A145654 Partial sums of A000918, starting from index 1.

0, 2, 8, 22, 52, 114, 240, 494, 1004, 2026, 4072, 8166, 16356, 32738, 65504, 131038, 262108, 524250, 1048536, 2097110, 4194260, 8388562, 16777168, 33554382, 67108812, 134217674, 268435400, 536870854, 1073741764, 2147483586

Offset: 1

Author

Vladimir Joseph Stephan Orlovsky, Oct 16 2008

Keywords

• nonn
• easy

Examples

			For n=7, a(7) = 6*2 + 5*2^2 + 4*2^3 + 3*2^4 + 2*2^5 + 1*2^6 = 240. - _Bruno Berselli_, Feb 10 2014
From _Bruno Berselli_, Jul 17 2018: (Start)
Row sums of the triangle:
   0   ......................................    0
   1,  1   ..................................    2
   3,  2,  3   ..............................    8
   6,  5,  5,  6   ..........................   22
  10, 11, 10, 11, 10   ......................   52
  15, 21, 21, 21, 21, 15   ..................  114
  21, 36, 42, 42, 42, 36, 21   ..............  240
  28, 57, 78, 84, 84, 78, 57, 28   ..........  494, etc.
(End)
		

Links

• Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
• Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Crossrefs

Cf. A000217, A000295, A000918, A005803, A121173.

Programs

Haskell

a145654 n = a145654_list !! (n-1)
a145654_list = scanl1 (+) $ tail a000918_list
-- Reinhard Zumkeller, Nov 06 2013

Mathematica

Accumulate[2^Range[30] - 2] (* or *) LinearRecurrence[{4, -5, 2}, {0, 2, 8}, 30] (* Harvey P. Dale, Jul 15 2017 *)

Formula

a(n) = Sum_{i=1..n} A000918(i).

a(n+1) - a(n) = A000918(n+1).

a(n) = A005803(n+1). - R. J. Mathar, Oct 21 2008

From Colin Barker, Jan 11 2012: (Start)

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

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

a(n+1) = A121173(2*n). - Reinhard Zumkeller, Nov 06 2013

a(n) = Sum_{i=1..n-1} (n-i)*2^i with a(1)=0. - Bruno Berselli, Feb 10 2014

a(n) = 2 * A000295(n). - Alois P. Heinz, May 28 2018

Extensions

Edited by R. J. Mathar, Oct 21 2008