Tetrahedron vertex degree recurrence

The positive harmonic sum in Tetrahedron vertex risk admits an explicit order-ten recurrence in the degree. Its polynomial coefficients have degree at most 69. The recurrence holds at every integer \(D\geq1\), has no exceptional integer indices, and determines the coefficient from its first ten values. This supplies a fixed-order recurrence; it does not evaluate the coefficient by a bounded number of classical special functions.

Theorem 37 (A certified degree recurrence)

Under the uniform-simplex assumptions of Theory of the skeleton risk, write \(a_D=a_0(4,D)\). The eleven integer polynomials recorded in tools/risk_research/examples/tetrahedron_vertex_recurrence_order10.json satisfy

\[\sum_{i=0}^{10}p_i(D)a_{D+i}=0\qquad(D\geq1).\]

The JSON rows give coefficients in ascending powers of \(D\). Their maximum degree is 69, and \(p_{10}(D)\ne0\) at every positive integer. The same artifact records the ten exact initial values, starting with \(a_1=8/5\) and \(a_2=16/35\).

The proof below constructs a rational telescoping certificate from these polynomials. The executable verifier rebuilds this certificate and checks four rational identities exactly. Finite sequence agreement is an additional regression check, rather than the proof of the recurrence.

Four polynomial difference equations

Recall the proved single sum

\[a_D=\frac{576}{d_D}\sum_{j=0}^{D-1} \binom{D-1}j\binom{D+1}j G_j, \qquad d_D=\frac{(2D+3)!}{(D-1)!(D+1)!},\]

where

\[G_j=\frac{H_j^2}{(j+1)(j+2)(j+3)} +\frac{H_j^{(2)}}{(j+1)(j+3)} -\frac{jH_j}{(j+1)(j+2)^2(j+3)} +\frac{2}{(j+1)^2(j+2)^2(j+3)}.\]

For a candidate order \(r\), put \(N=D+r-1\) and \(w_{N,j}=\binom Nj\binom{N+2}j\). With \(x^{\underline m}=x(x-1)\cdots(x-m+1)\), define

\[U_m(j)=(N-j)^{\underline m}(N+2-j)^{\underline m},\qquad B_m=N^{\underline m}(N+2)^{\underline m}.\]

The elementary shift identity is

\[w_{N,j}U_m(j)=B_m \binom{N-m}j\binom{N+2-m}j.\]

It holds on \(0\leq j\leq N\), including the zero terms outside the shorter support. For \(P(j)=\sum_{m=0}^r\lambda_m U_m(j)\), we seek a telescope for \(w_{N,j}P(j)G_j\).

Set \(A(j)=(N-j)(N+2-j)\) and introduce two polynomial operators over \(\mathbb Q(N)\):

\[\begin{split}L_2 q&=A(j)(j+1)q(j+1)-j^2(j+3)q(j),\\ L_3 q&=A(j)(j+1)^2q(j+1)-j^2(j+2)(j+3)q(j).\end{split}\]

Four certificate polynomials \(q_1,q_2,q_3,q_4\) must satisfy

(78)\[\begin{split}L_2q_1&=P,\\ L_2q_2&=(j+2)P,\\ L_3q_3&=-j(j+1)P -2A(j)(j+1)(j+2)q_1(j+1),\\ L_3q_4&=2P-A(j)(j+2)\bigl(q_1(j+1)+q_2(j+1)\bigr) -A(j)(j+1)q_3(j+1).\end{split}\]

These equations can be solved by polynomial reduction. The leading terms of \(L_2(j^m)\) and \(L_3(j^m)\) are respectively

\[(m-2N-4)j^{m+2},\qquad (m-2N-5)j^{m+3}.\]

Successively eliminating the highest powers therefore leaves remainders of degrees below \(2,2,3,3\). Their ten coefficients depend linearly on the \(r+1\) unknowns \(\lambda_m\). At \(r=10\) the resulting \(10\times11\) matrix has an exact nonzero kernel. The implementation computes that kernel over \(\mathbb Q(N)\) without using sequence values. This establishes an order bound for this certificate construction; it does not claim that order ten is minimal among all recurrences for \(a_D\).

The rational telescope and its boundaries

Let \(h_j=(H_j^2,H_j^{(2)},H_j,1)^T\) and let \(b(j)\) be the four rational coefficients of \(G_j=b(j)h_j\) above. Define the row

\[c(j)=j^2\left( \frac{q_1(j)}{(j+1)(j+2)}, \frac{q_2(j)}{(j+1)(j+2)}, \frac{q_3(j)}{(j+1)^2(j+2)}, \frac{q_4(j)}{(j+1)^2(j+2)}\right).\]

The harmonic transition \(h_{j+1}=T(j)h_j\) is

\[\begin{split}T(j)=\begin{pmatrix} 1&0&2/(j+1)&1/(j+1)^2\\ 0&1&0&1/(j+1)^2\\ 0&0&1&1/(j+1)\\ 0&0&0&1 \end{pmatrix}.\end{split}\]

Substitution of (78) gives the four rational identities

\[\frac{A(j)}{(j+1)^2}c(j+1)T(j)-c(j)=P(j)b(j).\]

Multiplication by \(w_{N,j}h_j\) telescopes the desired summand. The lower boundary vanishes because \(c(0)=0\). At the upper boundary \(w_{N,N+1}=0\), and \(c(N+1)\) is finite, so the sum is zero.

There are no hidden poles on the summation domain. Polynomial reduction uses indices \(m\leq2r-1\) and \(N=D+r-1\geq r\), hence \(m-2N-4<0\) and \(m-2N-5<0\). The factors \(B_m\) are positive. The normalizing ratio

\[\frac{d_{D+i}}{d_D} =\frac{(2D+4)_{2i}}{(D)_i(D+2)_i}\]

has positive factors as well. The certificate denominators in \(j\) involve only \(j+1,j+2\). The verifier additionally expands every certificate parameter denominator after \(N=D+r-1\) and checks that its coefficients are nonnegative and not all zero. This independently certifies positivity for \(D\geq1\).

Finally set

\[p_i(D)=\lambda_{r-i}(D+r-1)B_{r-i}(D+r-1) \frac{d_{D+i}}{d_D}.\]

Clearing the common denominator and integer content gives the recorded polynomials and proves the theorem. Conversely, the verifier starts with supplied \(p_i\), recovers \(\lambda_m\) through this formula, and checks the telescope directly. It therefore detects an incorrect coefficient array independently of its provenance metadata.

Integer domain and reproduction

The leading polynomial has one sign variation in its nonzero coefficients. Its exact evaluations satisfy \(p_{10}(4)<0<p_{10}(5)\). Descartes’ rule and the intermediate value theorem place its only positive real root strictly inside \((4,5)\). Thus it has no positive integer zero; recurrence evaluation needs no additional initial values beyond degrees one through ten.

With the optional Sage environment, reproduce both the derivation and the symbolic certificate using

python -m tools.risk_research.vertex_degree_recurrence \
  --output /tmp/tetrahedron-vertex.json --report /tmp/vertex-proof.json
python -m tools.risk_research.vertex_degree_recurrence \
  --verify tools/risk_research/examples/tetrahedron_vertex_recurrence_order10.json

Ordinary Python can evaluate the certified recurrence without Sage:

from fractions import Fraction
from tools.risk_research.vertex_degree_recurrence import recurrence_coefficients

values = recurrence_coefficients(300)
assert values[1] == Fraction(8, 5)

Regression checks compare recurrence values with the previously proved positive sum at degrees one through 48 and at 121, 241, 260, 271, 277, 290 and 300. The degrees above 260 were checked after the operator was frozen. These checks do not replace the symbolic certificate.