cp's OEIS Frontend

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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A098493 Triangle T(n,k) read by rows: difference between A098489 and A098490 at triangular rows.

Original entry on oeis.org

1, 0, -1, -1, -1, 1, -1, 1, 2, -1, 0, 3, 0, -3, 1, 1, 2, -5, -2, 4, -1, 1, -2, -7, 6, 5, -5, 1, 0, -5, 0, 15, -5, -9, 6, -1, -1, -3, 12, 9, -25, 1, 14, -7, 1, -1, 3, 15, -18, -29, 35, 7, -20, 8, -1, 0, 7, 0, -42, 14, 63, -42, -20, 27, -9, 1, 1, 4, -22, -24, 85, 14, -112, 42
Offset: 0

Views

Author

Ralf Stephan, Sep 12 2004

Keywords

Comments

Also, coefficients of polynomials that have values in A098495 and A094954.

Examples

			Triangle begins:
   1;
   0, -1;
  -1, -1, 1;
  -1,  1, 2, -1;
   0,  3, 0, -3, 1;
  ...
		

Crossrefs

Columns include A010892, -A076118. Diagonals include A033999, A038608, (-1)^n*A000096. Row sums are in A057077.
Cf. A098494 (diagonal polynomials), A085478, A244419.

Programs

  • Maple
    A098493 := proc (n, k)
    add((-1)^(k+binomial(n-j+1,2))*binomial(floor((1/2)*n+(1/2)*j),j)* binomial(j,k), j = k..n);
    end proc:
    seq(seq(A098493(n, k), k = 0..n), n = 0..10); # Peter Bala, Jul 13 2021
  • PARI
    T(n,k)=if(k>n||k<0||n<0,0,if(k>=n-1,(-1)^n*if(k==n,1,-k),if(n==1,0,if(k==0,T(n-1,0)-T(n-2,0),T(n-1,k)-T(n-2,k)-T(n-1,k-1)))))

Formula

T(n, k) = A098489[n(n+1)/2, k] - A098490[n(n+1)/2, k].
Recurrence: T(n, k) = T(n-1, k)-T(n-1, k-1)-T(n-2, k); T(n, k)=0 for n<0, k>n, k<0; T(n, n)=(-1)^n; T(n, n-1)=(-1)^n*(1-n).
G.f.: (1-x)/(1+(y-1)*x+x^2). [Vladeta Jovovic, Dec 14 2009]
From Peter Bala, Jul 13 2021: (Start)
Riordan array ( (1 - x)/(1 - x + x^2), -x/(1 - x + x^2) ).
T(n,k) = (-1)^k * the (n,k)-th entry of Q^(-1)*P = Sum_{j = k..n} (-1)^(k+binomial(n-j+1,2))*binomial(floor((1/2)*n+(1/2)*j),j)* binomial(j,k), where P denotes Pascal's triangle A007318 and Q denotes triangle A061554 (formed from P by sorting the rows into descending order). (End)
From Peter Bala, Jun 26 2025: (Start)
n-th row polynomial R(n, x) = Sum_{k = 0..n} (-1)^k * binomial(n+k, 2*k) * (1 + x)^k.
R(n, 2*x + 1) = (-1)^n * Dir(n, x), where Dir(n,x) denotes the n-th row polynomial of the triangle A244419.
R(n, -1 - x) = b(n, x), where b(n, x) denotes the n-th row polynomial of the triangle A085478. (End)

A332602 Tridiagonal matrix M read by antidiagonals: main diagonal is 1,2,2,2,2,..., two adjacent diagonals are 1,1,1,1,1,...

Original entry on oeis.org

1, 1, 1, 0, 2, 0, 0, 1, 1, 0, 0, 0, 2, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

N. J. A. Sloane, Mar 06 2020, following a suggestion from Gary W. Adamson

Keywords

Comments

From Gary W. Adamson, Mar 11 2020: (Start)
The upper left entry of M^n gives the Catalan numbers A000108. Extracting 2 X 2, 3 X 3, and 4 X 4 submatrices from M; then generating sequences from the upper left entries of M^n, we obtain the following sequences:
1, 1, 2, 5, 13, ... = A001519 and the convergent is 2.61803... = 2 + 2*cos(2*Pi/5) = (2*cos(Pi/5))^2.
1, 1, 2, 5, 14, 42, 131, ... = A080937 and the convergent is 3.24697... = 2 + 2*cos(2*Pi/7) = (2*cos(Pi/7))^2.
1, 1, 2, 5, 14, 42, 132, 429, 1429, ... = A080938 and the convergent is 3.53208... = 2 + 2*cos(2*Pi/9) = (2*cos(Pi/9))^2. (End)
The characteristic polynomial for the N X N main submatrix M_N is Phi(N, x) = S(N, 2-x) - S(N-1, 2-x), with Chebyshev's S polynomial (see A049310) evaluated at 2-x. Proof by determinant expansion, to obtain the recurrence Phi(N, x) - (x-2)*Phi(N-1, x) - Phi(N-2, x), for N >= 2, and Phi(0, x) = 1 and Phi(1, x) = 1 - x, that is Phi(-1, x) = 1. The trace is tr(M_N) = 1 + 2^(N-1) = A000051(N-1), and Det(M_N) = 1. - Wolfdieter Lang, Mar 13 2020
The explicit form of the characteristic polynomial for the N X N main submatrix M_N is Phi(N, x) := Det(M_N - x*1_N) = Sum_{k=0..N} binomial(N+k, 2*k)*(-x)^k = Sum_{k=0..N} A085478(N, k)*(-x)^k, for N >= 0, with Phi(0, x) := 1. Proof from the recurrence given in the preceding comment. - Wolfdieter Lang, Mar 25 2020
For the proofs of the 2 X 2, 3 X 3 and 4 X 4 conjectures, see the comments in the respective A-numbers A001519, A080937 and A080938. - Wolfdieter Lang, Mar 30 2020
Replace the main diagonal 1,2,2,2,... of the matrix M with 1,0,0,0,...; 1,1,1,1,...; 1,3,3,3,...; 1,2,1,2,...; 1,2,3,4,...; 1,0,1,0...; and 1,1,0,0,1,1,0,0,.... Take powers of M and extract the upper left terms, resulting in respectively: A001405, A001006, A033321, A176677, A006789, A090344, and A007902. - Gary W. Adamson, Apr 12 2022
The statement that the upper left entry of M^n is a Catalan number is equivalent to Exercise 41 of R. Stanley, "Catalan Numbers." - Richard Stanley, Feb 28 2023
If the upper left 1 in matrix M is replaced with 3, taking powers of the resulting matrix and extracting the upper left terms apparently results in sequence A001700. - Gary W. Adamson, Apr 03 2023

Examples

			The matrix begins:
  1, 1, 0, 0, 0, ...
  1, 2, 1, 0, 0, ...
  0, 1, 2, 1, 0, ...
  0, 0, 1, 2, 1, ...
  0, 0, 0, 1, 2, ...
  ...
The first few antidiagonals are:
  1;
  1, 1;
  0, 2, 0;
  0, 1, 1, 0;
  0, 0, 2, 0, 0;
  0, 0, 1, 1, 0, 0;
  0, 0, 0, 2, 0, 0, 0;
  0, 0, 0, 1, 1, 0, 0, 0;
  0, 0, 0, 0, 2, 0, 0, 0, 0;
  0, 0, 0, 0, 1, 1, 0, 0, 0, 0;
  ...
Characteristic polynomial of the 3 X 3 matrix M_3: Phi(3, x) = 1 - 6*x + 5*x^2 - x^3, from {A085478(3, k)}_{k=0..3} = {1, 6, 5, 1}. - _Wolfdieter Lang_, Mar 25 2020
		

References

  • Richard P. Stanley, "Catalan Numbers", Cambridge University Press, 2015.

Crossrefs

Cf. A001333 (permanent of the matrix M).
Cf. A054142, A053123, A011973 (characteristic polynomials of submatrices of M).
Cf. A001700.

A182432 Recurrence a(n)*a(n-2) = a(n-1)*(a(n-1) + 3) with a(0) = 1, a(1) = 4.

Original entry on oeis.org

1, 4, 28, 217, 1705, 13420, 105652, 831793, 6548689, 51557716, 405913036, 3195746569, 25160059513, 198084729532, 1559517776740, 12278057484385, 96664942098337, 761041479302308, 5991666892320124, 47172293659258681, 371386682381749321
Offset: 0

Views

Author

Peter Bala, Apr 30 2012

Keywords

Comments

The non-linear recurrence equation a(n)*a(n-2) = a(n-1)*(a(n-1) + r) with initial conditions a(0) = 1, a(1) = 1 + r has the solution a(n) = 1/2 + (1/2)*Sum_{k = 0..n} (2*r)^k*binomial(n+k,2*k) = 1/2 + b(n,2*r)/2, where b(n,x) are the Morgan-Voyce polynomials of A085478. The recurrence produces sequences A101265 (r = 1), A011900 (r = 2) and A054318 (r = 4), as well as signed versions of A133872 (r = -1), A109613 (r = -2), A146983 (r = -3) and A084159(r = -4).
Also the indices of centered pentagonal numbers (A005891) which are also centered triangular numbers (A005448). - Colin Barker, Jan 01 2015
Also positive integers y in the solutions to 3*x^2 - 5*y^2 - 3*x + 5*y = 0. - Colin Barker, Jan 01 2015

Crossrefs

Programs

  • Magma
    I:=[1, 4, 28]; [n le 3 select I[n] else 9*Self(n-1)-9*Self(n-2)+Self(n-3): n in [1..25]]; // Vincenzo Librandi, May 18 2012
    
  • Mathematica
    RecurrenceTable[{a[0]==1,a[1]==4,a[n]==(a[-1+n] (3+a[-1+n]))/a [-2+n]}, a[n],{n,30}] (* or *) LinearRecurrence[{9,-9,1},{1,4,28},30] (* Harvey P. Dale, May 14 2012 *)
  • PARI
    Vec((1-5*x+x^2)/((1-x)*(1-8*x+x^2)) + O(x^100)) \\ Colin Barker, Jan 01 2015

Formula

a(n) = 1/2 + (1/2)*Sum_{k = 0..n} 6^k*binomial(n+k,2*k).
a(n) = R(n,3) where R(n,x) denotes the row polynomials of A211955.
a(n) = (1/u)*T(n,u)*T(n+1,u) with u = sqrt(5/2) and T(n,x) the n-th Chebyshev polynomial of the first kind.
Recurrence equation: a(n) = 8*a(n-1) - a(n-2) - 3 with a(0) = 1 and a(1) = 4.
O.g.f.: (1 - 5*x + x^2)/((1 - x)*(1 - 8*x + x^2)) = 1 + 4*x + 28*x^2 + ....
Sum_{n >= 0} 1/a(n) = sqrt(5/3); 5 - 3*(Sum_{k = 0..2*n} 1/a(k))^2 = 2/A070997(n)^2.
a(0) = 1, a(1) = 4, a(2) = 28, a(n) = 9*a(n-1) - 9*a(n-2) + a(n-3). - Harvey P. Dale, May 14 2012

A143858 Number of pairwise disjoint unions of m integer-to-integer subintervals of [0,n]; a rectangular array by antidiagonals, n>=2m-1, m>=1.

Original entry on oeis.org

1, 3, 1, 6, 5, 1, 10, 15, 7, 1, 15, 35, 28, 9, 1, 21, 70, 84, 45, 11, 1, 28, 126, 210, 165, 66, 13, 1, 36, 210, 462, 495, 286, 91, 15, 1, 45, 330, 924, 1287, 1001, 455, 120, 17, 1, 55, 495, 1716, 3003, 3003, 1820, 680, 153, 19, 1, 66, 715, 3003, 6435, 8008, 6188, 3060
Offset: 1

Views

Author

Clark Kimberling, Sep 03 2008

Keywords

Comments

Main diagonal: A025174.

Examples

			R(2,4) counts these unions of 2 subintervals of [0,4]: [0,1]U[2,3], [0,1]U[2,4], [0,1]U[3,4], [0,2]U[3,4], [1,2]U[3,4].
   1    3    6   10   15   21   28   36   45   55   66   78
   0    0    1    5   15   35   70  126  210  330  495  715
   0    0    0    0    1    7   28   84  210  462  924 1716
   0    0    0    0    0    0    1    9   45  165  495 1287
   0    0    0    0    0    0    0    0    1   11   66  286
   0    0    0    0    0    0    0    0    0    0    1   13
		

Crossrefs

Programs

  • Haskell
    Seen as a triangle read by rows
    a143858 n k = a143858_tabl !! (n-1) !! k
    a143858_row n = a143858_tabl !! (n-1)
    a143858_tabl = map ((++ [1]) . tail) a258993_tabl
    -- Reinhard Zumkeller, Jun 22 2015
  • Maple
    A143858 := proc(m,n)
        binomial(n-1+2*m,2*m) ;
    end proc:
    seq(seq( A143858(n,d-n),n=1..d-1),d=2..8) ; # R. J. Mathar, Nov 16 2023

Formula

R(m,n) = C(n+1,2m), where n>=2m-1, m>=1. R is also given by the absolute values of terms in A109954.

A202023 Triangle T(n,k), read by rows, given by (1, 0, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, 1, -1, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 3, 0, 0, 1, 6, 1, 0, 0, 1, 10, 5, 0, 0, 0, 1, 15, 15, 1, 0, 0, 0, 1, 21, 35, 7, 0, 0, 0, 0, 1, 28, 70, 28, 1, 0, 0, 0, 0, 1, 36, 126, 84, 9, 0, 0, 0, 0, 0, 1, 45, 210, 210, 45, 1, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Philippe Deléham, Dec 10 2011

Keywords

Comments

Riordan array (1/(1-x), x^2/(1-x)^2).
A skewed version of triangular array A085478.
Mirror image of triangle in A098158.
Sum_{k, 0<=k<=n} T(n,k)*x^k = A138229(n), A006495(n), A138230(n),A087455(n), A146559(n), A000012(n), A011782(n), A001333(n),A026150(n), A046717(n), A084057(n), A002533(n), A083098(n),A084058(n), A003665(n), A002535(n), A133294(n), A090042(n),A125816(n), A133343(n), A133345(n), A120612(n), A133356(n), A125818(n) for x = -5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18 respectively.
Sum_{k, 0<=k<=n} T(n,k)*x^(n-k) = A009116(n), A000007(n), A011782(n), A006012(n), A083881(n), A081335(n), A090139(n), A145301(n), A145302(n), A145303(n), A143079(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively.
From Gus Wiseman, Jul 08 2025: (Start)
After the first row this is also the number of subsets of {1..n-1} with k maximal runs (sequences of consecutive elements increasing by 1) for k = 0..n. For example, row n = 5 counts the following subsets:
{} {1} {1,3} . . .
{2} {1,4}
{3} {2,4}
{4} {1,2,4}
{1,2} {1,3,4}
{2,3}
{3,4}
{1,2,3}
{2,3,4}
{1,2,3,4}
Requiring n-1 gives A202064.
For anti-runs instead of runs we have A384893.
(End)

Examples

			Triangle begins :
1
1, 0
1, 1, 0
1, 3, 0, 0
1, 6, 1, 0, 0
1, 10, 5, 0, 0, 0
1, 15, 15, 1, 0, 0, 0
1, 21, 35, 7, 0, 0, 0, 0
1, 28, 70, 28, 1, 0, 0, 0, 0
		

Crossrefs

Column k = 1 is A000217.
Column k = 2 is A000332.
Row sums are A011782 (or A000079 shifted right).
Removing all zeros gives A034839 (requiring n-1 A034867).
Last nonzero term in each row appears to be A093178, requiring n-1 A124625.
Reversing rows gives A098158, without zeros A109446.
Without the k = 0 column we get A210039.
Row maxima appear to be A214282.
A116674 counts strict partitions by number of maximal runs, for anti-runs A384905.
A268193 counts integer partitions by number of maximal runs, for anti-runs A384881.

Programs

  • Mathematica
    Table[Length[Select[Subsets[Range[n-1]],Length[Split[#,#2==#1+1&]]==k&]],{n,0,10},{k,0,n}] (* Gus Wiseman, Jul 08 2025 *)

Formula

T(n,k) = binomial(n,2k).
G.f.: (1-x)/((1-x)^2-y*x^2).
T(n,k)= Sum_{j, j>=0} T(n-1-j,k-1)*j with T(n,0)=1 and T(n,k)= 0 if k<0 or if n
T(n,k) = 2*T(n-1,k) + T(n-2,k-1) - T(n-2,k) for n>1, T(0,0) = T(1,0) = 1, T(1,1) = 0, T(n,k) = 0 if k>n or if k<0. - Philippe Deléham, Nov 10 2013

A204021 Triangle read by rows: row n shows the coefficients of the characteristic polynomial of the n-th principal submatrix of min(2i-1,2j-1) (A157454).

Original entry on oeis.org

1, 1, -1, 2, -4, 1, 4, -12, 9, -1, 8, -32, 40, -16, 1, 16, -80, 140, -100, 25, -1, 32, -192, 432, -448, 210, -36, 1, 64, -448, 1232, -1680, 1176, -392, 49, -1, 128, -1024, 3328, -5632, 5280, -2688, 672, -64, 1, 256, -2304, 8640, -17472, 20592
Offset: 0

Author

Clark Kimberling, Jan 11 2012

Keywords

Comments

Let p(n)=p(n,x) be the characteristic polynomial of the n-th principal submatrix. The zeros of p(n) are real, and they interlace the zeros of p(n+1). See A202605 and A204016 for guides to related sequences.
a(0)=1 by convention. - Philippe Deléham, Nov 17 2013
The n roots of the n-th polynomial are 1/(1+cos((2*k-1)*Pi/(2*n))) for k = 1..n. See my pdf in the link section for the proof. - Jianing Song, Dec 01 2023

Examples

			Top of the triangle:
  1
  1....-1
  2....-4.....1
  4....-12....9....-1
  8....-32....40...-16....1
  16...-80....140..-100...25....-1
  32...-192...432..-448...210...-36....1
  ...
-448=2*(-100)-2*140-(-32). - _Philippe Deléham_, Nov 17 2013
		

References

  • (For references regarding interlacing roots, see A202605.)

Crossrefs

Programs

  • Mathematica
    f[i_, j_] := Min[2 i - 1, 2 j - 1];
    m[n_] := Table[f[i, j], {i, 1, n}, {j, 1, n}]
    TableForm[m[6]] (* 6x6 principal submatrix *)
    Flatten[Table[f[i, n + 1 - i],
      {n, 1, 15}, {i, 1, n}]]   (* A157454 *)
    p[n_] := CharacteristicPolynomial[m[n], x];
    c[n_] := CoefficientList[p[n], x]
    TableForm[Flatten[Table[p[n], {n, 1, 10}]]]
    Table[c[n], {n, 1, 12}]
    Flatten[%]                  (* A204021 *)
    TableForm[Table[c[n], {n, 1, 10}]]

Formula

From Peter Bala, May 01 2012: (Start)
The triangle appears to be a signed version of the row reverse of A211957.
If true, then for 0 <= k <= n-1, T(n,k) = (-1)^k*n/(n-k)*2^(n-k-1)*binomial(2*n-k-1,k) and Sum_{k = 0..n} T(n,k)*x^(n-k) = 1/2*(-1)^n*(b(2*n,-2*x) + 1)/b(n,-2*x), where b(n,x) := Sum_{k = 0..n} binomial(n+k,2*k)*x^k are the Morgan-Voyce polynomials of A085478.
Conjectural o.g.f.: t*(1-x-x^2*t)/(1-2*t*(1-x)+t^2*x^2) = (1-x)*t + (2-4*x+x^2)*t^2 + .... (End)
T(n,k)=2*T(n-1,k)-2*T(n-1,k-1)-T(n-2,k-2), T(0,0)=T(1,0)=1, T(1,1)=-1, T(n,k)=0 of k<0 or if k>n. - Philippe Deléham, Nov 17 2013

A211956 Coefficients of a sequence of polynomials related to the Morgan-Voyce polynomials.

Original entry on oeis.org

1, 1, 1, 1, 1, 2, 1, 4, 2, 1, 6, 4, 1, 9, 12, 4, 1, 12, 20, 8, 1, 16, 40, 32, 8, 1, 20, 60, 56, 16, 1, 25, 100, 140, 80, 16, 1, 30, 140, 224, 144, 32, 1, 36, 210, 448, 432, 192, 32, 1, 42, 280, 672, 720, 352, 64, 1, 49, 392, 1176, 1680, 1232, 448, 64
Offset: 0

Author

Peter Bala, Apr 30 2012

Keywords

Comments

The row generating polynomials R(n,x) of A211955 factorize in the ring Z[x] as R(n,x) = P(n,x)*P(n+1,x) for n >= 1: explicitly, P(2*n,x) = 1/2*(b(2*n,2*x) + 1)/b(n,2*x) and P(2*n+1,x) = b(n,2*x), where b(n,x) := Sum_{k = 0..n} binomial(n+k,2*k)*x^k are the Morgan-Voyce polynomials of A085478. This triangle lists the coefficients in ascending powers of x of the polynomials P(n,x).
The odd numbered rows of the present triangle produce triangle A123519; the even numbered row entries are recorded separately in A211957 and appear to equal the unsigned and row reversed form of A204021. The even numbered rows with a factor of 2^(k-1) removed from the k-th column entries produce triangle A208513.

Examples

			Triangle begins
.n\k.|..0....1....2....3....4
= = = = = = = = = = = = = = =
..0..|..1
..1..|..1
..2..|..1....1
..3..|..1....2
..4..|..1....4....2
..5..|..1....6....4
..6..|..1....9...12....4
..7..|..1...12...20....8
..8..|..1...16...40...32....8
..9..|..1...20...60...56...16
...
		

Crossrefs

Formula

T(n,0) = 1; for k > 0, T(2*n,k) = 2^k * binomial(n+k,2*k) = A123519(n,k);
for k > 0, T(2*n-1,k) = n/(n+k)*(2^k)*binomial(n+k,2*k) = 2^(k-1)*A208513(n,k).
O.g.f.: ((1+t)*(1-t^2)-t^2*x)/((1-t^2)^2-2*t^2*x) = 1 + t + (1+x)*t^2 + (1+2*x)*t^3 + (1+4*x+2*x^2)*t^4 + ....
Row generating polynomials: P(2*n,x) := 1/2*(b(2*n,2*x)+1)/b(n,2*x) and P(2*n+1,x) := b(n,2*x), where b(n,x) := Sum_{k = 0..n} binomial(n+k,2*k)*x^k are the Morgan-Voyce polynomials of A085478.
The product P(n,x)*P(n+1,x) is the n-th row polynomial of A211955.
In terms of T(n,x), the Chebyshev polynomials of the first kind, we have P(2*n,x) = T(2*n,u) and P(2*n+1,x) = 1/u*T(2*n+1,u), where u = sqrt((x+2)/2).
Other representations for the row polynomials include
P(2*n,x) = 1/2*(1+x+sqrt(x^2+2*x))^n + 1/2*(1+x-sqrt(x^2+2*x))^n;
P(2*n,x) = n*Sum_{k = 0..n}(-1)^(n-k)/(n+k) * binomial(n+k,2*k) * (2*x+4)^k for n >= 1;
P(2*n+1,x) = (2*n+1)*Sum_{k=0..n} (-1)^(n-k)/(n+k+1) * binomial(n+k+1,2*k+1) * (2*x+4)^k.
Recurrence equation: P(n+1,x)*P(n-2,x) - P(n,x)*P(n-1,x) = x.
Row sums A005246(n+2).

A123519 Triangle read by rows: T(n,k) number of tilings of a 2n X 3 grid by dominoes, 2k of which are in a vertical position (0<=k<=n).

Original entry on oeis.org

1, 1, 2, 1, 6, 4, 1, 12, 20, 8, 1, 20, 60, 56, 16, 1, 30, 140, 224, 144, 32, 1, 42, 280, 672, 720, 352, 64, 1, 56, 504, 1680, 2640, 2112, 832, 128, 1, 72, 840, 3696, 7920, 9152, 5824, 1920, 256, 1, 90, 1320, 7392, 20592, 32032, 29120, 15360, 4352, 512, 1, 110, 1980, 13728, 48048, 96096, 116480, 87040, 39168, 9728, 1024
Offset: 0

Author

Emeric Deutsch, Oct 16 2006

Keywords

Comments

Sum of terms in row n = A001835(n+1). Sum(k*T(n,k), k=0..n)=A123520(n) (n>=1).

Examples

			T(1,1)=2 because a 2 X 3 grid can be tiled in 2 ways with dominoes so that exactly 2 dominoes are in vertical position: place a horizontal domino above or below two adjacent vertical dominoes.
		

Crossrefs

Programs

  • Maple
    T:=(n,k)->2^k*binomial(n+k,2*k): for n from 0 to 10 do seq(T(n,k),k=0..n) od; # yields sequence in triangular form
  • Mathematica
    Table[2^k*Binomial[n + k, 2*k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 14 2017 *)
    CoefficientList[Table[Sqrt[2] Cosh[(2 n + 1) ArcSinh[Sqrt[x/2]]]/Sqrt[2 + x], {n, 0, 10}] // FunctionExpand // Simplify, x] // Flatten (* Eric W. Weisstein, Apr 04 2018 *)
    CoefficientList[Table[ChebyshevT[2 n - 1, Sqrt[1 + x/2]]/Sqrt[1 + x/2], {n, 10}], x] (* Eric W. Weisstein, Apr 04 2018 *)
  • PARI
    for(n=0,10, for(k=0,n, print1(2^k*binomial(n+k,2*k), ", "))) \\ G. C. Greubel, Oct 14 2017

Formula

T(n,k) = 2^k * binomial(n+k,2*k).
G.f.: (1-z)/(1 - 2*z + z^2 - 2*t*z).
Sum_{k=0..n} k*T(n,k) = A123520(n) (n>=1).
Row polynomials are b(n,2*x), where b(n,x) := Sum_{k = 0..n} binomial(n+k,2*k) * x^k are the Morgan-Voyce polynomials of A085478. The triangle is made up of the odd-indexed rows of A211956. - Peter Bala, May 01 2012

Extensions

Terms a(57) onward added by G. C. Greubel, Oct 14 2017

A284938 Triangle read by rows: coefficients of the edge cover polynomial for the n-path graph P_n.

Original entry on oeis.org

0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 0, 0, 2, 1, 0, 0, 0, 1, 3, 1, 0, 0, 0, 0, 3, 4, 1, 0, 0, 0, 0, 1, 6, 5, 1, 0, 0, 0, 0, 0, 4, 10, 6, 1, 0, 0, 0, 0, 0, 1, 10, 15, 7, 1, 0, 0, 0, 0, 0, 0, 5, 20, 21, 8, 1, 0, 0, 0, 0, 0, 0, 1, 15, 35, 28, 9, 1, 0, 0, 0, 0, 0, 0, 0, 6, 35, 56, 36, 10, 1, 0, 0, 0, 0, 0, 0, 0, 1, 21, 70, 84, 45, 11, 1
Offset: 1

Author

Eric W. Weisstein, Apr 06 2017

Keywords

Examples

			0;
0,1;
0,0,1;
0,0,1,1;
0,0,0,2,1;
0,0,0,1,3,1;
0,0,0,0,3,4,1;
0,0,0,0,1,6,5,1;
0,0,0,0,0,4,10,6,1;
0,0,0,0,0,1,10,15,7,1;
0,0,0,0,0,0,5,20,21,8,1;
0,0,0,0,0,0,1,15,35,28,9,1;
0,0,0,0,0,0,0,6,35,56,36,10,1;
0,0,0,0,0,0,0,1,21,70,84,45,11,1;
...
		

Crossrefs

Unsigned version of A057094.
Row sums are A000045(n-1).

Programs

  • Mathematica
    Prepend[CoefficientList[Table[x^(n/2) Fibonacci[n - 1, Sqrt[x]], {n, 2, 14}], x], {0}] // Flatten (* Eric W. Weisstein, Apr 06 2017 *)
    Prepend[CoefficientList[LinearRecurrence[{x, x}, {0, x}, {2, 14}], x], {0}] // Flatten (* Eric W. Weisstein, Apr 07 2017 *)

Formula

a(n) = abs(A057094(n)).

A108366 L(n,n), where L is defined as in A108299.

Original entry on oeis.org

1, 0, 1, 13, 153, 2089, 33461, 620166, 13097377, 310957991, 8205571449, 238367471761, 7561422605881, 260127000028908, 9647591076297901, 383769576967012081, 16299953773597203585, 736281113282903567521, 35246262383544562907057, 1782495208063575448970418
Offset: 0

Author

Reinhard Zumkeller, Jun 01 2005

Keywords

Comments

A108367(n) = L(n,-n).

Crossrefs

Programs

  • Mathematica
    Join[{1,0,1}, Table[Sum[Binomial[n + k, 2*k] * (n-2)^k, {k,0,n}], {n,3,20}]] (* Vaclav Kotesovec, Jan 06 2021 *)
  • PARI
    a(n) = sum(k=0, n, binomial(n+k,2*k)*(n-2)^k); \\ Jinyuan Wang, Feb 25 2020

Formula

a(n) = Product_{k=1..n} (n - 2*cos((2*k-1)*Pi/(2*n+1))) with Pi = 3.14...
a(n) = Sum_{k=0..n} binomial(n+k,2*k)*(n-2)^k = b(n,n-2), where b(n,x) are the Morgan-Voyce polynomials of A085478. - Peter Bala, May 01 2012
a(n) ~ n^n * (1 - 2/n + 5/(2*n^2) - 31/(6*n^3) + 209/(24*n^4) - 173/(10*n^5) + ...). - Vaclav Kotesovec, Jan 06 2021

Extensions

More terms from Jinyuan Wang, Feb 25 2020
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