A167193 - OEIS

A167193

a(n) = (1/3)*(1 - (-2)^n + 3*(-1)^n ) = (-1)^(n+1)*A167030(n).

1, 0, 0, 2, -4, 10, -20, 42, -84, 170, -340, 682, -1364, 2730, -5460, 10922, -21844, 43690, -87380, 174762, -349524, 699050, -1398100, 2796202, -5592404, 11184810, -22369620, 44739242, -89478484, 178956970, -357913940, 715827882, -1431655764, 2863311530, -5726623060, 11453246122, -22906492244, 45812984490

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OFFSET

0,4

COMMENTS

This is the inverse binomial transform of 1, 1, 1, 3, 5, 11,.. (continued as in A001045 and conjectured to be equal to A152046).

Any sequence (like this one) which obeys a(n)= -2a(n-1)+a(n-2)+2a(n-3) also obeys a(n)=5a(n-2)-4a(n-4), proved by telescoping; see A101622.

LINKS

Vincenzo Librandi, Table of n, a(n) for n = 0..1000

Index entries for linear recurrences with constant coefficients, signature (-2,1,2).

FORMULA

a(2n) = (-1)^n* A084240(n). a(2n+1) = A020988(n).

G.f.: ( -1 - 2*x + x^2 ) / ( (x-1)*(1+2*x)*(1+x) ).

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

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

E.g.f.: (1/3)*(exp(x) + 3*exp(-x) - exp(-2*x)). - G. C. Greubel, Jun 04 2016

MATHEMATICA

LinearRecurrence[{-2, 1, 2}, {1, 0, 0}, 25] (* or *) Table[(1/3)*(1 + 3*(-1)^n - (-2)^n), {n, 0, 25}] (* G. C. Greubel, Jun 04 2016 *)

PROG

(Magma) [( 1-(-1)^n*2^n)/3+(-1)^n: n in [0..40] ]; // Vincenzo Librandi, Aug 06 2011

CROSSREFS

Sequence in context: A255386 A167030 A026644 * A026666 A325508 A238439

Adjacent sequences: A167190 A167191 A167192 * A167194 A167195 A167196

KEYWORD

easy,sign

AUTHOR

Paul Curtz, Oct 30 2009

STATUS

approved