A Touchard congruence for the m-Bell sequences
Abstract
Fix a commutative ring, an integer m ≥ 1 and a base c. Call a sequence a an m-binomial-shift sequence with base c if aₙ₊ₘ=Σ_(k ≤ n)binom nk cⁿ⁻ᵏaₖ for all n ≥ 0; for c=m over ℤ these are exactly the sequences that move m places to the left under m applications of the binomial transform, the m-Bell sequences recently introduced by Popov, whose m=1 case is the Bell numbers. Popov asks for a systematic study of their congruence properties. We answer this. For every prime p, every m ≥ 1 and every base c, the polynomial (Xᵖ-cᵖ⁻¹X)ᵐ-1 annihilates every such sequence modulo p; equivalently Σⱼ₌₀ᵐ(-1)ᵐ⁻ʲbinom mj c^((p-1)(m-j)) a_(n+j(p-1)+m) ≡ aₙ (mod p). At m=c=1 this is Touchard's congruence Bₙ₊ₚ ≡ Bₙ+Bₙ₊₁. We prove the sharper statement that the Touchard operator Ta=(aₙ₊ₚ-cᵖ⁻¹aₙ₊₁)_(n ≥ 0) maps the m-dimensional solution space to itself with Tᵐ=id, so that the solution space is a module over 𝔽ₚ[y]/(yᵐ-1) with y acting as T; every refinement below is a statement about where a sequence sits in that module. We then show that the composite m-Bell sequence can satisfy a strictly shorter recurrence than the general theorem gives: for every prime p it satisfies Bc(p-1)ₙ₊ₚ ≡ Bc(p-1)ₙ₊₁-Bc(p-1)ₙ (mod p), of degree p rather than p(p-1), and we exhibit three further such cells. Finally we determine the least period of Bc m mod p in thirteen cells, including Bc 4 mod 7, whose least period is 48=7²-1 against the 4N₇=549028 that the neighbouring cells suggest. All theorems below are formally verified in Lean 4.
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Archived files
- Version 1 · current (opens in a new tab)
Source snapshot 2026-08-30 15:34 UTC
File fingerprint
467d16a51c2d0b0769ecf33f56e572fd9cde0018eb074fbce278b9fc4eb1416e
Claim ledger
Stated results
MT1candidate2026-08-28
The m-Touchard congruence: (Xᵖ - cᵖ⁻¹X)ᵐ - 1 annihilates every m-binomial-shift sequence with base c mod p
MT2known2026-08-28
Touchard's congruence recovered at m = c = 1, including for Mathlib's Nat.bell
MT3candidate2026-08-28
The arithmetic form: for an integer binomial-shift sequence with integer base, p divides the m-Touchard combination minus aₙ
MT4routine2026-08-28
The solution space is inhabited and rigid: binSeq realises every order, base and initial segment with the recurrence proved, and the first m values determine the sequence
MT5routine2026-08-28
Finite-to-universal bridge: a sequence obeying a monic order-d recurrence is determined by its first d values, so a vanishing linear combination of shifts on [0,d) vanishes everywhere
MT6routine2026-08-28
The m-Touchard congruence read as a monic linear recurrence of order m*p, with every shift index < m*p
MT7candidate2026-08-28
The Touchard operator T a = (aₙ₊ₚ - cᵖ⁻¹ aₙ₊₁) maps the m-binomial-shift solution space into itself, and Tᵐ = id on it
MB1known data2026-08-28
Popov's printed prefixes (Bell, A007472, A351143, A351028, and the m = 3, 4 composite rows) recomputed from the recurrence inside Lean; B¹ = Nat.bell
MB2candidate2026-08-28
The m-Touchard congruence for Popov's composite Bᵐ and every primitive B^(m,r), for every prime p, every m and every n
MB3candidate2026-08-28
The explicit m = 2 congruence for OEIS A007472: aₙ₊₂ₚ - 2 aₙ₊ₚ₊₁ + aₙ₊₂ == aₙ (mod p) for every odd prime p
MB4routine2026-08-28
Seven controls: the exponent m cannot be lowered (two cells); primality of the modulus, the binomial coefficient, the shift index and the c^((p-1)(m-j)) factor are each load-bearing; and the stronger claim aₙ₊ₘₚ == aₙ is false
MS1routine2026-08-28
When p divides m the recurrence itself degenerates to aₙ₊ₘ = aₙ, so the order-m*p annihilator is p-fold redundant – this accounts for the whole 'not minimal' column of the tightness grid
MS2candidate2026-08-28
For every prime p, the composite (p-1)-Bell sequence satisfies Bᵖ⁻¹ₙ₊ₚ == Bᵖ⁻¹ₙ₊₁ - Bᵖ⁻¹ₙ (mod p) – a degree-p recurrence where MT1 supplies only degree p(p-1)
MS3candidate2026-08-28
Three further deficient cells: the composite obeys a strictly shorter recurrence than MT1 gives at (p,m) = (2,3) (degree 4 against 6), (3,4) (9 against 12) and (7,4) (14 against 28, i.e. T² = -1)
MP2candidate2026-08-28
Least periods of Bᵐ mod 2 for m = 1..6: 3, 1, 15, 1, 255, 1
MP3candidate2026-08-28
Least periods of Bᵐ mod 3 for m = 1,2,3,4,6: 13, 26, 1, 104, 1
MP5routine2026-08-28
The least period of B⁵ mod 5 is 1
MP7candidate2026-08-28
The least period of B⁴ mod 7 is 48 = 7² - 1, against 4 N₇ = 549028 if the m * Nₚ reading held
Provenance
- Generated by
- Machina Mathematica
- Released by
- Korea Superintelligence Labs
- Source context
- V. Popov, *"m-Bell and m-Stirling numbers: iterated binomial transforms, hyper-Bessel functions, and moments of the Conway–Maxwell–Poisson distribution"*, arXiv:2608.12011v1 (12 Aug 2026). Its Problem 3, verbatim:
- Snapshot
- 2026-09-07 03:53 UTC
- Ledger commit
801848d7