A063782 -id:A063782 - cp's OEIS frontend

A006138 a(n) = a(n-1) + 3*a(n-2).

1, 2, 5, 11, 26, 59, 137, 314, 725, 1667, 3842, 8843, 20369, 46898, 108005, 248699, 572714, 1318811, 3036953, 6993386, 16104245, 37084403, 85397138, 196650347, 452841761, 1042792802, 2401318085, 5529696491, 12733650746, 29322740219, 67523692457, 155491913114

Offset: 0

N. J. A. Sloane

The binomial transform of a(n) is b(n) = A006190(n+1), which satisfies b(n) = 3*b(n-1) + b(n-2). - Paul Barry, May 21 2006
Partial sums of A105476. - Paul Barry, Feb 02 2007
An elephant sequence, see A175654. For the corner squares four A[5] vectors, with decimal values 69, 261, 321 and 324, lead to this sequence. For the central square these vectors lead to the companion sequence A105476 (without the first leading 1). - Johannes W. Meijer, Aug 15 2010
Equals the INVERTi transform of A063782: (1, 3, 10, 32, 104, ...). - Gary W. Adamson, Aug 14 2010
Pisano period lengths: 1, 3, 1, 6, 24, 3, 24, 6, 3, 24, 120, 6, 156, 24, 24, 12, 16, 3, 90, 24, ... - R. J. Mathar, Aug 10 2012
The sequence is the INVERT transform of A016116: (1, 1, 2, 2, 4, 4, 8, 8, ...). - Gary W. Adamson, Jul 17 2015

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Vincenzo Librandi, Table of n, a(n) for n = 0..200
N. T. Gridgeman, A new look at Fibonacci generalization, Fib,. Quart., 11 (1973), 40-55.
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Index entries for linear recurrences with constant coefficients, signature (1,3).

Cf. A063782. - Gary W. Adamson, Aug 14 2010

Cf. A006130, A006190, A016116, A105476, A122950, A175654, A274977.

a:=[1,2];; for n in [3..40] do a[n]:=a[n-1]+3*a[n-2]; od; a; # G. C. Greubel, Nov 19 2019

[n le 2 select n else Self(n-1)+3*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Sep 15 2016

A006138:=-(1+z)/(-1+z+3*z**2); # Simon Plouffe in his 1992 dissertation

CoefficientList[Series[(1+z)/(1-z-3*z^2), {z,0,40}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 11 2011 *)
Table[(I*Sqrt[3])^(n-1)*(I*Sqrt[3]*ChebyshevU[n, 1/(2*I*Sqrt[3])] + ChebyshevU[n-1, 1/(2*I*Sqrt[3])]), {n, 0, 40}]//Simplify (* G. C. Greubel, Nov 19 2019 *)
LinearRecurrence[{1,3},{1,2},40] (* Harvey P. Dale, May 29 2025 *)

main(size)={my(v=vector(size),i);v[1]=1;v[2]=2;for(i=3,size,v[i]=v[i-1]+3*v[i-2]);return(v);} /* Anders Hellström, Jul 17 2015 */

def A006138_list(prec):
    P. = PowerSeriesRing(ZZ, prec)
    return P((1+x)/(1-x-3*x^2)).list()
A006138_list(40) # G. C. Greubel, Nov 19 2019

a(n) = Sum_{k=0..n+1} A122950(n+1,k)*2^(n+1-k). - Philippe Deléham, Jan 04 2008
G.f.: (1+x)/(1-x-3*x^2). - Paul Barry, May 21 2006
a(n) = Sum_{k=0..n} C(floor((2n-k)/2),n-k)*3^floor(k/2). - Paul Barry, Feb 02 2007
a(n) = A006130(n) + A006130(n-1). - R. J. Mathar, Jun 22 2011
a(n) = (i*sqrt(3))^(n-1)*(i*sqrt(3)*ChebyshevU(n, 1/(2*i*sqrt(3))) + ChebyshevU(n-1, 1/(2*i*sqrt(3)))), where i=sqrt(-1). - G. C. Greubel, Nov 19 2019

Typo in formula corrected by Johannes W. Meijer, Aug 15 2010

A215244 a(n) is the number of ways of writing the binary expansion of n as a product (or concatenation) of palindromes.

Original entry on oeis.org

1, 1, 1, 2, 2, 2, 2, 4, 4, 3, 3, 3, 4, 3, 4, 8, 8, 5, 4, 5, 5, 5, 4, 6, 8, 5, 5, 6, 8, 6, 8, 16, 16, 9, 7, 9, 8, 6, 6, 10, 10, 6, 8, 8, 7, 7, 7, 12, 16, 9, 7, 10, 8, 8, 8, 11, 16, 10, 10, 11, 16, 12, 16, 32, 32, 17, 13, 17, 13, 11, 11, 18, 15, 11, 10, 10, 12, 9, 11, 20, 20, 11, 10, 11, 13, 13, 10, 16, 14, 9, 10, 12, 13, 11, 13, 24, 32, 17, 13, 18, 14, 11, 12, 19, 16, 10, 13

Offset: 0

N. J. A. Sloane, Aug 07 2012

"Product" is used here is the usual sense of combinatorics on words.
The sequence may be arranged into a triangle in which row k contains the 2^k terms [a(2^k), ..., a(2^(k+1)-1)]: (1), (1), (1,2), (2,2,2,4), ... The row sums of this triangle appear to be A063782. - R. J. Mathar, Aug 09 2012. I have a proof that A063782 does indeed give the row sums. - N. J. A. Sloane, Aug 10 2012
a(A220654(n)) = n and a(m) < n for m < A220654(n). - Reinhard Zumkeller, Dec 17 2012

			a(4)=2 since 4 = 100, and 100 can be written as either 1.0.0 or 1.00.
a(5)=2 from 1.0.1, 101.
a(10)=3 from 1.0.1.0, 1.010, 101.1
Written as a triangle, the sequence is:
1,
1,
1, 2,
2, 2, 2, 4,
4, 3, 3, 3, 4, 3, 4, 8,
8, 5, 4, 5, 5, 5, 4, 6, 8, 5, 5, 6, 8, 6, 8, 16,
16, 9, 7, 9, 8, 6, 6, 10, 10, 6, 8, 8, 7, 7, 7, 12, 16, 9, 7, 10, 8, 8, 8, 11, 16, 10, 10, 11, 16, 12, 16, 32,
...

Cf. A050430, A215245, A215246, A215250, A215253, A215254, A063782.

Cf. A206925, A030308.

import Data.Map (Map, singleton, (!), insert)
import Data.List (inits, tails)
newtype Bin = Bin [Int] deriving (Eq, Show, Read)
instance Ord Bin where
   Bin us <= Bin vs | length us == length vs = us <= vs
                    | otherwise              = length us <= length vs
a215244 n = a215244_list !! n
a215244_list = 1 : f [1] (singleton (Bin [0]) 1) where
   f bs m | last bs == 1 = y : f (succ bs) (insert (Bin bs) y m)
          | otherwise    = f (succ bs) (insert (Bin bs) y m) where
     y = fromEnum (pal bs) +
         sum (zipWith (\us vs -> if pal us then m ! Bin vs else 0)
                      (init $ drop 1 $ inits bs) (drop 1 $ tails bs))
     pal ds = reverse ds == ds
     succ [] = [0]; succ (0:ds) = 1 : ds; succ (1:ds) = 0 : succ ds
-- Reinhard Zumkeller, Dec 17 2012

isPal := proc(L)
    local d ;
    for d from 1 to nops(L)/2 do
        if op(d,L) <> op(-d,L) then
            return false;
        end if;
    end do:
    return true;
end proc:
A215244L := proc(L)
    local a,c;
    a := 0 ;
    if isPal(L) then
        a := 1;
    end if;
    for c from 1 to nops(L)-1 do
        if isPal( [op(1..c,L)] ) then
            a := a+procname([op(c+1..nops(L),L)]) ;
        end if;
    end do:
    return a;
end proc:
A215244 := proc(n)
    if n = 1 then
        1;
    else
        convert(n,base,2) ;
        A215244L(%) ;
    end if;
end proc: # R. J. Mathar, Aug 07 2012
# Caution: the last procedure applies A215244L to the reverse of the binary expansion of n, which is perfectly OK but might cause a problem if the procedure was used in a different problem. - N. J. A. Sloane, Aug 11 2012

palQ[L_] := SameQ[L, Reverse[L]];
b[L_] := b[L] = Module[{a = palQ[L] // Boole, c}, For[c = 1, c < Length[L], c++, If[palQ[L[[;;c]]], a = a + b[L[[c+1;;]]]]]; a];
a[n_] := If[n == 1, 1, b[IntegerDigits[n, 2]]];
a /@ Range[0, 106] (* Jean-François Alcover, Oct 27 2019 *)

A179606 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 3x - 5x^2).

Original entry on oeis.org

1, 4, 17, 71, 298, 1249, 5237, 21956, 92053, 385939, 1618082, 6783941, 28442233, 119246404, 499950377, 2096083151, 8788001338, 36844419769, 154473265997, 647641896836, 2715292020493, 11384085545659, 47728716739442

Offset: 0

Johannes W. Meijer, Jul 28 2010

a(n) represents the number of n-move routes of a fairy chess piece starting in the central square (m = 5) on a 3 X 3 chessboard. This fairy chess piece behaves like a king on the eight side and corner squares but on the central square the king goes crazy and turns into a red king, see A179596.
The sequence above corresponds to 24 red king vectors, i.e., A[5] vectors, with decimal values 27, 30, 51, 54, 57, 60, 90, 114, 120, 147, 150, 153, 156, 177, 180, 210, 216, 240, 282, 306, 312, 402, 408 and 432. These vectors lead for the corner squares to A015523 and for the side squares to A152187.
This sequence belongs to a family of sequences with g.f. (1 + (k-4)*x)/(1 - 3*x - k*x^2). Red king sequences that are members of this family are A007483 (k= 2), A015521 (k=4), A179606 (k=5; this sequence), A154964 (k=6), A179603 (k=7) and A179599 (k=8). We observe that there is no red king sequence for k=3. Other members of this family are A006190 (k=1), A133494 (k=0) and A168616 (k=-2).
Inverse binomial transform of A052918.
The sequence b(n+1) = 6*a(n), n >= 0 with b(0)=1, is a berserker sequence, see A180147. The b(n) sequence corresponds to 16 A[5] vectors with decimal values between 111 and 492. These vectors lead for the corner squares to sequence c(n+1)=4*A179606(n), n >= 0 with c(0)=1, and for the side squares to A180140. - Johannes W. Meijer, Aug 14 2010
Equals the INVERT transform of A063782: (1, 3, 10, 32, 104, ...). Example: a(3) = 71 = (1, 1, 4, 7) dot (32, 10, 3, 1) = (32 + 10 + 12 + 17). - Gary W. Adamson, Aug 14 2010

Indranil Ghosh, Table of n, a(n) for n = 0..1603
Index entries for linear recurrences with constant coefficients, signature (3,5).

Cf. A179597 (central square).

Cf. A072263, A072264, A152187, A197189.

with(LinearAlgebra): nmax:=22; m:=5; A[1]:= [0,1,0,1,1,0,0,0,0]: A[2]:= [1,0,1,1,1,1,0,0,0]: A[3]:= [0,1,0,0,1,1,0,0,0]: A[4]:= [1,1,0,0,1,0,1,1,0]: A[5]:= [0,0,0,1,1,1,0,0,1]: A[6]:= [0,1,1,0,1,0,0,1,1]: A[7]:= [0,0,0,1,1,0,0,1,0]: A[8]:= [0,0,0,1,1,1,1,0,1]: A[9]:= [0,0,0,0,1,1,0,1,0]: A:=Matrix([A[1],A[2],A[3],A[4],A[5],A[6],A[7],A[8],A[9]]): for n from 0 to nmax do B(n):=A^n: a(n):= add(B(n)[m,k],k=1..9): od: seq(a(n), n=0..nmax);

CoefficientList[Series[(1+x)/(1-3*x-5*x^2), {x, 0, 22}],x] (* or *) LinearRecurrence[{3,5,0},{1,4},23] (* Indranil Ghosh, Mar 05 2017 *)

print(Vec((1 + x)/(1- 3*x - 5*x^2) + O(x^23))); \\ Indranil Ghosh, Mar 05 2017

G.f.: (1+x)/(1 - 3*x - 5*x^2).
a(n) = A015523(n) + A015523(n+1).
a(n) = 3*a(n-1) + 5*a(n-2) with a(0) = 1 and a(1) = 4.
a(n) = ((29 + 7*sqrt(29))*A^(-n-1) + (29-7*sqrt(29))*B^(-n-1))/290 with A = (-3+sqrt(29))/10 and B = (-3-sqrt(29))/10
Limit_{k->oo} a(n+k)/a(k) = (-1)^(n+1)*A000351(n)*A130196(n)/(A015523(n)*sqrt(29) - A072263(n)) for n >= 1.

A360571 Triangle read by rows: T(n,k) is the k-th Lie-Betti number of the path graph on n-vertices, n >= 1, 0 <= k <= 2*n - 1.

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 1, 3, 6, 6, 3, 1, 1, 4, 11, 16, 16, 11, 4, 1, 1, 5, 17, 33, 48, 48, 33, 17, 5, 1, 1, 6, 24, 58, 107, 140, 140, 107, 58, 24, 6, 1, 1, 7, 32, 92, 203, 329, 424, 424, 329, 203, 92, 32, 7, 1, 1, 8, 41, 136, 347, 668, 1039, 1280, 1280, 1039, 668, 347, 136, 41, 8, 1

Offset: 1

Samuel J. Bevins, Feb 12 2023

			Triangle begins:
      k=0  1  2   3   4    5    6    7     8     9   10   11   12  13 14 15
  n=1:  1  1
  n=2:  1  2  2   1
  n=3:  1  3  6   6   3    1
  n=4:  1  4 11  16  16   11    4    1
  n=5:  1  5 17  33  48   48   33   17     5     1
  n=6:  1  6 24  58 107  140  140  107    58    24    6    1
  n=7:  1  7 32  92 203  329  424  424   329   203   92   32    7   1
  n=8:  1  8 41 136 347  668 1039 1280  1280  1039  668  347  136  41  8  1

Marco Aldi and Samuel Bevins, L_oo-algebras and hypergraphs, arXiv:2212.13608 [math.CO], 2022. See page 9.
Meera Mainkar, Graphs and two step nilpotent Lie algebras, arXiv:1310.3414 [math.DG], 2013. See page 1.
Eric Weisstein's World of Mathematics, Path Graph.

Cf. A360572 (cycle graph), A088459 (star graph), A360625 (complete graph), A360938 (ladder graph), A360937 (wheel graph).

Cf. A063782 appears to be half the row sum.

from sage.algebras.lie_algebras.lie_algebra import LieAlgebra
def LieAlgebraFromGraph(G, Module = QQ):
    ''' Takes a graph and a module (optional) as an input.'''
    d = {}
    for edge in G.edges(): # this defines the relations among the generators of the Lie algebra
        key = ("x" + str(edge[0]), "x" + str(edge[1])) #[x_i, x_j]
        value = {"x_" + str(edge[0]) + "" + str(edge[1]): 1} #x{i,j}
        d[key] = value #appending to the dictionary d
    C = LieAlgebras(Module).WithBasis().Graded() #defines the category that we need to work with.
    C = C.FiniteDimensional().Stratified().Nilpotent() #specifies that the algebras we want should be finite, stratified, and nilpotent
    L = LieAlgebra(Module, d, nilpotent=True, category=C)
    def sort_generators_by_grading(lie_algebra, grading_operator): #this sorts the generators by their grading. In this case, V1 are vertices and V2
        generators = lie_algebra.gens()
        grading = [grading_operator(g) for g in generators] #using the grading operator to split the elements into their respective vector spaces
        sorted_generators = [g for _, g in sorted(zip(grading, generators))]
        grouped_generators = {}
        for g in sorted_generators:
            if grading_operator(g) in grouped_generators:
                grouped_generators[grading_operator(g)].append(g)
            else:
                grouped_generators[grading_operator(g)] = [g]
        return grouped_generators
    grading_operator = lambda g: g.degree() #defining the grading operator
    grouped_generators = sort_generators_by_grading(L, grading_operator) #evaluating the function to pull the generators apart
    V1 = grouped_generators[1] #elements from vertices
    V2 = grouped_generators[2] #elements from edges
    return L #, V1, V2 #returns the Lie algebra and the two vector spaces
def betti_numbers(lie_algebra): #this function will calculate the Lie theoretic Betti numbers and return them as a list
    dims = []
    H = lie_algebra.cohomology()
    for n in range(lie_algebra.dimension() + 1):
        dims.append(H[n].dimension())
    return dims
def A360571_row(n):
    if n == 1: return [1, 1]
    return betti_numbers(LieAlgebraFromGraph(graphs.PathGraph(n)))
for n in range(1,7): print(A360571_row(n))

A087206 a(n) = 2a(n-1) + 4a(n-2); with a(0)=1, a(1)=4.

Original entry on oeis.org

1, 4, 12, 40, 128, 416, 1344, 4352, 14080, 45568, 147456, 477184, 1544192, 4997120, 16171008, 52330496, 169345024, 548012032, 1773404160, 5738856448, 18571329536, 60098084864, 194481487872, 629355315200, 2036636581888

Offset: 0

Paul Barry, Aug 25 2003

Binomial transform of A056487. Unsigned version of A152174.
Number of words of length n over the alphabet {1,2,3,4} such that no odd letter is followed by an odd letter. - Armend Shabani, Feb 18 2017
From Sean A. Irvine, Jun 06 2025: (Start)
Also, the number of walks of length n starting at 0 in the following graph:
  1---2
  |\ /|
  | 0 |
  |/ \|
  4---3. (End)

Jens Christian Claussen, Time-evolution of the Rule 150 cellular automaton activity from a Fibonacci iteration, arXiv:math/0410429 [math.CO], 2004. See Table II, p. 4.
Sean A. Irvine, Walks on Graphs.
N. J. A. Sloane, On the Number of ON Cells in Cellular Automata, arXiv:1503.01168 [math.CO], 2015.
Index entries for linear recurrences with constant coefficients, signature (2,4).

Cf. A060925, A063782, A253064.

Equals (1/2) * A063727(n-1). Cf. A006483.

LinearRecurrence[{2, 4}, {1, 4}, 25] (* Jean-François Alcover, Sep 21 2017 *)

G.f.: (1+2x)/(1-2x-4x^2).
a(n) = (1-sqrt(5))^n*(1/2-3*sqrt(5)/10)+(1+sqrt(5))^n*(1/2+3*sqrt(5)/10).
a(n) = 2^n*Fibonacci(n+2). - Paul Barry, Mar 22 2004
a(n) = ((1+sqrt(5))^n-(1-sqrt(5))^n)/sqrt(80). Offset 2. a(4)=12. - Al Hakanson (hawkuu(AT)gmail.com), Apr 11 2009
G.f.: 1/(-2x-1/(-2x-1)). - Paul Barry, Mar 24 2010

Comment corrected by Philippe Deléham, Nov 27 2008

A209144 Triangle of coefficients of polynomials v(n,x) jointly generated with A209143; see the Formula section.

Original entry on oeis.org

1, 3, 6, 1, 12, 5, 24, 16, 1, 48, 44, 7, 96, 112, 30, 1, 192, 272, 104, 9, 384, 640, 320, 48, 1, 768, 1472, 912, 200, 11, 1536, 3328, 2464, 720, 70, 1, 3072, 7424, 6400, 2352, 340, 13, 6144, 16384, 16128, 7168, 1400, 96, 1, 12288, 35840, 39680, 20736

Offset: 1

Clark Kimberling, Mar 06 2012

Alternating row sums: 1,3,5,7,9,11,13,15,17,...
For a discussion and guide to related arrays, see A208510.
Subtriangle of the triangle given by (3,-1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1/3, -1/3, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 07 2012

			First five rows:
   1;
   3;
   6,  1;
  12,  5;
  24, 16, 1;
First three polynomials v(n,x): 1, 3, 6 + x.
(3,-1, 0, 0, 0, ...) DELTA (0, 1/3, -1/3, 0, 0, ...) begins:
    1;
    3,   0;
    6,   1,   0;
   12,   5,   0, 0;
   24,  16,   1, 0, 0;
   48,  44,   7, 0, 0, 0;
   96, 112,  30, 1, 0, 0, 0;
  192, 272, 104, 9, 0, 0, 0, 0;

Cf. A209143, A208510.

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + (x + 1)*v[n - 1, x];
v[n_, x_] := u[n - 1, x] + v[n - 1, x] + 1;
Table[Expand[u[n, x]], {n, 1, z/2}]
Table[Expand[v[n, x]], {n, 1, z/2}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]    (* A209143 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]    (* A209144 *)

u(n,x) = u(n-1,x) + (x+1)*v(n-1,x),
v(n,x) = u(n-1,x) + v(n-1,x) + 1,
where u(1,x)=1, v(1,x)=1.
From Philippe Deléham, Mar 07 2012: (Start)
As triangle T(n,k) with 0 <= k <= n:
T(n,k) = 2*T(n-1,k) + T(n-2,k-1), T(0,0) = 1, T(1,0) = 3, T(1,1) = 0 and T(n,k) = 0 if k < 0 or if k > n.
G.f.: (1+x)/(1-2*x-y*x^2).
Sum_{k=0..n} T(n,k)*x^k = A005408(n), A003945(n), A078057(n), A028859(n), A000244(n), A063782(n), A180168(n) for x = -1, 0, 1, 2, 3, 4, 5 respectively. (End)

A207543 Triangle read by rows, expansion of (1+yx)/(1-2y*x+y(y-1)x^2).

Original entry on oeis.org

1, 0, 3, 0, 1, 5, 0, 0, 5, 7, 0, 0, 1, 14, 9, 0, 0, 0, 7, 30, 11, 0, 0, 0, 1, 27, 55, 13, 0, 0, 0, 0, 9, 77, 91, 15, 0, 0, 0, 0, 1, 44, 182, 140, 17, 0, 0, 0, 0, 0, 11, 156, 378, 204, 19, 0, 0, 0, 0, 0, 1, 65, 450, 714, 285, 21, 0

Offset: 0

Philippe Deléham, Feb 24 2012

Previous name was: "A scaled version of triangle A082985."

Triangle, read by rows, given by (0, 1/3, -1/3, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (3, -4/3, 1/3, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

			Triangle begins :
1
0, 3
0, 1, 5
0, 0, 5, 7
0, 0, 1, 14, 9
0, 0, 0, 7, 30, 11
0, 0, 0, 1, 27, 55, 13
0, 0, 0, 0, 9, 77, 91, 15
0, 0, 0, 0, 1, 44, 182, 140, 17
0, 0, 0, 0, 0, 11, 156, 378, 204, 19
0, 0, 0, 0, 0, 1, 65, 450, 714, 285, 21
0, 0, 0, 0, 0, 0, 13, 275, 1122, 1254, 385, 23

R. Zielinski, Induction and Analogy in a Problem of Finite Sums arXiv:1608.04006 [math.GM], 2016.

Cf. A082985 which is another version of this triangle.

Cf. Diagonals : A005408, A000330, A005585, A050486, A054333, A057788. Cf. A119900.

s := (1+y*x)/(1-2*y*x+y*(y-1)*x^2): t := series(s,x,12):
seq(print(seq(coeff(coeff(t,x,n),y,m),m=0..n)),n=0..11); # Peter Luschny, Aug 17 2016

T(n,k) = 2*T(n-1,k-1) + T(n-2,k-1) - T(n-2,k-2), T(0,0) = 1, T(1,0) = 0, T(1,1) = 3.
G.f.: (1+y*x)/(1-2*y*x+y*(y-1)*x^2).
Sum_{i, i>=0} T(n+i,n) = A000204(2*n+1).
Sum_{k, 0<=k<=n} T(n,k)*x^k = A078069(n), A000007(n), A003945(n), A111566(n) for x = -1, 0, 1, 2 respectively.
Sum_{k, 0<=k<=n} T(n,k)*x^(n-k) = A090131(n), A005408(n), A003945(n), A078057(n), A028859(n), A000244(n), A063782(n), A180168(n) for x = -1, 0, 1, 2, 3, 4, 5, 6 respectively.
From Peter Bala, Aug 17 2016: (Start)
Let S(k,n) = Sum_{i = 1..n} i^k. Calculations in Zielinski 2016 suggest the following identity holds involving the p-th row elements of this triangle:
Sum_{k = 0..p} T(p,k)*S(2*k,n) = 1/2*(2*n + 1)*(n*(n + 1))^p.
For example, for row 6 we find S(6,n) + 27*S(8,n) + 55*S(10,n) + 13*S(12,n) = 1/2*(2*n + 1)*(n*(n + 1))^6.
There appears to be a similar result for the odd power sums S(2*k + 1,n) involving A119900. (End)

New name using a formula of the author from Peter Luschny, Aug 17 2016

A319053 a(n) is the exponent of the largest power of 2 that appears in the factorization of the entries in the matrix {{3,1},{1,-1}}^n.

Original entry on oeis.org

0, 1, 5, 3, 4, 8, 6, 7, 12, 9, 10, 15, 12, 13, 18, 15, 16, 20, 18, 19, 25, 21, 22, 28, 24, 25, 31, 27, 28, 32, 30, 31, 36, 33, 34, 39, 36, 37, 42, 39, 40, 44, 42, 43, 50, 45, 46, 53, 48, 49, 56, 51, 52, 56, 54, 55, 60, 57, 58, 63, 60, 61, 66, 63, 64, 68, 66, 67, 73, 69

Offset: 1

Greg Dresden, Sep 09 2018

nonn

a(n) appears to equal n-1 for n not a multiple of 3.
The matrix entries of M^n, with n >= 0, are M^n(1, 1) = 2^(n-1)*F(n+3) = A063782(n), M^n(2, 2) = 2^(n-1)*F(n-3) = A319196(n), M^n(1, 2) = M^n(2, 1) = 2^(n-1)*F(n) = A085449(n), where i = sqrt(-1), F = A000045, and F(-1) = 1, F(-2) = -1, F(-3) = 2. Proof by Cayley-Hamilton, with S(n, -i) = (-i)^n*F(n+1), where S(n, x) is given in A049310. - Wolfdieter Lang, Oct 08 2018
The above conjecture is true. From the preceding formulas for the elements of M^n this claims that the Fibonacci numbers F(n-3), F(n) and F(n+3) are always odd for n == 1 or 2 (mod 3). This is true because F(n) is even iff n == 0 (mod 3) (see e.g. Vajda, p.73), and each of the three indices is == 1 or 2 (mod 3) for n == 1 or 2 (mod 3), respectively. - Wolfdieter Lang, Oct 09 2018

			For n = 3, the matrix {{3,1},{1,-1}}^3 = {{32,8},{8,0}} and the largest power of 2 appearing in the factorization of any entry is 2^5 = 32. Hence, a(3) = 5.

S. Vajda, Fibonacci and Lucas numbers and the Golden Section, Ellis Horwood Ltd., Chichester, 1989, p. 73.

Cf. A000045, A049310, A063782, A085449, A130481, A319196.

Join[{0, 1, 5}, Table[Max[ IntegerExponent[Flatten[MatrixPower[{{3, 1}, {1, -1}}, n]], 2]], {n, 4, 40}]]

a(n) = vecmax(apply(x->if (x, valuation(x, 2), 0), [3,1;1,-1]^n)); \\ Michel Marcus, Sep 09 2018

A223211 3 X 3 X 3 triangular graph coloring a rectangular array: number of n X 1 0..5 arrays where 0..5 label nodes of a graph with edges 0,1 0,2 1,2 1,3 1,4 2,4 3,4 2,5 4,5 and every array movement to a horizontal or vertical neighbor moves along an edge of this graph.

Original entry on oeis.org

6, 18, 60, 192, 624, 2016, 6528, 21120, 68352, 221184, 715776, 2316288, 7495680, 24256512, 78495744, 254017536, 822018048, 2660106240, 8608284672, 27856994304, 90147127296, 291722231808, 944032972800, 3054954872832, 9886041636864

Offset: 1

R. H. Hardin, Mar 18 2013

nonn

Column 1 of A223218.

			Some solutions for n=3:
..4....4....0....2....1....4....2....3....2....2....0....5....1....3....4....5
..2....1....1....5....2....2....1....1....0....1....1....4....3....4....3....2
..0....4....4....2....0....4....2....0....2....0....0....2....1....1....1....1

R. H. Hardin, Table of n, a(n) for n = 1..210

Cf. A223218.

Empirical: a(n) = 2*a(n-1) + 4*a(n-2) = 6*A063782(n-1).
Conjectures from Colin Barker, Aug 17 2018: (Start)
G.f.: 6*x*(1 + x) / (1 - 2*x - 4*x^2).
a(n) = (3*((1-sqrt(5))^n*(-3+sqrt(5)) + (1+sqrt(5))^n*(3+sqrt(5)))) / (4*sqrt(5)).
(End)

A319196 a(n) = 2^(n-1)*Fibonacci(n-3), n >= 0.

Original entry on oeis.org

1, -1, 2, 0, 8, 16, 64, 192, 640, 2048, 6656, 21504, 69632, 225280, 729088, 2359296, 7634944, 24707072, 79953920, 258736128, 837287936, 2709520384, 8768192512, 28374466560, 91821703168, 297141272576, 961569357824, 3111703805952, 10069685043200, 32586185310208, 105451110793216

Offset: 0

Wolfdieter Lang, Oct 09 2018

This sequence gives the elements M^n(2, 2) of the matrix M = [[3, 1], [1, -1]].
Motivation to look into these matrix powers came from A319053. M^n[1, 1] = A063782 and M^n(1, 2) =  M^n(2, 1) = A085449(n). Proof by Cayley-Hamilton, using S(n, -I) = (-I)^n*F(n+1), and S(n, x) from A049310 and F = A000045.
For a similar signed sequence see A087205.

Index entries for linear recurrences with constant coefficients, signature (2,4).

Cf. A000045, A001622, A049310, A085449, A063782, A087205, A319053.

[2^(n-1)*Fibonacci(n-3): n in [0..30]]; // Vincenzo Librandi, Oct 09 2018

Table[2^(n-1) Fibonacci[n-3], {n, 0, 40}] (* Vincenzo Librandi, Oct 09 2018 *)
LinearRecurrence[{2,4},{1,-1},40] (* Harvey P. Dale, Mar 29 2020 *)

a(n) = 2^(n-1)*F(n-3), n >= 0, with F = A000045 with F(-1) = 1, F(-2) = -1 and F(-3) = 1.
G.f.: (1-3*x)/(1-2*x-(2*x)^2).
a(n) = 2*(a(n-1) + 2*a(n-2)), n >= 2, with a(0) = 1 and a(1) = -1.
a(n) = 2^(n-1)*(phi^(n-3) - (1 - phi)^(n-3))/(2*phi - 1) with the golden section phi =  A001622.

cp's OEIS Frontend

A006138 a(n) = a(n-1) + 3*a(n-2).

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A215244 a(n) is the number of ways of writing the binary expansion of n as a product (or concatenation) of palindromes.

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A179606 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 3*x - 5*x^2).

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A360571 Triangle read by rows: T(n,k) is the k-th Lie-Betti number of the path graph on n-vertices, n >= 1, 0 <= k <= 2*n - 1.

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A087206 a(n) = 2*a(n-1) + 4*a(n-2); with a(0)=1, a(1)=4.

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A209144 Triangle of coefficients of polynomials v(n,x) jointly generated with A209143; see the Formula section.

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A207543 Triangle read by rows, expansion of (1+y*x)/(1-2*y*x+y*(y-1)*x^2).

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A179606 Eight white kings and one red king on a 3 X 3 chessboard. G.f.: (1 + x)/(1 - 3x - 5x^2).

A087206 a(n) = 2a(n-1) + 4a(n-2); with a(0)=1, a(1)=4.

A207543 Triangle read by rows, expansion of (1+yx)/(1-2y*x+y(y-1)x^2).