0,2

A000466(n), A008586(n) and A053755(n) are Pythagorean triples. - Zak Seidov, Jan 16 2007

Sum_{n=1..infinity} (-1)^n*a(n)/n! = 1 - 1/e (A068996). - Gerald McGarvey (gerald.mcgarvey(AT)comcast.net), Nov 06 2007

Sum_(n=1,infinity,(-1)^n*a(n)/n!) = 1 - 1/e. - Gerald McGarvey (gerald.mcgarvey(AT)comcast.net), Nov 12 2007

Sequence arises from reading the line from -1, in the direction -1, 15,... and the same line from 3, in the direction 3, 35,..., in the square spiral whose nonnegative vertices are the squares A000290. - Omar E. Pol (info(AT)polprimos.com), May 24 2008

a(n+1)=A061037(4n+2)=4*(n+1)^2-1=(2n+1)*(2n+3), positive A000466(n). From Balmer spectrum of hydrogen. [From Paul Curtz (bpcrtz(AT)free.fr), Oct 09 2008]

a(n) is the product of the consecutive odd integers 2n-1 and 2n+1. [From Doug Bell (bell.doug(AT)gmail.com), Mar 08 2009]

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 3.

sum(n=1, inf, 1/a(n))=1/2 - Benoit Cloitre (benoit7848c(AT)orange.fr), Apr 05 2002

1 = Sum(1 through infinity) 2/a(n) = 2/3 + 2/15 + 2/35 + 2/63 + 2/99 + 2/143...; with partial sums: 2/3, 4/5, 6/7, 8/9, 10/11, 12/13, 14/15... - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 16 2003

1 = 2/(1)(3) + 2/(3)(5) + 2/(5)(7) + 2/(7)(9)... = 2/3 + 2/15 + 2/35 + 2/63 +...; & partial sums are 2/3, 4/5, 6/7, 8/9, ... - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 18 2003

1 = 1/3 + Sum(n=2, inf, 4/a(n)) = 1/3 + 4/15 + 4/35 + 4/63...; with partial sums 1/3, 3/5, 5/7, 7/9, 9/11...(2n+1)/(2n+3). - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 18 2003

Pi/4 = .7853981... = Sum(n=0, inf.): 2/a(2n+1) = 2/3 + 2/35 + 2/99...; = (1 - 1/3) + (1/5 - 2/7) + (1/9 - 1/11)...; = Sum(n=0, inf.): (-1)^n/(Nn+1), with N=2. 2. Pi/4 = DEF.INTEGRAL(0, 1, 1/(1 + x^2)dx)) - Gary W. Adamson (qntmpkt(AT)yahoo.com), Jun 22 2003

Product(1 through infinity, (a(n)+1)/a(n)) = pi/2 (Wallis formula). - Mohammed Bouayoun (mohammed.bouayoun(AT)sanef.com), Mar 03 2004

A053755(n)=a(n)+2. - Zak Seidov, Jan 16 2007

a(n)^2 + A008586(n)^2 = A053755(n)^2. - Zak Seidov, Jan 16 2007

lst={}; Do[AppendTo[lst, 4*n^2-1], {n, 0, 9^2}]; lst...or/and... lst={}; Do[AppendTo[lst, ChebyshevU[2, n]], {n, 0, 9^2}]; lst [From Vladimir Orlovsky (4vladimir(AT)gmail.com), Sep 11 2008]

Cf. A000290, A001539, A016286, A016742.

A002378 (0, 2, 6), A078371 (5, 21, 45). [From Paul Curtz (bpcrtz(AT)free.fr), Oct 09 2008]

Contribution from Johannes W. Meijer (meijgia(AT)hotmail.com), May 24 2009: (Start)

Factor of A160466.

(End)

Sequence in context: A086380 A009261 A102790 this_sequence A145949 A015809 A015715

Adjacent sequences: A000463 A000464 A000465 this_sequence A000467 A000468 A000469

sign,easy

skchan5(AT)hkein.ie.cuhk.hk (Chan Siu Kee)