1,1

Sums of squares of two consecutive elements multiplied (or divided) by 2 is always a perfect square. In general numbers represented by the quadratic form a(n)=(2*i*n+j)^2-2*i^2 for any i and j have 2(a(n)^2 + a(n+1)^2)) and (a(n)^2 + a(n+1)^2)/2 as perfect squares: in this case i=j=1.

The elements of this sequence may be seen to be 2 less than the odd squares. As such they run parallel to those in the square spiral as well as the Ulam square spiral. - Stuart M. Ellerstein (ellerstein(AT)aol.com), Oct 01 2002

a(n)=FrobeniusNumber[2n+1,2n+3] - Darrell Minor (dminor(AT)cscc.edu), Jul 29 2008

Second program 20% faster. [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Nov 18 2009]

a(n)=8*n+a(n-1), (with a(1)=7) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Oct 24 2009]

For n=2, a(2)=8*2+7=23; n=3, a(3)=8*3+23=47; n=4, a(4)=8*4+47=79 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Oct 24 2009]

Table[4*n^2+4*n-1, {n, 60}]..and/or..Table[n*(n+1)+(n-1), {n, 2, 120, 2}] [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Nov 18 2009]

Sequence in context: A097149 A139035 A002146 this_sequence A139830 A153210 A158035

Adjacent sequences: A073574 A073575 A073576 this_sequence A073578 A073579 A073580

nonn,new

M. N. Deshpande (dpratap(AT)nagpur.dot.net.in), Aug 27 2002

Edited and extended by Henry Bottomley (se16(AT)btinternet.com), = Oct 10 2002