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A143697 Least square k^2 such that n^2-k^2 = p*q with p and q odd primes and p<q for n>= 4. 3


%S 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,

%T 144,81,16,9,36,25,36,81,4,9,144,1,576,9,4,225,36,25,144,9,64,81,36,

%U 49,144,9,16,225,144,1,324,81,16,9,36,25,36,225,4,9,144,1,36,225

%N Least square k^2 such that n^2-k^2 = p*q with p and q odd primes and p<q for n>= 4.

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

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

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

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

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

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

%C 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]

%H Pierre CAMI, <a href="/A143697/b143697.txt">Table of n, a(n) for n = 4..60000</a>

%e 4*4-1=3*5 p=3 q=5

%e 5*5-4=3*7 p=3 q=7

%e 6*6-1=5*7 p=5 q=7

%e 7*7-16=3*11 p=3 q=11

%o (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

%Y Cf. A078587, A078496.

%K nonn

%O 4,2

%A _Pierre CAMI_, Aug 29 2008

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