# A conjecture about the degrees of special polynomials

Define the congruence "modulo m" on exponential Taylor series as $$\sum_{n=0}^\infty \frac{a_n}{n!}x^n \equiv \sum_{n=0}^\infty \frac{b_n}{n!} x^n \mod m \iff \forall n: \frac{a_n-b_n}{m}\in \mathbb{Z}$$ It's been conjectured, that if we define the sequence of polynomials $P_N(x)$ as $$P_N(x):=\ln^N(1+x)\mod (N+1)!$$ then $$\boxed{ \deg P_N(x)=N\cdot \mathrm{gpf}(N+1) }$$ (the greatest prime factor of $N+1$, see http://mathworld.wolfram.com/GreatestPrimeFactor.html)

It is obvious, that the coefficients of order $>N(N+1)$ vanish because of holding property $$a(x)b(x)\mod n=((a(x) \mod n)(b(x) \mod n)) \mod n$$

and the congruence $$\ln(1+x) \equiv 0!\frac{x}{1!}-1!\frac{x^2}{2!}+...+(-1)^{k-1}(k-1)!\frac{x^k}{k!} \mod k!$$ hence there is inequality $$N\leqslant\deg P_N(x)\leqslant N(N+1)$$ (Also, the conjecture is obviously true for prime $(N+1)$)

I haven't found the appropriate congruences of Stirling numbers of the first kind (the powers of natual logarithm are their exponential generating functions) to prove the whole conjecture and would be very grateful, if you could give a proof of this fact, share the ideas of why it might be true or post some references. Thank you.

Note: To see the list of these polynomials in Maple or Wolfram one should use the Laplace transform together with PolynomialMod command. Another interesting question is about the coefficients, but the regular behavior of them is observed only in the positions of vanishing coefficients of $P_{p-1}(x)$ and is unexplained too.

UPD. The equivalent form of the conjecture in terms of Stirling numbers states, that $$1.~~\forall k>N\cdot \mathrm{gpf}(N+1):~\left[ k \atop N\right]\equiv 0 \mod (N+1)\\ ~\\ 2.~~\left[ N\cdot\mathrm{gpf}(N+1)\atop N \right]\not\equiv 0 \mod (N+1)$$

where

$$\sum_{k=0}^n \left[ n \atop k \right]x^k=x(x-1)...(x-n+1)$$

• Can you specify the kind of congruence you are seeking for the Stirling numbers? It was not clear. – T. Amdeberhan Mar 23 '17 at 1:52
• @T.Amdeberhan It's fixed now. – Danil Krotkov Mar 23 '17 at 2:24