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Abstract:Let ${}_2F_1(1/3,1/3;1;27z)^3=\sum_{n\ge0}A_nz^n$. We prove the full prime-power supercongruence tower $A_{mp^r}\equiv A_{mp^{r-1}}\pmod{p^{4r}}$ for $p\ge5$, $m\ge1$, $r\ge1$. The level-three modular expansion used below was already recorded by Moy; the contribution here is the depth-preserving modulus $p^{4r}$, extending the previously known depth-one modulus $p^4$ case to all prime powers. The proof is modular. After replacing $A_n$ by $B_n=(-1)^nA_n$, the generating function is realized on $X_0(3)$ by $\sum_{n\ge0}B_nt^n=\eta(\tau)^9/\eta(3\tau)^3$ with $t=\eta(3\tau)^{12}/\eta(\tau)^{12}$. The logarithmic derivative $C=(\sum_{n\ge0}B_nt^n)\,(q/t)\,(dt/dq)$ is the Eisenstein series $3E_{5,\chi_0,\chi_3}$. Lagrange-Buermann gives $B_m=\mathrm{CT}_q(C(q)/t(q)^m)$. The new point is to replace the one-prime Hecke defect by the prime-power defect $T_{p^{s+1}}(C/t^{mp^{s+1}})-T_{p^s}(C/t^{mp^s})$. Its $i=0$ Hecke layer is the sparse Cartier defect controlling $A_{mp^{s+1}}-A_{mp^s}$, while all remaining Hecke layers are divisible by the required power of $p$ by induction. A Fricke involution argument on the two-dimensional space $M_5(\Gamma_0(3),\chi_3)=\mathrm{Span}\{C,tC\}$ then kills the low-order part exactly and the principal part modulo $p^{4(s+1)}$.

Submission history

From: Alex Shvets Mr [view email]
[v1] Sat, 13 Jun 2026 20:41:09 UTC (10 KB)