A125128 a(n) = 2^(n+1) - n - 2, or partial sums of main diagonal of array A125127 of k-step Lucas numbers.

1, 4, 11, 26, 57, 120, 247, 502, 1013, 2036, 4083, 8178, 16369, 32752, 65519, 131054, 262125, 524268, 1048555, 2097130, 4194281, 8388584, 16777191, 33554406, 67108837, 134217700, 268435427, 536870882, 1073741793, 2147483616, 4294967263, 8589934558

Offset: 1

Author

Jonathan Vos Post, Nov 22 2006

Keywords

• easy
• nonn

Comments

Essentially a duplicate of A000295: a(n) = A000295(n+1).

Partial sums of main diagonal of array A125127 = L(k,n): k-step Lucas numbers, read by antidiagonals.

Equals row sums of triangle A130128. - Gary W. Adamson, May 11 2007

Row sums of triangle A130330 which is composed of (1,3,7,15,...) in every column, thus: row sums of (1; 3,1; 7,3,1; ...). - Gary W. Adamson, May 24 2007

Row sums of triangle A131768. - Gary W. Adamson, Jul 13 2007

Convolution A130321 * (1, 2, 3, ...). Binomial transform of (1, 3, 4, 4, 4, ...). - Gary W. Adamson, Jul 27 2007

Row sums of triangle A131816. - Gary W. Adamson, Jul 30 2007

A000975 convolved with [1, 2, 2, 2, ...]. - Gary W. Adamson, Jun 02 2009

The eigensequence of a triangle with the triangular series as the left border and the rest 1's. - Gary W. Adamson, Jul 24 2010

Examples

			a(1) = 1 because "1-step Lucas number"(1) = 1.
a(2) = 4 = a(1) + [2-step] Lucas number(2) = 1 + 3.
a(3) = 11 = a(2) + 3-step Lucas number(3) = 1 + 3 + 7.
a(4) = 26 = a(3) + 4-step Lucas number(4) = 1 + 3 + 7 + 15.
a(5) = 57 = a(4) + 5-step Lucas number(5) = 1 + 3 + 7 + 15 + 31.
a(6) = 120 = a(5) + 6-step Lucas number(6) = 1 + 3 + 7 + 15 + 31 + 63.
G.f. = x + 4*x^2 + 11*x^3 + 26*x^4 + 57*x^5 + 120*x^6 + 247*x^7 + 502*x^8 + ...
		

Links

• Vincenzo Librandi, Table of n, a(n) for n = 1..1000
• Dillan Agrawal, Selena Ge, Jate Greene, Tanya Khovanova, Dohun Kim, Rajarshi Mandal, Tanish Parida, Anirudh Pulugurtha, Gordon Redwine, Soham Samanta, and Albert Xu, Chip-Firing on Infinite k-ary Trees, arXiv:2501.06675 [math.CO], 2025. See p. 18.
• Index entries for linear recurrences with constant coefficients, signature (4,-5,2).

Crossrefs

Cf. A000295, A125127, A125129, A130103.

Programs

GAP

List([1..40], n-> 2^(n+1) -n-2); # G. C. Greubel, Jul 26 2019

Magma

I:=[1, 4, 11]; [n le 3 select I[n] else 4*Self(n-1)-5*Self(n-2)+2*Self(n-3): n in [1..40]]; // Vincenzo Librandi, Jun 28 2012

Mathematica

CoefficientList[Series[1/((1-x)^2*(1-2*x)),{x,0,40}],x] (* Vincenzo Librandi, Jun 28 2012 *)
LinearRecurrence[{4,-5,2},{1,4,11},40] (* Harvey P. Dale, Nov 16 2014 *)
a[ n_] := With[{m = n + 1}, If[ m < 0, 0, 2^m - (1 + m)]]; (* Michael Somos, Aug 17 2015 *)

PARI

A125128(n)=2<M. F. Hasler, Jul 30 2015

PARI

{a(n) = n++; if( n<0, 0, 2^n - (1+n))}; /* Michael Somos, Aug 17 2015 */

Sage

[2^(n+1) -n-2 for n in (1..40)] # G. C. Greubel, Jul 26 2019

Formula

a(n) = A000295(n+1) = 2^(n+1) - n - 2 = Sum_{i=1..n} A125127(i,i) = Sum_{i=1..n} ((2^i)-1). [Edited by M. F. Hasler, Jul 30 2015]

From Colin Barker, Jun 17 2012: (Start)

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

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

a(n) = A000225(n) + A000325(n) - 1. - Miquel Cerda, Aug 07 2016

a(n) = A095151(n) - A000225(n). - Miquel Cerda, Aug 12 2016

E.g.f.: 2*exp(2*x) - (2+x)*exp(x). - G. C. Greubel, Jul 26 2019

Extensions

Edited by M. F. Hasler, Jul 30 2015