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This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A039599 Triangle formed from even-numbered columns of triangle of expansions of powers of x in terms of Chebyshev polynomials U_n(x).

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 5, 9, 5, 1, 14, 28, 20, 7, 1, 42, 90, 75, 35, 9, 1, 132, 297, 275, 154, 54, 11, 1, 429, 1001, 1001, 637, 273, 77, 13, 1, 1430, 3432, 3640, 2548, 1260, 440, 104, 15, 1, 4862, 11934, 13260, 9996, 5508, 2244, 663, 135, 17, 1
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

T(n,k) is the number of lattice paths from (0,0) to (n,n) with steps E = (1,0) and N = (0,1) which touch but do not cross the line x - y = k and only situated above this line; example: T(3,2) = 5 because we have EENNNE, EENNEN, EENENN, ENEENN, NEEENN. - Philippe Deléham, May 23 2005
The matrix inverse of this triangle is the triangular matrix T(n,k) = (-1)^(n+k)* A085478(n,k). - Philippe Deléham, May 26 2005
Essentially the same as A050155 except with a leading diagonal A000108 (Catalan numbers) 1, 1, 2, 5, 14, 42, 132, 429, .... - Philippe Deléham, May 31 2005
Number of Grand Dyck paths of semilength n and having k downward returns to the x-axis. (A Grand Dyck path of semilength n is a path in the half-plane x>=0, starting at (0,0), ending at (2n,0) and consisting of steps u=(1,1) and d=(1,-1)). Example: T(3,2)=5 because we have u(d)uud(d),uud(d)u(d),u(d)u(d)du,u(d)duu(d) and duu(d)u(d) (the downward returns to the x-axis are shown between parentheses). - Emeric Deutsch, May 06 2006
Riordan array (c(x),x*c(x)^2) where c(x) is the g.f. of A000108; inverse array is (1/(1+x),x/(1+x)^2). - Philippe Deléham, Feb 12 2007
The triangle may also be generated from M^n*[1,0,0,0,0,0,0,0,...], where M is the infinite tridiagonal matrix with all 1's in the super and subdiagonals and [1,2,2,2,2,2,2,...] in the main diagonal. - Philippe Deléham, Feb 26 2007
Inverse binomial matrix applied to A124733. Binomial matrix applied to A089942. - Philippe Deléham, Feb 26 2007
Number of standard tableaux of shape (n+k,n-k). - Philippe Deléham, Mar 22 2007
From Philippe Deléham, Mar 30 2007: (Start)
This triangle belongs to the family of triangles defined by: T(0,0)=1, T(n,k)=0 if k<0 or if k>n, T(n,0)=x*T(n-1,0)+T(n-1,1), T(n,k)=T(n-1,k-1)+y*T(n-1,k)+T(n-1,k+1) for k>=1. Other triangles arise by choosing different values for (x,y):
(0,0) -> A053121; (0,1) -> A089942; (0,2) -> A126093; (0,3) -> A126970
(1,0) -> A061554; (1,1) -> A064189; (1,2) -> A039599; (1,3) -> A110877;
(1,4) -> A124576; (2,0) -> A126075; (2,1) -> A038622; (2,2) -> A039598;
(2,3) -> A124733; (2,4) -> A124575; (3,0) -> A126953; (3,1) -> A126954;
(3,2) -> A111418; (3,3) -> A091965; (3,4) -> A124574; (4,3) -> A126791;
(4,4) -> A052179; (4,5) -> A126331; (5,5) -> A125906. (End)
The table U(n,k) = Sum_{j=0..n} T(n,j)*k^j is given in A098474. - Philippe Deléham, Mar 29 2007
Sequence read mod 2 gives A127872. - Philippe Deléham, Apr 12 2007
Number of 2n step walks from (0,0) to (2n,2k) and consisting of step u=(1,1) and d=(1,-1) and the path stays in the nonnegative quadrant. Example: T(3,0)=5 because we have uuuddd, uududd, ududud, uduudd, uuddud; T(3,1)=9 because we have uuuudd, uuuddu, uuudud, ududuu, uuduud, uduudu, uudduu, uduuud, uududu; T(3,2)=5 because we have uuuuud, uuuudu, uuuduu, uuduuu, uduuuu; T(3,3)=1 because we have uuuuuu. - Philippe Deléham, Apr 16 2007, Apr 17 2007, Apr 18 2007
Triangular matrix, read by rows, equal to the matrix inverse of triangle A129818. - Philippe Deléham, Jun 19 2007
Let Sum_{n>=0} a(n)*x^n = (1+x)/(1-mx+x^2) = o.g.f. of A_m, then Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n. Related expansions of A_m are: A099493, A033999, A057078, A057077, A057079, A005408, A002878, A001834, A030221, A002315, A033890, A057080, A057081, A054320, A097783, A077416, A126866, A028230, A161591, for m=-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15, respectively. - Philippe Deléham, Nov 16 2009
The Kn11, Kn12, Fi1 and Fi2 triangle sums link the triangle given above with three sequences; see the crossrefs. For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 20 2011
4^n = (n-th row terms) dot (first n+1 odd integer terms). Example: 4^4 = 256 = (14, 28, 20, 7, 1) dot (1, 3, 5, 7, 9) = (14 + 84 + 100 + 49 + 9) = 256. - Gary W. Adamson, Jun 13 2011
The linear system of n equations with coefficients defined by the first n rows solve for diagonal lengths of regular polygons with N= 2n+1 edges; the constants c^0, c^1, c^2, ... are on the right hand side, where c = 2 + 2*cos(2*Pi/N). Example: take the first 4 rows relating to the 9-gon (nonagon), N = 2*4 + 1; with c = 2 + 2*cos(2*Pi/9) = 3.5320888.... The equations are (1,0,0,0) = 1; (1,1,0,0) = c; (2,3,1,0) = c^2; (5,9,5,1) = c^3. The solutions are 1, 2.53208..., 2.87938..., and 1.87938...; the four distinct diagonal lengths of the 9-gon (nonagon) with edge = 1. (Cf. comment in A089942 which uses the analogous operations but with c = 1 + 2*cos(2*Pi/9).) - Gary W. Adamson, Sep 21 2011
Also called the Lobb numbers, after Andrew Lobb, are a natural generalization of the Catalan numbers, given by L(m,n)=(2m+1)*Binomial(2n,m+n)/(m+n+1), where n >= m >= 0. For m=0, we get the n-th Catalan number. See added reference. - Jayanta Basu, Apr 30 2013
From Wolfdieter Lang, Sep 20 2013: (Start)
T(n, k) = A053121(2*n, 2*k). T(n, k) appears in the formula for the (2*n)-th power of the algebraic number rho(N):= 2*cos(Pi/N) = R(N, 2) in terms of the odd-indexed diagonal/side length ratios R(N, 2*k+1) = S(2*k, rho(N)) in the regular N-gon inscribed in the unit circle (length unit 1). S(n, x) are Chebyshev's S polynomials (see A049310):
rho(N)^(2*n) = Sum_{k=0..n} T(n, k)*R(N, 2*k+1), n >= 0, identical in N > = 1. For a proof see the Sep 21 2013 comment under A053121. Note that this is the unreduced version if R(N, j) with j > delta(N), the degree of the algebraic number rho(N) (see A055034), appears.
For the odd powers of rho(n) see A039598. (End)
Unsigned coefficients of polynomial numerators of Eqn. 2.1 of the Chakravarty and Kodama paper, defining the polynomials of A067311. - Tom Copeland, May 26 2016
The triangle is the Riordan square of the Catalan numbers in the sense of A321620. - Peter Luschny, Feb 14 2023

Examples

			Triangle T(n, k) begins:
  n\k     0     1     2     3     4     5    6   7   8  9
  0:      1
  1:      1     1
  2:      2     3     1
  3:      5     9     5     1
  4:     14    28    20     7     1
  5:     42    90    75    35     9     1
  6:    132   297   275   154    54    11    1
  7:    429  1001  1001   637   273    77   13   1
  8:   1430  3432  3640  2548  1260   440  104  15   1
  9:   4862 11934 13260  9996  5508  2244  663 135  17  1
  ... Reformatted by _Wolfdieter Lang_, Dec 21 2015
From _Paul Barry_, Feb 17 2011: (Start)
Production matrix begins
  1, 1,
  1, 2, 1,
  0, 1, 2, 1,
  0, 0, 1, 2, 1,
  0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 1, 2, 1,
  0, 0, 0, 0, 0, 1, 2, 1 (End)
From _Wolfdieter Lang_, Sep 20 2013: (Start)
Example for rho(N) = 2*cos(Pi/N) powers:
n=2: rho(N)^4 = 2*R(N,1) + 3*R(N,3) + 1*R(N, 5) =
  2 + 3*S(2, rho(N)) + 1*S(4, rho(N)), identical in N >= 1. For N=4 (the square with only one distinct diagonal), the degree delta(4) = 2, hence R(4, 3) and R(4, 5) can be reduced, namely to R(4, 1) = 1 and R(4, 5) = -R(4,1) = -1, respectively. Therefore, rho(4)^4 =(2*cos(Pi/4))^4 = 2 + 3 -1 = 4. (End)

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 796.
  • T. Myers and L. Shapiro, Some applications of the sequence 1, 5, 22, 93, 386, ... to Dyck paths and ordered trees, Congressus Numerant., 204 (2010), 93-104.

Links

Crossrefs

Columns give: A000108, A000245, A000344, A000588, A001392, A000589, A000590, A000012, A005408, A014107(n>1).
Row sums: A000984.
Triangle sums (see the comments): A000958 (Kn11), A001558 (Kn12), A088218 (Fi1, Fi2).
Cf. A008313, A039598, A084938, A000007, A067311, A321620.

Programs

  • Magma
    /* As triangle */ [[Binomial(2*n, k+n)*(2*k+1)/(k+n+1): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Oct 16 2015
  • Maple
    T:=(n,k)->(2*k+1)*binomial(2*n,n-k)/(n+k+1): for n from 0 to 12 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form # Emeric Deutsch, May 06 2006
T := proc(n, k) option remember; if k = n then 1 elif k > n then 0 elif k = 0 then T(n-1, 0) + T(n-1,1) else T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1) fi end:
seq(seq(T(n, k), k = 0..n), n = 0..9) od; # Peter Luschny, Feb 14 2023
  • Mathematica
    Table[Abs[Differences[Table[Binomial[2 n, n + i], {i, 0, n + 1}]]], {n, 0,7}] // Flatten (* Geoffrey Critzer, Dec 18 2011 *)
Join[{1},Flatten[Table[Binomial[2n-1,n-k]-Binomial[2n-1,n-k-2],{n,10},{k,0,n}]]] (* Harvey P. Dale, Dec 18 2011 *)
Flatten[Table[Binomial[2*n,m+n]*(2*m+1)/(m+n+1),{n,0,9},{m,0,n}]] (* Jayanta Basu, Apr 30 2013 *)
  • PARI
    a(n, k) = (2*n+1)/(n+k+1)*binomial(2*k, n+k)
trianglerows(n) = for(x=0, n-1, for(y=0, x, print1(a(y, x), ", ")); print(""))
trianglerows(10) \\ Felix Fröhlich, Jun 24 2016
  • Sage
    # Algorithm of L. Seidel (1877)
# Prints the first n rows of the triangle
def A039599_triangle(n) :
    D = [0]*(n+2); D[1] = 1
    b = True ; h = 1
    for i in range(2*n-1) :
        if b :
            for k in range(h,0,-1) : D[k] += D[k-1]
            h += 1
        else :
            for k in range(1,h, 1) : D[k] += D[k+1]
        if b : print([D[z] for z in (1..h-1)])
        b = not b
A039599_triangle(10)  # Peter Luschny, May 01 2012

Formula

T(n,k) = C(2*n-1, n-k) - C(2*n-1, n-k-2), n >= 1, T(0,0) = 1.
From Emeric Deutsch, May 06 2006: (Start)
T(n,k) = (2*k+1)*binomial(2*n,n-k)/(n+k+1).
G.f.: G(t,z)=1/(1-(1+t)*z*C), where C=(1-sqrt(1-4*z))/(2*z) is the Catalan function. (End)
The following formulas were added by Philippe Deléham during 2003 to 2009: (Start)
Triangle T(n, k) read by rows; given by A000012 DELTA A000007, where DELTA is Deléham's operator defined in A084938.
T(n, k) = C(2*n, n-k)*(2*k+1)/(n+k+1). Sum(k>=0; T(n, k)*T(m, k) = A000108(n+m)); A000108: numbers of Catalan.
T(n, 0) = A000108(n); T(n, k) = 0 if k>n; for k>0, T(n, k) = Sum_{j=1..n} T(n-j, k-1)*A000108(j).
T(n, k) = A009766(n+k, n-k) = A033184(n+k+1, 2k+1).
G.f. for column k: Sum_{n>=0} T(n, k)*x^n = x^k*C(x)^(2*k+1) where C(x) = Sum_{n>=0} A000108(n)*x^n is g.f. for Catalan numbers, A000108.
T(0, 0) = 1, T(n, k) = 0 if n<0 or n=1, T(n, k) = T(n-1, k-1) + 2*T(n-1, k) + T(n-1, k+1).
a(n) + a(n+1) = 1 + A000108(m+1) if n = m*(m+3)/2; a(n) + a(n+1) = A039598(n) otherwise.
T(n, k) = A050165(n, n-k).
Sum_{j>=0} T(n-k, j)*A039598(k, j) = A028364(n, k).
Matrix inverse of the triangle T(n, k) = (-1)^(n+k)*binomial(n+k, 2*k) = (-1)^(n+k)*A085478(n, k).
Sum_{k=0..n} T(n, k)*x^k = A000108(n), A000984(n), A007854(n), A076035(n), A076036(n) for x = 0, 1, 2, 3, 4.
Sum_{k=0..n} (2*k+1)*T(n, k) = 4^n.
T(n, k)*(-2)^(n-k) = A114193(n, k).
Sum_{k>=h} T(n,k) = binomial(2n,n-h).
Sum_{k=0..n} T(n,k)*5^k = A127628(n).
Sum_{k=0..n} T(n,k)*7^k = A115970(n).
T(n,k) = Sum_{j=0..n-k} A106566(n+k,2*k+j).
Sum_{k=0..n} T(n,k)*6^k = A126694(n).
Sum_{k=0..n} T(n,k)*A000108(k) = A007852(n+1).
Sum_{k=0..floor(n/2)} T(n-k,k) = A000958(n+1).
Sum_{k=0..n} T(n,k)*(-1)^k = A000007(n).
Sum_{k=0..n} T(n,k)*(-2)^k = (-1)^n*A064310(n).
T(2*n,n) = A126596(n).
Sum_{k=0..n} T(n,k)*(-x)^k = A000007(n), A126983(n), A126984(n), A126982(n), A126986(n), A126987(n), A127017(n), A127016(n), A126985(n), A127053(n) for x=1,2,3,4,5,6,7,8,9,10 respectively.
Sum_{j>=0} T(n,j)*binomial(j,k) = A116395(n,k).
T(n,k) = Sum_{j>=0} A106566(n,j)*binomial(j,k).
T(n,k) = Sum_{j>=0} A127543(n,j)*A038207(j,k).
Sum_{k=0..floor(n/2)} T(n-k,k)*A000108(k) = A101490(n+1).
T(n,k) = A053121(2*n,2*k).
Sum_{k=0..n} T(n,k)*sin((2*k+1)*x) = sin(x)*(2*cos(x))^(2*n).
T(n,n-k) = Sum_{j>=0} (-1)^(n-j)*A094385(n,j)*binomial(j,k).
Sum_{j>=0} A110506(n,j)*binomial(j,k) = Sum_{j>=0} A110510(n,j)*A038207(j,k) = T(n,k)*2^(n-k).
Sum_{j>=0} A110518(n,j)*A027465(j,k) = Sum_{j>=0} A110519(n,j)*A038207(j,k) = T(n,k)*3^(n-k).
Sum_{k=0..n} T(n,k)*A001045(k) = A049027(n), for n>=1.
Sum_{k=0..n} T(n,k)*a(k) = (m+2)^n if Sum_{k>=0} a(k)*x^k = (1+x)/(x^2-m*x+1).
Sum_{k=0..n} T(n,k)*A040000(k) = A001700(n).
Sum_{k=0..n} T(n,k)*A122553(k) = A051924(n+1).
Sum_{k=0..n} T(n,k)*A123932(k) = A051944(n).
Sum_{k=0..n} T(n,k)*k^2 = A000531(n), for n>=1.
Sum_{k=0..n} T(n,k)*A000217(k) = A002457(n-1), for n>=1.
Sum{j>=0} binomial(n,j)*T(j,k)= A124733(n,k).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000984(n), A089022(n), A035610(n), A130976(n), A130977(n), A130978(n), A130979(n), A130980(n), A131521(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
Sum_{k=0..n} T(n,k)*A005043(k) = A127632(n).
Sum_{k=0..n} T(n,k)*A132262(k) = A089022(n).
T(n,k) + T(n,k+1) = A039598(n,k).
T(n,k) = A128899(n,k)+A128899(n,k+1).
Sum_{k=0..n} T(n,k)*A015518(k) = A076025(n), for n>=1. Also Sum_{k=0..n} T(n,k)*A015521(k) = A076026(n), for n>=1.
Sum_{k=0..n} T(n,k)*(-1)^k*x^(n-k) = A033999(n), A000007(n), A064062(n), A110520(n), A132863(n), A132864(n), A132865(n), A132866(n), A132867(n), A132869(n), A132897(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 respectively.
Sum_{k=0..n} T(n,k)*(-1)^(k+1)*A000045(k) = A109262(n), A000045:= Fibonacci numbers.
Sum_{k=0..n} T(n,k)*A000035(k)*A016116(k) = A143464(n).
Sum_{k=0..n} T(n,k)*A016116(k) = A101850(n).
Sum_{k=0..n} T(n,k)*A010684(k) = A100320(n).
Sum_{k=0..n} T(n,k)*A000034(k) = A029651(n).
Sum_{k=0..n} T(n,k)*A010686(k) = A144706(n).
Sum_{k=0..n} T(n,k)*A006130(k-1) = A143646(n), with A006130(-1)=0.
T(n,2*k)+T(n,2*k+1) = A118919(n,k).
Sum_{k=0..j} T(n,k) = A050157(n,j).
Sum_{k=0..2} T(n,k) = A026012(n); Sum_{k=0..3} T(n,k)=A026029(n).
Sum_{k=0..n} T(n,k)*A000045(k+2) = A026671(n).
Sum_{k=0..n} T(n,k)*A000045(k+1) = A026726(n).
Sum_{k=0..n} T(n,k)*A057078(k) = A000012(n).
Sum_{k=0..n} T(n,k)*A108411(k) = A155084(n).
Sum_{k=0..n} T(n,k)*A057077(k) = 2^n = A000079(n).
Sum_{k=0..n} T(n,k)*A057079(k) = 3^n = A000244(n).
Sum_{k=0..n} T(n,k)*(-1)^k*A011782(k) = A000957(n+1).
(End)
T(n,k) = Sum_{j=0..k} binomial(k+j,2j)*(-1)^(k-j)*A000108(n+j). - Paul Barry, Feb 17 2011
Sum_{k=0..n} T(n,k)*A071679(k+1) = A026674(n+1). - Philippe Deléham, Feb 01 2014
Sum_{k=0..n} T(n,k)*(2*k+1)^2 = (4*n+1)*binomial(2*n,n). - Werner Schulte, Jul 22 2015
Sum_{k=0..n} T(n,k)*(2*k+1)^3 = (6*n+1)*4^n. - Werner Schulte, Jul 22 2015
Sum_{k=0..n} (-1)^k*T(n,k)*(2*k+1)^(2*m) = 0 for 0 <= m < n (see also A160562). - Werner Schulte, Dec 03 2015
T(n,k) = GegenbauerC(n-k,-n+1,-1) - GegenbauerC(n-k-1,-n+1,-1). - Peter Luschny, May 13 2016
T(n,n-2) = A014107(n). - R. J. Mathar, Jan 30 2019
T(n,n-3) = n*(2*n-1)*(2*n-5)/3. - R. J. Mathar, Jan 30 2019
T(n,n-4) = n*(n-1)*(2*n-1)*(2*n-7)/6. - R. J. Mathar, Jan 30 2019
T(n,n-5) = n*(n-1)*(2*n-1)*(2*n-3)*(2*n-9)/30. - R. J. Mathar, Jan 30 2019

Extensions

Corrected by Philippe Deléham, Nov 26 2009, Dec 14 2009

A127864 Number of tilings of a 2 X n board with 1 X 1 and L-shaped tiles (where the L-shaped tiles cover 3 squares).

Original entry on oeis.org

1, 1, 5, 11, 33, 87, 241, 655, 1793, 4895, 13377, 36543, 99841, 272767, 745217, 2035967, 5562369, 15196671, 41518081, 113429503, 309895169, 846649343, 2313089025, 6319476735, 17265131521, 47169216511, 128868696065, 352075825151, 961889042433, 2627929735167
Offset: 0

Views

Author

Silvia Heubach (sheubac(AT)calstatela.edu), Feb 03 2007

Keywords

Comments

The signed version of this sequence appears as A077917.

Examples

			a(2) = 5 because the 2 X 2 board can be tiled either with 4 squares or with a single L-shaped tile (in four orientations) together with a single square tile.

Links

Crossrefs

Cf. A077917, A127865, A127866, A127867, A127868, A127869, A127870, A127871, A127872.
Column k=2 of A220054. - Alois P. Heinz, Dec 03 2012

Programs

  • Magma
    I:=[1,1,5]; [n le 3 select I[n] else Self(n-1) + 4*Self(n-2) + 2*Self(n-3): n in [1..41]]; // G. C. Greubel, Dec 08 2022
  • Mathematica
    CoefficientList[Series[1/(1-x-4*x^2-2*x^3), {x,0,30}], x]
  • SageMath
    A028860 = BinaryRecurrenceSequence(2,2,-1,1)
def A127864(n): return A028860(n+2) + (-1)^n
[A127864(n) for n in range(51)] # G. C. Greubel, Dec 08 2022

Formula

a(n) = a(n-1) + 4*a(n-2) + 2*a(n-3).
a(n) = (-1)^n + (1/sqrt(3)) * ((1+sqrt(3))^n - (1-sqrt(3))^n).
G.f.: 1/(1 - x - 4*x^2 - 2*x^3).
a(n) = A028860(n+2) + (-1)^n. - R. J. Mathar, Oct 29 2010
E.g.f.: exp(-x) + (2/sqrt(3))*exp(x)*sinh(sqrt(3)*x). - G. C. Greubel, Dec 08 2022
From Greg Dresden, Nov 10 2024: (Start)
a(n) = 1 + 4*a(n-2) + 6*Sum_{i=0..n-3} a(i) for n>1.
a(2*n) = a(n)^2 + 4*a(n-1)^2 + 4*a(n-1)*a(n-2) for n>1. (End)

A071035 Triangle read by rows giving successive states of cellular automaton generated by "Rule 126".

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1
Offset: 0

Views

Author

Hans Havermann, May 26 2002

Keywords

Comments

Row n has length 2n+1.

Examples

			Triangle begins:
                           1,
                        1, 1, 1,
                     1, 1, 0, 1, 1,
                  1, 1, 1, 1, 1, 1, 1,
               1, 1, 0, 0, 0, 0, 0, 1, 1,
            1, 1, 1, 1, 0, 0, 0, 1, 1, 1, 1,
         1, 1, 0, 0, 1, 1, 0, 1, 1, 0, 0, 1, 1,
      1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
   1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1,
1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1,
... - _N. J. A. Sloane_, Aug 24 2014

References

  • S. Wolfram, A New Kind of Science, Wolfram Media, 2002; Chapter 3.

Links

Crossrefs

See A071051 for number of ON states.
The right half is the triangle A127872.
Cf. A036987 (central column).

Programs

  • Mathematica
    A071035list[rowmax_]:=MapIndexed[ArrayPad[#1, #2-rowmax-1]&,CellularAutomaton[126,{{1},0},rowmax]];A071035list[10] (* Generates 11 rows *) (* Paolo Xausa, Jul 18 2023 *)

Extensions

Corrected by Hans Havermann, Jan 07 2012

A050614 Products of distinct terms of A001566: a(n) = Product_{i=0..floor(log_2(n+1))} L(2^(i+1))^bit(n,i).

Original entry on oeis.org

1, 3, 7, 21, 47, 141, 329, 987, 2207, 6621, 15449, 46347, 103729, 311187, 726103, 2178309, 4870847, 14612541, 34095929, 102287787, 228929809, 686789427, 1602508663, 4807525989, 10749959329, 32249877987, 75249715303
Offset: 0

Views

Author

Antti Karttunen, Dec 02 1999

Keywords

Comments

Each subset a(0..(2^k)-1) gives all the divisors of F(2^(k+1)) up to k=4 (F_32) and after that a subset of such divisors. E.g., the terms a(0)-a(7) are the divisors of F_16 = 987 (A018760).

Links

Crossrefs

Bisection of A075149 and A050613 (see there for the other Maple procedures), subset of A062877. Cf. also A050615.

Programs

  • Maple
    [seq(A050614(n),n=0..30)]; A050614 := n -> product('luc(2^(i+1))^bit_i(n,i)','i'=0..floor_log_2(n+1));
  • Mathematica
    Table[k = Floor[Log[2, n + 1]]; Product[j = 2^(i + 1); l = Fibonacci[j + 1] + Fibonacci[j - 1]; If[BitAnd[2^i, n] == 0, b = 0, b = 1]; l^b, {i, 0, k}], {n, 0, 200}] (* Robert Price, Feb 13 2017 *)

Formula

a(n) = Sum_{k=0..n} A127872(n,k)*Fibonacci(2*k+1), see A000045 and A001519. - Philippe Deléham, Aug 30 2007

A127865 Number of square tiles in all tilings of a 2 X n board with 1 X 1 and L-shaped tiles (where the L-shaped tiles cover 3 squares).

Original entry on oeis.org

2, 8, 30, 108, 354, 1152, 3614, 11204, 34170, 103176, 308598, 916236, 2702834, 7929872, 23155182, 67333140, 195082218, 563367960, 1622185958, 4658753564, 13347741666, 38160007200, 108881256414, 310108078116, 881761288154
Offset: 1

Views

Author

Silvia Heubach (sheubac(AT)calstatela.edu), Feb 03 2007

Keywords

Examples

			a(2) = 8 because the 2 X 2 board can be tiled either with 4 squares or with a single L-shaped tile (in four orientations) together with a single square tile and thus all the tilings of the 2 X 2 board contain 8 square tiles.

Links

Crossrefs

Cf. A127864, A127866, A127867, A127868, A127869, A127870, A127871, A127872.

Programs

  • Mathematica
    Table[(2n - 12)(-1)^n + (2/3)((9 - 5Sqrt[3])(1 + Sqrt[3])^n + (9 + 5Sqrt[3])(1 - Sqrt[3])^n) + (n/Sqrt[3])((Sqrt[3] - 1)( 1 + Sqrt[3])^n + (Sqrt[3] + 1)(1 - Sqrt[3])^n), {n, 1, 30}]

Formula

a(n) = (2*n - 12)*(-1)^n + (2/3)*((9-5*sqrt(3))*(1+sqrt(3))^n + (9+5*sqrt(3))*(1-sqrt(3))^n) + (n/sqrt(3))*((sqrt(3)-1)*(1+sqrt(3))^n+ (sqrt(3)+1)*(1-sqrt(3))^n).
G.f.: 2*x*(1+2*x)/((1+x)^2*(1-2*x-2*x^2)^2). - Colin Barker, Apr 30 2012

A127866 Number of L-shaped tiles in all tilings of a 2 X n board with 1 X 1 and L-shaped tiles (where the L-shaped tiles cover 3 squares).

Original entry on oeis.org

4, 12, 52, 172, 580, 1852, 5828, 17980, 54788, 165116, 493316, 1463036, 4312068, 12641276, 36887556, 107201532, 310427652, 896045052, 2579017732, 7403843580, 21205303300, 60604891132, 172872744964, 492233179132, 1399272374276
Offset: 2

Views

Author

Silvia Heubach (sheubac(AT)calstatela.edu), Feb 03 2007

Keywords

Examples

			a(2) = 4 because the 2 X 2 board can be tiled either with 4 squares or with a single L-shaped tile (in four orientations) together with a single square tile and thus all the tilings of the 2 X 2 board contain 4 L-shaped tiles.

Links

Crossrefs

Cf. A127864, A127865, A127867, A127868, A127869, A127870, A127871, A127872.

Programs

  • Mathematica
    Table[Coefficient[Normal[Series[4x^2/((1 + x)(1 - 2x - 2x^2)^2), {x, 0, 20}]], x, n], {n, 0, 20}]

Formula

a(n) = 4 (-1)^n - (2/9)[(9-5*Sqrt(3))(1+Sqrt(3))^n + (9+5*Sqrt(3))(1-Sqrt(3))^n] - (n/3)[(1-Sqrt(3))(1+Sqrt(3))^n+ (1+Sqrt(3))(1-Sqrt(3))^n].
G.f.: 4x^2/((1+x)(1-2x-2x^2)^2).

Extensions

G.f. proposed by Maksym Voznyy checked and corrected by R. J. Mathar, Sep 16 2009.

A062878 a(n) is the position of A050614(n) in A062877.

Original entry on oeis.org

1, 3, 6, 15, 24, 60, 102, 255, 384, 960, 1632, 4080, 6168, 15420, 26214, 65535, 98304, 245760, 417792, 1044480, 1579008, 3947520, 6710784, 16776960, 25166208, 62915520, 106956384, 267390960, 404232216, 1010580540, 1717986918, 4294967295, 6442450944
Offset: 0

Views

Author

Antti Karttunen, Jun 26 2001

Keywords

Comments

In binary this sequence looks like 1, 11, 110, 1111, 11000, 111100, 1100110, 11111111, 110000000, 1111000000, 11001100000, 111111110000, 1100000011000, 11110000111100, 110011001100110, ...
Sequence A282387 may be the same, but I cannot prove nor disprove this beyond a(22). - Robert Price, Feb 13 2017
Agrees with A282387 for at least 1000 terms. - Sean A. Irvine, Apr 14 2023

Programs

  • Mathematica
    A050614 = Table[k = Floor[Log[2, n + 1]]; Product[j = 2^(i + 1); l = Fibonacci[j + 1] + Fibonacci[j - 1]; If[BitAnd[2^i, n] == 0, b = 0, b = 1]; l^b, {i, 0, k}], {n, 0, 200}]; A062877 = Union[Total /@ Subsets[Fibonacci[Range[1, 46, 2]]]]; Flatten[Table[Position[ A062877, A050614[[i]] ] - 1, {i, 1, 25}]] (* Robert Price, Feb 13 2017 *)

Formula

a(2^n-1) = 2^(2^n) - 1. - Philippe Deléham, Apr 05 2007
a(n) = Sum_{k=0..n} A127872(n,k)*2^k. - Philippe Deléham, Oct 09 2007

Extensions

a(15)-a(22) from Robert Price, Feb 13 2017
More terms from Sean A. Irvine, Apr 14 2023

A302713 Decimal expansion of 2*sin(15*Pi/64).

Original entry on oeis.org

1, 3, 4, 3, 1, 1, 7, 9, 0, 9, 6, 9, 4, 0, 3, 6, 8, 0, 1, 2, 5, 0, 7, 5, 3, 7, 0, 0, 8, 5, 4, 8, 4, 3, 6, 0, 6, 4, 5, 7, 5, 0, 1, 2, 6, 4, 3, 9, 9, 5, 8, 8, 9, 9, 7, 7, 6, 6, 4, 2, 6, 1, 4, 6, 2, 1, 8, 9, 2, 9, 8, 2, 3, 7, 5, 8, 0, 0, 2, 8, 3, 0, 3, 3, 4, 5, 8, 0, 9, 8, 6, 3, 5, 6, 8, 0, 8, 3, 2, 1, 3, 2
Offset: 1

Views

Author

Wolfdieter Lang, Apr 28 2018

Keywords

Comments

This constant appears in a problem similar to a historic one posed by Adriaan van Roomen (Adrianus Romanus) in his Ideae mathematicae from 1593. See the Havil reference, pp. 69-74, problem 2 (not exemplum secundum of Romanus). See the comment on A302711, also for the Romanus link. In the Havil reference, problem 2, a further sqrt(2... is missing.
The present problem is equivalent to R(45, 2*sin(Pi/192)) =
2*sin(15*Pi/64), with the monic Chebyshev polynomial R from A127672, and for 2*sin(Pi/192) = 0.032723463252973563... see A302714. The general identity is R(2*k + 1, x) = x*(-1)^k*S(2*k, sqrt(4 - x^2)), with the Chebyshev S polynomials (see A049310 for the coefficients). Here k = 22, x = 2*sin(Pi/192).

Examples

			2*sin(15*Pi/64) = 1.3431179096940368012507537008548436064575012643995889977...

References

  • Julian Havil, The Irrationals, A Story of the Numbers You Can't Count On, Princeton University Press, Princeton and Oxford, 2012, pp. 69-74.

Links

Crossrefs

Cf. A049310, A127672, A302711, A302714.

Programs

  • Mathematica
    RealDigits[2*Sin[15 Pi/64],10,120][[1]] (* Harvey P. Dale, Sep 22 2019 *)

Formula

The constant is 2*sin(15*Pi/64) = sqrt(2-sqrt(2 - sqrt(2 + sqrt(2 + sqrt(2))))).
Root of the equation 2 + (-2 + x) * x^2 * (2 + x) * (-2 + x^2)^2 * (2 - 4*x^2 + x^4)^2 * (2 + x^2 * (-4 + x^2) * (-2 + x^2)^2)^2 = 0. - Vaclav Kotesovec, Apr 30 2018 [This is the polynomial R(32, x). See A127672 for all 32 roots. - Wolfdieter Lang, May 03 2018]
The constant also equals 2*cos(17*Pi/64), one of the roots of R(32, x) (the one for or k = 8 given in A127872). - Wolfdieter Lang, May 03 2018
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