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.

Showing 1-9 of 9 results.

A094829 Number of (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < 9 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+1, s(0) = 1, s(2n+1) = 6.

Original entry on oeis.org

1, 6, 27, 109, 417, 1548, 5644, 20349, 72846, 259579, 922209, 3269889, 11579032, 40967400, 144863001, 512050438, 1809503019, 6393427173, 22587086305, 79791176292, 281856708180, 995606748757, 3516721295214
Offset: 2

Views

Author

Herbert Kociemba, Jun 13 2004

Keywords

Comments

In general, a(n) = (2/m)*Sum_{r=1..m-1} sin(r*j*Pi/m)*sin(r*k*Pi/m)*(2*cos(r*Pi/m))^(2n+1) counts (s(0), s(1), ..., s(2n+1)) such that 0 < s(i) < m and |s(i)-s(i-1)| = 1 for i = 1,2,...,2n+1, s(0) = j, s(2n+1) = k.
a(n)/a(n-1) tends to 3.53208888...; = 2 + 2*cos(2*Pi/9) = A332438. - Gary W. Adamson, May 29 2008
From Wolfdieter Lang, Mar 27 2020: (Start)
The explicit form is written in terms of r = rho(9) = 2*cos(Pi/9) = A332437 as a Binet - de Moivre type formula a(n+2) = r^(2*(n+1))*(A(r) + Bp(r)*Cp(r)^(n+1)) + Bm(r)*Cm(r)^(n+1)), with A(r) = (1/9)*(2 + 5*r -r^2), approx. 0.87387081, Bp(r) = (1/18)*((14*r^2 - 5*r - 42)*sqrt(3*(3*r + 1)*(r - 1)) + r^2 - 5*r - 2) = (1/9)*(8 - r - 4*r^2), approx. -0.88974898, Cp(r) = (1/2)*(9*r^2 - 3*r - 26)*(3*r - 1 + sqrt(3*(3*r+1)*(r-1))) = 32 + 4*r - 11*r^2, approx. 0.66456322, Bm(r) = (1/18)*(-(14*r^2 - 5*r - 42)*sqrt(3*(3*r + 1)*(r - 1)) + r^2 - 5*r - 2) = (1/9)*(-10 -4*r + 5*r^2), approx. 0.01587816, and Cm(r) = (1/2)*(9*r^2 - 3*r - 26)*(3*r - 1 - sqrt(3*(3*r + 1)*(r - 1))) = 21 + 2*r - 7*r^2, approx. 0.03414828, for n >= 0.
Proof by partial fraction decomposition of the g.f. using the roots of 1 - 6*x + 9*x^2 - x^3 written in terms of r, which are X1(r) = 1/r^2 = 9 + r - 3*r^2, approx. 0.28311858, Xp(r) = (r/2)*(3*r - 1 + sqrt((3*(3*r+1))*(r-1))) = 1 + 2*r + r^2, approx. 8.29085937, Xm(r) = (r/2)*(3*r - 1 - sqrt((3*(3*r + 1))*(r - 1))) = -1 - 3*r + 2*r^2, approx. 0.42602205. Xp(r)*Xm(r) = r^2. The reduction with the minimal polynomial of r = rho(9), i.e., C(9, x) = x^3 - 3*x - 1 (see A187360) has been used to avoid all powers of r larger than 2. The reciprocal roots turn out to be the roots of the minimal polynomial of r^2, see A332438. 1/X1(r) = r^2, 1/Xp(r) = 2 - r, and 1/Xm(r) = 4 + r - r^2.
This proves the above stated limit of a(n+3)/a(n+2) for n to infinity, namely r^2 = A332438, approx. 3.53208889.
(End)

Crossrefs

Programs

  • Mathematica
    Drop[CoefficientList[Series[x^2/(1 - 6 x + 9 x^2 - x^3), {x, 0, 24}], x], 2] (* Michael De Vlieger, Aug 05 2021 *)

Formula

a(n) = (2/9)*Sum_{r=1..8} sin(r*Pi/9)*sin(2*r*Pi/3)*(2*cos(r*Pi/9))^(2*n+1), for n >= 2.
a(n) = 6*a(n-1) - 9*a(n-2) + a(n-3).
G.f.: x^2/(1 - 6x + 9x^2 - x^3).
For the explicit form of a(n+2), for n >= 0, see a comment above. - Wolfdieter Lang, Mar 26 2020

A332438 Decimal expansion of (2*cos(Pi/9))^2 = A332437^2.

Original entry on oeis.org

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

Views

Author

Wolfdieter Lang, Mar 31 2020

Keywords

Comments

This algebraic number rho(9)^2 of degree 3 is a root of its minimal polynomial x^3 - 6*x^2 + 9*x - 1.
The other two roots are x2 = (2*cos(5*Pi/9))^2 = (2*cos(4*Pi/9))^2 = (R(4,rho(9)))^2 = 2 - rho(9) = 0.120614758..., and x3 = (2*cos(7*Pi/9))^2 = (2*cos(7*Pi/9))^2 = (R(7,rho(9)))^2 = 4 + rho(9) - rho(9)^2 = 2.347296355... = A130880 + 2, with rho(9) = 2*cos(Pi/9) = A332437, the monic Chebyshev polynomials R (see A127672), and the computation is done modulo the minimal polynomial of rho(9) which is x^3 - 3*x - 1 (see A187360).
This gives the representation of these roots in the power basis of the simple field extension Q(rho(9)). See the linked W. Lang paper in A187360, sect. 4.
This number rho(9)^2 appears as limit of the quotient of consecutive numbers af various sequences, e.g., A094256 and A094829.
The algebraic number rho(9)^2 - 2 = 1.532088898... of Q(rho(9)) has minimal polynomial x^3 - 3*x + 1 over Q. The other roots are -rho(9) = -A332437 and 2 + rho(9) - rho(9)^2 = A130880. - Wolfdieter Lang, Sep 20 2022

Examples

			3.5320888862379560704047853011108333478716649...
		

Crossrefs

2 + 2*cos(2*Pi/n): A104457 (n = 5), A116425 (n = 7), A296184 (n = 10), A019973 (n = 12).

Programs

  • Mathematica
    RealDigits[(2*Cos[Pi/9])^2, 10, 100][[1]] (* Amiram Eldar, Mar 31 2020 *)
  • PARI
    (2*cos(Pi/9))^2 \\ Michel Marcus, Sep 23 2022

Formula

Equals (2*cos(Pi/9))^2 = rho(9)^2 = A332437^2.
Equals 2 + i^(4/9) - i^(14/9). - Peter Luschny, Apr 04 2020
Equals 2 + w1^(1/3) + w2^(1/3), where w1 = (-1 + sqrt(3)*i)/2 = exp(2*Pi*i/3) and w2 = (-1 - sqrt(3)*i)/2 are the complex roots of x^3 - 1. - Wolfdieter Lang, Sep 20 2022
Constant c = 2 + 2*cos(2*Pi/9). The linear fractional transformation z -> c - c/z has order 9, that is, z = c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(c - c/(z))))))))). - Peter Bala, May 09 2024
From Amiram Eldar, Nov 22 2024: (Start)
Equals 3 + sec(Pi/9)/2 = 3 + 1/(2*A019879).
Equals 3 + Product_{k>=3} (1 + (-1)^k/A063289(k)). (End)

A080938 Number of Catalan paths (nonnegative, starting and ending at 0, step +-1) of 2*n steps with all values less than or equal to 7.

Original entry on oeis.org

1, 1, 2, 5, 14, 42, 132, 429, 1429, 4846, 16645, 57686, 201158, 704420, 2473785, 8704089, 30664890, 108126325, 381478030, 1346396146, 4753200932, 16783118309, 59266297613, 209302921830, 739203970773, 2610763825782, 9221050139566, 32568630376132
Offset: 0

Views

Author

Henry Bottomley, Feb 25 2003

Keywords

Comments

From Wolfdieter Lang, Mar 27 2020: (Start)
a(n) also gives the upper left entry of the n-th power of the 4 X 4 tridiagonal matrix M_4, given in A332602: M_4 = Matrix([1,1,0,0], [1,2,1,0], [0,1,2,1], [0,0,1,2]): a(n) = (M_4)^n[1,1]. Proof from the formula for (M_4)^n, given in a comment in A094256, derived from the Cayley-Hamilton theorem, which leads to the recurrence. The formula for a(n) in terms of A094256 is given below.
For A094256(n+1)/A094256(n), like for A094829(n+1)/A094829(n), the limit for n -> infinity is rho(9)^2 = A332438 = 3.53208888..., with rho(9) = 2*cos(Pi/9) = A332437. Therefore the formula of a(n) in terms of A094256 shows that the same limit is reached for a(n+1)/a(n). See this conjecture by Gary W. Adamson in A332602.
(End)

Examples

			1 + x + 2*x^2 + 5*x^3 + 14*x^4 + 42*x^5 + 132*x^6 + 429*x^7 + ...
		

Crossrefs

Cf. A000007, A000012, A011782, A001519, A007051, A080937, A024175, A033191 which essentially provide the same sequence for different limits and tend to A000108.
Cf. A211216, A094826 (first differences), A094829, A094829, A332602, A332437, A332438.

Programs

  • Magma
    I:=[1,1,2,5]; [n le 4 select I[n] else 7*Self(n-1)-15*Self(n-2)+10*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Nov 30 2018
  • Mathematica
    CoefficientList[Series[(1 - 2 x) (2 x^2 - 4 x + 1) / ((x - 1) (x^3 - 9 x^2 + 6 x - 1)), {x, 0, 40}], x] (* Vincenzo Librandi, Nov 30 2018 *)
    LinearRecurrence[{7, -15, 10, -1}, {1, 1, 2, 5}, 30] (* Jean-François Alcover, Jan 07 2019 *)
  • PARI
    {a(n) = local(A); A = 1; for( i=1, 7, A = 1 / (1 - x*A)); polcoeff( A + x * O(x^n), n)} /* Michael Somos, May 12 2012 */
    

Formula

a(n) = A080934(n,7).
G.f.: -(2*x - 1)*(2*x^2 - 4*x + 1) / ( (x - 1)*(x^3 - 9*x^2 + 6*x - 1) ). - Ralf Stephan, May 13 2003
a(n) = 7*a(n-1) - 15*a(n-2) + 10*a(n-3) - a(n-4). - Herbert Kociemba, Jun 13 2004
G.f.: 1 / (1 - x / (1 - x / (1 - x / (1 - x / (1 - x / (1 - x / (1 - x))))))). - Michael Somos, May 12 2012
a(n) = 5*b(n-2) - 21*b(n-3) + 19*b(n-4) - 2*b(n-5), for n >= 0, with b(n) = A094256(n), for n >= -5. See a comment in A094256 for this offset, and the above comment. - Wolfdieter Lang, Mar 28 2020

A122588 Expansion of x/(1 - 9*x + 28*x^2 - 35*x^3 + 15*x^4 - x^5).

Original entry on oeis.org

1, 9, 53, 260, 1156, 4845, 19551, 76912, 297275, 1134705, 4292145, 16128061, 60304951, 224660626, 834641671, 3094322026, 11453607152, 42344301686, 156404021389, 577291806894, 2129654436910, 7853149169635, 28949515515376, 106692395098433, 393137817645838
Offset: 1

Views

Author

Roger L. Bagula and Gary W. Adamson, Sep 19 2006

Keywords

Comments

Essentially the same as A005025. - R. J. Mathar, Aug 02 2008

Crossrefs

Programs

  • Magma
    I:=[1,9,53,260,1156]; [n le 5 select I[n] else 9*Self(n-1) -28*Self(n-2) +35*Self(n-3) -15*Self(n-4) +Self(n-5): n in [1..30]]; // G. C. Greubel, Nov 29 2021
    
  • Mathematica
    m = 10; p[x_]:= ExpandAll[x^m*ChebyshevU[m, 1/x]]; Table[SeriesCoefficient[ Series[2^(n+m-1)*x/p[x], {x,0,30}], n], {n,1,30,2}]
  • Sage
    def A122588_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( x/(1-9*x+28*x^2-35*x^3+15*x^4-x^5) ).list()
    a=A122588_list(31); a[1:] # G. C. Greubel, Nov 29 2021

Formula

G.f.: x/(1 - 9*x + 28*x^2 - 35*x^3 + 15*x^4 - x^5).

Extensions

Generating function corrected by R. J. Mathar, Jul 09 2009
New name (using g.f.) and editing by Joerg Arndt, Feb 12 2015

A005023 Number of walks of length 2n+7 in the path graph P_8 from one end to the other.

Original entry on oeis.org

7, 34, 143, 560, 2108, 7752, 28101, 100947, 360526, 1282735, 4552624, 16131656, 57099056, 201962057, 714012495, 2523515514, 8916942687, 31504028992, 111295205284, 393151913464, 1388758662221, 4905479957435, 17327203698086, 61202661233823, 216176614077600
Offset: 1

Views

Author

Keywords

References

  • W. Feller, An Introduction to Probability Theory and its Applications, 3rd ed, Wiley, New York, 1968, p. 96.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A094829 (first differences), A094256 (essentially the same).

Programs

  • Magma
    I:=[7, 34, 143, 560]; [n le 4 select I[n] else 7*Self(n-1)-15*Self(n-2)+10*Self(n-3)-Self(n-4): n in [1..30]]; // Vincenzo Librandi, Jun 08 2013
  • Maple
    a:=k->sum(binomial(7+2*k,9*j+k-2),j=ceil((2-k)/9)..floor((9+k)/9))-sum(binomial(7+2*k,9*j+k-1),j=ceil((1-k)/9)..floor((8+k)/9)): seq(a(k),k=1..28);
    A005023:=-(-7+15*z-10*z**2+z**3)/(z-1)/(z**3-9*z**2+6*z-1); # Conjectured by Simon Plouffe in his 1992 dissertation.
  • Mathematica
    CoefficientList[Series[(-z^3 + 10 z^2 - 15 z + 7)/(z^4 - 10 z^3 + 15 z^2 - 7 z + 1), {z, 0, 100}], z] (* Vladimir Joseph Stephan Orlovsky, Jun 27 2011 *)
    LinearRecurrence[{7,-15,10,-1},{7,34,143,560},40] (* Harvey P. Dale, May 26 2013 *)
    CoefficientList[Series[(1 / x) (1 / (1 - 7 x + 15 x^2 - 10 x^3 + x^4) - 1), {x, 0, 50}], x] (* Vincenzo Librandi, Jun 08 2013 *)

Formula

G.f.: 1/(1-7x+15x^2-10x^3+x^4) - 1. a(n)=7a(n-1)-15a(n-2)+10a(n-3)-a(n-4). - Emeric Deutsch, Apr 02 2004
a(k) = sum(binomial(7+2k, 9j+k-2)-binomial(7+2k, 9j+k-1), j=-infinity..infinity) (a finite sum).

Extensions

Better definition from Emeric Deutsch, Apr 02 2004

A094828 Number of (s(0), s(1), ..., s(2n)) such that 0 < s(i) < 9 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = 1, s(2n) = 5.

Original entry on oeis.org

1, 5, 20, 75, 274, 988, 3536, 12597, 44745, 158632, 561683, 1987154, 7026408, 24835744, 87763945, 310088381, 1095490524, 3869911659, 13670143618, 48287147300, 170561502896, 602454835293, 2127962632993, 7516243783216
Offset: 2

Views

Author

Herbert Kociemba, Jun 13 2004

Keywords

Comments

In general, a(n) = (2/m)*Sum_{r=1..m-1} sin(r*j*Pi/m)*sin(r*k*Pi/m)*(2*cos(r*Pi/m))^(2*n) counts (s(0), s(1), ..., s(2n)) such that 0 < s(i) < m and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n, s(0) = j, s(2n) = k.

Programs

  • Mathematica
    LinearRecurrence[{7,-15,10,-1},{1,5,20,75},30] (* Harvey P. Dale, Apr 27 2020 *)

Formula

a(n) = (2/9)*Sum_{r=1..8} sin(r*Pi/9)*sin(5*r*Pi/9)*(2*cos(r*Pi/9))^(2*n).
a(n) = 7*a(n-1) - 15*a(n-2) + 10*a(n-3) - a(n-4).
G.f.: -x^2*(-1+2*x) / ( (x-1)*(x^3-9*x^2+6*x-1) ).
a(n) = A094256(n-1) - 2*A094256(n-2). - R. J. Mathar, Nov 14 2019
3*a(n) = A094829(n+2) -5*A094829(n+1)+7*A094829(n)-1. - R. J. Mathar, Nov 14 2019

A259475 Array read by antidiagonals: row n gives coefficients of Taylor series expansion of 1/F_{n+1}(t), where F_i(t) is a Fibonacci polynomial defined by F_0=1, F_1=1, F_{i+1} = F_i-t*F_{i-1}.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 1, 0, 1, 3, 4, 1, 0, 1, 4, 8, 8, 1, 0, 1, 5, 13, 21, 16, 1, 0, 1, 6, 19, 40, 55, 32, 1, 0, 1, 7, 26, 66, 121, 144, 64, 1, 0, 1, 8, 34, 100, 221, 364, 377, 128, 1, 0, 1, 9, 43, 143, 364, 728, 1093, 987, 256, 1, 0, 1, 10, 53, 196, 560, 1288, 2380, 3280, 2584, 512, 1, 0
Offset: 0

Views

Author

N. J. A. Sloane, Jul 03 2015

Keywords

Examples

			The first few antidiagonals are:
  1;
  1, 0;
  1, 1,  0;
  1, 2,  1,  0;
  1, 3,  4,  1,   0;
  1, 4,  8,  8,   1,   0;
  1, 5, 13, 21,  16,   1,  0;
  1, 6, 19, 40,  55,  32,  1, 0;
  1, 7, 26, 66, 121, 144, 64, 1, 0;
  ...
Square array starts:
  [0] 1, 0,  0,   0,    0,    0,     0,     0,      0,       0,       0, ...
  [1] 1, 1,  1,   1,    1,    1,     1,     1,      1,       1,       1, ...
  [2] 1, 2,  4,   8,   16,   32,    64,   128,    256,     512,    1024, ...
  [3] 1, 3,  8,  21,   55,  144,   377,   987,   2584,    6765,   17711, ...
  [4] 1, 4, 13,  40,  121,  364,  1093,  3280,   9841,   29524,   88573, ...
  [5] 1, 5, 19,  66,  221,  728,  2380,  7753,  25213,   81927,  266110, ...
  [6] 1, 6, 26, 100,  364, 1288,  4488, 15504,  53296,  182688,  625184, ...
  [7] 1, 7, 34, 143,  560, 2108,  7752, 28101, 100947,  360526, 1282735, ...
  [8] 1, 8, 43, 196,  820, 3264, 12597, 47652, 177859,  657800, 2417416, ...
  [9] 1, 9, 53, 260, 1156, 4845, 19551, 76912, 297275, 1134705, 4292145, ...
		

Crossrefs

The initial rows of the array are A000007, A000012, A000079, A001906, A003432, A005021, A094811, A094256.
A(n,n) gives A274969.
Cf. A309896.
A188843 is a variant without the first two rows and the first column, and the antidiagonals read in opposite direction.

Programs

  • Maple
    F:= proc(n) option remember;
          `if`(n<2, 1, expand(F(n-1)-t*F(n-2)))
        end:
    A:= (n, k)-> coeff(series(1/F(n+1), t, k+1), t, k):
    seq(seq(A(d-k, k), k=0..d), d=0..12);  # Alois P. Heinz, Jul 04 2015
  • Mathematica
    F[n_] := F[n] = If[n<2, 1, Expand[F[n-1] - t*F[n-2]]]; A[n_, k_] := SeriesCoefficient[1/F[n+1], { t, 0, k}]; Table[A[d-k, k], {d, 0, 12}, {k, 0, d}] // Flatten (* Jean-François Alcover, Feb 17 2016, after Alois P. Heinz *)
  • SageMath
    @cached_function
    def F(n, k):
        if k <  0: return 0
        if k == 0: return 1
        return sum((-1)^j*binomial(n-1-j,j+1)*F(n,k-2-2*j) for j in (0..(n-2)/2))
    def A(n, k): return F(n+1, 2*k)
    print([A(n-k, k) for n in (0..11) for k in (0..n)]) # Peter Luschny, Aug 21 2019

Formula

Let F(n, k) = Sum_{j=0..(n-2)/2} (-1)^j*binomial(n-1-j, j+1)*F(n, k-2-2*j) for k > 0; F(n, 0) = 1 and F(n, k) = 0 if k < 0. Then A(n, k) = F(n+1, 2*k). See [Shibukawa] and A309896. - Peter Luschny, Aug 21 2019

Extensions

More terms from Alois P. Heinz, Jul 04 2015

A122589 Expansion of 1/(1 - 11*x + 45*x^2 - 84*x^3 + 70*x^4 - 21*x^5 + x^6).

Original entry on oeis.org

1, 11, 76, 425, 2109, 9709, 42504, 179630, 740025, 2991495, 11920740, 46981740, 183579396, 712493461, 2750450981, 10572046555, 40495806764, 154683305139, 589504177384, 2242448706435, 8517201473375, 32309383853565
Offset: 0

Views

Author

Roger L. Bagula and Gary W. Adamson, Sep 19 2006

Keywords

Comments

Previous name was: Sum_{n >= 0} a(n)*x^(2n) / 4^(n+6) = 1/(4096 - 11264*x^2 + 11520*x^4 - 5376*x^6 + 1120*x^8 - 84*x^10 + x^12).
Suggested by study of polynomials associated with the regular 13-gon.

Crossrefs

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 30); Coefficients(R!( 1/(1-11*x+45*x^2 -84*x^3+70*x^4-21*x^5+x^6) )); // G. C. Greubel, Nov 29 2021
    
  • Maple
    A122589:= proc(n) coeftayl(1/(4096-11264*x^2+11520*x^4-5376*x^6+1120*x^8-84*x^10 +x^12), x=0,2*n); %*2^(2*n+12); end: seq(A122589(n), n=0..30); # R. J. Mathar, Sep 21 2007
  • Mathematica
    m=12; p[x_]:= ExpandAll[x^m*ChebyshevU[m, 1/x]]; Table[ SeriesCoefficient[ Series[2^(n+m-1)*x/p[x], {x,0,30}], n], {n,1,30,2}]
  • Sage
    def A122589_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( 1/(1-11*x+45*x^2-84*x^3+70*x^4-21*x^5+x^6) ).list()
    A122589_list(30) # G. C. Greubel, Nov 29 2021

Formula

G.f.: 1/(1 - 11*x + 45*x^2 - 84*x^3 + 70*x^4 - 21*x^5 + x^6). - Colin Barker, Oct 16 2013

Extensions

Edited by N. J. A. Sloane, Oct 02 2006
More terms from R. J. Mathar, Sep 21 2007
New name from Colin Barker, Oct 16 2013

A217315 Square array T, read by antidiagonals: T(n,k) = 0 if n-k >= 1 or if k-n >= 8, T(0,k)= 1 if 0<=k<=7, T(n,k) = T(n-1,k) + T(n,k-1).

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 2, 0, 0, 1, 3, 2, 0, 0, 1, 4, 5, 0, 0, 0, 1, 5, 9, 5, 0, 0, 0, 1, 6, 14, 14, 0, 0, 0, 0, 0, 7, 20, 28, 14, 0, 0, 0, 0, 0, 7, 27, 48, 42, 0, 0, 0, 0, 0, 0, 0, 34, 75, 90, 42, 0, 0, 0, 0, 0, 0, 0, 34, 109, 165, 132, 0, 0, 0, 0, 0, 0, 0, 0, 0, 143, 274, 297, 132, 0, 0, 0, 0, 0, 0, 0, 0, 0, 143, 417, 571, 429, 0, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Philippe Deléham, Mar 17 2013

Keywords

Comments

A hexagon arithmetic of E. Lucas.

Examples

			Square array begins:
1, 1, 1, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, ... row n=0
0, 1, 2, 3, 4, 5, 6, 7, 7, 0, 0, 0, 0, 0, 0, ... row n=1
0, 0, 2, 5, 9, 14, 20, 27, 34, 34, 0, 0, 0, ... row n=2
0, 0, 0, 5, 14, 28, 48, 75, 109, 143, 143, 0, 0, ... row n=3
0, 0, 0, 0, 14, 42, 90, 165, 274, 417, 560, 560, 0, ... row n=4
0, 0, 0, 0, 0, 42, 132, 297, 571, 988, 1548, 2108, 2108, 0, ... row n=5
...
		

Crossrefs

Cf. Similar sequence: A216230, A216228, A216226, A216238, A216054, A217257.

Programs

  • Mathematica
    t[0, k_ /; k <= 7] = 1; t[n_, k_] /; k < n || k > n+7 = 0; t[n_, k_] := t[n, k] = t[n-1, k] + t[n, k-1]; Table[t[n-k, k], {n, 0, 13}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Mar 18 2013 *)

Formula

T(n,n) = A080938(n).
T(n,n+1) = A080938(n+1).
T(n,n+2) = A094826(n+1).
T(n,n+3) = A094827(n+1).
T(n,n+4) = A094828(n+2).
T(n,n+5) = A094829(n+2).
T(n,n+6) = T(n,n+7) = A094256(n+1).
Sum_{k, 0<=k<=n} T(n-k,k) = A061551(n).
Showing 1-9 of 9 results.