2,1

Any number may be substituted for y to yield similar sequences. The number set used determines values given (i.e.- integer yields integer). All centered square integers in the set of integers may be found by this formula.

1/2 + 1/10 + 1/26 +...= (Pi/4)*tanh(Pi/2) [Jolley] - Gary W. Adamson (qntmpkt(AT)yahoo.com), Dec 21 2006

Except for the first term, a(n)=8*n+a(n-1), (with a(1)=10) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Oct 24 2009]

L. B. W. Jolley, "Summation of Series", Dover Publications, 1961, p. 176.

[y(2x + 1)]^2 + [y(2x^2 + 2x)]^2 = [y(2x^2 + 2x +1)]^2 where y = 2. If a^2 + b^2 = c^2, then c^2 = y^2(4x^4 + 8x^3 + 8x^2 + 4x + 1). Also 2*A001844.

a(n) = (2n+1)^2+1. [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Nov 10 2008, corrected R. J. Mathar, Sep 16 2009]

a(n)=8*n+a(n-1)-8 (with a(1)=2) [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 17 2009]

If y = 3, then 81 + 144 = 225; if y = 4, then 12^2 + 16^2 = 20^2; 7^2 + 24^2 = 25^2 = 15^2 + 20^2.

For n=2, a(2)=8*2+2-8=10; n=3, a(3)=8*3+10-8=26; n=4, a(4)=8*4+26-8=50 [From Vincenzo Librandi (vincenzo.librandi(AT)tin.it), Nov 17 2009]

Table[4n(n + 1) + 2, {n, 0, 45}]

lst={}; Do[AppendTo[lst, n^2+1], {n, 1, 2*4!, 2}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Nov 10 2008]

Cf. A001844.

Sequence in context: A058373 A167386 A027719 this_sequence A045605 A009307 A131130

Adjacent sequences: A069891 A069892 A069893 this_sequence A069895 A069896 A069897

nonn,new

Glenn B. Cox (igloos_r_us(AT)canada.com), Apr 10 2002

Edited by Robert G. Wilson v (rgwv(AT)rgwv.com), Apr 11 2002

Edited the equation 4n^2+4n+2=n^2+1 - R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 16 2009