A140096 a(n) = A000045(n) - A113405(n).

0, 1, 1, 1, 1, 1, 1, -1, -7, -23, -59, -139, -311, -677, -1443, -3031, -6295, -12967, -26543, -54073, -109743, -222071, -448323, -903411, -1817767, -3653245, -7335147, -14716663, -29508351, -59138095, -118472607

Offset: 0

Author

Paul Curtz, Jun 21 2008

Keywords

• sign
• easy
• less

Comments

Because the inverse binomial transform of A000045 and A113405 is individually the same as the original sequence up to a sign flip of each second term, the same is true for their difference here. (The inverse binomial transform is a linear transform.)

The sequence and its higher order differences in subsequent rows has zeros on the main diagonal:

0, 1, 1, 1, 1, 1, 1, -1, -7,-23, -59, -139, -311, -677, -1443, -3031

1, 0, 0, 0, 0, 0, -2, -6,-16,-36, -80, -172, -366, -766, -1588, -3264

-1, 0, 0, 0, 0,-2, -4,-10,-20,-44, -92, -194, -400, -822, -1676

1, 0, 0, 0,-2,-2, -6,-10,-24,-48,-102, -206, -422, -854, -1732

-1, 0, 0,-2, 0,-4, -4,-14,-24,-54,-104, -216, -432, -878, -1764

1, 0,-2, 2,-4, 0,-10,-10,-30,-50,-112, -216, -446, -886, -1790,

-1,-2, 4,-6, 4,-10, 0,-20,-20,-62,-104, -230, -440, -904, -1792

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Index entries for linear recurrences with constant coefficients, signature (3,-1,-3,3,-1,-2).

Crossrefs

Cf. A140428, A140413.

Programs

Mathematica

CoefficientList[Series[-x*(-1 + 2*x + x^2 - 2*x^3 + x^4)/((2*x - 1)*(1 + x)*(x^2 - x + 1)*(x^2 + x - 1)), {x, 0, 50}], x] (* G. C. Greubel, Jul 17 2017 *)

PARI

x='x+O('x^50); concat([0], Vec(-x*(-1+2*x+x^2-2*x^3+x^4)/( (2*x-1)*(1+x)*(x^2-x+1)*(x^2+x-1) ))) \\ G. C. Greubel, Jul 17 2017

Formula

a(n)= +3*a(n-1) -a(n-2) -3*a(n-3) +3*a(n-4) -a(n-5) -2*a(n-6).

G.f.: -x*(-1+2*x+x^2-2*x^3+x^4) / ( (2*x-1)*(1+x)*(x^2-x+1)*(x^2+x-1) ).

a(n+1)-2*a(n) = -A141325(n-2), n>2.