June 16, 2016 1. Super-Conjecture 1

Super-Conjecture 1.1. Let p ≥ 5 be a prime, and

A = ∑_{n=0}^{p−1} \binom{2n}{n}.

Then

A ≡ 1 mod p^{2} if p ≡ 1 mod 3,
A ≡ −1 mod p^{2} if p ≡ 2 mod 3.

Proof. Already from the paper we know

A = CT \frac{(1 + x)^{2p}}{x^{p−1}(1 + x + x^{2})} = CT \frac{(1 − x)(1 + x)^{2p}}{x^{p−1}(1 − x^{3})} .

So

(1)  A = ∑_{k≥0} (\binom{2p}{p − 1 − 3k} − \binom{2p}{p − 2 − 3k})

\underline{CASE 1} p ≡ 1 mod 3. In this case, the k = (p − 1)/3 term in (1) is 1, and all of the other terms are divisible by p,

A = 1 + pB

where

B = \frac{1}{p} ∑_{0≤k<(p−1)/3} (\binom{2p}{p − 1 − 3k} − \binom{2p}{p − 2 − 3k}) .

We’ll show that the integer B satsifies B ≡ 0 mod p, so that A ≡ 1 mod p^{2}. Since

\begin{aligned}\frac{1}{p}\binom{2p}{p − 1 − 3k} & ≡ 2\frac{(p − 1)!}{(p − 1 − 3k)!(3k + 1)!} mod p, \\ \frac{1}{p}\binom{2p}{p − 2 − 3k} & ≡ 2\frac{(p − 1)!}{(p − 2 − 3k)!(3k + 2)!} mod p \end{aligned}

we see that

B ≡ 2 ∑_{0≤k<(p−1)/3} (\frac{(p − 1)!}{(p − 1 − 3k)!(3k + 1)!} − \frac{(p − 1)!}{(p − 2 − 3k)!(3k + 2)!}) mod p

However, by trisecting the binomial theorem this sum of integers is 0

∑_{0≤k<(p−1)/3} (\frac{(p − 1)!}{(p − 1 − 3k)!(3k + 1)!} − \frac{(p − 1)!}{(p − 2 − 3k)!(3k + 2)!}) = 0,

so

B ≡ 0 mod p.

\underline{CASE 2} p ≡ 2 mod 3. In this case the second term with k = (p − 2)/3 gives −1. We proceed as in CASE 1, the trisection identity we need is

∑_{0≤k≤(p−2)/3} \frac{(p − 1)!}{(p − 1 − 3k)!(3k + 1)!} − ∑_{0≤k≤(p−5)/3} \frac{(p − 1)!}{(p − 2 − 3k)!(3k + 2)!} = 0,

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2. Super-Conjecture 1”

Super-Conjecture 2.1. Let p ≥ 5 be a prime, r be a positive integer, and

A = ∑_{n=0}^{rp−1} \binom{2n}{n}.

Then

A ≡ α_{r} mod p^{2} if p ≡ 1 mod 3,
A ≡ −α_{r} mod p^{2} if p ≡ 2 mod 3.

where

α_{r} = ∑_{n=0}^{r−1} \binom{2n}{n}.

Proof. (sketch, assuming a fact (3) you probably know, if true) This time

A = CT \frac{(1 + x)^{2rp}}{x^{rp−1}(1 + x + x^{2})} = CT \frac{(1 − x)(1 + x)^{2rp}}{x^{rp−1}(1 − x^{3})} .

So

(2)  A = ∑_{k≥0} (\binom{2rp}{rp − 1 − 3k} − \binom{2rp}{pr − 2 − 3k})

As before, we separate the terms which have a p for sure, namely those whose binomial denominator is not a multiple of p from those whose binomial denominator is a multiple of p. The non-multiples of p will sum to 0 as before, and we are left with the extra terms.

\underline{CASE 1} p ≡ 1 mod 3. Here the terms with denominator parameter a multiple of p are

EXT RA = ∑_{j≥0} (\binom{2rp}{(r − 1 − 3j)p} − \binom{2rp}{(r − 2 − 3j)p}) .

Assuming

(3)  \binom{Dp}{Ep} ≡ \binom{D}{E} mod p^{2}

then

EXT RA ≡ ∑_{j≥0} (\binom{2r}{r − 1 − 3j} − \binom{2r}{r − 2 − 3j}) = α_{r} mod p^{2}

by (1), which works for any positive integer p.

\underline{CASE 2} p ≡ 2 mod 3. Here the terms with denominator parameter being a multiple of p are

EXT RA = ∑_{j≥0} (\binom{2rp}{(r − 2 − 3j)p} − \binom{2rp}{(r − 1 − 3j)p}) .

then

EXT RA ≡ ∑_{j≥0} (\binom{2r}{r − 2 − 3j} − \binom{2r}{r − 1 − 3j}) = −α_{r} mod p^{2}

again by (1).

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School of Mathematics, University of Minnesota, Minneapolis, MN 55455

   E-mail address: stanton@math.umn.edu