0,4

Unsigned sequence (A008619) gives number of partitions of n in which the greatest part is 2. - Robert G. Wilson v (rgwv(AT)rgwv.com), Jan 11 2002

Go forwards and backwards with increasing step sizes. - Daniele Parisse and Franco Virga (daniele.parisse(AT)eads.com), Jun 06 2005

The partial sums of the divergent series 1 - 2 + 3 - 4 + ... give this sequence. Euler summed it to 1/4 which was one of the first examples of summing divergent series. - Michael Somos May 22 2007

Contribution from Peter Luschny (peter(AT)luschny.de), Jul 12 2009: (Start)

The general formula for alternating sums of powers is in terms of the Swiss-Knife polynomials P(n,x) A153641 2^(-n-1)(P(n,1)-(-1)^k P(n,2k+1)). Thus

a(k) = 2^(-2)(P(1,1)-(-1)^k P(1,2k+1)). (End)

G. Myerson and A. J. van der Poorten, Some problems concerning recurrence sequences, Amer. Math. Monthly, 102 (1995), 698-705.

T. D. Noe, Table of n, a(n) for n=0..1000

G. Myerson and A. J. van der Poorten, Some problems concerning recurrence sequences, Amer. Math. Monthly, 102 (1995), 698-705.

Wikipedia, 1 - 2 + 3 - 4 + ...

Euler transform of [ -1, 2] is sequence a(n+1).

G.f.: x/((1+x)(1-x^2)). E.g.f.: (exp(x)-(1-2x)exp(-x))/4.

a(n)=1-2*a(n-1)-a(n-2); a(2n)=-n, a(2n+1)=n+1.

a(n)=-a(n-1)+a(n-2)+a(n-3). a(n)=(-1)^(n+1)*floor((n+1)/2).

a(1) = 1, a(n) = a(n-1)+n or a(n-1)-n whichever is closer to 0 on the number line. Or a(n) = min{absolute[a(n-1) +n], absolute[a(n-1) -n]}. - Amarnath Murthy (amarnath_murthy(AT)yahoo.com), Jul 01 2003

a(n)=sum{k=0..n, k*(-1)^(k+1) } - Paul Barry (pbarry(AT)wit.ie), Aug 20 2003

a(n)=(1-(2n+1)(-1)^n)/4 - Paul Barry (pbarry(AT)wit.ie), Feb 02 2004

a(0)=0; a(n) = (-1)^(n-1) * (n-|a(n-1)|) for n >= 1. - Rick L. Shepherd (rshepherd2(AT)hotmail.com), Jul 14 2004

a(n)=a(n-1)-n*(-1)^n, a(0)=0; or a(n)=-a(n-1)+(1-(-1)^n)/2, a(0)=0. - Daniele Parisse and Franco Virga (daniele.parisse(AT)eads.com), Jun 06 2005

a := n -> (1-(-1)^n*(2*n+1))/4; [From Peter Luschny (peter(AT)luschny.de), Jul 12 2009]

(PARI) a(n)=if(n%2, n\2+1, -n/2)

A008619(n)=|a(n+1)|, A004526(n)=|a(n-1)|.

Cf. A166711, A166871. [From Jaume Oliver Lafont (joliverlafont(AT)gmail.com), Oct 30 2009]

Sequence in context: A111660 A127365 A065033 this_sequence A130472 A004526 A140106

Adjacent sequences: A001054 A001055 A001056 this_sequence A001058 A001059 A001060

sign,nice,easy

N. J. A. Sloane (njas(AT)research.att.com).

Thanks to Michael Somos for helpful comments.