-2\), a unit at every odd prime. The proof also does not extend unchanged to non-prime-power \(N\), whose interior binomial coefficients need not be divisible by the chosen prime. Neither observation is a counterexample to Nishibuchi's original eventual congruence.