A015565 -id:A015565 - cp's OEIS frontend

A015577 a(n+1) = 8a(n) + 9a(n-1), a(0) = 0, a(1) = 1.

0, 1, 8, 73, 656, 5905, 53144, 478297, 4304672, 38742049, 348678440, 3138105961, 28242953648, 254186582833, 2287679245496, 20589113209465, 185302018885184, 1667718169966657, 15009463529699912, 135085171767299209, 1215766545905692880, 10941898913151235921

Offset: 0

Olivier Gérard

Binomial transform is A011557, with a leading zero. - Paul Barry, Jul 09 2003
Number of walks of length n between any two distinct nodes of the complete graph K_10. Example: a(2) = 8 because the walks of length 2 between the nodes A and B of the complete graph ABCDEFGHIJ are: ACB, ADB, AEB, AFB, AGB, AHB, AIB and AJB. - Emeric Deutsch, Apr 01 2004
The ratio a(n+1)/a(n) converges to 9 as n approaches infinity. - Felix P. Muga II, Mar 09 2014

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Jean-Paul Allouche, Jeffrey Shallit, Zhixiong Wen, Wen Wu, Jiemeng Zhang, Sum-free sets generated by the period-k-folding sequences and some Sturmian sequences, arXiv:1911.01687 [math.CO], 2019.
Martin Burtscher, Igor Szczyrba, Rafał Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.
Index entries for linear recurrences with constant coefficients, signature (8,9).

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

[Round(9^n/10): n in [0..30]]; // Vincenzo Librandi, Jun 24 2011

seq(round(9^n/10),n=0..25); # Mircea Merca, Dec 28 2010

k=0;lst={k};Do[k=9^n-k;AppendTo[lst, k], {n, 0, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 11 2008 *)
Table[(9^n - (-1)^n)/10, {n,0,30}] (* or *) LinearRecurrence[{8,9}, {0,1}, 30] (* G. C. Greubel, Jan 06 2018 *)

a[0]:0$
a[n]:=9^(n-1)-a[n-1]$
A015577(n):=a[n]$
makelist(A015577(n),n,0,30); /* Martin Ettl, Nov 05 2012 */

A015577_vec(N=20)=Vec(O(x^N)+1/(1-8*x-9*x^2), -N-1) \\ M. F. Hasler, Jun 14 2008, edited Oct 25 2019

for(n=0,30, print1((9^n - (-1)^n)/10, ", ")) \\ G. C. Greubel, Jan 06 2018

apply( {A015577(n)=9^n\/10}, [0..25]) \\ M. F. Hasler, Oct 25 2019

[lucas_number1(n,8,-9) for n in range(0, 19)] # Zerinvary Lajos, Apr 25 2009

From Paul Barry, Jul 09 2003: (Start)
G.f.: x/((1+x)*(1-9*x)).
E.g.f. exp(4*x)*sinh(5*x)/5.
a(n) = (9^n - (-1)^n)/10. (End)
a(n) = 9^(n-1)-a(n-1). - Emeric Deutsch, Apr 01 2004
a(n) = round(9^n/10). - Mircea Merca, Dec 28 2010

Extended by T. D. Noe, May 23 2011

A062160 Square array T(n,k) = (n^k - (-1)^k)/(n+1), n >= 0, k >= 0, read by falling antidiagonals.

Original entry on oeis.org

0, 1, 0, -1, 1, 0, 1, 0, 1, 0, -1, 1, 1, 1, 0, 1, 0, 3, 2, 1, 0, -1, 1, 5, 7, 3, 1, 0, 1, 0, 11, 20, 13, 4, 1, 0, -1, 1, 21, 61, 51, 21, 5, 1, 0, 1, 0, 43, 182, 205, 104, 31, 6, 1, 0, -1, 1, 85, 547, 819, 521, 185, 43, 7, 1, 0, 1, 0, 171, 1640, 3277, 2604, 1111, 300, 57, 8, 1, 0, -1, 1, 341, 4921, 13107, 13021, 6665, 2101, 455, 73, 9, 1, 0

Offset: 0

Henry Bottomley, Jun 08 2001

For n >= 1, T(n, k) equals the number of walks of length k between any two distinct vertices of the complete graph K_(n+1). - Peter Bala, May 30 2024

			From _Seiichi Manyama_, Apr 12 2019: (Start)
Square array begins:
   0, 1, -1,  1,  -1,    1,    -1,      1, ...
   0, 1,  0,  1,   0,    1,     0,      1, ...
   0, 1,  1,  3,   5,   11,    21,     43, ...
   0, 1,  2,  7,  20,   61,   182,    547, ...
   0, 1,  3, 13,  51,  205,   819,   3277, ...
   0, 1,  4, 21, 104,  521,  2604,  13021, ...
   0, 1,  5, 31, 185, 1111,  6665,  39991, ...
   0, 1,  6, 43, 300, 2101, 14706, 102943, ... (End)

Seiichi Manyama, Antidiagonals n = 0..139, flattened
M. Dukes and C. D. White, Web Matrices: Structural Properties and Generating Combinatorial Identities, arXiv:1603.01589 [math.CO], 2016.

Rows include A062157, A000035, A001045, A015518, A015521, A015531, A015540, A015552, A015565, A015577, A015585, A015592, A015609.
Columns include A000004, A000012, A023443, A002061, A062158, A060884, A062159, A060888.
Related to repunits in negative bases (cf. A055129 for positive bases).
Main diagonal gives A081216.
Cf. A109502.

seq(print(seq((n^k - (-1)^k)/(n+1), k = 0..10)), n = 0..10); # Peter Bala, May 31 2024

T[n_,k_]:=(n^k - (-1)^k)/(n+1); Join[{0},Table[Reverse[Table[T[n-k,k],{k,0,n}]],{n,12}]]//Flatten (* Stefano Spezia, Feb 20 2024 *)

T(n, k) = n^(k-1) - n^(k-2) + n^(k-3) - ... + (-1)^(k-1) = n^(k-1) - T(n, k-1) = n*T(n, k-1) - (-1)^k = (n - 1)*T(n, k-1) + n*T(n, k-2) = round[n^k/(n+1)] for n > 1.
T(n, k) = (-1)^(k+1) * resultant( n*x + 1, (x^k-1)/(x-1) ). - Max Alekseyev, Sep 28 2021
G.f. of row n: x/((1+x) * (1-n*x)). - Seiichi Manyama, Apr 12 2019
E.g.f. of row n: (exp(n*x) - exp(-x))/(n+1). - Stefano Spezia, Feb 20 2024
From Peter Bala, May 31 2024: (Start)
Binomial transform of the m-th row: Sum_{k = 0..n} binomial(n, k)*T(m, k) = (m + 1)^(n-1) for n >= 1.
Let R(m, x) denote the g.f. of the m-th row of the square array. Then R(m_1, x) o R(m_2, x) = R(m_1 + m_2 + m_1*m_2, x),  where o denotes the black diamond product of power series as defined by Dukes and White. Cf. A109502.
T(m_1 + m_2 + m_1*m_2, k) = Sum_{i = 0..k} Sum_{j = i..k} binomial(k, i)* binomial(k-i, j-i)*T(m_1, j)*T(m_2, k-i).  (End)

A132805 A trisection of A024495.

Original entry on oeis.org

0, 3, 21, 171, 1365, 10923, 87381, 699051, 5592405, 44739243, 357913941, 2863311531, 22906492245, 183251937963, 1466015503701, 11728124029611, 93824992236885, 750599937895083, 6004799503160661, 48038396025285291, 384307168202282325, 3074457345618258603

Offset: 0

Paul Curtz, Nov 18 2007

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

Cf. A029898.

[-(1/3)*(-1)^n+(1/3)*8^n: n in [0..25]]; // Vincenzo Librandi, Aug 10 2011

LinearRecurrence[{7,8},{0,3},40] (* Harvey P. Dale, Feb 08 2015 *)

From Philippe Deléham, Nov 19 2007: (Start)
a(n) = A132804(n)/2.
G.f.: 3x/(1 - 7*x - 8*x^2).
a(n+1) = 7*a(n) + 8*a(n-1) for n >= 1, a(0)=0, a(1)=3. (End)
a(n) = 3*A015565(n). - R. J. Mathar, Aug 07 2017

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)

A033118 Base 8 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.

Original entry on oeis.org

1, 8, 65, 520, 4161, 33288, 266305, 2130440, 17043521, 136348168, 1090785345, 8726282760, 69810262081, 558482096648, 4467856773185, 35742854185480, 285942833483841, 2287542667870728, 18300341342965825, 146402730743726600

Offset: 1

Clark Kimberling

Partial sums of A015565. - Mircea Merca, Dec 28 2010

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (8,1,-8).

Pairwise sums are (8^n - 1)/7 (A023001).

[Round((8*8^n-8)/63): n in [1..30]]; // Vincenzo Librandi, Jun 25 2011

seq(1/7*floor(8^(n+1)/9),n=1..22); # Mircea Merca, Dec 27 2010

Table[FromDigits[PadRight[{},n,{1,0}],8],{n,20}] (* or *) LinearRecurrence[ {8,1,-8},{1,8,65},20] (* Harvey P. Dale, Jan 20 2021 *)

a(n) = 8*a(n-1) + a(n-2) - 8*a(n-3).
a(n) = 2^(3*n+3)/63 - 1/14 - (-1)^n/18. - R. J. Mathar, Jan 08 2011
From Paul Barry, Apr 04 2008: (Start)
G.f. x/((1-x^2)*(1-8*x));
a(n) = (1/3)*Sum_{k=0..n} A001045(3k). (End)
a(n) = floor(8^(n+1)/9)/7 = floor((8*8^n-1)/63) = round((8*8^n-8)/63) = round((16*8^n-9)/63) = ceiling((8*8^n-8)/63). a(n) = a(n-2) + 8^(n-1), n > 2. - Mircea Merca, Dec 28 2010

A132804 A trisection of A024495.

Original entry on oeis.org

0, 6, 42, 342, 2730, 21846, 174762, 1398102, 11184810, 89478486, 715827882, 5726623062, 45812984490, 366503875926, 2932031007402, 23456248059222, 187649984473770, 1501199875790166, 12009599006321322, 96076792050570582, 768614336404564650

Offset: 0

Paul Curtz, Nov 18 2007

Vincenzo Librandi, Table of n, a(n) for n = 0..300
Index entries for linear recurrences with constant coefficients, signature (7,8).

Cf. A132805, A078008, A015565.

[-(2/3)*(-1)^n+(2/3)*8^n: n in [0..25]]; // Vincenzo Librandi, Jun 08 2011

A132804:=n->-(2/3)*(-1)^n+(2/3)*8^n: seq(A132804(n), n=0..30); # Wesley Ivan Hurt, Apr 14 2017

LinearRecurrence[{7,8},{0,6},30] (* Harvey P. Dale, Mar 29 2018 *)

a(n)=2*(8^n-(-1)^n)/3 \\ Charles R Greathouse IV, Jun 08 2011

G.f.: 6*x/(1-7*x-8*x^2). a(n+1) = 7*a(n)+8*a(n-1) for n>=1, a(0)=0, a(1)=6. - Philippe Deléham, Nov 19 2007
a(n) = 2*A132805(n). - R. J. Mathar, Jun 07 2011
From Oboifeng Dira, Jun 05 2020: (Start)
a(n) = A078008(3n+1). Second trisection of A078008.
a(n) = 6*A015565(n).
a(n) = Sum_{k=0..n} binomial(3*n+1,3*k+2). (End)

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)

A189800 a(n) = 6a(n-1) + 8a(n-2), with a(0)=0, a(1)=1.

Original entry on oeis.org

0, 1, 6, 44, 312, 2224, 15840, 112832, 803712, 5724928, 40779264, 290475008, 2069084160, 14738305024, 104982503424, 747801460736, 5326668791808, 37942424436736, 270267896954880, 1925146777223168, 13713023838978048, 97679317251653632, 695780094221746176

Offset: 0

Vladimir Joseph Stephan Orlovsky, May 24 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (6,8).

Sequences of the form a(n) = c*a(n-1) + d*a(n-2), with a(0)=0, a(1)=1:
c/d...1.......2.......3.......4.......5.......6.......7.......8.......9......10
.A000045,A001045,A006130,A006131,A015440,A015441,A015442,A015443,A015445,A015446
.A000129,A002605,A015518,A063727,A002532,A083099,A015519,A003683,A002534,A083102
.A006190,A007482,A030195,A015521,A015523,A083858,A015524,A015525,A099012,A015528
.A001076,A090017,A015530,A057087,A015531,A085939,A015532,A180222,A015533,A180226
.A052918,A015535,A015536,A015537,A057088,A015540,A015541,A015544,A015545,A180250
.A005668,A135030,A090018,A135032,A015551,A057089,A015552,A189800,A189801,A190005
.A054413,A015555,A015559,A015561,A015562,A015564,A057090,A015565,A015566,A015568
.A041025,A190331,A015574,A190510,A015575,A190560,A015576,A057091,A015577,A190953
.A099371,A015579,A181353,A015580,A015581,A153191,A015583,A015584,A057092,A015585
A041041,A191014,A015588,A190954,A190955,A190956,A015589,A190957,A015591,A057093

I:=[0,1]; [n le 2 select I[n] else 6*Self(n-1)+8*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Nov 14 2011

LinearRecurrence[{6, 8}, {0, 1}, 50]
CoefficientList[Series[-(x/(-1+6 x+8 x^2)),{x,0,50}],x] (* Harvey P. Dale, Jul 26 2011 *)

a(n)=([0,1; 8,6]^n*[0;1])[1,1] \\ Charles R Greathouse IV, Oct 03 2016

G.f.: x/(1 - 2*x*(3+4*x)). - Harvey P. Dale, Jul 26 2011

A099322 An inverse Catalan transform of J(3n)/J(3).

Original entry on oeis.org

0, 1, 6, 43, 291, 1992, 13595, 92845, 633966, 4329023, 29560367, 201850896, 1378323999, 9411785201, 64267689006, 438847231427, 2996636337771, 20462312853336, 139725412120339, 954104794142789, 6515035056168654

Offset: 0

Paul Barry, Nov 17 2004

The g.f. is obtained from that of A015565 through the mapping g(x)->g(x(1-x)). A015565 may be retrieved through the mapping g(x)->g(xc(x)), where c(x) is the g.f. of A000108.

Index entries for linear recurrences with constant coefficients, signature (7,1,-16,8).

Cf. A001045.

LinearRecurrence[{7,1,-16,8},{0,1,6,43},30] (* Harvey P. Dale, Jul 19 2016 *)

G.f.: x(1-x)/(1-7x-x^2+16x^3-8x^4);
a(n) = 7a(n-1) + a(n-2) - 16a(n-3) + 8a(n-4);
a(n) = Sum_{k=0..floor(n/2)} binomial(n-k, k)*(-1)^k*J(3n-3k)/J(3).
a(n) = Sum_{k=0..n} A109466(n,k)*A015565(k). - Philippe Deléham, Oct 30 2008

A242563 a(n) = 2a(n-1) - a(n-3) + 2a(n-4), a(0)=a(1)=0, a(2)=2, a(3)=3.

Original entry on oeis.org

0, 0, 2, 3, 6, 10, 21, 42, 86, 171, 342, 682, 1365, 2730, 5462, 10923, 21846, 43690, 87381, 174762, 349526, 699051, 1398102, 2796202, 5592405, 11184810, 22369622, 44739243, 89478486, 178956970, 357913941, 715827882, 1431655766, 2863311531, 5726623062, 11453246122

Offset: 0

Paul Curtz, May 17 2014

Generally, a(n) is an autosequence if its inverse binomial transform is (-1)^n*a(n). It is of the first kind if the main diagonal is 0's and the first two upper diagonals (just above the main one) are the same. It is of the second kind if the main diagonal is equal to the first upper diagonal multiplied by 2. If the first upper diagonal is an autosequence, the sequence is a super autosequence. Example: A113405. The first upper diagonal is A001045(n). Another super autosequence: 0, 0, 0 followed by A059633(n). The first upper diagonal is A000045(n).
Difference table of a(n):
   0,  0,  2, 3, 6, 10, 21, 42, ...
   0,  2,  1, 3, 4, 11, 21, 44, ...
   2, -1,  2, 1, 7, 10, 23, 41, ...
  -3,  3, -1, 6, 3, 13, 18, 45, ... .
This is an autosequence of the second kind. The main diagonal is 2*A001045(n) = A078008(n). More precisely it is a super autosequence, companion of A113405(n).
a(n+1) mod 10 = period 12: repeat 0, 2, 3, 6, 0, 1, 2, 6, 1, 2, 2, 5.
It is shifted A081374(n+1) mod 10 =
    period 12: repeat 1, 2, 2, 5, 0, 2, 3, 6, 0, 1, 2, 6.
a(n) mod 9 = period 18:
repeat 0, 0, 2, 3, 6, 1, 3, 6, 5, 0, 0, 7, 6, 3, 8, 6, 3, 4 = c(n).
c(n) + c(n+9) = 0, 0, 9, 9, 9, 9, 9, 9, 9.

			G.f. = 2*x^2 + 3*x^3 + 6*x^4 + 10*x^5 + 21*x^6 + 42*x^7 + 86*x^8 + ...

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

Cf. A000032, 1/(n+1), A164555/A027642 (all autosequences of 2nd kind). A007283, A175805.

a[n_] := (m = Mod[n, 6]; 1/3*(2^n + (-1)^n + 1/120*(m-6)*(m+1)*(m^3-29*m+40))); Table[a[n], {n, 0, 35}] (* Jean-François Alcover, May 19 2014, a non-recursive formula, after Mathematica's RSolve *)
LinearRecurrence[{2, 0, -1, 2}, {0, 0, 2, 3},50] (* G. C. Greubel, Feb 21 2017 *)

concat([0,0], Vec(x^2*(x-2)/((x+1)*(2*x-1)*(x^2-x+1)) + O(x^100))) \\ Colin Barker, May 18 2014

a(n+3) = 3*2^n - a(n), a(0)=a(1)=0, a(2)=2.
a(n) = 2*A113405(n+1) - A113405(n).
a(n+1) = 2*a(n) + period 6: repeat 0, 2, -1, 0, -2, 1. a(0)=0.
a(n) = 2^n - A081374(n+1).
a(n+3) = a(n+1) + A130755(n).
G.f.: x^2*(x-2) / ((x+1)*(2*x-1)*(x^2-x+1)). - Colin Barker, May 18 2014
a(n) = A024495(n) + A131531(n).
a(n+6) = a(n) + 21*2^n, a(0)=a(1)=0, a(2)=2, a(3)=3, a(4)=6, a(5)=10.
a(n) = A001045(n) - A092220(n).
a(n+12) = a(n) + 1365*2^n. First 12 values in the Data. (A024495(n+12) = A024495(n) + 1365*2^n).
a(3n)   = A132805(n) = 3*A015565(n).
a(3n+1) = A132804(n) = 6*A015565(n).
a(3n+2) = A132397(n) = 2*A082311(n).
a(n) = 1/3*((-1)^n - 2*cos((n*Pi)/3) + 2^n). - Alexander R. Povolotsky,  Jun 02 2014

More terms from Colin Barker, May 18 2014

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A015577 a(n+1) = 8*a(n) + 9*a(n-1), a(0) = 0, a(1) = 1.

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A062160 Square array T(n,k) = (n^k - (-1)^k)/(n+1), n >= 0, k >= 0, read by falling antidiagonals.

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A132805 A trisection of A024495.

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A093134 A Jacobsthal trisection.

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A033118 Base 8 digits are, in order, the first n terms of the periodic sequence with initial period 1,0.

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A132804 A trisection of A024495.

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

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A015577 a(n+1) = 8a(n) + 9a(n-1), a(0) = 0, a(1) = 1.

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

A189800 a(n) = 6a(n-1) + 8a(n-2), with a(0)=0, a(1)=1.

A242563 a(n) = 2a(n-1) - a(n-3) + 2a(n-4), a(0)=a(1)=0, a(2)=2, a(3)=3.