A141516 The main diagonal of the array of A141425 and its higher order differences.

1, 2, 1, -7, -23, -1, 7, -103, -251, -133, -149, -1387, -3143, -3001, -4913, -19663, -42611, -55693, -101549, -291667, -612863, -960001, -1831433, -4460023, -9185771, -15980053, -31162949, -69500347, -141392183, -261261001

Offset: 0

Author

Paul Curtz, Aug 11 2008

Keywords

• sign
• easy

Comments

The sequence A141425 and higher order differences in subsequent rows starts (see A141533):

1, 2, 4, 5, 7, 8, 1, 2, 4, 5, 7, 8, 1, 2, 4,...

1, 2, 1, 2, 1,-7, 1, 2, 1, 2, 1,-7, 1, 2, 1, 2,...

1,-1, 1, -1, -8, 8, 1,-1, 1, -1, -8, 8, 1, -1,..

-2, 2,-2, -7, 16,-7,-2, 2,-2, -7, 16,-7, -2,..

4,-4,-5, 23,-23, 5, 4,-4,-5, 23,-23, 5, 4,..

-8,-1,28,-46, 28,-1,-8,-1,28,-46, 28,-1,..

Reading downwards the main diagonal of this array defines the sequence.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..200
• Index entries for linear recurrences with constant coefficients, signature (1,-1,3,6)

Programs

Maple

A108411 := proc(n) 3^floor(n/2) ; end proc:
A141516 := proc(n) if n = 0 then 1; else (-3*(-1)^n-2^n+3*(-1)^(floor((n-1)/2))*A108411(n))/2 ; end if; end proc: # R. J. Mathar, Mar 08 2011

Mathematica

LinearRecurrence[{1,-1,3,6},{1,2,1,-7,-23},30] (* Harvey P. Dale, Nov 23 2022 *)

Formula

a(n) = ( -3*(-1)^n -2^n +3*(-1)^(floor((n-1)/2))*A108411(n) )/2, n>0. - R. J. Mathar, Mar 08 2011

a(2n)+a(2n+1)= -A002023(n-1) = -3*A081294(n), n>0.

a(4n)+a(4n+1)+a(4n+2)+a(4n+3) = -120*16^(n-1), n>0.

a(4n+2)+a(4n+3)+a(4n+4)+a(4n+5) = -30*A001025(n).

G.f. x*(-2+x+6*x^2+21*x^3) / ( (2*x-1)*(1+x)*(3*x^2+1) ). - R. J. Mathar, Mar 08 2011