A characterization of A002450, A020988 and A080674.

				Peter Bala, Oct 2015

A002450 is the sequence 1/3*(4^n - 1). We show that A002450 gives the values of N such that

binomial(4*N + 1,N) is odd. The proof uses Lucas' theorem on the divisibility of binomial

coefficients by a prime p. We give similar characterizations for A020988 and A080674.

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We make use of the following congruence, which is an equivalent form of Lucas' theorem - see 

[Mestrovic, Section 2]:

	binomial(n*p + r, m*p + s) = binomial(n,m)*binomial(r,s) (mod p)	(1)
	
where n, m, r and s are nonnegative integers such that 0 <= r,s <= p-1.

The parity of numbers of the form binomial(4*N + 1,N) will be established on a case-by-case

basis by applying (1) when p = 2. 

PROPOSITION 1.  binomial(4*N + 1,N) is odd if and only if N = 1/3*(4^k - 1) for some k >= 0.

PROOF.

For N = 0, binomial(4*N + 1,N) = 1 is odd, and N = 1/3*(4^k - 1) with k = 0. 

From now on we assume N > 0.

Case 1. N is even, say, N = 2^a*(2*b + 1) for some a >= 1, b >= 0.

We have

	binomial(4*N + 1,N) = binomial( 2^(a+2)*(2*b + 1) + 1, 2^a*(2*b + 1) )

		    	    = binomial( 2^(a+1)(2*b + 1), 2^(a-1)*(2*b + 1) ) * binomial(1,0) (mod 2)  by (1)

		    	    = binomial( 2^(a+1)(2*b + 1), 2^(a-1)*(2*b + 1) ) (mod 2)

		   	    = binomial( 2^(2)(2*b + 1), (2*b + 1) ) (mod 2)  by repeated application of (1)

		   	    = binomial( 2*(2*b + 1), 2*b ) * binomial(0,1) (mod 2)  by (1) again

		    	    = 0 (mod 2).
		    	    
Thus if N is even then binomial(4*N + 1,N) is even.

Case 2. N is odd of the form N = 4*m + 3.

Then by (1)

	binomial(4*N + 1,N) = binomial(16*m + 13,4*m + 3)
	
			    = binomial(8*m + 6,2*m + 1) * binomial(1,1) (mod 2)
			    
			    = binomial(4*m + 3,m) * binomial(0,1) (mod 2)
			    
			    = 0 (mod 2).
			    
Thus if N is odd of the form 4*m + 3 then binomial(4*N + 1,N) is even.
			    
Case 3. N is odd of the form N = 4*m + 1.

By (1)

	binomial(4*N + 1,N) = binomial(16*m + 5,4*m + 1)
	
			    = binomial(8*m + 2,2*m) * binomial(1,1) (mod 2)
			    
			    = binomial(4*m + 1, m) * binomial(0,0) (mod 2)
			    
			    = binomial(4*m + 1,m) (mod 2).
			    
We have shown that if N = 4*m + 1 then

	binomial(4*N + 1,N) = binomial(4*m + 1,m) (mod 2).		(2)

We now use (2) in two reductio arguments to finish the proof of the proposition.

(i) Suppose N = 4*m + 1 is the minimal positive integer not of the form 1/3*(4^k - 1) such that

binomial(4*N + 1,N) is odd. Then by (2), binomial(4*m + 1,m) is also odd. Clearly, m is also

not of the form 1/3*(4^k - 1) and m < N. This contradicts the assumed minimality of N. We

conclude that if N = 4*m + 1 is not of the form 1/3*(4^k - 1) then binomial(4*N + 1,N) is even.


Thus the only possible values of N for which binomial(4*N + 1,N) may be odd are when N = 4*m + 1 and

also N is of the form 1/3*(4^k - 1) for some k >= 0. In fact, all such values of N make binomial(4*N + 1,N)

odd, as can be seen by a simple induction argument using (2) or the following reductio argument.


(ii) Suppose N = 4*m + 1 is the minimal number of the form 1/3*(4^k - 1) such that binomial(4*N + 1,N)

is even. Necessarily, m > 0 since binomial(5,1) = 5 is odd. Then by (2), binomial(4*m + 1,m) is

also even. But m = (N - 1)/ 4 = 1/3*(4^(k-1) - 1) and m < N. This contradicts the assumed minimality

of N. We conclude that if N = 4*m + 1 is of the form 1/3*(4^k - 1) then binomial(4*N + 1,N) is odd.

END PROOF.


We can find similar characterizations in terms of the parity of certain binomial coefficients for

A020988(n) = 2/3*(4^n - 1) and A080674(n) = 4/3*(4^n - 1).


PROPOSITION 2.  binomial(4*N + 2,N) is odd if and only if N = 2/3*(4^k - 1) = A020988(k) for some k >= 0.

PROOF

Case 1. Assume N = 2*m + 1 is odd.

Then by (1)

	 binomial(4*N + 2,N) = binomial(8*m + 6, 2*m + 1)
	 
	 		     = binomial(4*m + 3, m)*binomial(0,1) (mod 2) 
	 			
	 		     = 0 (mod 2)

Thus binomial(4*N + 2,N) is even when N is odd.

Case 2. Assume N = 2*m is even.

Then by (1)

	binomial(4*N + 2,N) = binomial(8*m + 2, 2*m )
	 
	 		    = binomial(4*m + 1, m) (mod 2) 
	 			
Hence by Proposition 1, binomial(4*N + 2,N) will be odd iff m = 1/3*(4^k - 1) for some k >= 0.

Thus binomial(4*N + 2,N) will be odd iff N = 2*m = 2/3*(4^k - 1) for some k.

END PROOF


PROPOSITION 3.  binomial(4*N + 4,N) is odd if and only if N = 4/3*(4^k - 1) = A080674(k) for some k >= 0.

PROOF

Case 1. Assume N = 2*m + 1 is odd.

Then by (1)

	 binomial(4*N + 4,N) = binomial(8*m + 8, 2*m + 1)
	 
	 		     = binomial(4*m + 4, m)*binomial(0,1) (mod 2) 
	 			
	 		     = 0 (mod 2)

Thus binomial(4*N + 4,N) is even when N is odd.

Case 2. Assume N = 2*m is even.

Then by (1)

	binomial(4*N + 4,N) = binomial(8*m + 4, 2*m )
	 
	 		    = binomial(4*m + 2, m) (mod 2) 
	 			
Hence by Proposition 2, binomial(4*N + 4,N) will be odd iff m = 2/3*(4^k - 1) for some k >= 0.

Thus binomial(4*N + 4,N) will be odd iff N = 2*m = 4/3*(4^k - 1) for some k.

END PROOF


	
REFERENCES
	
[1] ROMEO MESTROVIC, LUCAS’ THEOREM: ITS GENERALIZATIONS, EXTENSIONS AND APPLICATIONS (1878–2014),
    arXiv:1409.3820v1 [math.NT] available online at http://arxiv.org/abs/1409.3820