A006054 -id:A006054 - cp's OEIS frontend

A089068 a(n) = a(n-1)+a(n-2)+a(n-3)+2 with a(0)=0, a(1)=0 and a(2)=1.

0, 0, 1, 3, 6, 12, 23, 43, 80, 148, 273, 503, 926, 1704, 3135, 5767, 10608, 19512, 35889, 66011, 121414, 223316, 410743, 755475, 1389536, 2555756, 4700769, 8646063, 15902590, 29249424, 53798079, 98950095, 181997600, 334745776, 615693473

Offset: 0

Roger L. Bagula, Dec 03 2003

The a(n+2) represent the Kn12 and Kn22 sums of the square array of Delannoy numbers A008288. See A180662 for the definition of these knight and other chess sums. - Johannes W. Meijer, Sep 21 2010

Index entries for linear recurrences with constant coefficients, signature (2,0,0,-1).

Cf. A000931, A000073, A077939, A113300, A001057, A006054, A033505.

Cf. A000073 (Kn11 & Kn21), A089068 (Kn12 & Kn22), A180668 (Kn13 & Kn23), A180669 (Kn14 & Kn24), A180670 (Kn15 & Kn25). - Johannes W. Meijer, Sep 21 2010

Join[{a=0,b=0,c=1},Table[d=a+b+c+2;a=b;b=c;c=d,{n,50}]] (* Vladimir Joseph Stephan Orlovsky, Apr 19 2011 *)
RecurrenceTable[{a[0]==a[1]==0,a[2]==1,a[n]==a[n-1]+a[n-2]+a[n-3]+2}, a[n],{n,40}] (* or *) LinearRecurrence[{2,0,0,-1},{0,0,1,3},40] (* Harvey P. Dale, Sep 19 2011 *)

a(n) = A008937(n-2)+A008937(n-1). - Johannes W. Meijer, Sep 21 2010
a(n) = A018921(n-5)+A018921(n-4), n>4. - Johannes W. Meijer, Sep 21 2010
a(n) = A000073(n+2)-1. - R. J. Mathar, Sep 22 2010
From Johannes W. Meijer, Sep 22 2010: (Start)
a(n) = a(n-1)+A001590(n+1).
a(n) = Sum_{m=0..n} A040000(m)*A000073(n-m).
a(n+2) = Sum_{k=0..floor(n/2)} A008288(n-k+1,k+1).
G.f. = x^2*(1+x)/((1-x)*(1-x-x^2-x^3)). (End)
a(n) = 2*a(n-1)-a(n-4), a(0)=0, a(1)=0, a(2)=1, a(3)=3. - Bruno Berselli, Sep 23 2010

Corrected and information added by Johannes W. Meijer, Sep 22 2010, Oct 22 2010

Definition based on arbitrarily set floating-point precision removed by R. J. Mathar, Sep 30 2010

A187070 Let i be in {1,2,3}, let r >= 0 be an integer and n=2r+i-1. Then a(n)=a(2r+i-1) gives the quantity of H_(7,3,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt((2*cos(Pi/7))^2-1).

Original entry on oeis.org

0, 0, 1, 1, 1, 2, 3, 5, 6, 11, 14, 25, 31, 56, 70, 126, 157, 283, 353, 636, 793, 1429, 1782, 3211, 4004, 7215, 8997, 16212, 20216, 36428, 45425, 81853, 102069, 183922, 229347, 413269, 515338, 928607, 1157954, 2086561, 2601899

Offset: 0

L. Edson Jeffery, Mar 05 2011

Theory. (Start)
1. Definitions. Let T_(7,j,0) denote the rhombus with sides of unit length (=1), interior angles given by the pair (j*Pi/7,(7-j)*Pi/7) and Area(T_(7,j,0))=sin(j*Pi/7), j in {1,2,3}. Associated with T_(7,j,0) are its angle coefficients (j, 7-j) in which one coefficient is even while the other is odd. A half-tile is created by cutting T_(7,j,0) along a line extending between its two corners with even angle coefficient; let H_(7,j,0) denote this half-tile. Similarly, a T_(7,j,r) tile is a linearly scaled version of T_(7,j,0) with sides of length x^r and Area(T_(7,j,r))=x^(2*r)*sin(j*Pi/7), r>=0 an integer, where x is the positive, constant square root x=sqrt[(2*cos(j*Pi/7))^2 - 1]; likewise let H_(7,j,r) denote the corresponding half-tile. Often H_(7,i,r) (i in {1,2,3}) can be subdivided into an integral number of each equivalence class H_(7,j,0). But regardless of whether or not H_(7,j,r) subdivides, in theory such a proposed subdivision for each j can be represented by the matrix M=(m_(i,j)), i,j=1,2,3, in which the entry m_(i,j) gives the quantity of H_(7,j,0) tiles that should be present in a subdivided H_(7,i,r) tile. The number x^(2*r) (the square of the scaling factor) is an eigenvalue of M=(U_2)^r, where
  U_2= (0 0 1)
       (0 1 1)
       (1 1 1).
2. The sequence. Let r>=0, and let C_r be the r-th "block" defined by C_r={a(2*r),a(2*r+1),a(2*r+2)}. Note that C_r-2*C_(r-1)-C_(r-2)+C_(r-3)={0,0,0}. Let n=2*r+i-1. Then a(n)=a(2*r+i-1)=m_(i,3), where M=(m_(i,j))=(U_2)^r was defined above. Hence the block C_r corresponds component-wise to the third column of M, and a(n)=m_(i,3) gives the quantity of H_(7,3,0) tiles that should appear in a subdivided H_(7,i,r) tile. (End)
Combining blocks A_r, B_r and C_r, from A187068, A187069 and this sequence, respectively, as matrix columns [A_r,B_r,C_r] generates the matrix (U_2)^r, and a negative index (-1)*r yields the corresponding inverse [A_(-r),B_(-r),C_(-r)]=(U_2)^(-r) of (U_2)^r. Therefore, the three sequences need not be causal.
Since a(2*r+2)=a(2*(r+1)) for all r, this sequence arises by concatenation of third-column entries m_(1,3) and m_(2,3) from successive matrices M=(U_2)^r.
This sequence is a trivial extension of A038196.

			Suppose r=3. Then
C_r = C_3 = {a(2*r),a(2*r+1),a(2*r+2)} = {a(6),a(7),a(8)} = {3,5,6},
corresponding to the entries in the third column of
M = (U_2)^3 = (1 2 3)
              (2 4 5)
              (3 5 6).
Choose i=2 and set n=2*r+i-1. Then a(n) = a(2*r+i-1) = a(6+2-1) = a(7) = 5, which equals the entry in row 2 and column 3 of M. Hence a subdivided H_(7,2,3) tile should contain a(7) = m_(2,3) = 5 H_(7,3,0) tiles.

G. C. Greubel, Table of n, a(n) for n = 0..1000
L. Edson Jeffery, Unit-primitive matrices
Roman Witula, D. Slota and A. Warzynski, Quasi-Fibonacci Numbers of the Seventh Order, J. Integer Seq., 9 (2006), Article 06.4.3.
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1,0,-1).

Cf. A038196, A187068, A187069.

a[0] = a[1] = 0; a[2] = a[3] = a[4] = 1; a[?Negative] = 0; a[n] := a[n] = 2*a[n-2] + a[n-4] - a[n-6]; Table[a[n], {n, 0, 40}] (* Jean-François Alcover, Jan 02 2013 *)

x='x+O('x^50); concat([0,0], Vec(x^2*(1+x-x^2)/(1-2*x^2-x^4+x^6))) \\ G. C. Greubel, Jul 05 2017

Recurrence: a(n) = 2*a(n-2) + a(n-4) - a(n-6).
G.f.: x^2*(1+x-x^2)/(1-2*x^2-x^4+x^6).
a(2*n)=A106803(n); a(2*n+1)=A006054(n+1); a(2*n+2)=A077998(n).
Closed-form: a(n) = (1/14)*[[X_1+Y_1*(-1)^(n-1)]*[(w_2)^2-(w_3)^2]*(w_1)^(n-1)+[X_2+Y_2*(-1)^(n-1)]*[(w_3)^2-(w_1)^2]*(w_2)^(n-1)+[X_3+Y_3*(-1)^(n-1)]*[(w_1)^2-(w_2)^2]*(w_3)^(n-1)], where w_k = sqrt[(2cos(k*Pi/7))^2-1], X_k = (w_k)^3+(w_k)^2-w_k and Y_k = -(w_k)^3+(w_k)^2+w_k, k=1,2,3.

A231187 Decimal expansion of the length ratio (largest diagonal)/side in the regular 7-gon (or heptagon).

Original entry on oeis.org

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

Offset: 1

Wolfdieter Lang, Nov 21 2013

The length ratio (largest diagonal)/side in the regular 7-gon (heptagon) is sigma(7) = S(2, rho(7)) = -1 + rho(7)^2, with rho(7) = 2*cos(Pi/7), which is approx. 1.8019377358 (see A160389 for its decimal expansion, and A049310 for the Chebyshev S-polynomials). sigma(7), approx. 2.2469796, is also the reciprocal of one of the solutions of the minimal polynomial C(7, x) = x^3 - x^2 - 2*x + 1 of rho(7) (see A187360), namely 1/(2*cos(3*Pi/7)).
sigma(7) is the limit of a(n+1)/a(n) for n->infinity for the sequences A006054 and A077998 which can be considered as analogs of the Fibonacci sequence in the pentagon. Thus sigma(7) plays in the heptagon the role of the golden section in the pentagon.
See the Steinbach link.

			2.24697960371746706105000976800847962126454946179280421073109887819...

Peter Steinbach, Golden Fields: A Case for the Heptagon, Mathematics Magazine, Vol. 70, No. 1, Feb. 1997.
I. J. Zucker, G. S. Joyce, Special values of the hypergeometric series II, Math. Proc. Cam. Phil. Soc. 131 (2001) 309 eq. (8.10).
Index entries for algebraic numbers, degree 3.

Cf. A160389, A006054, A077998, A255240, A255241, A255249, A116425.

First[RealDigits[N[Csc[Pi/14]/2,104]]] (* Stefano Spezia, Jun 26 2022 *)

 5/cos(3*Pi/7) \\ Charles R Greathouse IV, Feb 04 2025

polrootsreal(x^3-2*x^2-x+1)[3] \\ Charles R Greathouse IV, Feb 04 2025

sigma(7) = -1 + (2*cos(Pi/7))^2 = 1/(2*cos(3*Pi/7)).
Equals A116425 -1.
From Geoffrey Caveney, Apr 23 2014: (Start)
sigma(7) = exp(asinh(cos(Pi/7))).
cos(Pi/7) + sqrt(1+cos(Pi/7)^2). (End)
From Peter Bala, Oct 12 2021: (Start)
Minimal polynomial x^3 - 2*x^2 - x + 1.
Equals 2*(cos(3*Pi/7) - cos(6*Pi/7)). The other zeros of the minimal polynomial are 2*(cos(Pi/7) - cos(2*Pi/7)) = A255240 and 2*(cos(5*Pi/7) - cos(10*Pi/7)) = 1 - A160389.
The quadratic mapping z -> z^2 - 2*z cyclically permutes the zeros of the minimal polynomial. The inverse cyclic permutation is given by the mapping z -> 2 + z - z^2.
Equals Product_{n >= 0} (7*n+3)*(7*n+4)/((7*n+1)*(7*n+6)) = 1 + Product_{n >= 0} (7*n+3)*(7*n+4)/((7*n+2)*(7*n+5)) = 1 + A255249 = 1/A255241. (End)
Equals 1/(2*sin(Pi/14)) = 1 + 2*sin(3*Pi/14). - Gary W. Adamson, Jun 25 2022
Equals (2*cos(Pi/7)) * (2*cos(2*Pi/7)) = (i^(2/7) + i^(-2/7)) * (i^(4/7) + i^(-4/7)) = 1 + i^(4/7) + i^(-4/7). - Gary W. Adamson, Jul 16 2022
Equals 2F1(1/7,2/7;1/2;1) [Zucker] - R. J. Mathar, Jun 24 2024

A033303 Expansion of (1 + x)/(1 - 2*x - x^2 + x^3).

Original entry on oeis.org

1, 3, 7, 16, 36, 81, 182, 409, 919, 2065, 4640, 10426, 23427, 52640, 118281, 265775, 597191, 1341876, 3015168, 6775021, 15223334, 34206521, 76861355, 172705897, 388066628, 871977798, 1959316327

Offset: 0

N. J. A. Sloane

Also the number of one-sided n-step prudent walks that avoid 3 or more consecutive east steps. - Shanzhen Gao, Apr 27 2011
Equivalently, number of ternary strings of length n with subwords (0,0) and (1,2) not allowed. - Olivier Gérard, Aug 28 2012
First differences are in A052534.
a(n) is the number of vertices of the Minkowski sum of n simplices with vertices e_(i+1), e_(i+2), e_(i+3) for i=0,...,n-1, where e_i is a standard basis vector. - Alejandro H. Morales, Oct 05 2022

R. P. Stanley, Enumerative Combinatorics I, p. 244.

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
L. Escobar, P. Gallardo, J. González-Anaya, J. L. González, G. Montúfar, and A. H. Morales, Enumeration of max-pooling responses with generalized permutohedra, arXiv:2209.14978 [math.CO], 2022. (See Ex. 4.7)
S. Gao and H. Niederhausen, Sequences Arising From Prudent Self-Avoiding Walks, 2010.
Index entries for linear recurrences with constant coefficients, signature (2,1,-1).

CoefficientList[Series[(1 + x)/(1 - 2*x - x^2 + x^3), {x, 0, 100}], x] (* Vincenzo Librandi, Oct 20 2012 *)
LinearRecurrence[{2,1,-1},{1,3,7},40] (* Harvey P. Dale, Oct 31 2013 *)

h(n):=sum(sum(binomial(k,j)*binomial(j,n-3*k+2*j)*2^(3*k-n-j)*(-1)^(k-j),j,0,k),k,1,n); a(n):=if n=0 then 1 else if n=2 then h(n) else h(n)+h(n-1); /* Vladimir Kruchinin, Sep 09 2010 */

a(n)=([0,1,0; 0,0,1; -1,1,2]^n*[1;3;7])[1,1] \\ Charles R Greathouse IV, Feb 19 2017

a(0)=1, a(1)=h(n), and a(n) = h(n) + h(n-1) for n >= 2, where h(n) = Sum_{k=1..n} Sum_{j=0..k} binomial(k, j) * binomial(j, n-3*k+2*j) * 2^(3*k-n-j) * (-1)^(k-j). - Vladimir Kruchinin, Sep 09 2010
a(0)=1, a(1)=3, a(2)=7, a(n) = 2*a(n-1) + a(n-2) - a(n-3). - Harvey P. Dale, Oct 31 2013
a(n) = A006054(n+1)+A006054(n+2). - R. J. Mathar, Jul 08 2022

A121442 Expansion of (1-x^2)/(1-x-9*x^2+x^3).

Original entry on oeis.org

1, 1, 9, 17, 97, 241, 1097, 3169, 12801, 40225, 152265, 501489, 1831649, 6192785, 22176137, 76079553, 269472001, 932011841, 3281180297, 11399814865, 39998425697, 139315579185, 487901595593, 1701743382561, 5953542163713, 20781331011169, 72661467102025

Offset: 0

Philippe Deléham, Sep 06 2006

From Roman Witula, Aug 08 2012: (Start)
We have a(n)=A(n;2), where A(n;2), B(n;2) and C(n;2) are the special cases of so-called quasi-Fibonacci numbers A(n;d), B(n;d), and C(n;d) for the value of argument d=2 - for details see Witula's comments to A121449 or the paper of Witula-Slota-Warzynski's. The sequences A(n;2), B(n;2) and C(n;2) are defined by the following system of recurrence equations:
A(0;2)=1, B(0;2)=C(0;2)=0,
A(n+1;2)=A(n;2)+4*B(n;2)-2*C(n;2), B(n+1;2)=2*A(n;2)+B(n;2), and C(n+1;2)=2*B(n;2)-C(n;2).
We note that A(n;1)=A077998(n), B(n;1)=A006054(n+1), and C(n;1)=A006054(n). We know (see formulas (3.61-63) in Witula et al.'s paper) that the sequences: (-2)^(-n)*(A(n;1)*(A(n;2)-C(n;2))-B(n;1)*(B(n;2)+C(n;2))+C(n;1)*B(n;2)), (-2)^(-n)*(-A(n;1)*C(n;2)+B(n;1)*(A(n;2)-C(n;2))-C(n;1)*(B(n;2)-C(n;2))), and (-2)^(-n)*(A(n;1)*(B(n;2)-C(n;2))-B(n;1)*B(n;2)+C(n;1)*(A(n;2)-B(n;2)+C(n;2))) are the binomial transforms of the sequences (-2)^(-n)*A(n;1), (-2)^(-n)*B(n;1), and (-2)^(-n)*C(n;1) respectively. Moreover the elements of the sequences A(n;1/2)=2^(-n)*A052975, B(n;1/2)=2^(-n)*A094789, and C(n;1/2) could be described by certain convolutions type identities for the elements of A(n;2), B(n;2), and C(n;2) (see identities (3.58-60) in Witula et al.'s paper). (End)

G. C. Greubel, Table of n, a(n) for n = 0..1000
Roman Witula, Damian Slota and Adam Warzynski, Quasi-Fibonacci Numbers of the Seventh Order, J. Integer Seq., 9 (2006), Article 06.4.3.
Index entries for linear recurrences with constant coefficients, signature (1, 9, -1).

Cf. A077998, A006054, A121449, A085810.

I:=[1,1,9]; [n le 3 select I[n] else Self(n-1)+9*Self(n-2)-Self(n-3): n in [1..30]]; // Vincenzo Librandi, Sep 18 2015

LinearRecurrence[{1,9,-1},{1,1,9},50] (* Roman Witula, Aug 08 2012 *)
CoefficientList[Series[(1 - x^2)/(1 - x - 9 x^2 + x^3), {x, 0, 40}], x] (* Vincenzo Librandi, Sep 18 2015 *)

Vec((1-x^2)/(1-x-9*x^2+x^3)+O(x^99)) \\ Charles R Greathouse IV, Sep 23 2012

a(0)=a(1)=1, a(2)=9, a(n+1) = a(n)+9*a(n-1)-a(n-2) for n>=2.

7*a(n) = (2-c(4))*(1-2*c(1))^n + (2-c(1))*(1-2*c(2))^n + (2-c(2))*(1-2*c(4))^n = (s(2))^2*(1-2*c(1))^n + (s(4))^2*(1-2*c(2))^n + (s(1))^2*(1-2*c(4))^n, where c(j):=2*Cos(2Pi*j/7) and s(j):=2*Sin(2Pi*j/7) - it is the special case, for d=2, of the Binet's formula for the respective quasi-Fibonacci number A(n;d) discussed in Witula-Slota-Warzynski's paper (see also A121449). - Roman Witula, Aug 08 2012

Corrected by T. D. Noe, Oct 25 2006

More terms from Vincenzo Librandi, Sep 18 2015

A106195 Riordan array (1/(1-2x), x(1-x)/(1-2*x)).

Original entry on oeis.org

1, 2, 1, 4, 3, 1, 8, 8, 4, 1, 16, 20, 13, 5, 1, 32, 48, 38, 19, 6, 1, 64, 112, 104, 63, 26, 7, 1, 128, 256, 272, 192, 96, 34, 8, 1, 256, 576, 688, 552, 321, 138, 43, 9, 1, 512, 1280, 1696, 1520, 1002, 501, 190, 53, 10, 1, 1024, 2816, 4096, 4048, 2972, 1683, 743, 253, 64, 11

Offset: 0

Gary W. Adamson, Apr 24 2005; Paul Barry, May 21 2006

Extract antidiagonals from the product P * A, where P = the infinite lower triangular Pascal's triangle matrix; and A = the Pascal's triangle array:
  1, 1,  1,  1, ...
  1, 2,  3,  4, ...
  1, 3,  6, 10, ...
  1, 4, 10, 20, ...
  ...
Row sums are Fibonacci(2n+2). Diagonal sums are A006054(n+2). Row sums of inverse are A105523. Product of Pascal triangle A007318 and A046854.
A106195 with an appended column of ones = A055587. Alternatively, k-th column (k=0, 1, 2) is the binomial transform of bin(n, k).
T(n,k) is the number of ideals in the fence Z(2n) having k elements of rank 1. - Emanuele Munarini, Mar 22 2011
Subtriangle of the triangle given by (0, 2, 0, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (1, 0, -1/2, 1/2, 0, 0, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Mar 22 2012

			Triangle begins
   1;
   2,   1;
   4,   3,   1;
   8,   8,   4,  1;
  16,  20,  13,  5,  1;
  32,  48,  38, 19,  6, 1;
  64, 112, 104, 63, 26, 7, 1;
(0, 2, 0, 0, 0, ...) DELTA (1, 0, -1/2, 1/2, 0, 0, ...) begins :
  1;
  0,  1;
  0,  2,   1;
  0,  4,   3,   1;
  0,  8,   8,   4,  1;
  0, 16,  20,  13,  5,  1;
  0, 32,  48,  38, 19,  6, 1;
  0, 64, 112, 104, 63, 26, 7, 1. - _Philippe Deléham_, Mar 22 2012

Reinhard Zumkeller, Rows n = 0..125 of triangle, flattened
E. Munarini, N. Zagaglia Salvi, On the Rank Polynomial of the Lattice of Order Ideals of Fences and Crowns, Discrete Mathematics 259 (2002), 163-177.

Column 0 = 1, 2, 4...; (binomial transform of 1, 1, 1...); column 1 = 1, 3, 8, 20...(binomial transform of 1, 2, 3...); column 2: 1, 4, 13, 38...= binomial transform of bin(n, 2): 1, 3, 6...

Cf. A001792, A001906, A002620, A007318, A029653, A049612, A055587, A078812, A208341.

a106195 n k = a106195_tabl !! n !! k
a106195_row n = a106195_tabl !! n
a106195_tabl = [1] : [2, 1] : f [1] [2, 1] where
   f us vs = ws : f vs ws where
     ws = zipWith (-) (zipWith (+) ([0] ++ vs) (map (* 2) vs ++ [0]))
                      ([0] ++ us ++ [0])
-- Reinhard Zumkeller, Dec 16 2013

[ (&+[Binomial(n-k, n-j)*Binomial(j, k): j in [0..n]]): k in [0..n], n in [0..10]]; // G. C. Greubel, Mar 15 2020

T := (n, k) -> hypergeom([-n+k, k+1],[1],-1):
seq(lprint(seq(simplify(T(n, k)), k=0..n)), n=0..7); # Peter Luschny, May 20 2015

u[1, x_] := 1; v[1, x_] := 1; z = 16;
u[n_, x_] := u[n - 1, x] + v[n - 1, x]
v[n_, x_] := u[n - 1, x] + (x + 1) v[n - 1, x]
Table[Factor[u[n, x]], {n, 1, z}]
Table[Factor[v[n, x]], {n, 1, z}]
cu = Table[CoefficientList[u[n, x], x], {n, 1, z}];
TableForm[cu]
Flatten[%]  (* A207605 *)
Table[Expand[v[n, x]], {n, 1, z}]
cv = Table[CoefficientList[v[n, x], x], {n, 1, z}];
TableForm[cv]
Flatten[%]  (* A106195 *)
(* Clark Kimberling, Feb 19 2012 *)
Table[Hypergeometric2F1[-n+k, k+1, 1, -1], {n, 0, 12}, {k, 0, n}]//Flatten (* G. C. Greubel, Mar 15 2020 *)

create_list(sum(binomial(i,k)*binomial(n-k,n-i),i,0,n),n,0,8,k,0,n); /* Emanuele Munarini, Mar 22 2011 */

from sympy import Poly, symbols
x = symbols('x')
def u(n, x): return 1 if n==1 else u(n - 1, x) + v(n - 1, x)
def v(n, x): return 1 if n==1 else u(n - 1, x) + (x + 1)*v(n - 1, x)
def a(n): return Poly(v(n, x), x).all_coeffs()[::-1]
for n in range(1, 13): print(a(n)) # Indranil Ghosh, May 28 2017

from mpmath import hyp2f1, nprint
def T(n, k): return hyp2f1(k - n, k + 1, 1, -1)
for n in range(13): nprint([int(T(n, k)) for k in range(n + 1)]) # Indranil Ghosh, May 28 2017, after formula from Peter Luschny

[[sum(binomial(n-k,n-j)*binomial(j,k) for j in (0..n)) for k in (0..n)] for n in (0..10)] # G. C. Greubel, Mar 15 2020

T(n,k) = Sum_{j=0..n} C(n-k,n-j)*C(j,k).
From Emanuele Munarini, Mar 22 2011: (Start)
T(n,k) = Sum_{i=0..n-k} C(k,i)*C(n-k,i)*2^(n-k-i).
T(n,k) = Sum_{i=0..n-k} C(k,i)*C(n-i,k)*(-1)^i*2^(n-k-i).
Recurrence: T(n+2,k+1) = 2*T(n+1,k+1)+T(n+1,k)-T(n,k). (End)
From Clark Kimberling, Feb 19 2012: (Start)
Define u(n,x) = u(n-1,x)+v(n-1,x), v(n,x) = u(n-1,x)+(x+1)*v(n-1,x),
where u(1,x)=1, v(1,x)=1. Then v matches A106195 and u matches A207605. (End)
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k-1). - Philippe Deléham, Mar 22 2012
T(n+k,k) is the coefficient of x^n y^k in 1/(1-2x-y+xy). - Ira M. Gessel, Oct 30 2012
T(n, k) = A208341(n+1,n-k+1), k = 0..n. - Reinhard Zumkeller, Dec 16 2013
T(n, k) = hypergeometric_2F1(-n+k, k+1, 1 , -1). - Peter Luschny, May 20 2015
G.f. 1/(1-2*x+x^2*y-x*y). - R. J. Mathar, Aug 11 2015
Sum_{k=0..n} T(n, k) = Fibonacci(2*n+2) = A088305(n+1). - G. C. Greubel, Mar 15 2020

Edited by N. J. A. Sloane, Apr 09 2007, merging two sequences submitted independently by Gary W. Adamson and Paul Barry

A187069 Let i be in {1,2,3}, let r >= 0 be an integer and n=2r+i-1. Then a(n)=a(2r+i-1) gives the quantity of H_(7,2,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt((2*cos(Pi/7))^2-1).

Original entry on oeis.org

0, 1, 0, 1, 1, 2, 2, 4, 5, 9, 11, 20, 25, 45, 56, 101, 126, 227, 283, 510, 636, 1146, 1429, 2575, 3211, 5786, 7215, 13001, 16212, 29213, 36428, 65641, 81853, 147494, 183922, 331416, 413269, 744685, 928607, 1673292, 2086561, 3759853, 4688460, 8448313, 10534874

Offset: 0

L. Edson Jeffery, Mar 06 2011

See A187070 for supporting theory. Define the matrix
U_2 = (0 0 1)
      (0 1 1)
      (1 1 1).
Let r>=0, and let B_r be the r-th "block" defined by B_r={a(2*r),a(2*r+1),a(2*r+2)}. Note that B_r-2*B_(r-1)-B_(r-2)+B_(r-3)={0,0,0}. Let n=2*r+i-1 and M=(m_(i,j))=(U_2)^r. Then B_r corresponds component-wise to the second column of M, and a(n)=a(2*r+i-1)=m_(i,2) gives the quantity of H_(7,2,0) tiles that should appear in a subdivided H_(7,i,r) tile.
Since a(2*r+2)=a(2*(r+1)) for all r, this sequence arises by concatenation of second-column entries m_(1,2) and m_(2,2) from successive matrices M=(U_2)^r.

			Suppose r=3.
Then B_r = B_3 = {a(2*r),a(2*r+1),a(2*r+2)} = {a(6),a(7),a(8)} = {2,4,5}, corresponding to the entries in the second column of
  M = (U_2)^3 = (1 2 3)
                (2 4 5)
                (3 5 6).
Suppose i=2. Setting n=2*r+i-1, then a(n) = a(2*r+i-1) = a(6+2-1) = a(7) = m_(2,2) = 4. Hence a subdivided H_(7,2,3) tile should contain a(7) = m_(2,2) = 4 H_(7,2,0) tiles.

G. C. Greubel, Table of n, a(n) for n = 0..1000
L. Edson Jeffery, Unit-primitive matrices
Roman Witula, Damian Slota and Adam Warzynski, Quasi-Fibonacci Numbers of the Seventh Order, J. Integer Seq., 9 (2006), Article 06.4.3.
Index entries for linear recurrences with constant coefficients, signature (0,2,0,1,0,-1).

Cf. A038196, A187068, A187070, A006054.

CoefficientList[Series[x*(1 - x^2 + x^3 - x^4)/(1 - 2*x^2 - x^4 + x^6), {x, 0, 50}], x] (* G. C. Greubel, Oct 20 2017 *)
LinearRecurrence[{0,2,0,1,0,-1},{0,1,0,1,1,2},50] (* Harvey P. Dale, Dec 16 2017 *)

my(x='x+O('x^50)); concat([0], Vec(x*(1-x^2+x^3-x^4)/(1-2*x^2-x^4+x^6))) \\ G. C. Greubel, Oct 20 2017

Recurrence: a(n) = 2*a(n-2) + a(n-4) - a(n-6).
G.f.: x*(1-x^2+x^3-x^4)/(1-2*x^2-x^4+x^6).
Closed-form: a(n) = (1/14)*[[X_1+Y_1*(-1)^(n-1)]*[(w_2)^2-(w_3)^2]*(w_1)^(n-1)+[X_2+Y_2*(-1)^(n-1)]*[(w_3)^2-(w_1)^2]*(w_2)^(n-1)+[X_3+Y_3*(-1)^(n-1)]*[(w_1)^2-(w_2)^2]*(w_3)^(n-1)], where w_k = sqrt[(2cos(k*Pi/7))^2-1], X_k = (w_k)^4-(w_k)^2+w_k-1 and Y_k = (w_k)^4+(w_k)^2-w_k-1, k=1,2,3.
a(2*n) = A006054(n), a(2*n+3) = A052534(n).

A215560 a(n) = 3a(n-1) + 46a(n-2) + a(n-3) with a(0)=a(1)=3, a(2)=101.

Original entry on oeis.org

3, 3, 101, 444, 5981, 38468, 390974, 2948431, 26868565, 216624495, 1888775906, 15657923053, 134074085330, 1124375492334, 9556192325235, 80523923708399, 682280993578341, 5760499663646612, 48746948619251921, 411906111379078256, 3483838470286469746, 29447943482916260935

Offset: 0

Roman Witula, Aug 16 2012

The Ramanujan-type sequence number 6 for the argument 2Pi/7 (see also A214683, A215112, A006053, A006054, A215076, A215100, A120757 for the numbers: 1, 1a, 2, 2a, 3, 4 and 5 respectively).
The sequence a(n) is one of the three special sequences (the remaining two are A215569 and A215572) connected with the following recurrence relation: T(n):=49^(1/3)*T(n-2)+T(n-3), with T(0)=3, T(1)=0, and T(2)=2*49^(1/3) - see the comments to A214683.
It can be proved that
T(n) = (c(1)^4/c(2))^(n/3) + (c(2)^4/c(4))^(n/3) + (c(4)^4/c(1))^(n/3), where c(j):=2*cos(2*Pi*j/7), and the following decomposition hold true:
T(n) = at(n) + bt(n)*7^(1/3) + ct(n)*49^(1/3), where sequences at(n), bt(n), and ct(n) satisfy the following system of recurrence equations: at(n)=7*bt(n-2)+at(n-3),
bt(n)=7*ct(n-2)+bt(n-3), ct(n)=at(n-2)+ct(n-3), with at(0)=3, at(1)=at(2)=bt(0)=bt(1)=bt(2)=ct(0)=ct(1)=0, ct(2)=2 - for details see the first Witula reference.
It follows that a(n)=at(3*n), bt(3*n)=ct(3*n)=0.
Every difference of the form a(n)-a(n-2)-a(n-3) is divisible by 3. Because the difference a(n+1)-a(n) is congruent to the difference a(n-4)-a(n-2) modulo 3 we easily deduce that a(6)-a(5) and a(7)-a(6)-2 are both divisible by 3.

R. Witula, E. Hetmaniok, D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the Fifteenth International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012

Roman Witula, Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2*Pi/7, Journal of Integer Sequences, Vol. 12 (2009), Article 09.8.5.
Roman Witula, Full Description of Ramanujan Cubic Polynomials, Journal of Integer Sequences, Vol. 13 (2010), Article 10.5.7.
Roman Witula, Ramanujan Cubic Polynomials of the Second Kind, Journal of Integer Sequences, Vol. 13 (2010), Article 10.7.5.
Roman Witula, Ramanujan Type Trigonometric Formulae, Demonstratio Math. 45 (2012) 779-796.
Index entries for linear recurrences with constant coefficients, signature (3,46,1).

Cf. A214683, A215112, A006053, A006054, A215076, A215100, A120757, A214699.

LinearRecurrence[{3, 46, 1}, {3, 3, 101}, 50]

Vec((3-6*x-46*x^2)/(1-3*x-46*x^2-x^3) + O(x^40)) \\ Michel Marcus, Apr 20 2016

a(n) = (c(1)^4/c(2))^n + (c(2)^4/c(4))^n + (c(4)^4/c(1))^n, where c(j) = 2*cos(2*Pi*j/7).

G.f.: (3-6*x-46*x^2)/(1-3*x-46*x^2-x^3).

More terms from Michel Marcus, Apr 20 2016

A215572 a(n) = 3a(n-1) + 46a(n-2) + a(n-3) with a(0)=2, a(1)=5, a(2)=106.

Original entry on oeis.org

2, 5, 106, 550, 6531, 44999, 435973, 3384404, 30252969, 246877464, 2135653370, 17793576423, 151867661753, 1276243154087, 10832435479322, 91356359187721, 773637352766062, 6534137016412674, 55281085635664595, 467187197014742851, 3951025667301212597, 33398969150217473532

Offset: 0

Roman Witula, Aug 16 2012

The Ramanujan-type sequence number 8 for the argument 2Pi/7 (see also A214683, A215112, A006053, A006054, A215076, A215100, A120757, A215560, A215569 for the numbers: 1, 1a, 2, 2a, 3-7 respectively). The sequence a(n) is one of the three special sequences (the remaining two are A215560 and A215569) connected with the following recurrence relation:
(c(1)^4/c(2))^(n/3) + (c(2)^4/c(4))^(n/3) + (c(4)^4/c(1))^(n/3) = at(n) + bt(n)*7^(1/3) + ct(n)*49^(1/3), where c(j):=2*cos(2*Pi*j/7), and the sequences at(n), bt(n), and ct(n) are defined in comments to A215560 (see also A215569). It follows that a(n)=ct(3*n+2), at(3*n+2)=bt(3*n+2)=0, which implies the first formula below.
We note that if a(n), a(n+1) and a(n+2) are all odd for some n in N then a(n+3) is even, a(n+4) is odd, a(n+5) and a(n+6) are both even, and the numbers a(n+7), a(n+8), a(n+9) are all odd again. In consequence, this situation hold for every n of the form 7*k+4, k=0,1,..., in the other words cyclical through all sequence a(n), n=4,5,... (from n=1 whenever we start from odd-even-even sequence).

			From 4*a(1)+5*a(2)=a(3) we obtain 4*((c(1)^4/c(2))^(5/3) + (c(2)^4/c(4))^(5/3) + (c(4)^4/c(1))^(5/3)) + 5*((c(1)^4/c(2))^(8/3) + (c(2)^4/c(4))^(8/3) + (c(4)^4/c(1))^(8/3)) = (4 + 5*c(1)^4/c(2))*((c(1)^4/c(2))^(5/3) + (4 + 5*c(2)^4/c(4))*((c(2)^4/c(4))^(5/3) + (4 + 5*c(4)^4/c(1))*((c(4)^4/c(1))^(5/3) = (c(1)^4/c(2))^(11/3) + (c(2)^4/c(4))^(11/3) + (c(4)^4/c(1))^(11/3) = 550*49^(1/3).

R. Witula, E. Hetmaniok, D. Slota, Sums of the powers of any order roots taken from the roots of a given polynomial, Proceedings of the Fifteenth International Conference on Fibonacci Numbers and Their Applications, Eger, Hungary, 2012.

G. C. Greubel, Table of n, a(n) for n = 0..1000
Roman Witula, Ramanujan Type Trigonometric Formulas: The General Form for the Argument 2*Pi/7, Journal of Integer Sequences, Vol. 12 (2009), Article 09.8.5.
Roman Witula, Full Description of Ramanujan Cubic Polynomials, Journal of Integer Sequences, Vol. 13 (2010), Article 10.5.7.
Roman Witula, Ramanujan Cubic Polynomials of the Second Kind, Journal of Integer Sequences, Vol. 13 (2010), Article 10.7.5.
Roman Witula, Ramanujan Type Trigonometric Formulae, Demonstratio Math. 45 (2012) 779-796.
Index entries for linear recurrences with constant coefficients, signature (3,46,1).

Cf. A214683, A215112, A006053, A006054, A215076, A215100, A120757, A215560, A215569.

LinearRecurrence[{3,46,1}, {2,5,106}, 50]
CoefficientList[Series[(2 - x - x^2)/(1 - 3*x - 46*x^2 - x^3), {x,0,50}], x] (* G. C. Greubel, Apr 16 2017 *)

Vec((2-x-x^2)/(1-3*x-46*x^2-x^3) + O(x^40)) \\ Michel Marcus, Apr 20 2016

49^(1/3)*a(n) = (c(1)^4/c(2))^(n+2/3) + (c(2)^4/c(4))^(n+2/3) + (c(4)^4/c(1))^(n+2/3) = (c(1)*(c(1)/c(2))^(1/3))^(3*n+2) + (c(2)*(c(2)/c(4))^(1/3))^(3*n+2) + (c(4)*(c(4)/c(1))^(1/3))^(3*n+2).

G.f.: (2-x-x^2)/(1-3*x-46*x^2-x^3).

More terms from Michel Marcus, Apr 20 2016

A059594 Convolution triangle based on A008619 (positive integers repeated).

Original entry on oeis.org

1, 1, 1, 2, 2, 1, 2, 5, 3, 1, 3, 8, 9, 4, 1, 3, 14, 19, 14, 5, 1, 4, 20, 39, 36, 20, 6, 1, 4, 30, 69, 85, 60, 27, 7, 1, 5, 40, 119, 176, 160, 92, 35, 8, 1, 5, 55, 189, 344, 376, 273, 133, 44, 9, 1, 6, 70, 294, 624, 820, 714, 434

Offset: 0

Wolfdieter Lang, Feb 02 2001

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as a matrix, belongs to the Bell-subgroup of the Riordan-group.
The G.f. for the row polynomials p(n,x) = Sum_{m=0..n} a(n,m)*x^m is 1/((1-z^2)*(1-z)-x*z).
The column sequences are A008619(n); A006918(n); A038163(n-2), n >= 2; A038164(n-3), n >= 3; A038165(n-4), n >= 4; A038166(n-5), n >= 5; A059595(n-6), n >= 6; A059596(n-7), n >= 7; A059597(n-8), n >= 8; A059598(n-9), n >= 9; A059625(n-10), n >= 10 for m=0..10.
The sequence of row sums is A006054(n+2).
From Gary W. Adamson, Aug 14 2016: (Start)
The sequence can be generated by extracting the descending antidiagonals of an array formed by taking powers of the natural integers with repeats, (1, 1, 2, 2, 3, 3, ...), as follows:
  1, 1,  2,  2,  3,   3, ...
  1, 2,  5,  8, 14,  20, ...
  1, 3,  9, 19, 39,  69, ...
  1, 4, 14, 36, 85, 176, ...
  ...
Row sums of the triangle = (1, 2, 5, 11, 25, 56, ...), the INVERT transform of (1, 1, 2, 2, 3, 3, ...). (End)

			{1}; {1,1}; {2,2,1}; {2,5,3,1}; ...
Fourth row polynomial (n=3): p(3,x)= 2 + 5*x + 3*x^2 + x^3.

t[n_, m_] := Sum[Sum[Binomial[j, n-m-3*k+2*j]*(-1)^(j-k)*Binomial[k, j], {j, 0, k}]*Binomial[m+k, m], {k, 0, n-m}]; Table[t[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, May 27 2013, after Vladimir Kruchinin *)

T(n,m):=sum((sum(binomial(j,n-m-3*k+2*j)*(-1)^(j-k)*binomial(k,j),j,0,k)) *binomial(m+k,m),k,0,n-m); /* Vladimir Kruchinin, Dec 14 2011 */

a(n, m) := a(n-1, m) + (-(n-m+1)*a(n, m-1) + 3*(n+2*m)*a(n-1, m-1))/(8*m), n >= m >= 1; a(n, 0) := floor((n+2)/2) = A008619(n), n >= 0; a(n, m) := 0 if n < m.
G.f.for column m >= 0: ((x/((1-x^2)*(1-x)))^m)/((1-x^2)*(1-x)).
T(n,m) = Sum_{k=0..n-m} (Sum_{j=0..k} binomial(j, n-m-3*k+2*j)*(-1)^(j-k)*binomial(k,j))*binomial(m+k,m). - Vladimir Kruchinin, Dec 14 2011
Recurrence: T(n,k) = T(n-1,k) + T(n-1,k-1) + T(n-2,k) - T(n-3,k) with T(0,0) = 1. - Philippe Deléham, Feb 23 2012

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A089068 a(n) = a(n-1)+a(n-2)+a(n-3)+2 with a(0)=0, a(1)=0 and a(2)=1.

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A187070 Let i be in {1,2,3}, let r >= 0 be an integer and n=2*r+i-1. Then a(n)=a(2*r+i-1) gives the quantity of H_(7,3,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt((2*cos(Pi/7))^2-1).

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A231187 Decimal expansion of the length ratio (largest diagonal)/side in the regular 7-gon (or heptagon).

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A033303 Expansion of (1 + x)/(1 - 2*x - x^2 + x^3).

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A121442 Expansion of (1-x^2)/(1-x-9*x^2+x^3).

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A106195 Riordan array (1/(1-2*x), x*(1-x)/(1-2*x)).

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A187070 Let i be in {1,2,3}, let r >= 0 be an integer and n=2r+i-1. Then a(n)=a(2r+i-1) gives the quantity of H_(7,3,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt((2*cos(Pi/7))^2-1).

A106195 Riordan array (1/(1-2x), x(1-x)/(1-2*x)).

A187069 Let i be in {1,2,3}, let r >= 0 be an integer and n=2r+i-1. Then a(n)=a(2r+i-1) gives the quantity of H_(7,2,0) tiles in a subdivided H_(7,i,r) tile after linear scaling by the factor x^r, where x=sqrt((2*cos(Pi/7))^2-1).

A215560 a(n) = 3a(n-1) + 46a(n-2) + a(n-3) with a(0)=a(1)=3, a(2)=101.

A215572 a(n) = 3a(n-1) + 46a(n-2) + a(n-3) with a(0)=2, a(1)=5, a(2)=106.