Understanding the Divisibility Property of the Central Trinomial Coefficient

The central trinomial coefficient, denoted as ( T_n ), is a well-known sequence in combinatorics. This article delves into the divisibility property of the central trinomial coefficient, presented with a simplified and detailed proof that does not rely on Fermat's Little Theorem.

What is a Central Trinomial Coefficient?

A central trinomial coefficient ( T_n ) is the coefficient of ( x^n ) in the expansion of the polynomial ( (1 x x^2)^n ). It can also be defined as the sum of multinomial coefficients involving the terms ( (1, x, x^2) ).

Divisibility Property

The main focus of this discussion is a particular divisibility property of the central trinomial coefficient, which can be expressed as:

( T_{2p} k cdot p )

where ( k ) is some integer.

The Simplified Proof

Let's explore why this property holds using a detailed and accessible proof.

Definition and Initial Expression

Let ( T_n ) be the central trinomial coefficient, which is the coefficient of ( x^n ) in the expansion of ( (1 x x^2)^n ).

The binomial theorem can be applied twice to simplify this expression:

( (1 x x^2)^n sum_{k0}^{n} binom{n}{k} (x x^2)^{n-k} )

Let's further break this down using another application of the binomial theorem:

( (x x^2)^{n-k} sum_{j0}^{n-k} binom{n-k}{j} x^{n-k-j} )

Combining these results, we get:

( (1 x x^2)^n sum_{k0}^{n} binom{n}{k} sum_{j0}^{n-k} binom{n-k}{j} x^{n-k-j} )

Since we are interested in the coefficient of ( x^n ), we need ( k j ). Therefore:

( T_n sum_{k0}^{n} binom{n}{k} binom{n-k}{k} )

Simplifying the Expression

Notice that ( binom{n-k}{k} binom{n-k}{n-2k} ), which is a symmetric property of binomial coefficients. This allows us to rewrite the expression as:

( T_n sum_{k0}^{n} binom{n}{k} binom{n-k}{n-2k} )

Now, we need to use the identity:

( binom{n}{k} binom{n-k}{n-2k} binom{2k}{k} binom{n}{2k} )

Let's verify this identity:

The left hand side is equal to:

( binom{n}{k} binom{n-k}{n-2k} frac{n!}{k!(n-k)!} cdot frac{(n-k)!}{(n-2k)!k!} frac{n!}{k!^2 (n-2k)!} )

The right hand side is equal to:

( binom{2k}{k} binom{n}{2k} frac{(2k)!}{k!^2} cdot frac{n!}{2k!(n-2k)!} frac{n!}{k!^2 (n-2k)!} )

Both expressions are identical, confirming the identity.

Modulo ( n ) Consideration

Now, let's consider the expression modulo ( n ). Notice that:

( binom{n}{2k} frac{n!}{2k!(n-2k)!} )

This term is divisible by ( n ) if ( k > 0 ). Therefore, all terms in the sum except for ( k 0 ) will be zero modulo ( n ).

Thus:

( T_{2p} equiv binom{n}{0} binom{0}{0} equiv 1 pmod{n} )

Conclusion

The divisibility property ( T_{2p} k cdot p ) holds true, showing that ( T_{2p} equiv 1 pmod{p} ). This does not depend on ( n ) being prime, as initially discussed on MathWorld.

Understanding these proofs and properties of the central trinomial coefficients can provide valuable insights in combinatorics and number theory.