A015518 -id:A015518 - cp's OEIS frontend

A202064 Triangle T(n,k), read by rows, given by (2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

1, 2, 0, 3, 1, 0, 4, 4, 0, 0, 5, 10, 1, 0, 0, 6, 20, 6, 0, 0, 0, 7, 35, 21, 1, 0, 0, 0, 8, 56, 56, 8, 0, 0, 0, 0, 9, 84, 126, 36, 1, 0, 0, 0, 0, 10, 120, 252, 120, 10, 0, 0, 0, 0, 0, 11, 165, 462, 330, 55, 1, 0, 0, 0, 0, 0

Offset: 0

Philippe Deléham, Dec 10 2011

Riordan array (x/(1-x)^2, x^2/(1-x)^2).
Mirror image of triangle in A119900.
A203322*A130595 as infinite lower triangular matrices. - Philippe Deléham, Jan 05 2011
From Gus Wiseman, Jul 07 2025: (Start)
Also the number of subsets of {1..n} containing n with k maximal runs (sequences of consecutive elements increasing by 1). For example, row n = 5 counts the following subsets:
  {5}          {1,5}      {1,3,5}
  {4,5}        {2,5}
  {3,4,5}      {3,5}
  {2,3,4,5}    {1,2,5}
  {1,2,3,4,5}  {1,4,5}
               {2,3,5}
               {2,4,5}
               {1,2,3,5}
               {1,2,4,5}
               {1,3,4,5}
For anti-runs instead of runs we have A053538.
Without requiring n see A210039, A202023, reverse A098158, A109446.
(End)

			Triangle begins :
1
2, 0
3, 1, 0
4, 4, 0, 0
5, 10, 1, 0, 0
6, 20, 6, 0, 0, 0
7, 35, 21, 1, 0, 0, 0
8, 56, 56, 8, 0, 0, 0, 0

Cf. A007318, A005314 (antidiagonal sums), A119900, A084938, A130595, A203322.
Column k = 1 is A000027.
Row sums are A000079.
Column k = 2 is A000292.
Without zeros we have A034867.
Last nonzero term in each row appears to be A124625.
A034839 counts subsets by number of maximal runs, for anti-runs A384893.
A116674 counts strict partitions by number of maximal runs, for anti-runs A384905.
Cf. A000045, A000071, A001629, A010027, A053538, A208342, A210034, A245563, A268193, A384177, A384890.
Cf. A098158, A109446, A202023, A210039.

Table[Length[Select[Subsets[Range[n]],MemberQ[#,n]&&Length[Split[#,#2==#1+1&]]==k&]],{n,12},{k,n}] (* Gus Wiseman, Jul 07 2025 *)

G.f.: 1/((1-x)^2-y*x^2).
Sum_{k, 0<=k<=n} T(n,k)*x^k = A000027(n+1), A000079(n), A000129(n+1), A002605(n+1), A015518(n+1), A063727(n), A002532(n+1), A083099(n+1), A015519(n+1), A003683(n+1), A002534(n+1), A083102(n), A015520(n+1), A091914(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 10, 11, 12, 13 respectively.
T(n,k) = binomial(n+1,2k+1).
T(n,k) = 2*T(n-1,k) + T(n-2,k-1) - T(n-2,k), T(0,0) = 1, T(1,0) = 2, T(1,1) = 0 and T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Mar 15 2012

A238801 Triangle T(n,k), read by rows, given by T(n,k) = C(n+1, k+1)*(1-(k mod 2)).

Original entry on oeis.org

1, 2, 0, 3, 0, 1, 4, 0, 4, 0, 5, 0, 10, 0, 1, 6, 0, 20, 0, 6, 0, 7, 0, 35, 0, 21, 0, 1, 8, 0, 56, 0, 56, 0, 8, 0, 9, 0, 84, 0, 126, 0, 36, 0, 1, 10, 0, 120, 0, 252, 0, 120, 0, 10, 0, 11, 0, 165, 0, 462, 0, 330, 0, 55, 0, 1, 12, 0, 220, 0, 792, 0, 792, 0, 220, 0, 12, 0

Offset: 0

Philippe Deléham, Mar 05 2014

Row sums are powers of 2.

			Triangle begins:
1;
2, 0;
3, 0, 1;
4, 0, 4, 0;
5, 0, 10, 0, 1;
6, 0, 20, 0, 6, 0;
7, 0, 35, 0, 21, 0, 1;
8, 0, 56, 0, 56, 0, 8, 0;
9, 0, 84, 0, 126, 0, 36, 0, 1;
10, 0, 120, 0, 252, 0, 120, 0, 10, 0; etc.

G. C. Greubel, Table of n, a(n) for the first 100 rows, flattened

Cf. Columns: A000027, A000292, A000389, A000580, A000582, A001288, A010966, A010968, A010970, A010972, A010974, A010976, A010980, A010982, A010984, A010986, A010988, A010990, A010992, A010994, A010996, A010998, A011000, A017713, A017715, A017717, A017719, A017721, A017723, A017725, A017727, A017729, A017731, A017733, A017735, A017737, A017739, A017741, A017743, A017745, A017747, A017749, A017751, A017753, A017755, A017757, A017759, A017761, A017763.

Cf. A095704, A178616.

Table[Binomial[n + 1, k + 1]*(1 - Mod[k , 2]), {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Nov 22 2017 *)

T(n,k) = binomial(n+1, k+1)*(1-(k % 2));
tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n,k), ", ")); print); \\ Michel Marcus, Nov 23 2017

G.f.: 1/((1+(y-1)*x)*(1-(y+1)*x)).
T(n,k) = 2*T(n-1,k) + T(n-2,k-2) - T(n-2,k), T(0,0) = 1, T(1,0) = 2, T(1,1) = 0, T(n,k) = 0 if k<0 or if k>n.
Sum_{k=0..n} T(n,k)*x^k = A000027(n+1), A000079(n), A015518(n+1), A003683(n+1), A079773(n+1), A051958(n+1), A080920(n+1), A053455(n), A160958(n+1) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively.

A093134 A Jacobsthal trisection.

Original entry on oeis.org

1, 0, 8, 56, 456, 3640, 29128, 233016, 1864136, 14913080, 119304648, 954437176, 7635497416, 61083979320, 488671834568, 3909374676536, 31274997412296, 250199979298360, 2001599834386888, 16012798675095096, 128102389400760776, 1024819115206086200, 8198552921648689608

Offset: 0

Paul Barry, Mar 23 2004

Counts closed walks at a vertex of the complete graph on 9 nodes K_9.

Second binomial transform is A047855.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (7,8).

Other sequences with a(n+1) = 8^n - a(n) are A001045, A078008, A097073, A115341, A015518, A054878, A015521, A109499, A015531, A109500, A109501, A015552, A015565. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008

Cf. A047855.

[(8^n/9+8*(-1)^n/9): n in [0..20]]; // Vincenzo Librandi, Oct 11 2011

k=0;lst={1, k};Do[k=8^n-k;AppendTo[lst, k], {n, 1, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 11 2008 *)
Table[(8^n + 8*(-1)^n)/9, {n,0,30}] (* or *) LinearRecurrence[{7,8}, {1,0}, 30] (* G. C. Greubel, Jan 06 2018 *)

for(n=0,30, print1((8^n + 8*(-1)^n)/9, ", ")) \\ G. C. Greubel, Jan 06 2018

G.f.: (1-7*x)/(1 - 7*x - 8*x^2).
a(n) = (8^n + 8*(-1)^n)/9.
a(n) = 8*A001045(3*n-3)/3.
From Elmo R. Oliveira, Aug 17 2024: (Start)
E.g.f.: exp(-x)*(exp(9*x) + 8)/9.
a(n) = 7*a(n-1) + 8*a(n-2) for n > 1. (End)

A154692 Triangle read by rows: T(n, k) = (2^(n-k)3^k + 2^k3^(n-k))*binomial(n, k).

Original entry on oeis.org

2, 5, 5, 13, 24, 13, 35, 90, 90, 35, 97, 312, 432, 312, 97, 275, 1050, 1800, 1800, 1050, 275, 793, 3492, 7020, 8640, 7020, 3492, 793, 2315, 11550, 26460, 37800, 37800, 26460, 11550, 2315, 6817, 38064, 97776, 157248, 181440, 157248, 97776, 38064, 6817

Offset: 0

Roger L. Bagula and Gary W. Adamson, Jan 14 2009

			Triangle begins
     2;
     5,     5;
    13,    24,    13;
    35,    90,    90,     35;
    97,   312,   432,    312,     97;
   275,  1050,  1800,   1800,   1050,    275;
   793,  3492,  7020,   8640,   7020,   3492,   793;
  2315, 11550, 26460,  37800,  37800,  26460, 11550,  2315;
  6817, 38064, 97776, 157248, 181440, 157248, 97776, 38064, 6817;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
A. Lakhtakia, R. Messier, V. K. Varadan, and V. V. Varadan, Use of combinatorial algebra for diffusion on fractals, Physical Review A, volume 34, Number 3 (1986) p. 2502, Fig. 3.

Cf. A000225, A007482, A013620, A088137, A119309.
Sums include: A010673 (alternating sign row), A020699 (row), A020729 (row).
Related sequences: A007318, A154690,

A154692:= func< n,k | (2^(n-k)*3^k + 2^k*3^(n-k))*Binomial(n,k) >;
[A154692(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jan 18 2025

A154692 := proc(n,m)
        (2^(n-m)*3^m+2^m*3^(n-m))*binomial(n,m) ;
end proc:
seq(seq(A154692(n,m),m=0..n),n=0..10) ; # R. J. Mathar, Oct 24 2011

p=2; q=3;
T[n_, m_]= (p^(n-m)*q^m + p^m*q^(n-m))*Binomial[n,m];
Table[T[n,m], {n,0,10}, {m,0,n}]//Flatten

from sage.all import *
def A154692(n,k): return (pow(2,n-k)*pow(3,k)+pow(2,k)*pow(3,n-k))*binomial(n,k)
print(flatten([[A154692(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Jan 18 2025

Sum_{k=0..n} T(n, k) = A020729(n) = A020699(n+1).
T(n,m) = A013620(n,m) + A013620(m,n). - R. J. Mathar, Oct 24 2011
From G. C. Greubel, Jan 18 2025: (Start)
T(2*n, n) = A119309(n).
Sum_{k=0..n} (-1)^k*T(n, k) = A010673(n+1).
Sum_{k=0..floor(n/2)} T(n-k, k) = A015518(n+1) + A007482(n).
Sum_{k=0..floor(n/2)} (-1)^k*T(n-k, k) = A088137(n+1) + A000225(n+1). (End)

A203362 T(n,k)=Number of nXk binary arrays with every 1 immediately preceded by 0 to the left or above.

Original entry on oeis.org

1, 2, 2, 3, 7, 3, 5, 20, 20, 5, 8, 61, 102, 61, 8, 13, 182, 565, 565, 182, 13, 21, 547, 3042, 5677, 3042, 547, 21, 34, 1640, 16538, 55517, 55517, 16538, 1640, 34, 55, 4921, 89610, 547897, 984703, 547897, 89610, 4921, 55, 89, 14762, 486103, 5390325, 17631496

Offset: 1

R. H. Hardin Dec 31 2011

Table starts
..1....2......3........5..........8...........13.............21
..2....7.....20.......61........182..........547...........1640
..3...20....102......565.......3042........16538..........89610
..5...61....565.....5677......55517.......547897........5390325
..8..182...3042....55517.....984703.....17631496......314688054
.13..547..16538...547897...17631496....572818585....18550312526
.21.1640..89610..5390325..314688054..18550312526..1089970014327
.34.4921.486103.53087833.5622720187.601393482628.64114204398028

			Some solutions for n=5 k=3
..0..1..0....0..0..1....0..1..0....0..1..0....0..0..1....0..0..0....0..0..1
..0..0..0....0..0..0....1..0..0....0..0..1....0..0..1....1..0..0....1..0..0
..0..0..0....0..0..0....0..0..1....0..0..0....0..1..0....0..1..1....0..0..0
..0..0..0....0..1..1....1..0..0....1..1..1....0..0..1....0..0..1....0..1..1
..1..1..0....0..1..0....0..0..1....0..0..0....0..1..0....1..0..1....0..0..1

R. H. Hardin, Table of n, a(n) for n = 1..1011

Column 1 is A000045(n+1)

Column 2 is A015518(n+1)

A152011 a(0) = 1 and a(n) = (3^n - (-1)^n)/2 for n >= 1.

Original entry on oeis.org

1, 2, 4, 14, 40, 122, 364, 1094, 3280, 9842, 29524, 88574, 265720, 797162, 2391484, 7174454, 21523360, 64570082, 193710244, 581130734, 1743392200, 5230176602, 15690529804, 47071589414, 141214768240, 423644304722

Offset: 0

Roger L. Bagula and Gary W. Adamson, Nov 19 2008

Previous name was: A product form based on the Fibonacci product form: f(n) = 2^n*Product_{k=1..floor((n-1)/2)} (1 + 3*cos(k*Pi/n)^2).
Gary W. Adamson found this article, I experimented. Based on the paper Fibonacci identity of: f(n) = Product_{k=1..floor((n-1)/2)} (1 + 4*cos(k*Pi/n)^2). I changed the 4 to a 3 and used 2^n to get rid of the rational terms. The product comes down slow in Mathematica: I tried 30 but no luck.
For n > 0, Select an odd size subset S of {1, 2, ..., n}, then select a subset of S. - Geoffrey Critzer, Mar 03 2010
It appears that if s(n) is a first order rational sequence of the form s(1) = 2, s(n) = (s(n-1) + 2)/(2*s(n-1) + 1), n > 1 then s(n) = a(n)/(a(n) + (-1)^n). - Gary Detlefs, Nov 16 2010
For n >= 1, a(n) counts closed walks of length n + 1 on the vertex of a triangle to which two loops have been added to one of remaining vertices. - David Neil McGrath, Sep 04 2014

			G.f. = 1 + 2*x + 4*x^2 + 14*x^3 + 40*x^4 + 122*x^5 + 364*x^6 + 1094*x^7 + ...

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Dhroova Aiylam and Tanya Khovanova, Weighted Mediants and Fractals, arXiv:1711.01475 [math.NT], 2017. See p. 18.
M. H. Albert and R. Brignall, Enumerating indices of Schubert varieties defined by inclusions, arXiv:1301.3188 [math.CO], 2013. - From _N. J. A. Sloane_, Feb 11 2013
N. Garnier and O. Ramaré, Fibonacci numbers and trigonometric identities
N. Garnier and O. Ramaré, Fibonacci numbers and trigonometric identities, Fibonacci Quart. 46/47 (2008/2009), no. 1, 56-61.
Index entries for linear recurrences with constant coefficients, signature (2,3).

Cf. A000045.

A152011 = 2*A015518, except for the first term. [From Geoffrey Critzer, Mar 03 2010; corrected by M. F. Hasler, Nov 16 2010]

[1] cat [(3^n-(-1)^n)/2: n in [1..30]]; // Vincenzo Librandi, Sep 15 2014

f[n_] = 2^n Product[(1 + 3 Cos[k Pi/n]^2), {k, Floor[(n - 1)/2]}]; Table[FullSimplify[ExpandAll[f[n]]], {n, 0, 15}]
(* Second program: *)
CoefficientList[Series[(1-3x^2)/((1+x)(1-3x)), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 15 2014 *)
Join[{1}, LinearRecurrence[{2, 3}, {2, 4}, 30]] (* Jean-François Alcover, Jan 07 2019 *)

a(n)=floor(2^n*prod(k=1,floor((n-1)/2),1+3*cos(k*Pi/n)^2)+1/2) \\ Edward Jiang, Sep 08 2014

a(n)=if(n,(3^n-(-1)^n)/2,1) \\ Charles R Greathouse IV, Sep 15 2014

def A152011(n) :
    if n == 0 : return 1
    return add(2^(n-k)*binomial(n,k) for k in range(n)[::2])  # Peter Luschny, Jul 30 2012

a(n) = 2^n * Product_{k = 1..floor((n-1)/2)} (1 + 3 * cos(k * Pi/n)^2).
From Geoffrey Critzer, Mar 03 2010: (Start)
For n > 0, a(n) = Sum_{k = 1, 3, 5, ...} C(n, k)* 2^k.
E.g.f.: 1 + sinh(2*x)*exp(x). (End)
From R. J. Mathar, Mar 11 2010: (Start)
a(n) = (3^n - (-1)^n)/2, n > 0.
G.f.: (1 - 3*x^2)/((1 + x)*(1 - 3*x)). (End)
a(n) = 2*a(n-1) + 3*a(n-2) for n >= 2, a(1) = 2, and a(2) = 4. - David Neil McGrath, Sep 04 2014
a(n) = M^n[1,2] = M^n[2,1] for n>0, where M = [1,2;2,1]. - Rigoberto Florez, May 05 2020

Terms a(16)-a(25) from Peter Luschny, Jul 30 2012

New name (using R. J. Mathar's formula) by Joerg Arndt, Sep 09 2014

A054880 a(n) = 3*(9^n - 1)/4.

Original entry on oeis.org

0, 6, 60, 546, 4920, 44286, 398580, 3587226, 32285040, 290565366, 2615088300, 23535794706, 211822152360, 1906399371246, 17157594341220, 154418349070986, 1389765141638880, 12507886274749926, 112570976472749340, 1013138788254744066, 9118249094292696600, 82064241848634269406, 738578176637708424660

Offset: 0

Paolo Dominici (pl.dm(AT)libero.it), May 23 2000

Number of walks of length 2n+1 along the edges of a (3 dimensional) cube between two opposite vertices.

Urn A initially contains 3 labeled balls while urn B is empty. A ball is randomly selected and switched from one urn to the other. a(n)/3^(2n+1) is the probability that urn A is empty after 2n+1 switches. - Geoffrey Critzer, May 23 2013

G. C. Greubel, Table of n, a(n) for n = 0..1000
G. Benkart and D. Moon, A Schur-Weyl Duality Approach to Walking on Cubes, arXiv preprint arXiv:1409.8154 [math.RT], 2014 and Ann. Combin. 20 (3) (2016) 397-417
R. J. Mathar, Counting Walks on Finite Graphs, Nov 2020, Section 5.
Index entries for linear recurrences with constant coefficients, signature (10,-9).

Cf. A002452, A015518, A046717, A066443.

List([0..30], n-> 3*(9^n -1)/4); # G. C. Greubel, Jul 14 2019

[3*(9^n -1)/4: n in [0..30]]; // G. C. Greubel, Jul 14 2019

Table[(2 n + 1)! Coefficient[Series[Sinh[x]^3, {x, 0, 2 n + 1}],
x^(2 n + 1)], {n, 0, 30}]  (* Geoffrey Critzer, May 23 2013 *)
LinearRecurrence[{10,-9},{0,6},30] (* Harvey P. Dale, Sep 17 2024 *)

vector(30, n, n--; 3*(9^n -1)/4) \\ G. C. Greubel, Jul 14 2019

[3*(9^n -1)/4 for n in (0..30)] # G. C. Greubel, Jul 14 2019

G.f.: (3/4)/(1 - 9*x) - (3/4)/(1 - x).
a(n) = 6*A002452(n).
sin(x)^3 = Sum_{k>=0} (-1)^(k+1)*a(k)*x^(2k+1)/(2k+1)!. - Dan Fux (dan.fux(AT)OpenGaia.com or danfux(AT)OpenGaia.com), Apr 08 2001
a(n) = A015518(2n+1) - 1 = (A046717(2n+1) - 1)/2. - M. F. Hasler, Mar 20 2008
a(n) = 9*a(n-1) + 6 with n > 0, a(0) = 0. - Vincenzo Librandi, Aug 07 2010
a(n) = A066443(n) - 1. - Georg Fischer, Nov 25 2018
E.g.f.: 3*(exp(9*x) - exp(x))/4. - G. C. Greubel, Jul 14 2019
a(n) = 10*a(n-1) - 9*a(n-2) with a(0) = 0 and a(1) = 6. - Miquel A. Fiol, Mar 09 2024

A137241 Number triples (k,3-k,2-2k), concatenated for k=0, 1, 2, 3,...

Original entry on oeis.org

0, 3, 2, 1, 2, 0, 2, 1, -2, 3, 0, -4, 4, -1, -6, 5, -2, -8, 6, -3, -10, 7, -4, -12, 8, -5, -14, 9, -6, -16, 10, -7, -18, 11, -8, -20, 12, -9, -22, 13, -10, -24, 14, -11, -26, 15, -12, -28, 16, -13, -30, 17, -14, -32, 18, -15, -34, 19, -16, -36, 20, -17, -38, 21, -18, -40

Offset: 0

Paul Curtz, Mar 09 2008

The entries are the coefficients in a family of Jacobsthal recurrences: a(n)=k*a(n-1)+(3-k)*a(n-2)+(2-2k)*a(n-3).
Examples for k=0 are in A001045 and A113954. Examples for k=1 are A001045, A078008.
Examples for k=2 are A000975, A087288, A084639, A000012 and A001045.
Examples for k=3 are A045883, A059570. Examples for k=4 are A094705 and A015518.

			The triples (k,3-k,2-2k) are (0,3,2), (1,2,0), (2,1,-2), (3,0,-4),...

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

CoefficientList[Series[x*(3 + 2*x + x^2 - 4*x^3 - 4*x^4)/((x - 1)^2*(1 + x + x^2)^2), {x, 0, 50}], x] (* G. C. Greubel, Sep 28 2017 *)
Table[{n,3-n,2-2n},{n,0,30}]//Flatten (* or *) LinearRecurrence[ {0,0,2,0,0,-1},{0,3,2,1,2,0},100] (* Harvey P. Dale, Jun 23 2019 *)

x='x+O('x^50); Vec(x*(3+2*x+x^2-4*x^3-4*x^4)/((x-1)^2*(1+x +x^2 )^2)) \\ G. C. Greubel, Sep 28 2017

From R. J. Mathar, Feb 25 2009: (Start)
a(n) = 2*a(n-3) - a(n-6).
G.f.: x*(3+2*x+x^2-4*x^3-4*x^4)/((x-1)^2*(1+x+x^2)^2). (End)

Edited by R. J. Mathar, Jun 28 2008

A015592 a(n) = 10a(n-1) + 11a(n-2).

Original entry on oeis.org

0, 1, 10, 111, 1220, 13421, 147630, 1623931, 17863240, 196495641, 2161452050, 23775972551, 261535698060, 2876892678661, 31645819465270, 348104014117971, 3829144155297680, 42120585708274481, 463326442791019290, 5096590870701212191, 56062499577713334100

Offset: 0

Olivier Gérard

Number of walks of length n between any two distinct nodes of the complete graph K_12. Example: a(2)=10 because the walks of length 2 between the nodes A and B of the complete graph ABCDEFGHIJKL are ACB, ADB, AEB, AFB, AGB, AHB, AIB, AJB, AKB and ALB. - Emeric Deutsch, Apr 01 2004

General form: k=11^n-k. Also: A001045, A078008, A097073, A115341, A015518, A054878, A015521, A109499, A015531, A109500, A109501, A015552, A093134, A015565, A015577, A015585. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008

Vincenzo Librandi, Table of n, a(n) for n = 0..900
Index entries for linear recurrences with constant coefficients, signature (10,11).

Cf. A001045, A078008, A097073, A115341, A015518, A054878, A015521, A109499, A015531, A109500, A109501, A015552, A093134, A015565, A015577, A015585. - Vladimir Joseph Stephan Orlovsky, Dec 11 2008

[-(1/12)*(-1)^n+(1/12)*11^n: n in [0..20]]; // Vincenzo Librandi, Oct 11 2011

k=0;lst={k};Do[k=11^n-k;AppendTo[lst, k], {n, 0, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 11 2008 *)

[lucas_number1(n,10,-11) for n in range(0, 18)] # Zerinvary Lajos, Apr 26 2009

a(n) = 11^(n-1) - a(n-1). G.f.: x/(1 - 10x - 11x^2). - Emeric Deutsch, Apr 01 2004
From Elmo R. Oliveira, Aug 17 2024: (Start)
E.g.f.: exp(5*x)*sinh(6*x)/6.
a(n) = (11^n - (-1)^n)/12. (End)

A051068 Partial sums of A014578.

Original entry on oeis.org

0, 1, 2, 2, 3, 4, 4, 5, 6, 7, 8, 9, 9, 10, 11, 11, 12, 13, 14, 15, 16, 16, 17, 18, 18, 19, 20, 20, 21, 22, 22, 23, 24, 24, 25, 26, 27, 28, 29, 29, 30, 31, 31, 32, 33, 34, 35, 36, 36, 37, 38, 38, 39, 40, 40, 41, 42, 42, 43, 44, 44, 45, 46, 47, 48, 49, 49

Offset: 0

N. J. A. Sloane, Gary W. Adamson

nonn

Duplicate of A050294? [Joerg Arndt, Apr 27 2013]
From Michel Dekking, Feb 10 2019: (Start)
The answer to Joerg Arndt's question is: yes (modulo an offset).  To see this, it suffices to prove that the two sequences of first differences Da and Db of a= A051068 and b:=A050294 are equal. Clearly the sequence Da of first differences of a is the sequence A014578. According to Philippe Deleham (2004), Da equals 0x = 0110110111110..., where x is the fixed point of the morphism 0->111, 1->110.
From Vladimir Shevelev (2011) we know a formula for b=A050294: b(n) = n-b(floor(n/3)). This gives that the sequence of first differences Db:=(b(n+1)-b(n)) of b satisfies
      Db(3m+1) = Db(3m+2) = 1, and Db(3m+3) = 1 - Db(m).
This implies that Db = x, the fixed point of 0->111, 1->110.
(End)

Cf. A014578, A051069, A050294.

a(3^n) = A015518(n+1) = -(-1)^n*A014983(n+1). - Philippe Deléham, Mar 31 2004

cp's OEIS Frontend

A202064 Triangle T(n,k), read by rows, given by (2, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1/2, -1/2, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

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A238801 Triangle T(n,k), read by rows, given by T(n,k) = C(n+1, k+1)*(1-(k mod 2)).

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A093134 A Jacobsthal trisection.

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A154692 Triangle read by rows: T(n, k) = (2^(n-k)*3^k + 2^k*3^(n-k))*binomial(n, k).

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A203362 T(n,k)=Number of nXk binary arrays with every 1 immediately preceded by 0 to the left or above.

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A152011 a(0) = 1 and a(n) = (3^n - (-1)^n)/2 for n >= 1.

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A054880 a(n) = 3*(9^n - 1)/4.

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A154692 Triangle read by rows: T(n, k) = (2^(n-k)3^k + 2^k3^(n-k))*binomial(n, k).

A015592 a(n) = 10a(n-1) + 11a(n-2).