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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A005993 Expansion of (1+x^2)/((1-x)^2*(1-x^2)^2).

Original entry on oeis.org

1, 2, 6, 10, 19, 28, 44, 60, 85, 110, 146, 182, 231, 280, 344, 408, 489, 570, 670, 770, 891, 1012, 1156, 1300, 1469, 1638, 1834, 2030, 2255, 2480, 2736, 2992, 3281, 3570, 3894, 4218, 4579, 4940, 5340, 5740, 6181, 6622, 7106, 7590, 8119, 8648, 9224, 9800
Offset: 0

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Author

N. J. A. Sloane, Winston C. Yang (yang(AT)math.wisc.edu)

Keywords

Comments

Alkane (or paraffin) numbers l(6,n).
Dimension of the space of homogeneous degree n polynomials in (x1, y1, x2, y2) invariant under permutation of variables x1<->y1, x2<->y2.
Also multidigraphs with loops on 2 nodes with n arcs (see A138107). - Vladeta Jovovic, Dec 27 1999
Euler transform of finite sequence [2,3,0,-1]. - Michael Somos, Mar 17 2004
a(n-2) is the number of plane partitions with trace 2. - Michael Somos, Mar 17 2004
With offset 4, a(n) is the number of bracelets with n beads, 3 of which are red, 1 of which is blue. For odd n, a(n) = C(n-1,3)/2. For even n, a(n) = C(n-1,3)/2 +(n-2)/4. For n >= 6, with K = (n-1)(n-2)/((n-5)(n-4)), for odd n, a(n) = K*a(n-2). For even n, a(n) = K*a(n-2) -(n-2)/(n-5). - Washington Bomfim, Aug 05 2008
Equals (1,2,3,4,...) convolved with (1,0,3,0,5,...). - Gary W. Adamson, Feb 16 2009
Equals row sums of triangle A177878.
Equals (1/2)*((1, 4, 10, 20, 35, 56, ...) + (1, 0, 2 0, 3, 0, 4, ...)).
From Ctibor O. Zizka, Nov 21 2014: (Start)
With offset 4, a(n) is the number of different patterns of the 2-color 4-partition of n.
P(n)_(k;t) gives the number of different patterns of the t-color, k-partition of n.
P(n;i;j) = Sum(r=1..m) c_(i,j)*v_r*F_r(X_1,...,X_i).
m partition number of i.
c_(i,j) number of different coloring patterns on the r-th form (X_1,...,X_i) of i-partition with j-colors.
v_r number of i-partitions of n of the r-th form (X_1,...,X_i).
F_r(X_1,...,X_i) number of different patterns of the r-th form i-partition of n.
Some simple results:
P(1)(k;t)=1, P(2)(k;t)=2, P(3)(k;t)=4, P(4)(k;t)=11, etc.
P(n;1;1) = P(n;n;n) = 1 for all n;
P(n;2;2) = floor(n/2) (A004526);
P(n;3;2) = (n*n - 2*n + n mod 2)/4 (A002620).
This sequence is a(n) = P(n;4;2).
2-coloring of 4-partition is (A,B,A,B) or (B,A,B,A).
Each 4-partition of n has one of the form (X_1,X_1,X_1,X_1),(X_1,X_1,X_1,X_2), (X_1,X_1,X_2,X_2),(X_1,X_1,X_2,X_3),(X_1,X_2,X_3,X_4).
The number of forms is m=5 which is the partition number of k=4.
Partition form (X_1,X_1,X_1,X_1) gives 1 pattern ((X_1A,X_1B,X_1A,X_1B), (X_1,X_1,X_1,X_2) gives 2 patterns, (X_1,X_1,X_2,X_2) gives 4 patterns, (X_1,X_1,X_2,X_3) gives 6 patterns and (X_1,X_2,X_3,X_4) gives 12 patterns.
Thus a(n) = P(n;4;2) = 1*1*v_1 + 1*2*v_2 + 1*4*v_3 + 1*6*v_4 + 1*12*v_5 where v_r is the number of different 4-partitions of the r-th form (X_1,X_2,X_3,X_4) for a given n.
Example:
The 4-partitions of 8 are (2,2,2,2), (1,1,1,5), (1,1,3,3), (1,1,2,4), and (1,2,2,3):
(2,2,2,2) 1 pattern
(1,1,1,5), (1,1,5,1) 2 patterns
(1,1,3,3), (1,3,3,1), (3,1,1,3), (1,3,1,3) 4 patterns
(1,1,2,4), (1,1,4,2), (1,2,1,4), (1,2,4,1), (1,4,1,2), (2,1,1,4) 6 patterns
(2,2,1,3), (2,2,3,1), (2,1,2,3), (2,1,3,2), (2,3,2,1), (1,2,2,3) 6 patterns
Thus a(8) = P(8,4,2) = 1 + 2 + 4 + 6 + 6 = 19. (End)
a(n) = length of run n+2 of consecutive 1's in A254338. - Reinhard Zumkeller, Feb 27 2015
Take a chessboard of (n+2) X (n+2) unit squares in which the a1 square is black. a(n) is the number of composite squares having black unit squares on their vertices. - Ivan N. Ianakiev, Jul 19 2018
a(n) is the number of 1423-avoiding odd Grassmannian permutations of size n+2. Avoiding any of the patterns 2314 or 3412 gives the same sequence. - Juan B. Gil, Mar 09 2023

Examples

			a(2) = 6, since ( x1*y1, x2*y2, x1*x1+y1*y1, x2*x2+y2*y2, x1*x2+y1*y2, x1*y2+x2*y1 ) are a basis for homogeneous quadratic invariant polynomials.
		

References

  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
  • L. Smith, Polynomial Invariants of Finite Groups, A K Peters, 1995, p. 96.

Crossrefs

Cf. A177878.
Partial sums of A008794 (without 0). - Bruno Berselli, Aug 30 2013

Programs

  • Haskell
    Following Gary W. Adamson.
    import Data.List (inits, intersperse)
    a005993 n = a005994_list !! n
    a005993_list = map (sum . zipWith (*) (intersperse 0 [1, 3 ..]) . reverse) $
                       tail $ inits [1..]
    -- Reinhard Zumkeller, Feb 27 2015
    
  • Magma
    I:=[1,2,6,10,19,28]; [n le 6 select I[n] else 2*Self(n-1)+Self(n-2)-4*Self(n-3)+Self(n-4)+2*Self(n-5)-Self(n-6): n in [1..60]]; // Vincenzo Librandi, Jul 19 2015
    
  • Maple
    g := proc(n) local i; add(floor(i/2)^2,i=1..n+1) end: # Joseph S. Riel (joer(AT)k-online.com), Mar 22 2002
    a:= n-> (Matrix([[1, 0$3, -1, -2]]).Matrix(6, (i,j)-> if (i=j-1) then 1 elif j=1 then [2, 1, -4, 1, 2, -1][i] else 0 fi)^n)[1,1]; seq (a(n), n=0..44); # Alois P. Heinz, Jul 31 2008
  • Mathematica
    CoefficientList[Series[(1+x^2)/((1-x)^2*(1-x^2)^2),{x,0,44}],x]  (* Jean-François Alcover, Apr 08 2011 *)
    LinearRecurrence[{2,1,-4,1,2,-1},{1,2,6,10,19,28},50] (* Harvey P. Dale, Feb 20 2012 *)
  • PARI
    a(n)=polcoeff((1+x^2)/(1-x)^2/(1-x^2)^2+x*O(x^n),n)
    
  • PARI
    a(n) = (binomial(n+3, n) + (1-n%2)*binomial((n+2)/2, n>>1))/2 \\ Washington Bomfim, Aug 05 2008
    
  • PARI
    a = vector(50); a[1]=1; a[2]=2;
    for(n=3, 50, a[n] = ((n+2)*a[n-2]+2*a[n-1]-n)/(n-2)); a \\ Gerry Martens, Jun 03 2018
    
  • Sage
    def A005993():
        a, b, to_be = 0, 0, True
        while True:
            yield (a*(a*(2*a+9)+13)+b*(b+1)*(2*b+1)+6)//6
            if to_be: b += 1
            else: a += 1
            to_be = not to_be
    a = A005993()
    [next(a) for  in range(48)] # _Peter Luschny, May 04 2016

Formula

l(c, r) = 1/2 C(c+r-3, r) + 1/2 d(c, r), where d(c, r) is C((c + r - 3)/2, r/2) if c is odd and r is even, 0 if c is even and r is odd, C((c + r - 4)/2, r/2) if c is even and r is even, C((c + r - 4)/2, (r - 1)/2) if c is odd and r is odd.
G.f.: (1+x^2)/((1-x)^2*(1-x^2)^2) = (1+x^2)/((1+x)^2*(x-1)^4) = (1/(1-x)^4 +1/(1-x^2)^2)/2.
a(2n) = (n+1)(2n^2+4n+3)/3, a(2n+1) = (n+1)(n+2)(2n+3)/3. a(-4-n) = -a(n).
From Yosu Yurramendi, Sep 12 2008: (Start)
a(n+1) = a(n) + A008794(n+3) with a(1)=1.
a(n) = A027656(n) + 2*A006918(n).
a(n+2) = a(n) + A000982(n+2) with a(1)=1, a(2)=2. (End)
a(n) = 2*a(n-1) + a(n-2) - 4*a(n-3) + a(n-4) + 2*a(n-5) - a(n-6). - Jaume Oliver Lafont, Dec 05 2008
a(n) = (n^3 + 6*n^2 + 11*n + 6)/12 + ((n+2)/4)[n even] (the bracket means that the second term is added if and only if n is even). - Benoit Jubin, Mar 31 2012
a(n) = (1/12)*n*(n+1)*(n+2) + (1/4)*(n+1)*(1/2)*(1-(-1)^n), with offset 1. - Yosu Yurramendi, Jun 20 2013
a(n) = Sum_{i=0..n+1} ceiling(i/2) * round(i/2) = Sum_{i=0..n+2} floor(i/2)^2. - Bruno Berselli, Aug 30 2013
a(n) = (n + 2)*(3*(-1)^n + 2*n^2 + 8*n + 9)/24. - Ilya Gutkovskiy, May 04 2016
Recurrence formula: a(n) = ((n+2)*a(n-2)+2*a(n-1)-n)/(n-2), a(1)=1, a(2)=2. - Gerry Martens, Jun 10 2018
E.g.f.: exp(-x)*(6 - 3*x + exp(2*x)*(18 + 39*x + 18*x^2 + 2*x^3))/24. - Stefano Spezia, Feb 23 2020
a(n) = Sum_{j=0..n/2} binomial(c+2*j-1,2*j)*binomial(c+n-2*j-1,n-2*j) where c=2. For other values of c we have: A008619 (c=1), A005995 (c=3), A018211 (c=4), A018213 (c=5), A062136 (c=6). - Miquel A. Fiol, Sep 24 2024

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

A282011 Number T(n,k) of k-element subsets of [n] having an even sum; triangle T(n,k), n>=0, 0<=k<=n, read by rows.

Original entry on oeis.org

1, 1, 0, 1, 1, 0, 1, 1, 1, 1, 1, 2, 2, 2, 1, 1, 2, 4, 6, 3, 0, 1, 3, 6, 10, 9, 3, 0, 1, 3, 9, 19, 19, 9, 3, 1, 1, 4, 12, 28, 38, 28, 12, 4, 1, 1, 4, 16, 44, 66, 60, 40, 20, 5, 0, 1, 5, 20, 60, 110, 126, 100, 60, 25, 5, 0, 1, 5, 25, 85, 170, 226, 226, 170, 85, 25, 5, 1, 1, 6, 30, 110, 255, 396, 452, 396, 255, 110, 30, 6, 1
Offset: 0

Views

Author

Alois P. Heinz, Feb 04 2017

Keywords

Comments

Row n is symmetric if and only if n mod 4 in {0,3} (or if T(n,n) = 1).

Examples

			T(5,0) = 1: {}.
T(5,1) = 2: {2}, {4}.
T(5,2) = 4: {1,3}, {1,5}, {2,4}, {3,5}.
T(5,3) = 6: {1,2,3}, {1,2,5}, {1,3,4}, {1,4,5}, {2,3,5}, {3,4,5}.
T(5,4) = 3: {1,2,3,4}, {1,2,4,5}, {2,3,4,5}.
T(5,5) = 0.
T(7,7) = 1: {1,2,3,4,5,6,7}.
Triangle T(n,k) begins:
  1;
  1, 0;
  1, 1,  0;
  1, 1,  1,   1;
  1, 2,  2,   2,   1;
  1, 2,  4,   6,   3,   0;
  1, 3,  6,  10,   9,   3,   0;
  1, 3,  9,  19,  19,   9,   3,   1;
  1, 4, 12,  28,  38,  28,  12,   4,   1;
  1, 4, 16,  44,  66,  60,  40,  20,   5,   0;
  1, 5, 20,  60, 110, 126, 100,  60,  25,   5,  0;
  1, 5, 25,  85, 170, 226, 226, 170,  85,  25,  5, 1;
  1, 6, 30, 110, 255, 396, 452, 396, 255, 110, 30, 6, 1;
		

Crossrefs

Columns k=0..10 give (offsets may differ): A000012, A004526, A002620, A005993, A005994, A032092, A032093, A018211, A018212, A282077, A282078.
Row sums give A011782.
Main diagonal gives A133872(n+1).
Lower diagonals T(n+j,n) for j=1..10 give: A004525(n+1), A282079, A228705, A282080, A282081, A282082, A282083, A282084, A282085, A282086.
T(2n,n) gives A119358.

Programs

  • Maple
    b:= proc(n, s) option remember; expand(
          `if`(n=0, s, b(n-1, s)+x*b(n-1, irem(s+n, 2))))
        end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n, 1)):
    seq(T(n), n=0..16);
  • Mathematica
    Flatten[Table[Sum[Binomial[Ceiling[n/2],2j]Binomial[Floor[n/2],k-2j],{j,0,Floor[(n+1)/4]}],{n,0,10},{k,0,n}]] (* Indranil Ghosh, Feb 26 2017 *)
  • PARI
    a(n,k)=sum(j=0,floor((n+1)/4),binomial(ceil(n/2),2*j)*binomial(floor(n/2),k-2*j));
    tabl(nn)={for(n=0,nn,for(k=0,n,print1(a(n,k),", "););print(););} \\ Indranil Ghosh, Feb 26 2017

Formula

T(n,k) = Sum_{j=0..floor((n+1)/4)} C(ceiling(n/2),2*j) * C(floor(n/2),k-2*j).
T(n,k) = A007318(n,k) - A159916(n,k).
Sum_{k=0..n} k * T(n,k) = A057711(n-1) for n>0.
Sum_{k=0..n} (k+1) * T(n,k) = A087447(n) + [n=2].

A062135 Odd-numbered columns of Losanitsch triangle A034851 formatted as triangle with an additional first column.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 2, 1, 0, 2, 6, 3, 1, 0, 3, 10, 12, 4, 1, 0, 3, 19, 28, 20, 5, 1, 0, 4, 28, 66, 60, 30, 6, 1, 0, 4, 44, 126, 170, 110, 42, 7, 1, 0, 5, 60, 236, 396, 365, 182, 56, 8, 1, 0, 5, 85, 396, 868, 1001, 693, 280, 72, 9, 1
Offset: 0

Views

Author

Wolfdieter Lang, Jun 19 2001

Keywords

Comments

Because the sequence of column m=2*k, k >= 1, of A034851 is the partial sum sequence of the one of column m=2*k-1 the present triangle is essentially Losanitsch's triangle A034851.
Row sums give A051450 with A051450(0) := 1. Column sequences (without leading zeros) are for m=0..6: A000007, A008619, A005993, A005995, A018211, A018213, A062136.

Examples

			Triangle begins:
  {1};
  {0,1};
  {0,1,1};
  {0,2,2,1};
  ...
Pe(4,x^2)=1+6*x^2+x^4.
		

Crossrefs

Programs

  • Mathematica
    t[n_?EvenQ, k_?OddQ] := Binomial[n, k]/2; t[n_, k_] := (Binomial[n, k] + Binomial[Quotient[n, 2], Quotient[k, 2]])/2; Flatten[Table[t[n - 1 + m, n - m], {n, 0, 12}, {m, 0, n}]] (* Michael De Vlieger, Sep 28 2024, after Jean-François Alcover at A034851  *)

Formula

T(n, m) = A034851(n-1+m, n-m), n >= m >= 0; A034851(n-1, n) := 0, n >= 1, A034851(-1, 0) := 1.
T(n, m) = 0 if n= 1; T(n, m) = T(n-1, m)+sum(T(k, m-1), k=m-1..n-1) if n+m even and T(n, m) = T(n-1, m)+sum(T(k, m-1), k=m-1..n-1)-binomial((n+m-3)/2, m-1) if n+m odd, n >= m >= 1.
G.f. for column m: x^m*Pe(m, x^2)/(((1-x)^(2*m))*(1+x)^m), m >= 0, with Pe(m, x^2)= sum(A034839(m, k)*x^(2*k), k=0..floor(n/2)), the row polynomial of array A034839 (even-indexed entries of the rows of Pascal's triangle).

Extensions

More terms from Michael De Vlieger, Sep 28 2024

A141783 Number of bracelets (turn over necklaces) with n beads: 1 blue, 12 green, and r = n - 13 red.

Original entry on oeis.org

1, 7, 49, 231, 924, 3108, 9324, 25236, 63090, 147070, 323554, 676270, 1352540, 2600612, 4829708, 8692788, 15212379, 25949469, 43249115, 70562765, 112900424, 177412664, 274183208, 417232088, 625848132, 926250780, 1353751140
Offset: 13

Views

Author

Washington Bomfim, Aug 17 2008

Keywords

Crossrefs

Programs

  • Maple
    A141783:=n->(1/2)*(binomial(n - 1, 12) + binomial((n - 2 + (n mod 2))/2, 6)); seq(A141783(n), n=13..50); # Wesley Ivan Hurt, Jan 30 2014
  • Mathematica
    Table[(1/2) (Binomial[n - 1, 12] + Binomial[(n - 2 + Mod[n, 2])/2, 6]), {n, 13, 50}] (* Wesley Ivan Hurt, Jan 30 2014 *)

Formula

a(n) = 1/2*(binomial(n-1,12) + binomial((n-2+n mod 2)/2, 6)).
a(n) = (1/(2*12!))*(n+2)*(n+4)*(n+6)*(n+8)*(n+10)*(n+12)*((n+1)*(n+3)*(n+5)*(n+7)*(n+9)*(n+11) + 1*3*5*7*9*11) - (1/15)*(1/2^10)*(n^5+(65/2)*n^4+400*n^3+(4615/2)*n^2+6154*n+(11895/2))*(1/2)*(1-(-1)^n) [Yosu Yurramendi, Jun 24 2013]

Extensions

Revised by Washington Bomfim, Jul 24 2012
Showing 1-5 of 5 results.