A129339 -id:A129339 - cp's OEIS frontend

A057083 Scaled Chebyshev U-polynomials evaluated at sqrt(3)/2; expansion of 1/(1 - 3x + 3x^2).

1, 3, 6, 9, 9, 0, -27, -81, -162, -243, -243, 0, 729, 2187, 4374, 6561, 6561, 0, -19683, -59049, -118098, -177147, -177147, 0, 531441, 1594323, 3188646, 4782969, 4782969, 0, -14348907, -43046721, -86093442, -129140163, -129140163, 0

Offset: 0

Wolfdieter Lang, Aug 11 2000

With different sign pattern, see A000748.

Conjecture: Let M be any endomorphism on any vector space, such that M^3 = 1 (identity). Then (1-M)^n = A057681(n) - A057682(n)*M + z(n)*M^2, where z(0) = z(1) = 0 and, apparently, z(n+2) = a(n). - Stanislav Sykora, Jun 10 2012

Robert Israel, Table of n, a(n) for n = 0..4170
T. Alden Gassert, Discriminants of simplest 3^n-tic extensions, arXiv preprint arXiv:1409.7829 [math.NT], 2014.
A. F. Horadam, Special properties of the sequence W_n(a,b; p,q), Fib. Quart., 5.5 (1967), 424-434. Case n->n+1, a=0,b=1; p=3, q=-3.
Vladimir Kruchinin, Composition of ordinary generating functions, arXiv:1009.2565 [math.CO], 2010.
Wolfdieter Lang, On polynomials related to powers of the generating function of Catalan's numbers, Fib. Quart. 38 (2000) 408-419. Eqs. (38) and (45),lhs, m=3.
Vladimir Shevelev, Combinatorial identities generated by difference analogs of hyperbolic and trigonometric functions of order n, arXiv:1706.01454 [math.CO], 2017.
Index entries for sequences related to Chebyshev polynomials.
Index entries for linear recurrences with constant coefficients, signature (3,-3).

Cf. A000748, A010892, A049310, A057078, A057681, A057682, A129339.

I:=[1,3]; [n le 2 select I[n] else 3*Self(n-1) - 3*Self(n-2): n in [1..30]]; // G. C. Greubel, Oct 23 2018

seq(3^(n/2)*orthopoly[U](n,sqrt(3)/2),n=0..100); # Robert Israel, Nov 21 2016

Join[{a=1,b=3},Table[c=3*b-3*a;a=b;b=c,{n,100}]] (* Vladimir Joseph Stephan Orlovsky, Jan 17 2011 *)
CoefficientList[Series[1/(1 - 3 x + 3 x^2), {x, 0, 35}], x] (* Michael De Vlieger, Jul 30 2017 *)

a(n)=([0,1; -3,3]^n*[1;3])[1,1] \\ Charles R Greathouse IV, Apr 08 2016

[lucas_number1(n,3,3) for n in range(1, 37)] # Zerinvary Lajos, Apr 23 2009

a(n) = S(n, sqrt(3))*(sqrt(3))^n with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310.
a(2*n) = A057078(n)*3^n; a(2*n+1)= A010892(n)*3^(n+1).
G.f.: 1/(1-3*x+3*x^2).
Binomial transform of A057079. a(n) = Sum_{k=0..n} 2*binomial(n, k)*cos((k-1)Pi/3). - Paul Barry, Aug 19 2003
For n > 5, a(n) = -27*a(n-6) - Gerald McGarvey, Apr 21 2005
a(n) = Sum_{k=0..n} A109466(n,k)*3^k. - Philippe Deléham, Nov 12 2008
a(n) = Sum_{k=1..n} binomial(k,n-k) * 3^k *(-1)^(n-k) for n>0; a(0)=1. - Vladimir Kruchinin, Feb 07 2011
By the conjecture: Start with x(0)=1, y(0)=0, z(0)=0 and set x(n+1) = x(n) - z(n), y(n+1) = y(n) - x(n), z(n+1) = z(n) - y(n). Then a(n) = z(n+2). This recurrence indeed ends up in a repetitive cycle of length 6 and multiplicative factor -27, confirming G. McGarvey's observation. - Stanislav Sykora, Jun 10 2012
G.f.: Q(0) where Q(k) =  1 + k*(3*x+1) + 9*x - 3*x*(k+1)*(k+4)/Q(k+1) ; (continued fraction). - Sergei N. Gladkovskii, Mar 15 2013
G.f.: G(0)/(2-3*x), where G(k)= 1 + 1/(1 - x*(k+3)/(x*(k+4) + 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 16 2013
a(n) = Sum_{k = 0..floor(n/3)} (-1)^k*binomial(n+2,3*k+2). Sykora's conjecture in the Comments section follows easily from this. - Peter Bala, Nov 21 2016
From Vladimir Shevelev, Jul 30 2017: (Start)
a(n) = 2*3^(n/2)*cos(Pi*(n-2)/6);
a(n) = K_2(n+2) - K_1(n+2);
For m,n>=1, a(n+m) = a(n-1)*K_1(m+1) + K_2(n+1)*K_2(m+1) + K_1(n+1)*a(m-1) where K_1 = A057681, K_2 = A057682. (End)

A131022 Triangular array T read by rows: T(j,1) = 1 for ((j-1) mod 6) < 3, else 0; T(j,k) = T(j-1,k-1) + T(j,k-1) for 2 <= k <= j.

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 0, 1, 3, 7, 0, 0, 1, 4, 11, 0, 0, 0, 1, 5, 16, 1, 1, 1, 1, 2, 7, 23, 1, 2, 3, 4, 5, 7, 14, 37, 1, 2, 4, 7, 11, 16, 23, 37, 74, 0, 1, 3, 7, 14, 25, 41, 64, 101, 175, 0, 0, 1, 4, 11, 25, 50, 91, 155, 256, 431, 0, 0, 0, 1, 5, 16, 41, 91, 182, 337, 593, 1024

Offset: 1

Klaus Brockhaus, following a suggestion of Paul Curtz, Jun 10 2007

All columns are periodic with period length 6. The (3+6*i)-th row equals the first (3+6*i) terms of main diagonal (i >= 0).

			First seven rows of T are
[ 1 ]
[ 1, 2 ]
[ 1, 2, 4 ]
[ 0, 1, 3, 7 ]
[ 0, 0, 1, 4, 11 ]
[ 0, 0, 0, 1, 5, 16 ]
[ 1, 1, 1, 1, 2, 7, 23 ].

Michel Marcus, Rows n = 1..100 of triangle, flattened

Cf. A129339 (main diagonal of T), A131023 (first subdiagonal of T), A131024 (row sums of T), A131025 (antidiagonal sums of T). First through sixth column of T are in A088911, A131026, A131027, A131028, A131029, A131030 resp.

m:=13; M:=ZeroMatrix(IntegerRing(), m, m); for j:=1 to m do if (j-1) mod 6 lt 3 then M[j, 1]:=1; end if; end for; for k:=2 to m do for j:=k to m do M[j, k]:=M[j-1, k-1]+M[j, k-1]; end for; end for; &cat[ [ M[j, k]: k in [1..j] ]: j in [1..m] ];

T[j_, 1] := If[Mod[j-1, 6]<3, 1, 0]; T[j_, k_] := T[j, k] = T[j-1, k-1]+T[j, k-1]; Table[T[j, k], {j, 1, 13}, {k, 1, j}] // Flatten (* Jean-François Alcover, Mar 06 2014 *)

{m=13; M=matrix(m, m); for(j=1, m, M[j, 1]=if((j-1)%6<3, 1, 0)); for(k=2, m, for(j=k, m, M[j, k]=M[j-1, k-1]+M[j, k-1])); for(j=1, m, for(k=1, j, print1(M[j, k], ",")))}

From Werner Schulte, Jul 22 2017: (Start)
T(n,k) = 2^(k-2) + 2*sqrt(3)^(k-3) * sin(Pi/6*(2*n-k)) for 1 < k <= n, and
T(n,1) = 1 - floor((n-1)/3) mod 2 for n >= 1. (End)

A130781 Sequence is identical to its third differences: a(n+3) = 3a(n+2) - 3a(n+1) + 2*a(n), with a(0)=a(1)=1, a(2)=2.

Original entry on oeis.org

1, 1, 2, 5, 11, 22, 43, 85, 170, 341, 683, 1366, 2731, 5461, 10922, 21845, 43691, 87382, 174763, 349525, 699050, 1398101, 2796203, 5592406, 11184811, 22369621, 44739242, 89478485, 178956971, 357913942, 715827883, 1431655765, 2863311530

Offset: 0

Paul Curtz, Jul 14 2007, Jul 18 2007

The inverse binomial transform is 1,0,1,... repeated with period 3, essentially A011655. - R. J. Mathar, Aug 28 2023

Index entries for linear recurrences with constant coefficients, signature (3,-3,2).

See A130750, A130752, A130755, A129339.

Essentially a duplicate of A024493.

a[n_] := a[n] = 3 a[n - 1] - 3 a[n - 2] + 2 a[n - 3]; a[0] = a[1] = 1; a[2] = 2; Table[a@n, {n, 0, 33}] (* Or *)
CoefficientList[ Series[(1 - 2 x + 2 x^2)/(1 - 3 x + 3 x^2 - 2 x^3), {x, 0, 33}], x] (* Robert G. Wilson v, Sep 08 2007 *)
LinearRecurrence[{3,-3,2},{1,1,2},40] (* Harvey P. Dale, Sep 17 2013 *)

3*a(n) = 2^(n+1) + A087204(n+1).
Also first differences of A024494.
G.f.: (1-2x+2x^2)/(1-3x+3x^2-2x^3).
Binomial transform of [1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, ...]; i.e., ones in positions 2, 5, 8, 11, ... and the rest zeros. [Corrected by Gary W. Adamson, Jan 07 2008]

Edited by N. J. A. Sloane, Jul 28 2007

A129961 Main diagonal of triangular array T: T(j,1) = 1 for ((j-1) mod 8) < 4, else 0; T(j,k) = T(j-1,k-1)+T(j,k-1) for 2 <= k <= j.

Original entry on oeis.org

1, 2, 4, 8, 15, 26, 42, 64, 94, 140, 232, 464, 1092, 2744, 6840, 16384, 37384, 81296, 169120, 338240, 654192, 1232288, 2280864, 4194304, 7761376, 14635712, 28384384, 56768768, 116566080, 243472256, 511907712, 1073741824, 2232713344

Offset: 1

Paul Curtz, Jun 10 2007

nonn

First column is periodically 1 1 1 1 0 0 0 0 (see A131078).
First subdiagonal is 1, 2, 4, 7, 11, 16, 22, ... (see A131075); next subdiagonals are 1, 2, 3, 4, 5, 6, 8, 16, 46, 140, ..., 1, 1, 1, 1, 1, 2, 8, 30, 94, 256, ..., 0, 0, 0, 0, 1, 6, 22, 64, 162, 372, ..., 0, 0, 0, 1, 5, 16, 42, 98, 210, 420, ...., 0, 0, 1, 4, 11, 26, 56, 112, 210, 372, ..., 0, 1, 3, 7, 15, 30, 56, 98, 162, 256, ...,1, 2, 4, 8, 15, 26, 42, 64, 94, 140, ... . Main diagonal and eighth subdiagonal agree; generally j-th subdiagonal equals (j+8)-th subdiagonal.
Antidiagonal sums are 1, 1, 3, 3, 6, 5, 11, ... (see A131077).

			First seven rows of T are
[ 1 ]
[ 1, 2 ]
[ 1, 2, 4 ]
[ 1, 2, 4, 8 ]
[ 0, 1, 3, 7, 15 ]
[ 0, 0, 1, 4, 11, 26 ]
[ 0, 0, 0, 1, 5, 16, 42 ].

Cf. A129339, A131074 (T read by rows), A131075 (first subdiagonal of T), A131076 (row sums of T), A131077 (antidiagonal sums of T). First through sixth column of T are in A131078, A131079, A131080, A131081, A131082, A131083 resp.

m:=33; M:=ZeroMatrix(IntegerRing(), m, m); for j:=1 to m do if (j-1) mod 8 lt 4 then M[j, 1]:=1; end if; end for; for k:=2 to m do for j:=k to m do M[j, k]:=M[j-1, k-1]+M[j, k-1]; end for; end for; [ M[n, n]: n in [1..m] ]; // Klaus Brockhaus, Jun 14 2007

m:=33; S:=[ [1, 1, 1, 1, 0, 0, 0, 0][(n-1) mod 8 + 1]: n in [1..m] ]; [ &+[ Binomial(i-1, k-1)*S[k]: k in [1..i] ]: i in [1..m] ]; // Klaus Brockhaus, Jun 17 2007

{m=33; v=concat([1, 2, 4, 8, 15], vector(m-5)); for(n=6, m, v[n]=6*v[n-1]-14*v[n-2]+16*v[n-3]-10*v[n-4]+4*v[n-5]); v} \\ Klaus Brockhaus, Jun 14 2007

G.f.: x*(1-x)^4/((1-2*x)*(1-4*x+6*x^2-4*x^3+2*x^4)).
a(1) = 1, a(2) = 2, a(3) = 4, a(4) = 8, a(5) = 15; for n > 5, a(n) = 6*a(n-1)-14*a(n-2)+16*a(n-3)-10*a(n-4)+4*a(n-5).
Binomial transform of A131078. - Klaus Brockhaus, Jun 17 2007

Edited and extended by Klaus Brockhaus, Jun 14 2007

G.f. corrected by Klaus Brockhaus, Oct 15 2009

A131025 Antidiagonal sums of triangular array T: T(j,1) = 1 for ((j-1) mod 6) < 3, else 0; T(j,k) = T(j-1,k-1) + T(j-1,k) for 2 <= k <= j.

Original entry on oeis.org

1, 1, 3, 2, 5, 3, 9, 6, 16, 11, 27, 22, 50, 50, 101, 114, 215, 255, 471, 552, 1024, 1145, 2169, 2290, 4460, 4460, 8921, 8556, 17477, 16383, 33861, 31674, 65536, 62255, 127791, 124510, 252302, 252302, 504605, 514446, 1019051, 1048575, 2067627

Offset: 1

Klaus Brockhaus, following a suggestion of Paul Curtz, Jun 10 2007

nonn

			For first seven rows of T see A131022 or A129339.

Michael De Vlieger, Table of n, a(n) for n = 1..6644
Scott M. Bailey and Donald M. Larson, The A(1)-module structure of the homology of Brown-Gitler spectra, arXiv:2107.01316 [math.AT], 2021.

Cf. A131022 (T read by rows), A129339 (main diagonal of T), A131023 (first subdiagonal of T), A131024 (row sums of T). First through sixth column of T are in A088911, A131026, A131027, A131028, A131029, A131030 resp.

m:=43; M:=ZeroMatrix(IntegerRing(), m, m); for j:=1 to m do if (j-1) mod 6 lt 3 then M[j, 1]:=1; end if; end for; for k:=2 to m do for j:=k to m do M[j, k]:=M[j-1, k-1]+M[j, k-1]; end for; end for; [ &+[ M[j-k+1, k]: k in [1..(j+1) div 2] ]: j in [1..m] ];

CoefficientList[Series[(1 - 3 x^2 + 2 x^4 + 2 x^6 - 2 x^8 + x^9)/((1 - x)*(1 + x)*(1 - x + x^2)*(1 - 2 x^2)*(1 - 3 x^2 + 3 x^4)), {x, 0, 42}], x] (* Michael De Vlieger, Oct 26 2021 *)

{m=43; M=matrix(m, m); for(j=1, m, M[j, 1]=if((j-1)%6<3, 1, 0)); for(k=2, m, for(j=k, m, M[j, k]=M[j-1, k-1]+M[j, k-1])); for(j=1, m, print1(sum(k=1, (j+1)\2, M[j-k+1, k]), ","))}

G.f.: (1-3*x^2+2*x^4+2*x^6-2*x^8+x^9)/((1-x)*(1+x)*(1-x+x^2)*(1-2*x^2)*(1-3*x^2+3*x^4)).

A131023 First subdiagonal of triangular array T: T(j,1) = 1 for ((j-1) mod 6) < 3, else 0; T(j,k) = T(j-1,k-1) + T(j-1,k) for 2 <= k <= j.

Original entry on oeis.org

1, 2, 3, 4, 5, 7, 14, 37, 101, 256, 593, 1267, 2534, 4825, 8921, 16384, 30581, 58975, 117950, 242461, 504605, 1048576, 2156201, 4371451, 8742902, 17308657, 34085873, 67108864, 132623405, 263652487, 527304974, 1059392917, 2133134741

Offset: 1

Klaus Brockhaus, following a suggestion of Paul Curtz, Jun 10 2007

Also first differences of main diagonal A129339.

			For first seven rows of T see A131022 or A129339.

Index entries for linear recurrences with constant coefficients, signature (5,-9,6).

Cf. A131022 (T read by rows), A129339 (main diagonal of T), A131024 (row sums of T), A131025 (antidiagonal sums of T). First through sixth column of T are in A088911, A131026, A131027, A131028, A131029, A131030 resp.

m:=34; M:=ZeroMatrix(IntegerRing(), m, m); for j:=1 to m do if (j-1) mod 6 lt 3 then M[j, 1]:=1; end if; end for; for k:=2 to m do for j:=k to m do M[j, k]:=M[j-1, k-1]+M[j, k-1]; end for; end for; [ M[n+1, n]: n in [1..m-1] ];

{m=33; v=concat([1, 2, 3, 4],vector(m-4)); for(n=5, m, v[n]=5*v[n-1]-9*v[n-2]+6*v[n-3]); v}

a(1) = 1, a(2) = 2, a(3) = 3, a(4) = 4; for n > 4, a(n) = 5*a(n-1) - 9*a(n-2) + 6*a(n-3).
G.f.: x*(1-3*x+2*x^2+x^3)/((1-2*x)*(1-3*x+3*x^2)).
a(n) = A057681(n-1) + 2^(n-2), a(1) = 1. - Bruno Berselli, Feb 17 2011

A131024 Row sums of triangular array T: T(j,1) = 1 for ((j-1) mod 6) < 3, else 0; T(j,k) = T(j-1,k-1) + T(j-1,k) for 2 <= k <= j.

Original entry on oeis.org

1, 3, 7, 11, 16, 22, 36, 73, 175, 431, 1024, 2290, 4824, 9649, 18571, 34955, 65536, 124510, 242460, 484921, 989527, 2038103, 4194304, 8565754, 17308656, 34617313, 68703187, 135812051, 268435456, 532087942, 1059392916, 2118785833, 4251920575, 8546887871

Offset: 1

Klaus Brockhaus, following a suggestion of Paul Curtz, Jun 10 2007

Sum of n-th row equals (n+1)-th term of main diagonal minus (n+1)-th term of first column. A088911 has offset 0, so a(n) = A129339(n+1) - A088911(n).

			For first seven rows of T see A131022 or A129339.

Index entries for linear recurrences with constant coefficients, signature (6,-14,14,0,-14,15,-6).

Cf. A131022 (T read by rows), A129339 (main diagonal of T), A131023 (first subdiagonal of T), A131025 (antidiagonal sums of T). First through sixth column of T are in A088911, A131026, A131027, A131028, A131029, A131030 resp.

m:=32; M:=ZeroMatrix(IntegerRing(), m, m); for j:=1 to m do if (j-1) mod 6 lt 3 then M[j, 1]:=1; end if; end for; for k:=2 to m do for j:=k to m do M[j, k]:=M[j-1, k-1]+M[j, k-1]; end for; end for; [ &+[ M[j, k]: k in [1..j] ]: j in [1..m] ];

lista(m) = my(M=matrix(m, m)); for(j=1, m, M[j, 1]=if((j-1)%6<3, 1, 0)); for(k=2, m, for(j=k, m, M[j, k]=M[j-1, k-1]+M[j, k-1])); for(j=1, m, print1(sum(k=1, j, M[j, k]), ", "))

G.f.: x*(1-3*x+3*x^2-3*x^3+6*x^4-4*x^5+x^6)/((1-x)*(1+x)*(1-2*x)*(1-x+x^2)*(1-3*x+3*x^2)).

A132357 a(n) = 3a(n-1) - a(n-3) + 3a(n-4), with initial values 1,4,14,41.

Original entry on oeis.org

1, 4, 14, 41, 122, 364, 1093, 3280, 9842, 29525, 88574, 265720, 797161, 2391484, 7174454, 21523361, 64570082, 193710244, 581130733, 1743392200, 5230176602, 15690529805, 47071589414, 141214768240, 423644304721

Offset: 0

Paul Curtz, Nov 24 2007

Paolo Xausa, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-1,3).

First differences of A132353.

Cf. A129339.

LinearRecurrence[{3,0,-1,3},{1,4,14,41},50] (* Paolo Xausa, Dec 05 2023 *)

a(n)=([0,1,0,0; 0,0,1,0; 0,0,0,1; 3,-1,0,3]^n*[1;4;14;41])[1,1] \\ Charles R Greathouse IV, Oct 08 2016

O.g.f.: -(1+x+2*x^2)/((3*x-1)*(x+1)*(x^2-x+1)) = -(3/2)/(3*x-1)+(1/3)*(x-2)/(x^2-x+1)+(1/ 6)/(x+1). - R. J. Mathar, Nov 28 2007
a(n) = (1/2)*3^(n+1) + (1/6)*(-1)^n - (2/3)*cos(Pi*n/3). Or, a(n) = (1/2)*3^(n+1) + (1/2)*[ -1; -1; 1; 1; 1; -1]. - Richard Choulet, Jan 02 2008
a(n+1) - 3a(n) = A132367(n+1). - Paul Curtz, Dec 02 2007
6*a(n) = (-1)^n +3^(n+2) -2*A057079(n+1). - R. J. Mathar, Oct 03 2021

A132868 a(n) = 3a(n-1) - a(n-3) + 3a(n-4), with initial values 1,3,7,20.

Original entry on oeis.org

1, 3, 7, 20, 60, 182, 547, 1641, 4921, 14762, 44286, 132860, 398581, 1195743, 3587227, 10761680, 32285040, 96855122, 290565367, 871696101, 2615088301, 7845264902, 23535794706, 70607384120, 211822152361, 635466457083, 1906399371247, 5719198113740

Offset: 0

Paul Curtz, Nov 22 2007

Andrew Howroyd, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,0,-1,3).

Cf. A129339, A132951.

LinearRecurrence[{3,0,-1,3},{1,3,7,20},50] (* Harvey P. Dale, Jan 21 2012 *)

4*a(n) = 3^(n+1) + A132951(n).
O.g.f.: (-1+2*x^2)/((3*x-1)*(x+1)*(x^2-x+1)) = -(3/4)/(3*x-1)-(1/12)/(x+1)+(1/3)*(x+1)/(x^2-x+1). - R. J. Mathar, Nov 28 2007
a(n) = (1/12)*(3^(n+2) - 4*cos((n+1)*Pi/3) + cos((n+1)*Pi) + 4*sqrt(3) * sin(((n+1)*Pi)/3) + I*sin((n+1)*Pi)). - Harvey P. Dale, Jan 21 2012
12*a(n) = -(-1)^n +3^(n+2) +4*A057079(n). - R. J. Mathar, Oct 03 2021

A135343 a(n) = 3a(n-1) + 4a(n-2) - a(n-3) + 3a(n-4) + 4a(n-5).

Original entry on oeis.org

1, 3, 12, 51, 205, 820, 3277, 13107, 52428, 209715, 838861, 3355444, 13421773, 53687091, 214748364, 858993459, 3435973837, 13743895348, 54975581389, 219902325555, 879609302220, 3518437208883, 14073748835533, 56294995342132, 225179981368525, 900719925474099

Offset: 0

Paul Curtz, Dec 05 2007

See A129339 comments. Third sequence after b(n) = 3*b(n-1) - 3*b(n-2) + 2*b(n-3) and c(n) = 3*c(n-1) - c(n-3) + 3*c(n-4). The first is for every sequence identical to its third differences. What characterizes the two others?

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3, 4, -1, 3, 4).

Cf. A129339, A135345.

LinearRecurrence[{3,4,-1,3,4},{1,3,12,51,205},30] (* Harvey P. Dale, Jun 03 2013 *)

Vec((1-x+4*x^3)/((1+x)*(1-4*x)*(1-x+x^2)) + O(x^30)) \\ Colin Barker, Oct 11 2016

a(n+1) - 4*a(n) = hexaperiodic -1, 0, 3, 1, 0, -3.
a(n) = (1/15)*( 3*4^(n+1) - 2*(-1)^n + 5*cos(Pi*n/3) - 5*sqrt(3)*cos(Pi*n/3) ). - Richard Choulet, Jan 04 2008
G.f.: (1-x+4*x^3) / ((1+x)*(1-4*x)*(1-x+x^2)). - Colin Barker, Oct 11 2016

More terms from Harvey P. Dale, Jun 03 2013

Removed incorrect formula, Joerg Arndt, Oct 11 2016

cp's OEIS Frontend

A057083 Scaled Chebyshev U-polynomials evaluated at sqrt(3)/2; expansion of 1/(1 - 3*x + 3*x^2).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Sage

Formula

A131022 Triangular array T read by rows: T(j,1) = 1 for ((j-1) mod 6) < 3, else 0; T(j,k) = T(j-1,k-1) + T(j,k-1) for 2 <= k <= j.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A130781 Sequence is identical to its third differences: a(n+3) = 3*a(n+2) - 3*a(n+1) + 2*a(n), with a(0)=a(1)=1, a(2)=2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

A129961 Main diagonal of triangular array T: T(j,1) = 1 for ((j-1) mod 8) < 4, else 0; T(j,k) = T(j-1,k-1)+T(j,k-1) for 2 <= k <= j.

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Magma

Magma

PARI

Formula

Extensions

A131025 Antidiagonal sums of triangular array T: T(j,1) = 1 for ((j-1) mod 6) < 3, else 0; T(j,k) = T(j-1,k-1) + T(j-1,k) for 2 <= k <= j.

Views

Author

Keywords

Examples

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A131023 First subdiagonal of triangular array T: T(j,1) = 1 for ((j-1) mod 6) < 3, else 0; T(j,k) = T(j-1,k-1) + T(j-1,k) for 2 <= k <= j.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Magma

PARI

Formula

A131024 Row sums of triangular array T: T(j,1) = 1 for ((j-1) mod 6) < 3, else 0; T(j,k) = T(j-1,k-1) + T(j-1,k) for 2 <= k <= j.

Views

Author

Keywords

A057083 Scaled Chebyshev U-polynomials evaluated at sqrt(3)/2; expansion of 1/(1 - 3x + 3x^2).

A130781 Sequence is identical to its third differences: a(n+3) = 3a(n+2) - 3a(n+1) + 2*a(n), with a(0)=a(1)=1, a(2)=2.

A132357 a(n) = 3a(n-1) - a(n-3) + 3a(n-4), with initial values 1,4,14,41.

A132868 a(n) = 3a(n-1) - a(n-3) + 3a(n-4), with initial values 1,3,7,20.

A135343 a(n) = 3a(n-1) + 4a(n-2) - a(n-3) + 3a(n-4) + 4a(n-5).