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A143697
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Least square k^2 such that n^2-k^2 = p*q with p and q odd primes and p<q for n>= 4.
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3
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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
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OFFSET
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4,2
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COMMENTS
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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]
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LINKS
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EXAMPLE
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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
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PROG
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(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
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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