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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A059260 Triangle read by rows giving coefficient T(i,j) of x^i y^j in 1/(1-y-x*y-x^2) = 1/((1+x)(1-x-y)) for (i,j) = (0,0), (1,0), (0,1), (2,0), (1,1), (0,2), ...

Original entry on oeis.org

1, 0, 1, 1, 1, 1, 0, 2, 2, 1, 1, 2, 4, 3, 1, 0, 3, 6, 7, 4, 1, 1, 3, 9, 13, 11, 5, 1, 0, 4, 12, 22, 24, 16, 6, 1, 1, 4, 16, 34, 46, 40, 22, 7, 1, 0, 5, 20, 50, 80, 86, 62, 29, 8, 1, 1, 5, 25, 70, 130, 166, 148, 91, 37, 9, 1, 0, 6, 30, 95, 200, 296, 314, 239, 128, 46, 10, 1
Offset: 0

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Author

N. J. A. Sloane, Jan 23 2001

Keywords

Comments

Coefficients of the (left, normalized) shifted cyclotomic polynomial. Or, coefficients of the basic n-th q-series for q=-2. Indeed, let Y_n(x) = Sum_{k=0..n} x^k, having as roots all the n-th roots of unity except for 0; then coefficients in x of (-1)^n Y_n(-x-1) give exactly the n-th row of A059260 and a practical way to compute it. - Olivier Gérard, Jul 30 2002
The maximum in the (2n)-th row is T(n,n), which is A026641; also T(n,n) ~ (2/3)*binomial(2n,n). The maximum in the (2n-1)-th row is T(n-1,n), which is A014300 (but T does not have the same definition as in A026637); also T(n-1,n) ~ (1/3)*binomial(2n,n). Here is a generalization of the formula given in A026641: T(i,j) = Sum_{k=0..j} binomial(i+k-x,j-k)*binomial(j-k+x,k) for all x real (the proof is easy by induction on i+j using T(i,j) = T(i-1,j) + T(i,j-1)). - Claude Morin, May 21 2002
The second greatest term in the (2n)-th row is T(n-1,n+1), which is A014301; the second greatest term in the (2n+1)-th row is T(n+1,n) = 2*T(n-1,n+1), which is 2*A014301. - Claude Morin
Diagonal sums give A008346. - Paul Barry, Sep 23 2004
Riordan array (1/(1-x^2), x/(1-x)). As a product of Riordan arrays, factors into the product of (1/(1+x),x) and (1/(1-x),1/(1-x)) (binomial matrix). - Paul Barry, Oct 25 2004
Signed version is A239473 with relations to partial sums of sequences. - Tom Copeland, Mar 24 2014
From Robert Coquereaux, Oct 01 2014: (Start)
Columns of the triangle (cf. Example below) give alternate partial sums along nw-se diagonals of the Pascal triangle, i.e., sequences A000035, A004526, A002620 (or A087811), A002623 (or A173196), A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808, etc.
The dimension of the space of closed currents (distributional forms) of degree p on Gr(n), the Grassmann algebra with n generators, equivalently, the dimension of the space of Gr(n)-valued symmetric multilinear forms with vanishing graded divergence, is V(n,p) = 2^n T(p,n-1) - (-1)^p.
If p is odd V(n,p) is also the dimension of the cyclic cohomology group of order p of the Z2 graded algebra Gr(n).
If p is even the dimension of this cohomology group is V(n,p)+1.
Cf. A193844. (End)
From Peter Bala, Feb 07 2024: (Start)
The following remarks assume the row indexing starts at n = 1.
The sequence of row polynomials R(n,x), beginning R(1,x) = 1, R(2,x) = x, R(3,x) = 1 + x + x^2 , ..., is a strong divisibility sequence of polynomials in the ring Z[x]; that is, for all positive integers n and m, poly_gcd( R(n,x), R(m,x)) = R(gcd(n, m), x) - apply Norfleet (2005), Theorem 3. Consequently, the polynomial sequence {R(n,x): n >= 1} is a divisibility sequence; that is, if n divides m then R(n,x) divides R(m,x) in Z[x]. (End)
From Miquel A. Fiol, Oct 04 2024: (Start)
For j>=1, T(i,j) is the independence number of the (i-j)-supertoken graph FF_(i-j)(S_j) of the star graph S_j with j points.
(Given a graph G on n vertices and an integer k>=1, the k-supertoken (or reduced k-th power) FF_k(G) of G has vertices representing configurations of k indistinguishable tokens in the (not necessarily different) vertices of G, with two configurations being adjacent if one can be obtained from the other by moving one token along an edge. See an example below.)
Following the suggestion of Peter Munn, the k-supertoken graph FF_k(S_j) can also be defined as follows: Consider the Lattice graph L(k,j), whose vertices are the k^j j-vectors with elements in the set {0,..,k-1}, two being adjacent if they differ in just one coordinate by one unity. Then, FF_k(S_j) is the subgraph of L(k+1,j) induced by the vertices at distance at most k from (0,..,0). (End)

Examples

			Triangle begins
  1;
  0,  1;
  1,  1,  1;
  0,  2,  2,  1;
  1,  2,  4,  3,  1;
  0,  3,  6,  7,  4,  1;
  1,  3,  9, 13, 11,  5,  1;
  0,  4, 12, 22, 24, 16,  6,  1;
  1,  4, 16, 34, 46, 40, 22,  7,  1;
  0,  5, 20, 50, 80, 86, 62, 29,  8,  1;
Sequences obtained with _Miquel A. Fiol_'s Sep 30 2024 formula of A(n,c1,c2) for other values of (c1,c2). (In the table, rows are indexed by c1=0..6 and columns by c2=0..6):
A000007  A000012  A000027  A025747  A000292* A000332* A000389*
A059841  A008619  A087811* A002623  A001752  A001753  A001769
A193356  A008794* A005993  A005994  -------  -------  -------
-------  -------  -------  A005995  A018210  -------  A052267
-------  -------  -------  -------  A018211  A018212  -------
-------  -------  -------  -------  -------  A018213  A018214
-------  -------  -------  -------  -------  -------  A062136
*requires offset adjustment.
The 2-supertoken FF_2(S_3) of the star graph S_3 with central vertex 1 and peripheral vertices 2,3,4. (The vertex `ij' of FF_2(S_3) represents the configuration of one token in `ì' and the other token in `j'). The T(5,3)=7 independent vertices are 22, 24, 44, 23, 11, 34, and 33.
     22--12---24---14---44
          | \    / |
         23   11   34
            \  |  /
              13
               |
              33
		

Crossrefs

Cf. A059259. Row sums give A001045.
Seen as a square array read by antidiagonals this is the coefficient of x^k in expansion of 1/((1-x^2)*(1-x)^n) with rows A002620, A002623, A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808 etc. (allowing for signs). A058393 would then effectively provide the table for nonpositive n. - Henry Bottomley, Jun 25 2001

Programs

  • Maple
    read transforms; 1/(1-y-x*y-x^2); SERIES2(%,x,y,12); SERIES2TOLIST(%,x,y,12);
  • Mathematica
    t[n_, k_] := Sum[ (-1)^(n-j)*Binomial[j, k], {j, 0, n}]; Flatten[ Table[t[n, k], {n, 0, 12}, {k, 0, n}]] (* Jean-François Alcover, Oct 20 2011, after Paul Barry *)
  • PARI
    T(n, k) = sum(j=0, n, (-1)^(n - j)*binomial(j, k));
    for(n=0, 12, for(k=0, n, print1(T(n, k),", ");); print();) \\ Indranil Ghosh, Apr 11 2017
    
  • Python
    from sympy import binomial
    def T(n, k): return sum((-1)**(n - j)*binomial(j, k) for j in range(n + 1))
    for n in range(13): print([T(n, k) for k in range(n + 1)]) # Indranil Ghosh, Apr 11 2017
  • Sage
    def A059260_row(n):
        @cached_function
        def prec(n, k):
            if k==n: return 1
            if k==0: return 0
            return -prec(n-1,k-1)-sum(prec(n,k+i-1) for i in (2..n-k+1))
        return [(-1)^(n-k+1)*prec(n+1, n-k+1) for k in (1..n)]
    for n in (1..9): print(A059260_row(n)) # Peter Luschny, Mar 16 2016
    

Formula

G.f.: 1/(1-y-x*y-x^2) = 1 + y + x^2 + xy + y^2 + 2x^2y + 2xy^2 + y^3 + ...
E.g.f: (exp(-t)+(x+1)*exp((x+1)*t))/(x+2). - Tom Copeland, Mar 19 2014
O.g.f. (n-th row): ((-1)^n+(x+1)^(n+1))/(x+2). - Tom Copeland, Mar 19 2014
T(i, 0) = 1 if i is even or 0 if i is odd, T(0, i) = 1 and otherwise T(i, j) = T(i-1, j) + T(i, j-1); also T(i, j) = Sum_{m=j..i+j} (-1)^(i+j+m)*binomial(m, j). - Robert FERREOL, May 17 2002
T(i, j) ~ (i+j)/(2*i+j)*binomial(i+j, j); more precisely, abs(T(i, j)/binomial(i+j, j) - (i+j)/(2*i+j) )<=1/(4*(i+j)-2); the proof is by induction on i+j using the formula 2*T(i, j) = binomial(i+j, j)+T(i, j-1). - Claude Morin, May 21 2002
T(n, k) = Sum_{j=0..n} (-1)^(n-j)binomial(j, k). - Paul Barry, Aug 25 2004
T(n, k) = Sum_{j=0..n-k} binomial(n-j, j)*binomial(j, n-k-j). - Paul Barry, Jul 25 2005
Equals A097807 * A007318. - Gary W. Adamson, Feb 21 2007
Equals A128173 * A007318 as infinite lower triangular matrices. - Gary W. Adamson, Feb 17 2007
Equals A130595*A097805*A007318 = (inverse Pascal matrix)*(padded Pascal matrix)*(Pascal matrix) = A130595*A200139. Inverse is A097808 = A130595*(padded A130595)*A007318. - Tom Copeland, Nov 14 2016
T(i, j) = binomial(i+j, j)-T(i-1, j). - Laszlo Major, Apr 11 2017
Recurrence for row polynomials (with row indexing starting at n = 1): R(n,x) = x*R(n-1,x) + (x + 1)*R(n-2,x) with R(1,x) = 1 and R(2,x) = x. - Peter Bala, Feb 07 2024
From Miquel A. Fiol, Sep 30 2024: (Start)
The triangle can be seen as a slice of a 3-dimensional table that links it to well-known sequences as follows.
The j-th column of the triangle, T(i,j) for i >= j, equals A(n,c1,c2) = Sum_{k=0..floor(n/2)} binomial(c1+2*k-1,2*k)*binomial(c2+n-2*k-1,n-2*k) when c1=1, c2=j, and n=i-j.
This gives T(i,j) = Sum_{k=0..floor((i-j)/2)} binomial(i-2*k-1, j-1). For other values of (c1,c2), see the example below. (End)

Extensions

Formula corrected by Philippe Deléham, Jan 11 2014

A001753 Expansion of 1/((1+x)*(1-x)^6).

Original entry on oeis.org

1, 5, 16, 40, 86, 166, 296, 496, 791, 1211, 1792, 2576, 3612, 4956, 6672, 8832, 11517, 14817, 18832, 23672, 29458, 36322, 44408, 53872, 64883, 77623, 92288, 109088, 128248, 150008, 174624, 202368, 233529
Offset: 0

Views

Author

Keywords

Comments

Number of symmetric nonnegative integer 5 X 5 matrices with sum of elements equal to 4*n under action of dihedral group D_4.
a(n) = A108561(n+6,n) for n>0. - Reinhard Zumkeller, Jun 10 2005

Examples

			There are 5 symmetric nonnegative integer 5 X 5 matrices with sum of elements equal to 4 under action of D_4:
[1 0 0 0 1] [0 0 1 0 0] [0 0 0 0 0] [0 0 0 0 0] [0 0 0 0 0]
[0 0 0 0 0] [0 0 0 0 0] [0 1 0 1 0] [0 0 1 0 0] [0 0 0 0 0]
[0 0 0 0 0] [1 0 0 0 1] [0 0 0 0 0] [0 1 0 1 0] [0 0 4 0 0]
[0 0 0 0 0] [0 0 0 0 0] [0 1 0 1 0] [0 0 1 0 0] [0 0 0 0 0]
[1 0 0 0 1] [0 0 1 0 0] [0 0 0 0 0] [0 0 0 0 0] [0 0 0 0 0].
		

Crossrefs

Cf. A000217, A002620, A008804, A038163, A054343, A001769 (partial sums), A001752 (first differences), A169793 (binomial transf).

Programs

  • Magma
    [(4*n^5+70*n^4+460*n^3+1400*n^2+1936*n+945)/960+(-1)^n/64: n in [0..40]]; // Vincenzo Librandi, Aug 15 2011
    
  • Mathematica
    CoefficientList[Series[1/((1+x)*(1-x)^6), {x, 0, 50}], x] (* G. C. Greubel, Nov 22 2017 *)
    LinearRecurrence[{5,-9,5,5,-9,5,-1},{1,5,16,40,86,166,296},40] (* Harvey P. Dale, Jun 05 2021 *)
  • PARI
    a(n)=(4*n^5+70*n^4+460*n^3+1400*n^2+1936*n)\/960+1 \\ Charles R Greathouse IV, Apr 17 2012

Formula

a(n) = Sum{k=0..n} (-1)^(n-k)*binomial(k+5, 5); a(n) = (4*n^5 + 70*n^4 + 460*n^3 + 1400*n^2 + 1936*n + 945)/960 + (-1)^n/64. - Paul Barry, Jul 01 2003
a(n) = a(n-2) + (n*(n + 1)*(n + 2)*(n - 1))/24, a(1) = 0, a(2) = 1; (15*(-1)^n - 15*(-1)^(2*n) + 96*n - 160*(-1)^(2*n)*n + 200*n^2 - 200*(-1)^(2*n)*n^2 + 140*n^3 - 80*(-1)^(2*n)*n^3 + 40*n^4 - 10*(-1)^(2*n)*n^4 + 4*n^5)/960. - Cecilia Rossiter (cecilia(AT)noticingnumbers.net), Dec 14 2004
a(n) + a(n+1) = A000389(n+6). - R. J. Mathar, Mar 14 2011

Extensions

Comment and example from Vladeta Jovovic, May 14 2000

A158454 Riordan array (1/(1-x^2), x/(1+x)^2).

Original entry on oeis.org

1, 0, 1, 1, -2, 1, 0, 4, -4, 1, 1, -6, 11, -6, 1, 0, 9, -24, 22, -8, 1, 1, -12, 46, -62, 37, -10, 1, 0, 16, -80, 148, -128, 56, -12, 1, 1, -20, 130, -314, 367, -230, 79, -14, 1, 0, 25, -200, 610, -920, 771, -376, 106, -16, 1, 1, -30, 295, -1106, 2083, -2232, 1444, -574, 137, -18, 1
Offset: 0

Views

Author

Paul Barry, Mar 19 2009

Keywords

Comments

Coefficient table of the square of Chebyshev S-polynomials. For the S-polynomials see A049310, and for a proof see the array A181878, where the odd numbered rows are shifted by one to the left. - Wolfdieter Lang, Dec 15 2010
Image of the Catalan numbers A000108 by this matrix is the all 1's sequence.
Image of the central binomial numbers A000984 by this matrix is the counting numbers A000027.
Inverse array is the Riordan array (1-x^2*c(x)^4, xc(x)^2), where c(x) is the g.f. of A000108.
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k belong to the class of Boas-Buck polynomials. Hence they satisfy the Boas-Buck identity: (E_x - n*1)*R(n, x) = -Sum_{p=0..n-1} ((1 - (-1)^p)*1 + 2*(-1)^(p+1)*E_x) R(n-1-p, x) for n >= 0. See the Boas-Buck comments and references in A046521. The ensuing recurrence for the column sequences is given in the formula section. - Wolfdieter Lang, Aug 10 2017

Examples

			The triangle T(n,k) begins:
  n\k  0   1    2     3     4      5     6     7    8    9  10...
  0:   1
  1:   0   1
  2:   1  -2    1
  3:   0   4   -4     1
  4:   1  -6   11    -6     1
  5:   0   9  -24    22    -8      1
  6:   1 -12   46   -62    37    -10     1
  7:   0  16  -80   148  -128     56   -12     1
  8:   1 -20  130  -314   367   -230    79   -14    1
  9:   0  25 -200   610  -920    771  -376   106  -16    1
  10:  1 -30  295 -1106  2083  -2232  1444  -574  137  -18   1
  ... Reformatted and extended by _Wolfdieter Lang_, Nov 24 2012
Recurrences (from A- and Z-sequences):
  1 = T(6,0) = 0*0 + 1*9 +2*(-24) + 5*22 + 14*(-8)+ 42*1.
-80 = T(7,2) = 1*(-12) -2*(46) -1*(-62) -2*37 -5*(-10) -14*1. - _Wolfdieter Lang_, Dec 20 2010
		

References

  • Kenneth Edwards, Michael A. Allen, A new combinatorial interpretation of the Fibonacci numbers squared, Part II, Fib. Q., 58:2 (2020), 169-177.

Crossrefs

From Wolfdieter Lang, Aug 10 2017: (Start)
Row sums A011655(n+1), alternating row sums A007598(n+1)*(-1)^(n+1).
Column sequences k=0..5: A059841, A002620(n+2)*(-1)^(n), A001752(n)*(-1)^n, A001769(n)*(-1)^n, A001780(n)*(-1)^n, A001786(n)*(-1)^n. (End)

Programs

  • GAP
    T:=Flat(List([0..10], n->List([0..n], k->Sum([0..n], j-> (-1)^(j-k)*Binomial(k+j, 2*k))))); # G. C. Greubel, Dec 15 2018
  • Magma
    [[(&+[(-1)^(j-k)*Binomial(k+j, 2*k): j in [0..n]]): k in [0..n]]: n in [0..10]]; // G. C. Greubel, Dec 15 2018
    
  • Maple
    A158454 := proc(n,k) (-1)^(n+k)*add(binomial(n+k-1-2*j,2*k-1),j=0..floor(n/2)) ; end proc;
    seq(seq(A158454(n,k),k=0..n),n=0..10) ; # R. J. Mathar, Dec 17 2010
  • Mathematica
    nmax = 10; t[n_, k_] := (-1)^(n+k)* Sum[Binomial[n+k-1-2*j, 2*k-1], {j, 0, Floor[n/2]}]; t[n_?EvenQ, 0] = 1; Flatten[ Table[ t[n, k], {n, 0, nmax}, {k, 0, n}]] (* Jean-François Alcover, Nov 08 2011, after Maple *)
    With[{m = 15}, CoefficientList[CoefficientList[Series[(1+x)/((1-x)*(1 + x)^2 -t*x*(1-x)), {x, 0, m}, {t, 0, m}], x], t]]//Flatten (* G. C. Greubel, Dec 15 2018 *)
    T[n_, 0] := Boole[EvenQ[n]]; T[n_, k_] := (-1)^(n - k) Binomial[k+n-1, 2*k-1] HypergeometricPFQ[{1, (k - n)/2, (1 + k - n)/2}, {(1 - k - n)/2, (2 - k - n)/2}, 1]; Table[T[n, k], {n, 0, 9}, {k, 0, n}] // TableForm  (* Peter Luschny, Aug 20 2022 *)
  • PARI
    {T(n,k) = sum(j=0,n, (-1)^(j-k)*binomial(k+j, 2*k))};
    for(n=0, 10, for(k=0,n, print1(T(n,k), ", "))) \\ G. C. Greubel, Dec 15 2018
    
  • Sage
    [[sum((-1)^(j-k)*binomial(k+j, 2*k) for j in range(n+1)) for k in range(n+1)] for n in range(10)] # G. C. Greubel, Dec 15 2018
    

Formula

Number triangle T(n, k) = Sum_{j=0..n} (-1)^(j-k)*binomial(k+j, 2*k) = Sum_{j=0..n-k} (-1)^(n-k-j)*binomial(n+k-j, 2*k).
O.g.f. column k with leading zeros (Riordan array, see NAME): (1/(1-x^2))*(x/(1+x)^2)^k, k >= 0. - Wolfdieter Lang, Dec 15 2010
T(n, k) = (-1)^(n-k)*Sum_{j=0..floor(n/2)} binomial(n+k-1-2*j, 2*k-1), 0 <= k <= n, else 0. From the o.g.f. for column k after convolution. - Wolfdieter Lang, Dec 17 2010
O.g.f. row polynomials (rising powers in y): ((1+x)/(1-x))/(1+(2-y)*x+x^2) = Sum_{n>=0} S(n,sqrt(y))^2*x^n, with Chebyshev S-polynomials from A049310. - Wolfdieter Lang, Dec 15 2010
Recurrences from the A- and Z-sequences for Riordan arrays. See the W. Lang link under A006232 for details and references.
T(n, 0) = Sum_{j=0..n-1} Z(j)*T(n-1, j), n >= 1.
T(n, k) = Sum_{j=0..n-k} A(j)*T(n-1, k-1+j), n >= k >= 1.
Here Z(0)=0 and Z(j) = A000108(j), j >= 1, (o.g.f. -1 + c(x), with the Catalan o.g.f. c(x)), and A(j) = A115141(j) = [1,-2,-1,-2,-5,-14,...], j >= 0, with o.g.f. 1/c(x)^2. - Wolfdieter Lang, Dec 20 2010
T(n, k) = Sum_{m=0..n} A129818(m, k), 0 <= k <= n. - Wolfdieter Lang, Dec 15 2010
Boas-Buck recurrence for column k: R(n, k) = (1/(n-k))*Sum_{p=k..n-1}((-1)^(n-p)*(2*k+1) + 1) * R(p, k), for n > k >= 0, with input R(k, k) = 1. See a comment above. - Wolfdieter Lang, Aug 10 2017
G.f.: (1 + x)/((1 - x)*(1 + x)^2 - t*x*(1 - x)). - G. C. Greubel, Dec 15 2018
T(n, k) = (-1)^(n - k)*binomial(k + n - 1, 2*k-1)*hypergeom([1, (k - n)/2, (1 + k - n)/2], [(1 - k - n)/2, (2 - k - n)/2], 1) for k >= 1 . - Peter Luschny, Aug 20 2022

A112465 Riordan array (1/(1+x), x/(1-x)).

Original entry on oeis.org

1, -1, 1, 1, 0, 1, -1, 1, 1, 1, 1, 0, 2, 2, 1, -1, 1, 2, 4, 3, 1, 1, 0, 3, 6, 7, 4, 1, -1, 1, 3, 9, 13, 11, 5, 1, 1, 0, 4, 12, 22, 24, 16, 6, 1, -1, 1, 4, 16, 34, 46, 40, 22, 7, 1, 1, 0, 5, 20, 50, 80, 86, 62, 29, 8, 1, -1, 1, 5, 25, 70, 130, 166, 148, 91, 37, 9, 1, 1, 0, 6, 30, 95, 200, 296, 314, 239, 128, 46, 10, 1
Offset: 0

Views

Author

Paul Barry, Sep 06 2005

Keywords

Comments

Inverse is A112466. Note that C(n,k) = Sum_{j = 0..n-k} C(j+k-1, j).

Examples

			Triangle starts
   1;
  -1, 1;
   1, 0, 1;
  -1, 1, 1,  1;
   1, 0, 2,  2,  1;
  -1, 1, 2,  4,  3,  1;
   1, 0, 3,  6,  7,  4,  1;
  -1, 1, 3,  9, 13, 11,  5, 1;
   1, 0, 4, 12, 22, 24, 16, 6, 1;
Production matrix begins
  -1, 1;
   0, 1, 1;
   0, 0, 1, 1;
   0, 0, 0, 1, 1;
   0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 0, 1, 1;
   0, 0, 0, 0, 0, 0, 0, 1, 1; - _Paul Barry_, Apr 08 2011
		

Crossrefs

Columns: A033999(n) (k=0), A000035(n) (k=1), A004526(n) (k=2), A002620(n-1) (k=3), A002623(n-4) (k=4), A001752(n-5) (k=5), A001753(n-6) (k=6), A001769(n-7) (k=7), A001779(n-8) (k=8), A001780(n-9) (k=9), A001781(n-10) (k=10), A001786(n-11) (k=11), A001808(n-12) (k=12).
Diagonals: A000012(n) (k=n), A023443(n) (k=n-1), A152947(n-1) (k=n-2), A283551(n-3) (k=n-3).
Main diagonal: A072547.
Sums: A078008 (row), A078024 (diagonal), A092220 (signed diagonal), A280560 (signed row).

Programs

  • Haskell
    a112465 n k = a112465_tabl !! n !! k
    a112465_row n = a112465_tabl !! n
    a112465_tabl = iterate f [1] where
       f xs'@(x:xs) = zipWith (+) ([-x] ++ xs ++ [0]) ([0] ++ xs')
    -- Reinhard Zumkeller, Jan 03 2014
    
  • Magma
    A112465:= func< n,k | (-1)^(n+k)*(&+[(-1)^j*Binomial(j+k-1,j): j in [0..n-k]]) >;
    [A112465(n,k): k in [0..n], n in [0..13]]; // G. C. Greubel, Apr 18 2025
    
  • Mathematica
    T[n_, k_]:= Sum[Binomial[j+k-1, j]*(-1)^(n-k-j), {j, 0, n-k}];
    Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* Jean-François Alcover, Jul 23 2018 *)
  • SageMath
    def A112465(n,k): return (-1)^(n+k)*sum((-1)^j*binomial(j+k-1,j) for j in range(n-k+1))
    print(flatten([[A112465(n,k) for k in range(n+1)] for n in range(13)])) # G. C. Greubel, Apr 18 2025

Formula

Number triangle T(n, k) = Sum_{j=0..n-k} (-1)^(n-k-j)*C(j+k-1, j).
T(2*n, n) = A072547(n) (main diagonal). - Paul Barry, Apr 08 2011
From Reinhard Zumkeller, Jan 03 2014: (Start)
T(n, k) = T(n-1, k-1) + T(n-1, k), 0 < k < n, with T(n, 0) = (-1)^n and T(n, n) = 1.
T(n, k) = A108561(n, n-k). (End)
T(n, k) = T(n-1, k-1) + T(n-2, k) + T(n-2, k-1), T(0, 0) = 1, T(1, 0) = -1, T(1, 1) = 1, T(n, k) = 0 if k < 0 or if k > n. - Philippe Deléham, Jan 11 2014
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(-1 + x + x^2/2! + x^3/3!) = -1 + 2*x^2/2! + 6*x^3/3! + 13*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014

A001779 Expansion of 1/((1+x)(1-x)^8).

Original entry on oeis.org

1, 7, 29, 91, 239, 553, 1163, 2269, 4166, 7274, 12174, 19650, 30738, 46782, 69498, 101046, 144111, 201993, 278707, 379093, 508937, 675103, 885677, 1150123, 1479452, 1886404, 2385644, 2993972
Offset: 0

Views

Author

Keywords

Comments

a(n) is the number of positive terms in the expansion of (a_1 + a_2 + a_3 + a_4 + a_5 + a_6 + a_7 - z)^n. Also the convolution of A001769 and A000012; A001753 and A001477; A001752 and A000217; A002623 and A000292; A002620 and A000332; A004526 and A000389. - Sergio Falcon (sfalcon(AT)dma.ulpgc.es), Feb 13 2007

Crossrefs

Cf. A001769 (first differences), A169795 (binomial transf.)

Programs

  • Magma
    [1/80640*(2*n+9) *(4*n^6 +108*n^5 +1138*n^4 +5904*n^3 +15628*n^2 +19638*n +8925)+(-1)^n/256 : n in [0..30]]; // Vincenzo Librandi, Oct 08 2011
    
  • Maple
    A001779 := proc(n) 1/80640*(2*n+9) *(4*n^6 +108*n^5 +1138*n^4 +5904*n^3 +15628*n^2 +19638*n +8925)+(-1)^n/256 ; end proc:
    seq(A001779(n),n=0..50) ; # R. J. Mathar, Mar 22 2011
  • Mathematica
    CoefficientList[Series[1/((1 + x) (1 - x)^8), {x, 0, 50}], x] (* G. C. Greubel, Nov 24 2017 *)
    LinearRecurrence[{7,-20,28,-14,-14,28,-20,7,-1},{1,7,29,91,239,553,1163,2269,4166},30] (* Harvey P. Dale, Jan 21 2023 *)
  • PARI
    a(n)=(2*n+9)*(4*n^6+108*n^5+1138*n^4+5904*n^3+15628*n^2+19638*n + 8925)/80640 +(-1)^n/256 \\ Charles R Greathouse IV, Apr 17 2012

Formula

a(n) = (-1)^{7-n}*Sum_{i=0..n} ((-1)^(7-i)*binomial(7+i,i)). - Sergio Falcon, Feb 13 2007
a(n)+a(n+1) = A000580(n+8). - R. J. Mathar, Jan 06 2021

A001780 Expansion of 1/((1+x)(1-x)^9).

Original entry on oeis.org

1, 8, 37, 128, 367, 920, 2083, 4352, 8518, 15792, 27966, 47616, 78354, 125136, 194634, 295680, 439791, 641784, 920491, 1299584, 1808521, 2483624, 3369301, 4519424, 5998876, 7885280, 10270924, 13264896
Offset: 0

Views

Author

Keywords

Crossrefs

Cf. A001769, A158454 (signed column k=4), A001779 (first differences), A169796 (binomial trans.).

Programs

Formula

a(n) = 431*n/168 + (-1)^n/512 + 391*n^3/288 + 26011*n^2/10080 + 797*n^4/1920 + 11*n^5/144 + n^6/120 + n^7/2016 + n^8/80640 + 511/512. - R. J. Mathar, Mar 15 2011
Boas-Buck recurrence: a(n) = (1/n)*Sum_{p=0..n-1} (9 + (-1)^(n-p))*a(p), n >= 1, a(0) = 1. See the Boas-Buck comment in A046521 (here for the unsigned column k = 4 with offset 0). - Wolfdieter Lang, Aug 10 2017
a(n)+a(n+1) = A000581(n+9) . - R. J. Mathar, Jan 06 2021

A185957 Second accumulation array of the array min{n,k}, by antidiagonals.

Original entry on oeis.org

1, 3, 3, 6, 10, 6, 10, 21, 21, 10, 15, 36, 46, 36, 15, 21, 55, 81, 81, 55, 21, 28, 78, 126, 146, 126, 78, 28, 36, 105, 181, 231, 231, 181, 105, 36, 45, 136, 246, 336, 371, 336, 246, 136, 45, 55, 171, 321, 461, 546, 546, 461, 321, 171, 55, 66, 210, 406, 606, 756, 812, 756
Offset: 1

Views

Author

Clark Kimberling, Feb 07 2011

Keywords

Comments

A member of the accumulation chain
... < A003982 < A003783 < A115262 < A185957 <...,
where A003783(n,k)=min{n,k}. See A144112 for the definition of accumulation array.
A185957 also gives the symmetric matrix based on the triangular numbers s=(1,3,6,10,15,....; viz, let T be the infinite square matrix whose n-th row is formed by putting n-1 zeros before the terms of s. Let T' be the transpose of T. Then A185957 represents the matrix product M=T'*T. M is the self-fusion matrix of s, as defined at A193722. See A202678 for characteristic polynomials of principal submatrices of M.

Examples

			Northwest corner:
1....3....6....10...15
3....10...21...36...55
6....21...46...81...126
10...36...81...146..231
		

Crossrefs

Programs

  • Mathematica
    U = NestList[Most[Prepend[#, 0]] &, #, Length[#] - 1] &[Table[k (k + 1)/2, {k, 1, 15}]];
    L = Transpose[U]; M = L.U; TableForm[M]
    m[i_, j_] := M[[i]][[j]];
    Flatten[Table[m[i, n + 1 - i], {n, 1, 12}, {i, 1, n}]]
    f[n_] := Sum[m[i, n], {i, 1, n}] + Sum[m[n, j], {j, 1, n - 1}]
    Table[f[n], {n, 1, 12}]
    Table[Sqrt[f[n]], {n, 1, 12}] (* A000292 *)
    Table[m[1, j], {j, 1, 12}] (* A000217 *)
    Table[m[2, j], {j, 1, 12}] (* A014105 *)
    Table[m[j, j], {j, 1, 12}] (* A024166 *)
    Table[m[j, j + 1], {j, 1, 12}] (* A112851 *)
    Table[Sum[m[i, n + 1 - i], {i, 1, n}], {n, 1, 12}] (* A001769 *)

A299337 Expansion of 1 / ((1 - x)^7*(1 + x)^5).

Original entry on oeis.org

1, 2, 8, 14, 35, 56, 112, 168, 294, 420, 672, 924, 1386, 1848, 2640, 3432, 4719, 6006, 8008, 10010, 13013, 16016, 20384, 24752, 30940, 37128, 45696, 54264, 65892, 77520, 93024, 108528, 128877, 149226, 175560, 201894, 235543, 269192, 311696, 354200, 407330
Offset: 0

Views

Author

Colin Barker, Feb 07 2018

Keywords

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{2, 4, -10, -5, 20, 0, -20, 5, 10, -4, -2, 1}, {1, 2, 8, 14, 35, 56, 112, 168, 294, 420, 672, 924}, 41] (* Michael De Vlieger, Dec 19 2024 *)
  • PARI
    Vec(1 / ((1 - x)^7*(1 + x)^5) + O(x^40))

Formula

a(n) = (2*n^6 + 72*n^5 + 1040*n^4 + 7680*n^3 + 30368*n^2 + 60288*n + 46080) / 46080 for n even.
a(n) = (2*n^6 + 72*n^5 + 1010*n^4 + 6960*n^3 + 24278*n^2 + 39048*n + 20790) / 46080 for n odd.
a(n) = 2*a(n-1) + 4*a(n-2) - 10*a(n-3) - 5*a(n-4) + 20*a(n-5) - 20*a(n-7) + 5*a(n-8) + 10*a(n-9) - 4*a(n-10) - 2*a(n-11) + a(n-12) for n>11.

A128494 Coefficient table for sums of Chebyshev's S-Polynomials.

Original entry on oeis.org

1, 1, 1, 0, 1, 1, 0, -1, 1, 1, 1, -1, -2, 1, 1, 1, 2, -2, -3, 1, 1, 0, 2, 4, -3, -4, 1, 1, 0, -2, 4, 7, -4, -5, 1, 1, 1, -2, -6, 7, 11, -5, -6, 1, 1, 1, 3, -6, -13, 11, 16, -6, -7, 1, 1, 0, 3, 9, -13, -24, 16, 22, -7, -8, 1, 1, 0, -3, 9, 22, -24, -40, 22, 29, -8, -9, 1, 1, 1, -3, -12, 22, 46, -40, -62, 29, 37, -9, -10, 1, 1, 1, 4, -12
Offset: 0

Views

Author

Wolfdieter Lang, Apr 04 2007

Keywords

Comments

See A049310 for the coefficient table of Chebyshev's S(n,x)=U(n,x/2) polynomials.
This is a 'repetition triangle' based on a signed version of triangle A059260: a(2*p,2*k) = a(2*p+1,2*k) = A059260(p+k,2*k)*(-1)^(p+k) and a(2*p+1,2*k+1) = a(2*p+2,2*k+1) = A059260(p+k+1,2*k+1)*(-1)^(p+k), k >= 0.

Examples

			The triangle a(n,m) begins:
  n\m  0   1   2   3   4   5   6   7   8   9  10
   0:  1
   1:  1   1
   2:  0   1   1
   3:  0  -1   1   1
   4:  1  -1  -2   1   1
   5:  1   2  -2  -3   1   1
   6:  0   2   4  -3  -4   1   1
   7:  0  -2   4   7  -4  -5   1   1
   8:  1  -2  -6   7  11  -5  -6   1   1
   9:  1   3  -6 -13  11  16  -6  -7   1   1
  10:  0   3   9 -13 -24  16  22  -7  -8   1   1
... reformatted by _Wolfdieter Lang_, Oct 16 2012
Row polynomial S(1;4,x) = 1 - x - 2*x^2 + x^3 + x^4 = Sum_{k=0..4} S(k,x).
S(4,y)*S(5,y)/y = 3 - 13*y^2 + 16*y^4 - 7*y^6 + y^8, with y=sqrt(2+x) this becomes S(1;4,x).
From _Wolfdieter Lang_, Oct 16 2012: (Start)
S(1;4,x) = (1 - (S(5,x) - S(4,x)))/(2-x) = (1-x)*(2-x)*(1+x)*(1-x-x^2)/(2-x) = (1-x)*(1+x)*(1-x-x^2).
S(5,x) - S(4,x) = R(11,sqrt(2+x))/sqrt(2+x) = -1 + 3*x + 3*x^2 - 4*x^3 - x^4 + x^5. (End)
		

Crossrefs

Row sums (signed): A021823(n+2). Row sums (unsigned): A070550(n).
Cf. A128495 for S(2; n, x) coefficient table.
The column sequences (unsigned) are, for m=0..4: A021923, A002265, A008642, A128498, A128499.
For m >= 1 the column sequences (without leading zeros) are of the form a(m, 2*k) = a(m, 2*k+1) = ((-1)^k)*b(m, k) with the sequences b(m, k), given for m=1..11 by A008619, A002620, A002623, A001752, A001753, A001769, A001779, A001780, A001781, A001786, A001808.

Formula

S(1;n,x) = Sum_{k=0..n} S(k,x) = Sum_{m=0..n} a(n,m)*x^m, n >= 0.
a(n,m) = [x^m](S(n,y)*S(n+1,y)/y) with y:=sqrt(2+x).
G.f. for column m: (x^m)/((1-x)*(1+x^2)^(m+1)), which shows that this is a lower diagonal matrix of the Riordan type, named (1/((1+x^2)*(1-x)), x/(1+x^2)).
From Wolfdieter Lang, Oct 16 2012: (Start)
a(n,m) = [x^m](1- (S(n+1,x) - S(n,x)))/(2-x). From the Binet - de Moivre formula for S and use of the geometric sum.
a(n,m) = [x^m](1- R(2*n+3,sqrt(2+x))/sqrt(2+x))/(2-x) with the monic integer T-polynomials R with coefficient triangle given in A127672. From the odd part of the bisection of the T-polynomials. (End)

A299335 Expansion of 1 / ((1 - x)^7*(1 + x)^2).

Original entry on oeis.org

1, 5, 17, 45, 103, 211, 399, 707, 1190, 1918, 2982, 4494, 6594, 9450, 13266, 18282, 24783, 33099, 43615, 56771, 73073, 93093, 117481, 146965, 182364, 224588, 274652, 333676, 402900, 483684, 577524, 686052, 811053, 954465, 1118397, 1305129, 1517131, 1757063
Offset: 0

Views

Author

Colin Barker, Feb 07 2018

Keywords

Crossrefs

Programs

  • Mathematica
    CoefficientList[Series[1/((1 - x)^7 (1 + x)^2), {x, 0, 40}], x] (* or *) LinearRecurrence[{5, -8, 0, 14, -14, 0, 8, -5, 1}, {1, 5, 17, 45, 103, 211, 399, 707, 1190}, 41] (* Robert G. Wilson v, Feb 07 2018 *)
  • PARI
    Vec(1 / ((1 - x)^7*(1 + x)^2) + O(x^40))

Formula

a(n) = (2*n^6 + 54*n^5 + 575*n^4 + 3060*n^3 + 8468*n^2 + 11376*n + 5760) / 5760 for n even.
a(n) = (2*n^6 + 54*n^5 + 575*n^4 + 3060*n^3 + 8468*n^2 + 11286*n + 5355) / 5760 for n odd.
a(n) = 5*a(n-1) - 8*a(n-2) + 14*a(n-4) - 14*a(n-5) + 8*a(n-7) - 5*a(n-8) + a(n-9) for n>8.
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