A125129 -id:A125129 - cp's OEIS frontend

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

Jonathan Vos Post, Nov 22 2006

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

			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 + ...

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).

Cf. A000295, A125127, A125129, A130103.

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

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

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
```

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

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

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

Edited by M. F. Hasler, Jul 30 2015

A125127 Array L(k,n) read by antidiagonals: k-step Lucas numbers.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 3, 4, 1, 1, 3, 7, 7, 1, 1, 3, 7, 11, 11, 1, 1, 3, 7, 15, 21, 18, 1, 1, 3, 7, 15, 26, 39, 29, 1, 1, 3, 7, 15, 31, 51, 71, 47, 1, 1, 3, 7, 15, 31, 57, 99, 131, 76, 1, 1, 3, 7, 15, 31, 63, 113, 191, 241, 123, 1

Offset: 1

Jonathan Vos Post, Nov 21 2006

			Table begins:
1 | 1  1  1   1   1   1    1    1    1    1
2 | 1  3  4   7  11  18   29   47   76  123
3 | 1  3  7  11  21  39   71  131  241  443
4 | 1  3  7  15  26  51   99  191  367  708
5 | 1  3  7  15  31  57  113  223  439  863
6 | 1  3  7  15  31  63  120  239  475  943
7 | 1  3  7  15  31  63  127  247  493  983
8 | 1  3  7  15  31  63  127  255  502 1003
9 | 1  3  7  15  31  63  127  255  511 1013

Freddy Barrera, Table of n, a(n) for n = 1..5050
C. A. Charalambides, Lucas numbers and polynomials of order k and the length of the longest circular success run, The Fibonacci Quarterly, 29 (1991), 290-297.
Eric Weisstein's World of Mathematics, Lucas n-Step Number

n-step Lucas number analog of A092921 Array F(k, n) read by antidiagonals: k-generalized Fibonacci numbers (and see related A048887, A048888). L(1, n) = "1-step Lucas numbers" = A000012. L(2, n) = 2-step Lucas numbers = A000204. L(3, n) = 3-step Lucas numbers = A001644. L(4, n) = 4-step Lucas numbers = A001648 Tetranacci numbers A073817 without the leading term 4. L(5, n) = 5-step Lucas numbers = A074048 Pentanacci numbers with initial conditions a(0)=5, a(1)=1, a(2)=3, a(3)=7, a(4)=15. L(6, n) = 6-step Lucas numbers = A074584 Esanacci ("6-anacci") numbers. L(7, n) = 7-step Lucas numbers = A104621 Heptanacci-Lucas numbers. L(8, n) = 8-step Lucas numbers = A105754. L(9, n) = 9-step Lucas numbers = A105755. See A000295, A125129 for comments on partial sums of diagonals.

Cf. A000012, A000032, A000204, A001644, A001648, A048887, A048888, A074048, A074584, A092921, A104621, A105754, A105755, A125129.

def L(k, n):
    if n < 0:
        return -1
    a = [-1]*(k-1) + [k] # [-1, -1, ..., -1, k]
    for i in range(1, n+1):
        a[:] = a[1:] + [sum(a)]
    return a[-1]
[L(k, n) for d in (1..12) for k, n in zip((d..1, step=-1), (1..d))] # Freddy Barrera, Jan 10 2019

L(k,n) = L(k,n-1) + L(k,n-2) + ... + L(k,n-k); L(k,n) = -1 for n < 0, and L(k,0) = k.

G.f. for row k: x*(dB(k,x)/dx)/(1-B(k,x)), where B(k,x) = x + x^2 + ... + x^k. - Petros Hadjicostas, Jan 24 2019

Corrected by Freddy Barrera, Jan 10 2019

cp's OEIS Frontend

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

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