A143697 Least square k^2 such that n^2-k^2 = p*q with p and q odd primes and p= 4.

1, 4, 1, 16, 9, 4, 9, 36, 1, 36, 9, 4, 9, 36, 1, 144, 9, 4, 81, 36, 25, 36, 9, 16, 81, 144, 1, 144, 81, 16, 9, 36, 25, 36, 81, 4, 9, 144, 1, 576, 9, 4, 225, 36, 25, 144, 9, 64, 81, 36, 49, 144, 9, 16, 225, 144, 1, 324, 81, 16, 9, 36, 25, 36, 225, 4, 9, 144, 1, 36, 225

Offset: 4

Author

Pierre CAMI, Aug 29 2008

Keywords

• nonn

Comments

The product p*q is the sum of p consecutive odd numbers with 2*n-1 the greatest.

For n=4 p*q=3*5=15, 15=7+5+3

For n=5 p*q=3*7=21, 21=9+7+5

For n=6 p*q=5*7=35, 35=11+9+7+5+3

For n=7 p*q=3*11=33, 33=13+11+9

k^2 is the sum of the k first consecutive odd numbers p=n-k and q=n+k.

Assuming a strong version of the Goldbach conjecture, every term exists and we have a(n)=A082467(n)^2, p(n)=A078587(n) and q(n)=A078496(n). [T. D. Noe, Jan 22 2009]

Examples

			4*4-1=3*5 p=3 q=5
5*5-4=3*7 p=3 q=7
6*6-1=5*7 p=5 q=7
7*7-16=3*11 p=3 q=11
		

Links

• Pierre CAMI, Table of n, a(n) for n = 4..60000

Crossrefs

Cf. A078587, A078496.

Programs

PARI

a(n) = {for (k=1, n-1, my(x=n^2-k^2); if ((omega(x)==2) && (bigomega(x)==2) && (x%2), return(k^2);););} \\ Michel Marcus, Sep 23 2019