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 1-10 of 14 results. Next

A005899 Number of points on surface of octahedron; also coordination sequence for cubic lattice: a(0) = 1; for n > 0, a(n) = 4n^2 + 2.

Original entry on oeis.org

1, 6, 18, 38, 66, 102, 146, 198, 258, 326, 402, 486, 578, 678, 786, 902, 1026, 1158, 1298, 1446, 1602, 1766, 1938, 2118, 2306, 2502, 2706, 2918, 3138, 3366, 3602, 3846, 4098, 4358, 4626, 4902, 5186, 5478, 5778, 6086, 6402, 6726, 7058, 7398, 7746, 8102, 8466
Offset: 0

Views

Author

Keywords

Comments

Also, the number of regions the plane can be cut into by two overlapping concave (2n)-gons. - Joshua Zucker, Nov 05 2002
If X is an n-set and Y_i (i=1,2,3) are mutually disjoint 2-subsets of X then a(n-5) is equal to the number of 5-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Aug 26 2007
Binomial transform of a(n) is A055580(n). - Wesley Ivan Hurt, Apr 15 2014
The identity (4*n^2+2)^2 - (n^2+1)*(4*n)^2 = 4 can be written as a(n)^2 - A002522(n)*A008586(n)^2 = 4. - Vincenzo Librandi, Jun 15 2014
Also the least number of unit cubes required, at the n-th iteration, to surround a 3D solid built from unit cubes, in order to hide all its visible faces, starting with a unit cube. - R. J. Cano, Sep 29 2015
Also, coordination sequence for "tfs" 3D uniform tiling. - N. J. A. Sloane, Feb 10 2018
Also, the number of n-th order specular reflections arriving at a receiver point from an emitter point inside a cuboid with reflective faces. - Michael Schutte, Sep 18 2018

References

  • H. S. M. Coxeter, "Polyhedral numbers," in R. S. Cohen et al., editors, For Dirk Struik. Reidel, Dordrecht, 1974, pp. 25-35.
  • Gmelin Handbook of Inorg. and Organomet. Chem., 8th Ed., 1994, TYPIX search code (225) cF8
  • B. Grünbaum, Uniform tilings of 3-space, Geombinatorics, 4 (1994), 49-56. See tilings #16 and #22.
  • R. W. Marks and R. B. Fuller, The Dymaxion World of Buckminster Fuller. Anchor, NY, 1973, p. 46.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Partial sums give A001845.
Column 2 * 2 of array A188645.
The 28 uniform 3D tilings: cab: A299266, A299267; crs: A299268, A299269; fcu: A005901, A005902; fee: A299259, A299265; flu-e: A299272, A299273; fst: A299258, A299264; hal: A299274, A299275; hcp: A007899, A007202; hex: A005897, A005898; kag: A299256, A299262; lta: A008137, A299276; pcu: A005899, A001845; pcu-i: A299277, A299278; reo: A299279, A299280; reo-e: A299281, A299282; rho: A008137, A299276; sod: A005893, A005894; sve: A299255, A299261; svh: A299283, A299284; svj: A299254, A299260; svk: A010001, A063489; tca: A299285, A299286; tcd: A299287, A299288; tfs: A005899, A001845; tsi: A299289, A299290; ttw: A299257, A299263; ubt: A299291, A299292; bnn: A007899, A007202. See the Proserpio link in A299266 for overview.
Row 3 of A035607, A266213, A343599.
Column 3 of A113413, A119800, A122542.

Programs

Formula

G.f.: ((1+x)/(1-x))^3. - Simon Plouffe in his 1992 dissertation
Binomial transform of [1, 5, 7, 1, -1, 1, -1, 1, ...]. - Gary W. Adamson, Nov 02 2007
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), with a(0)=1, a(1)=6, a(2)=18, a(3)=38. - Harvey P. Dale, Nov 08 2011
Recurrence: n*a(n) = (n-2)*a(n-2) + 6*a(n-1), a(0)=1, a(1)=6. - Fung Lam, Apr 15 2014
For n > 0, a(n) = A001844(n-1) + A001844(n) = (n-1)^2 + 2n^2 + (n+1)^2. - Doug Bell, Aug 18 2015
For n > 0, a(n) = A010014(n) - A195322(n). - R. J. Cano, Sep 29 2015
For n > 0, a(n) = A000384(n+1) + A014105(n-1). - Bruce J. Nicholson, Oct 08 2017
a(n) = A008574(n) + A008574(n-1) + a(n-1). - Bruce J. Nicholson, Dec 18 2017
a(n) = 2*d*Hypergeometric2F1(1-d, 1-n, 2, 2) where d=3, n>0. - Shel Kaphan, Feb 16 2023
a(n) = A035597(n)*3/n, for n>0. - Shel Kaphan, Feb 26 2023
E.g.f.: exp(x)*(2 + 4*x + 4*x^2) - 1. - Stefano Spezia, Mar 08 2023
Sum_{n>=0} 1/a(n) = 3/4 + Pi *sqrt(2)*coth( Pi/sqrt 2)/8 = 1.31858... - R. J. Mathar, Apr 27 2024

A119258 Triangle read by rows: T(n,0) = T(n,n) = 1 and for 0

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 5, 7, 1, 1, 7, 17, 15, 1, 1, 9, 31, 49, 31, 1, 1, 11, 49, 111, 129, 63, 1, 1, 13, 71, 209, 351, 321, 127, 1, 1, 15, 97, 351, 769, 1023, 769, 255, 1, 1, 17, 127, 545, 1471, 2561, 2815, 1793, 511, 1, 1, 19, 161, 799, 2561, 5503, 7937, 7423, 4097, 1023, 1
Offset: 0

Views

Author

Reinhard Zumkeller, May 11 2006

Keywords

Comments

From Richard M. Green, Jul 26 2011: (Start)
T(n,n-k) is the (k-1)-st Betti number of the subcomplex of the n-dimensional half cube obtained by deleting the interiors of all half-cube shaped faces of dimension at least k.
T(n,n-k) is the (k-2)-nd Betti number of the complement of the k-equal real hyperplane arrangement in R^n.
T(n,n-k) gives a lower bound for the complexity of the problem of determining, given n real numbers, whether some k of them are equal.
T(n,n-k) is the number of nodes used by the Kronrod-Patterson-Smolyak cubature formula in numerical analysis. (End)

Examples

			Triangle begins as:
  1;
  1, 1;
  1, 3,  1;
  1, 5,  7,  1;
  1, 7, 17, 15,  1;
  1, 9, 31, 49, 31, 1;
		

Crossrefs

A145661, A119258 and A118801 are all essentially the same (see the Shattuck and Waldhauser paper). - Tamas Waldhauser, Jul 25 2011

Programs

  • Haskell
    a119258 n k = a119258_tabl !! n !! k
    a119258_row n = a119258_tabl !! n
    a119258_list = concat a119258_tabl
    a119258_tabl = iterate (\row -> zipWith (+)
       ([0] ++ init row ++ [0]) $ zipWith (+) ([0] ++ row) (row ++ [0])) [1]
    -- Reinhard Zumkeller, Nov 15 2011
    
  • Magma
    function T(n,k)
      if k eq 0 or k eq n then return 1;
      else return 2*T(n-1,k-1) + T(n-1,k);
      end if;
      return T;
    end function;
    [T(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Nov 18 2019
    
  • Maple
    # Case m = 2 of the more general:
    A119258 := (n,k,m) -> (1-m)^(-n+k)-m^(k+1)*pochhammer(n-k, k+1) *hypergeom([1,n+1],[k+2],m)/(k+1)!;
    seq(seq(round(evalf(A119258(n,k,2))),k=0..n), n=0..10); # Peter Luschny, Jul 25 2014
  • Mathematica
    T[n_, k_] := Binomial[n, k] Hypergeometric2F1[-k, n-k, n-k+1, -1];
    Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Sep 10 2017 *)
  • PARI
    T(n,k) = if(k==0 || k==n, 1, 2*T(n-1, k-1) + T(n-1,k) ); \\ G. C. Greubel, Nov 18 2019
    
  • Sage
    @CachedFunction
    def T(n, k):
        if (k==0 or k==n): return 1
        else: return 2*T(n-1, k-1) + T(n-1, k)
    [[T(n, k) for k in (0..n)] for n in (0..12)] # G. C. Greubel, Nov 18 2019

Formula

T(2*n,n-1) = T(2*n-1,n) for n > 0;
central terms give A119259; row sums give A007051;
T(n,0) = T(n,n) = 1;
T(n,1) = A005408(n-1) for n > 0;
T(n,2) = A056220(n-1) for n > 1;
T(n,n-4) = A027608(n-4) for n > 3;
T(n,n-3) = A055580(n-3) for n > 2;
T(n,n-2) = A000337(n-1) for n > 1;
T(n,n-1) = A000225(n) for n > 0.
T(n,k) = [k<=n]*(-1)^k*Sum_{i=0..k} (-1)^i*C(k-n,k-i)*C(n,i). - Paul Barry, Sep 28 2007
From Richard M. Green, Jul 26 2011: (Start)
T(n,k) = [k<=n] Sum_{i=n-k..n} (-1)^(n-k-i)*2^(n-i)*C(n,i).
T(n,k) = [k<=n] Sum_{i=n-k..n} C(n,i)*C(i-1,n-k-1).
G.f. for T(n,n-k): x^k/(((1-2x)^k)*(1-x)). (End)
T(n,k) = R(n,k,2) where R(n, k, m) = (1-m)^(-n+k)-m^(k+1)*Pochhammer(n-k,k+1)* hyper2F1([1,n+1], [k+2], m)/(k+1)!. - Peter Luschny, Jul 25 2014
From Peter Bala, Mar 05 2018: (Start)
The n-th row polynomial R(n,x) equals the n-th degree Taylor polynomial of the function (1 + 2*x)^n/(1 + x) about 0. For example, for n = 4 we have (1 + 2*x)^4/(1 + x) = 1 + 7*x + 17*x^2 + 15*x^3 + x^4 + O(x^5).
Row reverse of A112857. (End)

A119259 Central terms of the triangle in A119258.

Original entry on oeis.org

1, 3, 17, 111, 769, 5503, 40193, 297727, 2228225, 16807935, 127574017, 973168639, 7454392321, 57298911231, 441739706369, 3414246490111, 26447737520129, 205272288591871, 1595964714385409, 12427568655368191, 96905907580960769, 756583504975757311, 5913649000782757889
Offset: 0

Views

Author

Reinhard Zumkeller, May 11 2006

Keywords

Comments

The Gauss congruences a(n*p^k) == a(n^p^(k-1)) (mod p^k) hold for prime p and positive integers n and k. - Peter Bala, Jan 06 2022

References

  • R. P. Stanley, Enumerative Combinatorics Volume 2, Cambridge Univ. Press, 1999, Theorem 6.33, p. 197.

Crossrefs

Programs

  • Haskell
    a119259 n = a119258 (2 * n) n  -- Reinhard Zumkeller, Aug 06 2014
    
  • Mathematica
    Table[Binomial[2k - 1, k] Hypergeometric2F1[-2k, -k, 1 - 2k, -1], {k, 0, 10}] (* Vladimir Reshetnikov, Feb 16 2011 *)
  • Python
    from itertools import count, islice
    def A119259_gen(): # generator of terms
        yield from (1,3)
        a, c = 2, 1
        for n in count(1):
            yield (a<>1
    A119259_list = list(islice(A119259_gen(),20)) # Chai Wah Wu, Apr 26 2023

Formula

a(n) = A119258(2*n,n).
a(n) = Sum_{k=0..n} C(2*n,k)*C(2*n-k-1,n-k). - Paul Barry, Sep 28 2007
a(n) = Sum_{k=0..n} C(n+k-1,k)*2^k. - Paul Barry, Sep 28 2007
2*a(n) = A064062(n)+A178792(n). - Joseph Abate, Jul 21 2010
G.f.: (4*x^2+3*sqrt(1-8*x)*x-5*x)/(sqrt(1-8*x)*(2*x^2+x-1)-8*x^2-7*x+1). - Vladimir Kruchinin, Aug 19 2013
a(n) = (-1)^n - 2^(n+1)*binomial(2*n,n-1)*hyper2F1([1,2*n+1],[n+2],2). - Peter Luschny, Jul 25 2014
a(n) = (-1)^n + 2^(n+1)*A014300(n). - Peter Luschny, Jul 25 2014
a(n) = [x^n] ( (1 + x)^2/(1 - x) )^n. Exp( Sum_{n >= 1} a(n)*x^n/n ) = 1 + 3*x + 13*x^2 + 67*x^3 + ... is essentially the o.g.f. for A064062. - Peter Bala, Oct 01 2015
The o.g.f. is the diagonal of the bivariate rational function 1/(1 - t*(1 + x)^2/(1 - x)) and hence is algebraic by Stanley 1999, Theorem 6.33, p.197. - Peter Bala, Aug 21 2016
n*(3*n-4)*a(n) +(-21*n^2+40*n-12)*a(n-1) -4*(3*n-1)*(2*n-3)*a(n-2)=0. - R. J. Mathar, Aug 09 2017
From Peter Bala, Mar 23 2020: (Start)
a(p) == 3 ( mod p^3 ) for prime p >= 5. Cf. A002003, A103885 and A156894.
More generally, we conjecture that a(n*p^k) == a(n*p^(k-1)) ( mod p^(3*k) ) for prime p >= 5 and positive integers n and k. (End)
G.f.: (8*x)/(sqrt(1-8*x)*(1+4*x)-1+8*x). - Fabian Pereyra, Jul 20 2024
a(n) = 2^(n+1)*binomial(2*n,n) - A178792(n). - Akiva Weinberger, Dec 06 2024
a(n) = Sum_{k=0..n} 2^k * (-1)^(n-k) * binomial(2*n,k). - Seiichi Manyama, Jul 31 2025

A055252 Triangle of partial row sums (prs) of triangle A055249.

Original entry on oeis.org

1, 4, 1, 13, 5, 1, 38, 18, 6, 1, 104, 56, 24, 7, 1, 272, 160, 80, 31, 8, 1, 688, 432, 240, 111, 39, 9, 1, 1696, 1120, 672, 351, 150, 48, 10, 1, 4096, 2816, 1792, 1023, 501, 198, 58, 11, 1, 9728, 6912, 4608, 2815, 1524, 699, 256, 69, 12, 1, 22784, 16640, 11520
Offset: 0

Views

Author

Wolfdieter Lang, May 26 2000

Keywords

Comments

In the language of the Shapiro et al. reference (given in A053121) such a lower triangular (ordinary) convolution array, considered as matrix, belongs to the Riordan-group. The G.f. for the row polynomials p(n,x) (increasing powers of x) is (((1-z)^2)/(1-2*z)^3)/(1-x*z/(1-z)).
This is the third member of the family of Riordan-type matrices obtained from A007318(n,m) (Pascal's triangle read as lower triangular matrix) by repeated application of the prs-procedure.
The column sequences appear as A049611(n+1), A001793, A001788, A055580, A055581, A055582, A055583 for m=0..6.

Examples

			[0] 1
[1] 4, 1
[2] 13, 5, 1
[3] 38, 18, 6, 1
[4] 104, 56, 24, 7, 1
[5] 272, 160, 80, 31, 8, 1
[6] 688, 432, 240, 111, 39, 9, 1
[7] 1696, 1120, 672, 351, 150, 48, 10, 1
Fourth row polynomial (n = 3): p(3, x) = 38 + 18*x + 6*x^2 + x^3.
		

Crossrefs

Cf. A007318, A055248, A055249. Row sums: A049612(n+1)= A055584(n, 0).

Programs

  • Maple
    T := (n, k) -> binomial(n, k)*hypergeom([3, k - n], [k + 1], -1):
    for n from 0 to 7 do seq(simplify(T(n, k)), k = 0..n) od; # Peter Luschny, Sep 23 2024

Formula

a(n, m)=sum(A055249(n, k), k=m..n), n >= m >= 0, a(n, m) := 0 if n
Column m recursion: a(n, m)= sum(a(j, m), j=m..n-1)+ A055249(n, m), n >= m >= 0, a(n, m) := 0 if n
G.f. for column m: (((1-x)^2)/(1-2*x)^3)*(x/(1-x))^m, m >= 0.
T(n, k) = binomial(n, k)*hypergeom([3, k - n], [k + 1], -1). - Peter Luschny, Sep 23 2024

A112857 Triangle T(n,k) read by rows: number of Green's R-classes in the semigroup of order-preserving partial transformations (of an n-element chain) consisting of elements of height k (height(alpha) = |Im(alpha)|).

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 5, 1, 1, 15, 17, 7, 1, 1, 31, 49, 31, 9, 1, 1, 63, 129, 111, 49, 11, 1, 1, 127, 321, 351, 209, 71, 13, 1, 1, 255, 769, 1023, 769, 351, 97, 15, 1, 1, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1, 1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1
Offset: 0

Author

Abdullahi Umar, Aug 25 2008

Keywords

Comments

Sum of rows of T(n, k) is A007051; T(n,k) = |A118801(n,k)|.
Row-reversed variant of A119258. - R. J. Mathar, Jun 20 2011
Pairwise sums of row terms starting from the right yields triangle A038207. - Gary W. Adamson, Feb 06 2012
Riordan array (1/(1 - x), x/(1 - 2*x)). - Philippe Deléham, Jan 17 2014
Appears to coincide with the triangle T(n,m) (n >= 1, 1 <= m <= n) giving number of set partitions of [n], avoiding 1232, with m blocks [Crane, 2015]. See also A250118, A250119. - N. J. A. Sloane, Nov 25 2014
(A007318)^2 = A038207 = T*|A167374|. See A118801 for other relations to the Pascal matrix. - Tom Copeland, Nov 17 2016

Examples

			T(3,2) = 5 because in a regular semigroup of transformations the Green's R-classes coincide with convex partitions of subsets of {1,2,3} with convex classes (modulo the subsets): {1}, {2}/{1}, {3}/{2}, {3}/{1,2}, {3}/{1}, {2,3}
Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
  1;
  1,    1;
  1,    3,    1;
  1,    7,    5,    1;
  1,   15,   17,    7,    1;
  1,   31,   49,   31,    9,    1;
  1,   63,  129,  111,   49,   11,    1;
  1,  127,  321,  351,  209,   71,   13,   1;
  1,  255,  769, 1023,  769,  351,   97,  15,   1;
  1,  511, 1793, 2815, 2561, 1471,  545, 127,  17,  1;
  1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1;
  ...
As to matrix M, top row of M^3 = (1, 7, 5, 1, 0, 0, 0, ...)
		

Programs

  • Maple
    A112857 := proc(n,k) if k=0 or k=n then 1; elif k <0 or k>n then 0; else 2*procname(n-1,k)+procname(n-1,k-1) ; end if; end proc: # R. J. Mathar, Jun 20 2011
  • Mathematica
    Table[Abs[1 + (-1)^k*2^(n - k + 1)*Sum[ Binomial[n - 2 j - 2, k - 2 j - 1], {j, 0, Floor[k/2]}]] - 4 Boole[And[n == 1, k == 0]], {n, 0, 10}, {k, 0, n}] // Flatten (* Michael De Vlieger, Nov 24 2016 *)

Formula

T(n,k) = Sum_{j = k..n} C(n,j)*C(j-1,k-1).
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) for n >= 2 and 1 <= k <= n-1 with T(n,0) = 1 = T(n,n) for n >= 0.
n-th row = top row of M^n, deleting the zeros, where M is an infinite square production matrix with (1,1,1,...) as the superdiagonal and (1,2,2,2,...) as the main diagonal. - Gary W. Adamson, Feb 06 2012
From Peter Bala, Mar 05 2018 (Start):
The following remarks are particular cases of more general results for Riordan arrays of the form (f(x), x/(1 - k*x)).
Let R(n,x) denote the n-th row polynomial of this triangle. The polynomial R(n,2*x) has the e.g.f. Sum_{k = 0..n} T(n,k)*(2*x)^k/k!. The e.g.f. for the n-th diagonal of the triangle (starting at n = 0 for the main diagonal) equals exp(x) * the e.g.f. for the polynomial R(n,2*x). For example, when n = 3 we have exp(x)*(1 + 7*(2*x) + 5*(2*x)^2/2! + (2*x)^3/3!) = 1 + 15*x + 49*x^2/2! + 111*x^3/3! + 209*x^4/4! + ....
Let P(n,x) = Sum_{k = 0..n} T(n,k)*x^(n-k) denote the n-th row polynomial in descending powers of x. Then P(n,x) is the n-th degree Taylor polynomial of the function (1 + 2*x)^n/(1 + x) about 0. For example, for n = 4 we have (1 + 2*x)^4/(1 + x) = x^4 + 15*x^3 + 17*x^2 + 7*x + 1 + O(x^5).
See A118801 for a signed version of this triangle and A145661 for a signed version of the row reversed triangle. (End)
Bivariate o.g.f.: Sum_{n,k>=0} T(n,k)*x^n*y^k = (1 - 2*x)/((1 - x)*(1 - 2*x - x*y)). - Petros Hadjicostas, Feb 14 2021
The matrix inverse of the Lucas triangle A029635 is -T(n, k)/(-2)^(n-k+1). - Peter Luschny, Dec 22 2024

A055581 Fifth column of triangle A055252.

Original entry on oeis.org

1, 8, 39, 150, 501, 1524, 4339, 11762, 30705, 77808, 192495, 466926, 1114093, 2621420, 6094827, 14024682, 31981545, 72351720, 162529255, 362807270, 805306341, 1778384868, 3909091299, 8556380130, 18656264161, 40533753824
Offset: 0

Author

Wolfdieter Lang, May 26 2000

Keywords

Comments

a(n) = number of directed column-convex polyominoes of area n+5 having along the lower contour exactly two reentrant corners. - Emeric Deutsch, May 21 2003

Crossrefs

Cf. A055252, A055249, A045889, partial sums of A055580.

Programs

  • Mathematica
    Table[(n^2-n+4)2^(n+1)-7-n,{n,0,30}] (* or *) LinearRecurrence[ {8,-25,38,-28,8},{1,8,39,150,501},30] (* Harvey P. Dale, Nov 07 2011 *)

Formula

G.f.: 1/(((1-2*x)^3)*(1-x)^2).
a(n) = A055252(n+4, 4). a(n) = sum(a(j), j=0..n-1)+A045889(n), n >= 1.
a(n) = (n^2-n+4)2^(n+1)-7-n - Emeric Deutsch, May 21 2003
a(0)=1, a(1)=8, a(2)=39, a(3)=150, a(4)=501, a(n) = 8*a(n-1)- 25*a(n-2)+ 38*a(n-3)-28*a(n-4)+8*a(n-5). [Harvey P. Dale, Nov 07 2011]

A193844 Triangular array: the fission of ((x+1)^n) by ((x+1)^n); i.e., the self-fission of Pascal's triangle.

Original entry on oeis.org

1, 1, 3, 1, 5, 7, 1, 7, 17, 15, 1, 9, 31, 49, 31, 1, 11, 49, 111, 129, 63, 1, 13, 71, 209, 351, 321, 127, 1, 15, 97, 351, 769, 1023, 769, 255, 1, 17, 127, 545, 1471, 2561, 2815, 1793, 511, 1, 19, 161, 799, 2561, 5503, 7937, 7423, 4097, 1023
Offset: 0

Author

Clark Kimberling, Aug 07 2011

Keywords

Comments

See A193842 for the definition of fission of two sequences of polynomials or triangular arrays.
A193844 is also the fission of (p1(n,x)) by (q1(n,x)), where p1(n,x)=x^n+x^(n-1)+...+x+1 and q1(n,x)=(x+2)^n.
Essentially A119258 but without the main diagonal. - Peter Bala, Jul 16 2013
From Robert Coquereaux, Oct 02 2014: (Start)
This is also a rectangular array A(n,p) read down the antidiagonals:
1 1 1 1 1 1 1 1 1
3 5 7 9 11 13 15 17 19
7 17 31 49 71 97 127 161 199
15 49 111 209 351 545 799 1121 1519
31 129 351 769 1471 2561 4159 6401 9439
...
Calling Gr(n) the Grassmann algebra with n generators, A(n,p) is the dimension of the space of Gr(n)-valued symmetric multilinear forms with vanishing graded divergence. If p is odd A(n,p) is 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 A(n,p)+1. A(n,p) = 2^n*A059260(p,n-1)-(-1)^p.
(End)
The n-th row are also the coefficients of the polynomial P=sum_{k=0..n} (X+2)^k (in falling order, i.e., that of X^n first). - M. F. Hasler, Oct 15 2014

Examples

			First six rows:
1
1....3
1....5....7
1....7....17....15
1....9....31....49....31
1....11...49....111...129...63
		

Crossrefs

A145661 is an essentially identical triangle.

Programs

  • Maple
    A193844 := (n,k) -> 2^k*binomial(n+1,k)*hypergeom([1,-k],[-k+n+2],1/2);
    for n from 0 to 5 do seq(round(evalf(A193844(n,k))),k=0..n) od; # Peter Luschny, Jul 23 2014
    # Alternatively
    p := (n,x) -> add(x^k*(1+2*x)^(n-k), k=0..n): for n from 0 to 7 do [n], PolynomialTools:-CoefficientList(p(n,x), x) od; # Peter Luschny, Jun 18 2017
  • Mathematica
    z = 10;
    p[n_, x_] := (x + 1)^n;
    q[n_, x_] := (x + 1)^n
    p1[n_, k_] := Coefficient[p[n, x], x^k];
    p1[n_, 0] := p[n, x] /. x -> 0;
    d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
    h[n_] := CoefficientList[d[n, x], {x}]
    TableForm[Table[Reverse[h[n]], {n, 0, z}]]
    Flatten[Table[Reverse[h[n]], {n, -1, z}]]  (* A193844 *)
    TableForm[Table[h[n], {n, 0, z}]]
    Flatten[Table[h[n], {n, -1, z}]]  (* A193845 *)
  • Sage
    # uses[fission from A193842]
    p = lambda n,x: (x+1)^n
    A193844_row = lambda n: fission(p, p, n)
    for n in range(7): print(A193844_row(n)) # Peter Luschny, Jul 23 2014

Formula

From Peter Bala, Jul 16 2013: (Start)
T(n,k) = sum {i = 0..k} (-1)^i*binomial(n+1,k-i)*2^(k-i).
O.g.f.: 1/( (1 - x*t)*(1 - (2*x + 1)*t) ) = 1 + (1 + 3*x)*t + (1 + 5*x + 7*x^2)*t^2 + ....
The n-th row polynomial R(n,x) = 1/(x+1)*( (2*x+1)^(n+1) - x^(n+1) ). (End)
T(n,k) = T(n-1,k) + 3*T(n-1,k-1) - T(n-2,k-1) - 2*T(n-2,k-2), T(0,0) = 1, T(1,0) = 1, T(1,1) = 3, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 17 2014
T(n,k) = 2^k*binomial(n+1,k)*hyper2F1(1,-k,-k+n+2, 1/2). - Peter Luschny, Jul 23 2014

A058395 Square array read by antidiagonals. Based on triangular numbers (A000217) with each term being the sum of 2 consecutive terms in the previous row.

Original entry on oeis.org

1, 0, 1, 3, 1, 1, 0, 3, 2, 1, 6, 3, 4, 3, 1, 0, 6, 6, 6, 4, 1, 10, 6, 9, 10, 9, 5, 1, 0, 10, 12, 15, 16, 13, 6, 1, 15, 10, 16, 21, 25, 25, 18, 7, 1, 0, 15, 20, 28, 36, 41, 38, 24, 8, 1, 21, 15, 25, 36, 49, 61, 66, 56, 31, 9, 1, 0, 21, 30, 45, 64, 85, 102, 104, 80, 39, 10, 1, 28, 21, 36, 55, 81, 113, 146, 168, 160, 111, 48, 11, 1
Offset: 0

Author

Henry Bottomley, Nov 24 2000

Keywords

Comments

Changing the formula by replacing T(2n, 0) = T(n, 3) with T(2n, 0) = T(n, m) for some other value of m would change the generating function to the coefficient of x^n in expansion of (1 + x)^k / (1 - x^2)^m. This would produce A058393, A058394, A057884 (and effectively A007318).

Examples

			The array T(n, k) starts:
[0] 1, 0,  3,   0,   6,   0,  10,    0,   15,    0, ...
[1] 1, 1,  3,   3,   6,   6,  10,   10,   15,   15, ...
[2] 1, 2,  4,   6,   9,  12,  16,   20,   25,   30, ...
[3] 1, 3,  6,  10,  15,  21,  28,   36,   45,   55, ...
[4] 1, 4,  9,  16,  25,  36,  49,   64,   81,  100, ...
[5] 1, 5, 13,  25,  41,  61,  85,  113,  145,  181, ...
[6] 1, 6, 18,  38,  66, 102, 146,  198,  258,  326, ...
[7] 1, 7, 24,  56, 104, 168, 248,  344,  456,  584, ...
[8] 1, 8, 31,  80, 160, 272, 416,  592,  800, 1040, ...
[9] 1, 9, 39, 111, 240, 432, 688, 1008, 1392, 1840, ...
		

Crossrefs

Rows are A000217 with zeros, A008805, A002620, A000217, A000290, A001844, A005899.
Columns are A000012, A001477, A016028.
The triangle A055252 also appears in half of the array.

Programs

  • Maple
    gf := n -> (1 + x)^n / (1 - x^2)^3: ser := n -> series(gf(n), x, 20):
    seq(lprint([n], seq(coeff(ser(n), x, k), k = 0..9)), n = 0..9); # Peter Luschny, Apr 12 2023
  • Mathematica
    T[0, k_] := If[OddQ[k], 0, (k+2)(k+4)/8];
    T[n_, k_] := T[n, k] = If[k == 0, 1, T[n-1, k-1] + T[n-1, k]];
    Table[T[n-k, k], {n, 0, 12}, {k, n, 0, -1}] // Flatten (* Jean-François Alcover, Apr 13 2023 *)

Formula

T(n, k) = T(n-1, k-1) + T(n, k-1) with T(0, k) = 1, T(2*n, 0) = T(n, 3) and T(2*n + 1, 0) = 0. Coefficient of x^n in expansion of (1 + x)^k / (1 - x^2)^3.

A193845 Mirror of the triangle A193844.

Original entry on oeis.org

1, 3, 1, 7, 5, 1, 15, 17, 7, 1, 31, 49, 31, 9, 1, 63, 129, 111, 49, 11, 1, 127, 321, 351, 209, 71, 13, 1, 255, 769, 1023, 769, 351, 97, 15, 1, 511, 1793, 2815, 2561, 1471, 545, 127, 17, 1, 1023, 4097, 7423, 7937, 5503, 2561, 799, 161, 19, 1
Offset: 0

Author

Clark Kimberling, Aug 07 2011

Keywords

Comments

This triangle is obtained by reversing the rows of the triangle A193844.
From Philippe Deléham, Jan 17 2014: (Start)
Subtriangle of the triangle in A112857.
T(n,0) = A000225(n+1).
T(n,1) = A000337(n).
T(n+2,2) = A055580(n).
T(n+3,3) = A027608(n).
T(n+4,4) = A211386(n).
T(n+5,5) = A211388(n).
T(n,n) = A000012(n).
T(n+1,n) = A005408(n).
T(n+2,n) = A056220(n+2).
T(n+3,n) = A199899(n+1).
Row sums are A003462(n+1).
Diagonal sums are A048739(n).
Riordan array (1/((1-2*x)*(1-x)), x/(1-2*x)). (End)
Consider the transformation 1 + x + x^2 + x^3 + ... + x^n = A_0*(x-2)^0 + A_1*(x-2)^1 + A_2*(x-2)^2 + ... + A_n*(x-2)^n. This sequence gives A_0, ... A_n as the entries in the n-th row of this triangle, starting at n = 0. - Derek Orr, Oct 14 2014
The n-th row lists the coefficients of the polynomial sum_{k=0..n} (X+2)^k, in order of increasing powers. - M. F. Hasler, Oct 15 2014

Examples

			First six rows:
1
3....1
7....5....1
15...17...7....1
31...49...31...9...1
63...129..111..49..11..1
		

Crossrefs

Programs

  • Mathematica
    z = 10;
    p[n_, x_] := (x + 1)^n;
    q[n_, x_] := (x + 1)^n
    p1[n_, k_] := Coefficient[p[n, x], x^k];
    p1[n_, 0] := p[n, x] /. x -> 0;
    d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
    h[n_] := CoefficientList[d[n, x], {x}]
    TableForm[Table[Reverse[h[n]], {n, 0, z}]]
    Flatten[Table[Reverse[h[n]], {n, -1, z}]]  (* A193844 *)
    TableForm[Table[h[n], {n, 0, z}]]
    Flatten[Table[h[n], {n, -1, z}]]  (* A193845 *)
    Table[2^k*Binomial[n + 1, k]*Hypergeometric2F1[1, -k, -k + n + 2, 1/2], {n, 0, 9}, {k, n, 0, -1}] // Flatten (* Michael De Vlieger, Nov 09 2021 *)
  • PARI
    for(n=0,20,for(k=0,n,print1(1/k!*sum(i=0,n,(2^(i-k)*prod(j=0,k-1,i-j))),", "))) \\ Derek Orr, Oct 14 2014

Formula

T(n,k) = A193844(n,n-k).
T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 2*T(n-2,k) - T(n-2,k-1), T(0,0) = 1, T(1,0) = 3, T(1,1) = 1, T(n,k) = 0 if k<0 or if k>n. - Philippe Deléham, Jan 17 2014

A248917 a(n) = 2^n * n^2 + 1.

Original entry on oeis.org

1, 3, 17, 73, 257, 801, 2305, 6273, 16385, 41473, 102401, 247809, 589825, 1384449, 3211265, 7372801, 16777217, 37879809, 84934657, 189267969, 419430401, 924844033, 2030043137, 4437573633, 9663676417, 20971520001, 45365592065, 97844723713, 210453397505, 451508436993
Offset: 0

Author

Paul Curtz, Oct 22 2014

Keywords

Comments

Binomial transform of A118239 (Engel expansion of cosh(1)).
Table of successive differences of a(n):
1, 3, 17, 73, 257, 801, 2305,...
2, 14, 56, 184, 544, 1504,...
12, 42, 128, 360, 960,...
30, 86, 232, 600,...
56, 146, 368,...
90, 222,...
132,...
etc.
Via b(n) = 0, 0, 0 followed by A055580(n), i.e., 0, 0, 0, 1, 7, 31, 111, ... (the main sequence for the recurrence), a link can be found between a(n) and A002064(n): c(n) = b(n+1) - 2*b(n) = 0, 0, 1, 5, 17, 49, 129, ... (the main sequence for the signature (5, -8, 4)).

Examples

			a(3) = 9 * 8 + 1 = 73.
a(4) = 16 * 16 + 1 = 257.
a(5) = 25 * 32 + 1 = 801.
		

Crossrefs

Cf. A000225, A002064 (Cullen numbers), A006784, A007758, A055580, A118239, A168298.

Programs

  • Magma
    [2^n*n^2+1: n in [0..30]]; // Vincenzo Librandi, Oct 29 2016
  • Mathematica
    Table[n^2 * 2^n + 1, {n, 0, 31}] (* Alonso del Arte, Oct 22 2014 *)
    LinearRecurrence[{7,-18,20,-8}, {1,3,17,73}, 25] (* G. C. Greubel, Oct 28 2016 *)
  • PARI
    Vec(-(12*x^3-14*x^2+4*x-1)/((x-1)*(2*x-1)^3) + O(x^100)) \\ Colin Barker, Oct 22 2014
    
  • PARI
    a(n)=n^2<Charles R Greathouse IV, Oct 22 2014
    

Formula

a(n) = 4*a(n-1) - 4*a(n-2) + 2^(n+1) + 1.
a(n) = A007758(n) + 1.
a(n) = 7*a(n-1) - 18*a(n-2) + 20*a(n-3) - 8*a(n-4). - Jean-François Alcover, Oct 22 2014
G.f.: -(12*x^3-14*x^2+4*x-1) / ((x-1)*(2*x-1)^3). - Colin Barker, Oct 22 2014
E.g.f.: exp(x) + 2*x*(1 + 2*x)*exp(2*x). - G. C. Greubel, Oct 28 2016
Showing 1-10 of 14 results. Next