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 19 results. Next

A124182 A skewed version of triangular array A081277.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 0, 3, 4, 0, 0, 1, 8, 8, 0, 0, 0, 5, 20, 16, 0, 0, 0, 1, 18, 48, 32, 0, 0, 0, 0, 7, 56, 112, 64, 0, 0, 0, 0, 1, 32, 160, 256, 128, 0, 0, 0, 0, 0, 9, 120, 432, 576, 256, 0, 0, 0, 0, 0, 1, 50, 400, 1120, 1280, 512
Offset: 0

Views

Author

Philippe Deléham, Dec 05 2006

Keywords

Comments

Triangle T(n,k), 0 <= k <= n, read by rows given by [0, 1, -1, 0, 0, 0, 0, 0, 0, ...] DELTA [1, 1, 0, 0, 0, 0, 0, 0, 0,...] where DELTA is the operator defined in A084938. Falling diagonal sums in A052980.

Examples

			Triangle begins:
  1;
  0, 1;
  0, 1, 2;
  0, 0, 3, 4;
  0, 0, 1, 8,  8;
  0, 0, 0, 5, 20, 16;
  0, 0, 0, 1, 18, 48,  32;
  0, 0, 0, 0,  7, 56, 112,  64;
  0, 0, 0, 0,  1, 32, 160, 256,  128;
  0, 0, 0, 0,  0,  9, 120, 432,  576,  256;
  0, 0, 0, 0,  0,  1,  50, 400, 1120, 1280, 512;
		

Crossrefs

Cf. A025192 (column sums). Diagonals include A011782, A001792, A001793, A001794, A006974, A006975, A006976.

Formula

T(0,0)=T(1,1)=1, T(n,k)=0 if n < k or if k < 0, T(n,k) = T(n-2,k-1) + 2*T(n-1,k-1).
Sum_{k=0..n} x^k*T(n,k) = (-1)^n*A090965(n), (-1)^n*A084120(n), (-1)^n*A006012(n), A033999(n), A000007(n), A001333(n), A084059(n) for x = -4, -3, -2, -1, 0, 1, 2 respectively.
Sum_{k=0..floor(n/2)} T(n-k,k) = Fibonacci(n-1) = A000045(n-1).
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A011782(n), A001333(n), A026150(n), A046717(n), A084057(n), A002533(n), A083098(n), A084058(n), A003665(n), A002535(n), A133294(n), A090042(n), A125816(n), A133343(n), A133345(n), A120612(n), A133356(n), A125818(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17 respectively. - Philippe Deléham, Dec 26 2007
Sum_{k=0..n} T(n,k)*(-x)^(n-k) = A011782(n), A000012(n), A146559(n), A087455(n), A138230(n), A006495(n), A138229(n) for x= 0,1,2,3,4,5,6 respectively. - Philippe Deléham, Nov 14 2008
G.f.: (1-y*x)/(1-2y*x-y*x^2). - Philippe Deléham, Dec 04 2011
Sum_{k=0..n} T(n,k)^2 = A002002(n) for n > 0. - Philippe Deléham, Dec 04 2011

A005900 Octahedral numbers: a(n) = n*(2*n^2 + 1)/3.

Original entry on oeis.org

0, 1, 6, 19, 44, 85, 146, 231, 344, 489, 670, 891, 1156, 1469, 1834, 2255, 2736, 3281, 3894, 4579, 5340, 6181, 7106, 8119, 9224, 10425, 11726, 13131, 14644, 16269, 18010, 19871, 21856, 23969, 26214, 28595, 31116, 33781, 36594, 39559, 42680
Offset: 0

Views

Author

Keywords

Comments

Series reversion of g.f.: A(x) is Sum_{n>0} - A066357(n)(-x)^n.
Partial sums of centered square numbers A001844. - Paul Barry, Jun 26 2003
Also as a(n) = (1/6)*(4n^3 + 2n), n>0: structured tetragonal diamond numbers (vertex structure 5) (cf. A000447 - structured diamonds); and structured trigonal anti-prism numbers (vertex structure 5) (cf. A100185 - structured anti-prisms). Cf. A100145 for more on structured polyhedral numbers. - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Schlaefli symbol for this polyhedron: {3,4}.
If X is an n-set and Y and Z are disjoint 2-subsets of X then a(n-4) is equal to the number of 5-subsets of X intersecting both Y and Z. - Milan Janjic, Aug 26 2007
Starting with 1 = binomial transform of [1, 5, 8, 4, 0, 0, 0, ...] where (1, 5, 8, 4) = row 3 of the Chebyshev triangle A081277. - Gary W. Adamson, Jul 19 2008
a(n) = largest coefficient of (1 + ... + x^(n-1))^4. - R. H. Hardin, Jul 23 2009
Convolution square root of (1 + 6x + 19x^3 + ...) = (1 + 3x + 5x^2 + 7x^3 + ...) = A005408(x). - Gary W. Adamson, Jul 27 2009
Starting with offset 1 = the triangular series convolved with [1, 3, 4, 4, 4, ...]. - Gary W. Adamson, Jul 28 2009
One of the 5 Platonic polyhedral (tetrahedral, cube, octahedral, dodecahedral, and icosahedral) numbers (cf. A053012). - Daniel Forgues, May 14 2010
Let b be any product of four different primes. Then the divisor lattice of b^n is of width a(n+1). - Jean Drabbe, Oct 13 2010
Arises in Bezdek's proof on contact numbers for congruent sphere packings (see preprint). - Jonathan Vos Post, Feb 08 2011
Euler transform of length 2 sequence [6, -2]. - Michael Somos, Mar 27 2011
a(n+1) is the number of 2 X 2 matrices with all terms in {0,1,...,n} and (sum of terms) = 2n. - Clark Kimberling, Mar 19 2012
a(n) is the number of semistandard Young tableaux over all partitions of 3 with maximal element <= n. - Alois P. Heinz, Mar 22 2012
Self convolution of the odd numbers. - Reinhard Zumkeller, Apr 04 2012
a(n) is the number of (w,x,y,z) with all terms in {1,...,n} and w+x=y+z; also the number of (w,x,y,z) with all terms in {0,...,n} and |w-x|<=y. - Clark Kimberling, Jun 02 2012
The sequence is the third partial sum of (0, 1, 3, 4, 4, 4, ...). - Gary W. Adamson, Sep 11 2015
a(n) is the number of join-irreducible elements in the Weyl group of type B_n with respect to the strong Bruhat order. - Rafael Mrden, Aug 26 2020
Number of unit octahedra contained in an n-scale octahedron composed of a tetrahedral-octahedral honeycomb. The number of unit tetrahedra in it is 8*A000292(n-1) = 4*(n^3 - n)/3. Also, the number of unit tetrahedra and unit octahedra contained in an n-scale tetrahedron composed of a tetrahedral-octahedral honeycomb is respectively A006527(n) = (n^3 + 2*n)/3 and A000292(n-1) = (n^3 - n)/6. - Jianing Song, Feb 24 2025

Examples

			G.f. = x + 6*x^2 + 19*x^3 + 44*x^4 + 85*x^5 + 146*x^6 + 231*x^7 + ...
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 50.
  • H. S. M. Coxeter, Polyhedral numbers, pp. 25-35 of R. S. Cohen, J. J. Stachel and M. W. Wartofsky, eds., For Dirk Struik: Scientific, historical and political essays in honor of Dirk J. Struik, Reidel, Dordrecht, 1974.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Sums of 2 consecutive terms give A001845. Cf. A001844.
1/12*t*(n^3-n)+n for t = 2, 4, 6, ... gives A004006, A006527, A006003, A005900, A004068, A000578, A004126, A000447, A004188, A004466, A004467, A007588, A062025, A063521, A063522, A063523.
Cf. A022521.
Cf. A081277.
Row n=3 of A210391. - Alois P. Heinz, Mar 22 2012
Cf. A005408.
Cf. A002061.
Cf. A000292 (tetrahedral numbers), A000578 (cubes), A006566 (dodecahedral numbers), A006564 (icosahedral numbers).
Similar sequence: A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193(m=7), A099195 (m=8), A099196 (m=9), A099197 (m=10).

Programs

  • Haskell
    a005900 n = sum $ zipWith (*) odds $ reverse odds
                where odds = take n a005408_list
    a005900_list = scanl (+) 0 a001844_list
    -- Reinhard Zumkeller, Jun 16 2013, Apr 04 2012
    
  • Magma
    [n*(2*n^2+1)/3: n in [0..50]]; // Wesley Ivan Hurt, Sep 11 2015
    
  • Magma
    I:=[0,1,6,19]; [n le 4 select I[n] else 4*Self(n-1)-6*Self(n-2)+4*Self(n-3)-Self(n-4): n in [1..50]]; // Vincenzo Librandi, Sep 12 2015
    
  • Maple
    al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n), r=0..d-1); end; [seq(be(3,n), n=0..100)];
    A005900:=(z+1)**2/(z-1)**4; # Simon Plouffe in his 1992 dissertation
    with(combinat): seq(fibonacci(4,2*n)/12, n=0..40); # Zerinvary Lajos, Apr 21 2008
  • Mathematica
    Table[(2n^3+n)/3, {n,0,40}] (* or *) LinearRecurrence[{4,-6,4,-1}, {0,1,6,19},50] (* Harvey P. Dale, Oct 10 2013 *)
    CoefficientList[Series[x (1 + x)^2/(1 - x)^4, {x, 0, 45}], x] (* Vincenzo Librandi, Sep 12 2015 *)
  • Maxima
    makelist(n*(2*n^2+1)/3, n, 0, 20); /* Martin Ettl, Jan 07 2013 */
    
  • PARI
    {a(n) = n*(2*n^2+1)/3};
    
  • PARI
    concat([0],Vec(x*(1 + x)^2/(1 - x)^4 + O(x^50))) \\ Indranil Ghosh, Mar 16 2017
    
  • Python
    def a(n): return n*(2*n*n + 1)//3
    print([a(n) for n in range(41)]) # Michael S. Branicky, Sep 03 2021

Formula

a(n) = 1^2 + 2^2 + ... + (n-1)^2 + n^2 + (n-1)^2 + ... + 2^2 + 1^2. - Amarnath Murthy, May 28 2001
G.f.: x * (1 + x)^2 / (1 - x)^4. a(n) = -a(-n) = (2*n^3 + n) / 3.
a(n) = ( ((n+1)^5-n^5) - (n^5-(n-1)^5) )/30. - Xavier Acloque, Oct 17 2003
a(n) is the sum of the products pq, where p and q are both positive and odd and p + q = 2n, e.g., a(4) = 7*1 + 5*3 + 3*5 + 1*7 = 44. - Jon Perry, May 17 2005
a(n) = 4*binomial(n,3) + 4*binomial(n,2) + binomial(n,1). - Mitch Harris, Jul 06 2006
a(n) = binomial(n+2,3) + 2*binomial(n+1,3) + binomial(n,3), (this pair generalizes; see A014820, the 4-cross polytope numbers).
Sum_{n>=1} 1/a(n) = 3*gamma + 3*Psi((I*(1/2))*sqrt(2)) - (1/2)*(3*I)*Pi*coth((1/2)*Pi*sqrt(2)) - (1/2)*(3*I)*sqrt(2) = A175577, where I=sqrt(-1). - Stephen Crowley, Jul 14 2009
a(n) = A035597(n)/2. - J. M. Bergot, Jun 11 2012
a(n) = A000578(n) - 2*A000292(n-1) for n>0. - J. M. Bergot, Apr 05 2014
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), n>3. - Wesley Ivan Hurt, Sep 11 2015
E.g.f.: (1/3)*x*(3 + 6*x + 2*x^2)*exp(x). - Ilya Gutkovskiy, Mar 16 2017
a(n) = (A002061(A002061(n+1)) - A002061(A002061(n)))/6. - Daniel Poveda Parrilla, Jun 10 2017
a(n) = 6*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018
Sum_{n >= 1} (-1)^(n+1)/(a(n)*a(n+1)) = 6*log(2) - 4 = 1/(6 + 2/(6 + 6/(6 + ... + n*(n-1)/(6 + ...)))). See A142983. - Peter Bala, Mar 06 2024

A193722 Triangular array: the fusion of (x+1)^n and (x+2)^n; see Comments for the definition of fusion.

Original entry on oeis.org

1, 1, 2, 1, 5, 6, 1, 8, 21, 18, 1, 11, 45, 81, 54, 1, 14, 78, 216, 297, 162, 1, 17, 120, 450, 945, 1053, 486, 1, 20, 171, 810, 2295, 3888, 3645, 1458, 1, 23, 231, 1323, 4725, 10773, 15309, 12393, 4374, 1, 26, 300, 2016, 8694, 24948, 47628, 58320, 41553, 13122
Offset: 0

Views

Author

Clark Kimberling, Aug 04 2011

Keywords

Comments

Suppose that p = p(n)*x^n + p(n-1)*x^(n-1) + ... + p(1)*x + p(0) is a polynomial and that Q is a sequence of polynomials
...
q(k,x)=t(k,0)*x^k+t(k,1)*x^(k-1)+...+t(k,k-1)*x+t(k,k),
...
for k=0,1,2,... The Q-upstep of p is the polynomial given by
...
U(p) = p(n)*q(n+1,x) + p(n-1)*q(n,x) + ... + p(0)*q(1,x); note that q(0,x) does not appear.
...
Now suppose that P=(p(n,x)) and Q=(q(n,x)) are sequences of polynomials, where n indicates degree. The fusion of P by Q, denoted by P**Q, is introduced here as the sequence W=(w(n,x)) of polynomials defined by w(0,x)=1 and w(n+1,x)=U(p(n,x)).
...
Strictly speaking, ** is an operation on sequences of polynomials. However, if P and Q are regarded as numerical triangles (e.g., coefficients of polynomials), then ** can be regarded as an operation on numerical triangles. In this case, row (n+1) of P**Q, for n >= 0, is given by the matrix product P(n)*QQ(n), where P(n)=(p(n,n)...p(n,n-1)......p(n,1), p(n,0)) and QQ(n) is the (n+1)-by-(n+2) matrix given by
...
q(n+1,0) .. q(n+1,1)........... q(n+1,n) .... q(n+1,n+1)
0 ......... q(n,0)............. q(n,n-1) .... q(n,n)
0 ......... 0.................. q(n-1,n-2) .. q(n-1,n-1)
...
0 ......... 0.................. q(2,1) ...... q(2,2)
0 ......... 0 ................. q(1,0) ...... q(1,1);
here, the polynomial q(k,x) is taken to be
q(k,0)*x^k + q(k,1)x^(k-1) + ... + q(k,k)*x+q(k,k-1); i.e., "q" is used instead of "t".
...
If s=(s(1),s(2),s(3),...) is a sequence, then the infinite square matrix indicated by
s(1)...s(2)...s(3)...s(4)...s(5)...
..0....s(1)...s(2)...s(3)...s(4)...
..0......0....s(1)...s(2)...s(3)...
..0......0.......0...s(1)...s(2)...
is the self-fusion matrix of s; e.g., A202453, A202670.
...
Example: let p(n,x)=(x+1)^n and q(n,x)=(x+2)^n. Then
...
w(0,x) = 1 by definition of W
w(1,x) = U(p(0,x)) = U(1) = p(0,0)*q(1,x) = 1*(x+2) = x+2;
w(2,x) = U(p(1,x)) = U(x+1) = q(2,x) + q(1,x) = x^2+5x+6;
w(3,x) = U(p(2,x)) = U(x^2+2x+1) = q(3,x) + 2q(2,x) + q(1,x) = x^3+8x^2+21x+18;
...
From these first 4 polynomials in the sequence P**Q, we can write the first 4 rows of P**Q when P, Q, and P**Q are regarded as triangles:
1;
1, 2;
1, 5, 6;
1, 8, 21, 18;
...
Generally, if P and Q are the sequences given by p(n,x)=(ax+b)^n and q(n,x)=(cx+d)^n, then P**Q is given by (cx+d)(bcx+a+bd)^n.
...
In the following examples, r(P**Q) is the mirror of P**Q, obtained by reversing the rows of P**Q.
...
..P...........Q.........P**Q.......r(P**Q)
(x+1)^n.....(x+1)^n.....A081277....A118800 (unsigned)
(x+1)^n.....(x+2)^n.....A193722....A193723
(x+2)^n.....(x+1)^n.....A193724....A193725
(x+2)^n.....(x+2)^n.....A193726....A193727
(x+2)^n.....(2x+1)^n....A193728....A193729
(2x+1)^n....(x+1)^n.....A038763....A136158
(2x+1)^n....(2x+1)^n....A193730....A193731
(2x+1)^n,...(x+1)^n.....A193734....A193735
...
Continuing, let u denote the polynomial x^n+x^(n-1)+...+x+1, and let Fibo[n,x] denote the n-th Fibonacci polynomial.
...
P.............Q.........P**Q.......r(P**Q)
Fib[n+1,x]...(x+1)^n....A193736....A193737
u.............u.........A193738....A193739
u**u..........u**u......A193740....A193741
...
Regarding A193722:
col 1 ..... A000012
col 2 ..... A016789
col 3 ..... A081266
w(n,n) .... A025192
w(n,n-1) .. A081038
...
Associated with "upstep" as defined above is "downstep" defined at A193842 in connection with fission.

Examples

			First six rows:
  1;
  1,   2;
  1,   5,   6;
  1,   8,  21,  18;
  1,  11,  45,  81,  54;
  1,  14,  78, 216, 297, 162;
		

Crossrefs

Programs

  • GAP
    Flat(List([0..10], n-> List([0..n], k-> 3^(k-1)*( Binomial(n-1,k) + 2*Binomial(n,k) ) ))); # G. C. Greubel, Feb 18 2020
  • Magma
    [3^(k-1)*( Binomial(n-1,k) + 2*Binomial(n,k) ): k in [0..n], n in [0..10]]; // G. C. Greubel, Feb 18 2020
    
  • Maple
    fusion := proc(p, q, n) local d, k;
    p(n-1,0)*q(n,x)+add(coeff(p(n-1,x),x^k)*q(n-k,x), k=1..n-1);
    [1,seq(coeff(%,x,n-1-k), k=0..n-1)] end:
    p := (n, x) -> (x + 1)^n; q := (n, x) -> (x + 2)^n;
    A193722_row := n -> fusion(p, q, n);
    for n from 0 to 5 do A193722_row(n) od; # Peter Luschny, Jul 24 2014
  • Mathematica
    (* First program *)
    z = 9; a = 1; b = 1; c = 1; d = 2;
    p[n_, x_] := (a*x + b)^n ; q[n_, x_] := (c*x + d)^n
    t[n_, k_] := Coefficient[p[n, x], x^k]; t[n_, 0] := p[n, x] /. x -> 0;
    w[n_, x_] := Sum[t[n, k]*q[n + 1 - k, x], {k, 0, n}]; w[-1, x_] := 1
    g[n_] := CoefficientList[w[n, x], {x}]
    TableForm[Table[Reverse[g[n]], {n, -1, z}]]
    Flatten[Table[Reverse[g[n]], {n, -1, z}]] (* A193722 *)
    TableForm[Table[g[n], {n, -1, z}]]
    Flatten[Table[g[n], {n, -1, z}]] (* A193723 *)
    (* Second program *)
    Table[3^(k-1)*(Binomial[n-1,k] +2*Binomial[n,k]), {n,0,10}, {k,0,n}]//Flatten (* G. C. Greubel, Feb 18 2020 *)
  • PARI
    T(n,k) = 3^(k-1)*(binomial(n-1,k) +2*binomial(n,k)); \\ G. C. Greubel, Feb 18 2020
    
  • Sage
    def fusion(p, q, n):
        F = p(n-1,0)*q(n,x)+add(expand(p(n-1,x)).coefficient(x,k)*q(n-k,x) for k in (1..n-1))
        return [1]+[expand(F).coefficient(x,n-1-k) for k in (0..n-1)]
    A193842_row = lambda k: fusion(lambda n,x: (x+1)^n, lambda n,x: (x+2)^n, k)
    for n in range(7): A193842_row(n) # Peter Luschny, Jul 24 2014
    

Formula

Triangle T(n,k), read by rows, given by [1,0,0,0,0,0,0,0,...] DELTA [2,1,0,0,0,0,0,0,0,0,...] where DELTA is the operator defined in A084938. - Philippe Deléham, Oct 04 2011
T(n,k) = 3*T(n-1,k-1) + T(n-1,k) with T(0,0)=T(1,0)=1 and T(1,1)=2. - Philippe Deléham, Oct 05 2011
T(n, k) = 3^(k-1)*( binomial(n-1,k) + 2*binomial(n,k) ). - G. C. Greubel, Feb 18 2020

A025192 a(0)=1; a(n) = 2*3^(n-1) for n >= 1.

Original entry on oeis.org

1, 2, 6, 18, 54, 162, 486, 1458, 4374, 13122, 39366, 118098, 354294, 1062882, 3188646, 9565938, 28697814, 86093442, 258280326, 774840978, 2324522934, 6973568802, 20920706406, 62762119218, 188286357654, 564859072962, 1694577218886, 5083731656658, 15251194969974
Offset: 0

Views

Author

Keywords

Comments

Warning: there is considerable overlap between this entry and the essentially identical A008776.
Shifts one place left when plus-convolved (PLUSCONV) with itself. a(n) = 2*Sum_{i=0..n-1} a(i). - Antti Karttunen, May 15 2001
Let M = { 0, 1, ..., 2^n-1 } be the set of all n-bit numbers. Consider two operations on this set: "sum modulo 2^n" (+) and "bitwise exclusive or" (XOR). The results of these operations are correlated.
To give a numerical measure, consider the equations over M: u = x + y, v = x XOR y and ask for how many pairs (u,v) is there a solution? The answer is exactly a(n) = 2*3^(n-1) for n >= 1. The fraction a(n)/4^n of such pairs vanishes as n goes to infinity. - Max Alekseyev, Feb 26 2003
Number of (s(0), s(1), ..., s(2n+2)) such that 0 < s(i) < 6 and |s(i) - s(i-1)| = 1 for i = 1,2,...,2n+2, s(0) = 3, s(2n+2) = 3. - Herbert Kociemba, Jun 10 2004
Number of compositions of n into parts of two kinds. For a string of n objects, before the first, choose first kind or second kind; before each subsequent object, choose continue, first kind, or second kind. For example, compositions of 3 are 3; 2,1; 1,2; and 1,1,1. Using parts of two kinds, these produce respectively 2, 4, 4 and 8 compositions, 2+4+4+8 = 18. - Franklin T. Adams-Watters, Aug 18 2006
In the compositions the kinds of parts are ordered inside a run of identical parts, see example. Replacing "ordered" by "unordered" gives A052945. - Joerg Arndt, Apr 28 2013
Number of permutations of {1, 2, ..., n+1} such that no term is more than 2 larger than its predecessor. For example, a(3) = 18 because all permutations of {1, 2, 3, 4} are valid except 1423, 1432, 2143, 3142, 2314, 3214, in which 1 is followed by 4. Proof: removing (n + 1) gives a still-valid sequence. For n >= 2, can insert (n + 1) either at the beginning or immediately following n or immediately following (n - 1), but nowhere else. Thus the number of such permutations triples when we increase the sequence length by 1. - Joel B. Lewis, Nov 14 2006
Antidiagonal sums of square array A081277. - Philippe Deléham, Dec 04 2006
Equals row sums of triangle A160760. - Gary W. Adamson, May 25 2009
Let M = a triangle with (1, 2, 4, 8, ...) as the left border and all other columns = (0, 1, 2, 4, 8, ...). A025192 = lim_{n->oo} M^n, the left-shifted vector considered as a sequence. - Gary W. Adamson, Jul 27 2010
Number of nonisomorphic graded posets with 0 and uniform hasse graph of rank n with no 3-element antichain. ("Uniform" used in the sense of Retakh, Serconek and Wilson. By "graded" we mean that all maximal chains have the same length n.) - David Nacin, Feb 13 2012
Equals partial sums of A003946 prefaced with a 1: (1, 1, 4, 12, 36, 108, ...). - Gary W. Adamson, Feb 15 2012
Number of vertices (or sides) of the (n-1)-th iteration of a Gosper island. - Arkadiusz Wesolowski, Feb 07 2013
Row sums of triangle in A035002. - Jon Perry, May 30 2013
a(n) counts walks (closed) on the graph G(1-vertex; 1-loop, 1-loop, 2-loop, 2-loop, 3-loop, 3-loop, ...). - David Neil McGrath, Jan 01 2015
From Tom Copeland, Dec 03 2015: (Start)
For n > 0, a(n) are the traces of the even powers of the adjacency matrix M of the simple Lie algebra B_3, tr(M^(2n)) where M = Matrix(row 1; row 2; row 3) = Matrix[0,1,0; 1,0,2; 0,1,0], same as the traces of Matrix[0,2,0; 1,0,1; 0,1,0] (cf. Damianou). The traces of the odd powers vanish.
The characteristic polynomial of M equals determinant(x*I - M) = x^3 - 3x = A127672(3,x), so 1 - 3*x^2 = det(I - x M) = exp(-Sum_{n>=1} tr(M^n) x^n / n), implying Sum_{n>=1} a(n+1) x^(2n) / (2n) = -log(1 - 3*x^2), giving a logarithmic generating function for the aerated sequence, excluding a(0) and a(1).
a(n+1) = tr(M^(2n)), where tr(M^n) = 3^(n/2) + (-1)^n * 3^(n/2) = 2^n*(cos(Pi/6)^n + cos(5*Pi/6)^n) = n-th power sum of the eigenvalues of M = n-th power sum of the zeros of the characteristic polynomial.
The relation det(I - x M) = exp(-Sum_{n>=1} tr(M^n) x^n / n) = Sum_{n>=0} P_n(-tr(M), -tr(M^2), ..., -tr(M^n)) x^n/n! = exp(P.(-tr(M), -tr(M^2), ...)x), where P_n(x(1), ..., x(n)) are the partition polynomials of A036039 implies that with x(2n) = -tr(M^(2n)) = -a(n+1) for n > 0 and x(n) = 0 otherwise, the partition polynomials evaluate to zero except for P_2(x(1), x(2)) = P_2(0,-6) = -6.
Because of the inverse relation between the partition polynomials of A036039 and the Faber polynomials F_k(b1,b2,...,bk) of A263916, F_k(0,-3,0,0,...) = tr(M^k) gives aerated a(n), excluding n=0,1. E.g., F_2(0,-3) = -2(-3) = 6, F_4(0,-3,0,0) = 2 (-3)^2 = 18, and F_6(0,-3,0,0,0,0) = -2(-3)^3 = 54. (Cf. A265185.)
(End)
Number of permutations of length n > 0 avoiding the partially ordered pattern (POP) {1>2, 1>3, 1>4} of length 4. That is, number of length n permutations having no subsequences of length 4 in which the first element is the largest. - Sergey Kitaev, Dec 08 2020
For n > 0, a(n) is the number of 3-colorings of the grid graph P_2 X P_(n-1). More generally, for q > 1, the number of q-colorings of the grid graph P_2 X P_n is given by q*(q - 1)*((q - 1)*(q - 2) + 1)^(n - 1). - Sela Fried, Sep 25 2023
For n > 1, a(n) is the largest solution to the equation phi(x) = a(n-1). - M. Farrokhi D. G., Oct 25 2023
Number of dotted compositions of degree n. - Diego Arcis, Feb 01 2024

Examples

			There are a(3)=18 compositions of 3 into 2 kinds of parts. Here p:s stands for "part p of sort s":
01:  [ 1:0  1:0  1:0  ]
02:  [ 1:0  1:0  1:1  ]
03:  [ 1:0  1:1  1:0  ]
04:  [ 1:0  1:1  1:1  ]
05:  [ 1:0  2:0  ]
06:  [ 1:0  2:1  ]
07:  [ 1:1  1:0  1:0  ]
08:  [ 1:1  1:0  1:1  ]
09:  [ 1:1  1:1  1:0  ]
10:  [ 1:1  1:1  1:1  ]
11:  [ 1:1  2:0  ]
12:  [ 1:1  2:1  ]
13:  [ 2:0  1:0  ]
14:  [ 2:0  1:1  ]
15:  [ 2:1  1:0  ]
16:  [ 2:1  1:1  ]
17:  [ 3:0  ]
18:  [ 3:1  ]
- _Joerg Arndt_, Apr 28 2013
G.f. = 1 + 2*x + 6*x^2 + 18*x^3 + 54*x^4 + 162*x^5 + 486*x^6 + 1458*x^7 + ...
		

References

  • Richard P. Stanley, Enumerative combinatorics, Vol. 1, Cambridge University Press, Cambridge, 1997, pp. 96-100.

Crossrefs

First differences of 3^n (A000244). Other self-convolved sequences: A000108, A007460, A007461, A007462, A007463, A007464, A061922.
Apart from initial term, same as A008776.

Programs

  • Haskell
    a025192 0 = 1
    a025192 n = 2 * 3 ^ (n -1)
    a025192_list = 1 : iterate (* 3) 2  -- Reinhard Zumkeller, Nov 27 2012
  • Maple
    A025192 := proc(n): if n=0 then 1 else 2*3^(n-1) fi: end: seq(A025192(n),n=0..26);
  • Mathematica
    Join[{1},2*3^(Range[30]-1)]  (* Harvey P. Dale, Mar 22 2011 *)
  • PARI
    a(n)=max(1,2*3^(n-1)) \\ Charles R Greathouse IV, Jul 25 2011
    
  • PARI
    Vec((1-x)/(1-3*x) + O(x^100)) \\ Altug Alkan, Dec 05 2015
    
  • Python
    [1]+[2*3**(n-1) for n in range(1,30)] # David Nacin, Mar 04 2012
    

Formula

G.f.: (1-x)/(1-3*x).
E.g.f.: (2*exp(3*x) + exp(0))/3. - Paul Barry, Apr 20 2003
a(n) = phi(3^n) = A000010(A000244(n)). - Labos Elemer, Apr 14 2003
a(0) = 1, a(n) = Sum_{k=0..n-1} (a(k) + a(n-k-1)). - Benoit Cloitre, Jun 24 2003
a(n) = A002326((3^n-1)/2). - Vladimir Shevelev, May 26 2008
a(1) = 2, a(n) = 3*a(n-1). - Vincenzo Librandi, Jan 01 2011
a(n) = lcm(a(n-1), Sum_{k=1..n-1} a(k)) for n >= 3. - David W. Wilson, Sep 27 2011
a(n) = ((2*n-1)*a(n-1) + (3*n-6)*a(n-2))/(n-1); a(0)=1, a(1)=2. - Sergei N. Gladkovskii, Jul 16 2012
From Sergei N. Gladkovskii, Jul 17 2012: (Start)
For the e.g.f. E(x) = (2/3)*exp(3*x) + exp(0)/3 we have
E(x) = 2*G(0)/3 where G(k) = 1 + k!/(3*(9*x)^k - 3*(9*x)^(2*k+1)/((9*x)^(k+1) + (k+1)!/G(k+1))); (continued fraction, 3rd kind, 3-step).
E(x) = 1+2*x/(G(0)-3*x) where G(k) = 3*x + 1 + k - 3*x*(k+1)/G(k+1); (continued fraction, Euler's 1st kind, 1-step). (End)
a(n) = A114283(0,0). - Reinhard Zumkeller, Nov 27 2012
G.f.: 1 + ((1/2)/G(0) - 1)/x where G(k) = 1 - 2^k/(2 - 4*x/(2*x - 2^k/G(k+1) )); (recursively defined continued fraction). - Sergei N. Gladkovskii, Dec 22 2012
G.f.: 1 + x*W(0), where W(k) = 1 + 1/(1 - x*(2*k+3)/(x*(2*k+4) + 1/W(k+1))); (continued fraction). - Sergei N. Gladkovskii, Aug 28 2013
G.f.: 1 / (1 - 2*x / (1 - x)). - Michael Somos, Apr 03 2014
Construct the power matrix T(n,j) = [A(n)^*j]*[S(n)^*(j-1)] where A(n)=(2,2,2,...) and S(n)=(0,1,0,0,...). (* is convolution operation.) Then a(n) = Sum_{j=1..n} T(n,j). - David Neil McGrath, Jan 01 2015
G.f.: 1 + 2*x/(1 + 2*x)*( 1 + 5*x/(1 + 5*x)*( 1 + 8*x/(1 + 8*x)*( 1 + 11*x/(1 + 11*x)*( 1 + .... - Peter Bala, May 27 2017
Sum_{n>=0} 1/a(n) = 7/4. - Bernard Schott, Oct 02 2021
From Amiram Eldar, May 08 2023: (Start)
Sum_{n>=0} (-1)^n/a(n) = 5/8.
Product_{n>=1} (1 - 1/a(n)) = A132019. (End)

Extensions

Additional comments from Barry E. Williams, May 27 2000
a(22) corrected by T. D. Noe, Feb 08 2008
Maple programs simplified by Johannes W. Meijer, Jun 02 2011

A002492 Sum of the first n even squares: 2*n*(n+1)*(2*n+1)/3.

Original entry on oeis.org

0, 4, 20, 56, 120, 220, 364, 560, 816, 1140, 1540, 2024, 2600, 3276, 4060, 4960, 5984, 7140, 8436, 9880, 11480, 13244, 15180, 17296, 19600, 22100, 24804, 27720, 30856, 34220, 37820, 41664, 45760, 50116, 54740, 59640, 64824, 70300, 76076, 82160
Offset: 0

Views

Author

Keywords

Comments

Total number of possible bishop moves on an n+1 X n+1 chessboard, if the bishop is placed anywhere. E.g., on a 3 X 3-Board: bishop has 8 X 2 moves and 1 X 4 moves, so a(2)=20. - Ulrich Schimke (ulrschimke(AT)aol.com)
Let M_n denote the n X n matrix M_n(i,j)=(i+j)^2; then the characteristic polynomial of M_n is x^n - a(n)x^(n-1) - .... - Michael Somos, Nov 14 2002
Partial sums of A016742. - Lekraj Beedassy, Jun 19 2004
0,4,20,56,120 gives the number of electrons in closed shells in the double shell periodic system of elements. This is a new interpretation of the periodic system of the elements. The factor 4 in the formula 4*n(n+1)(2n+1)/6 plays a significant role, since it designates the degeneracy of electronic states in this system. Closed shells with more than 120 electrons are not expected to exist. - Karl-Dietrich Neubert (kdn(AT)neubert.net)
Inverse binomial transform of A240434. - Wesley Ivan Hurt, Apr 13 2014
Atomic number of alkaline-earth metals of period 2n. - Natan Arie Consigli, Jul 03 2016
a(n) are the negative cubic coefficients in the expansion of sin(kx) into powers of sin(x) for the odd k: sin(kx) = k sin(x) - c(k) sin^3(x) + O(sin^5(x)); a(n) = c(2n+1) = A000292(2n). - Mathias Zechmeister, Jul 24 2022
Also the number of distinct series-parallel networks under series-parallel reduction on three unlabeled edges of n element kinds. - Michael R. Hayashi, Aug 02 2023

References

  • A. O. Barut, Group Structure of the Periodic System, in Wybourne, Ed., The Structure of Matter, University of Canterbury Press, Christchurch, 1972, p. 126.
  • Edward G. Mazur, Graphic Representation of the Periodic System during One Hundred Years, University of Alabama Press, Alabama, 1974.
  • W. Permans and J. Kemperman, "Nummeringspribleem van S. Dockx, Mathematisch Centrum. Amsterdam," Rapport ZW; 1949-005, 4 leaves, 19.8 X 34 cm.
  • N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A033586 (King), A035005 (Queen), A035006 (Rook), A035008 (Knight) and A049450 (Pawn).

Programs

  • Magma
    [2*n*(n+1)*(2*n+1)/3: n in [0..40]]; // Vincenzo Librandi, Jun 16 2011
  • Maple
    A002492:=n->2*n*(n+1)*(2*n+1)/3; seq(A002492(n), n=0..50); # Wesley Ivan Hurt, Apr 04 2014
  • Mathematica
    Table[2n(n+1)(2n+1)/3, {n,0,40}] (* or *) Binomial[2*Range[0,40]+2,3] (* or *) LinearRecurrence[{4,-6,4,-1}, {0,4,20,56},40] (* Harvey P. Dale, Aug 15 2012 *)
    Accumulate[(2*Range[0,40])^2] (* Harvey P. Dale, Jun 04 2019 *)
  • PARI
    a(n)=2*n*(n+1)*(2*n+1)/3
    

Formula

G.f.: 4*x*(1+x)/(1-x)^4. - Simon Plouffe in his 1992 dissertation
a(-1-n) = -a(n).
a(n) = 4*A000330(n) = 2*A006331(n) = A000292(2*n).
a(n) = (-1)^(n+1)*A053120(2*n+1,3) (fourth unsigned column of Chebyshev T-triangle, zeros omitted).
a(n) = binomial(2*n+2, 3). - Lekraj Beedassy, Jun 19 2004
A035005(n+1) = a(n) + A035006(n+1) since Queen = Bishop + Rook. - Johannes W. Meijer, Feb 04 2010
a(n) - a(n-1) = 4*n^2. - Joerg Arndt, Jun 16 2011
a(n) = 4*a(n-1)-6*a(n-2)+4*a(n-3)-a(n-4) for n>3. - Harvey P. Dale, Aug 15 2012
a(n) = Sum_{k=0..3} C(n-2+k,n-2)*C(n+3-k,n), for n>2. - J. M. Bergot, Jun 14 2014
a(n) = 2*A006331(n). - R. J. Mathar, May 28 2016
From Natan Arie Consigli Jul 03 2016: (Start)
a(n) = A166464(n) - 1.
a(n) = A168380(2*n). (End)
a(n) = Sum_{i=0..n} A005408(i)*A005408(i-1)+1 with A005408(-1):=-1. - Bruno Berselli, Jan 09 2017
a(n) = A002412(n) + A016061(n). - Bruce J. Nicholson, Nov 12 2017
From Amiram Eldar, Jan 04 2022: (Start)
Sum_{n>=1} 1/a(n) = 9/2 - 6*log(2).
Sum_{n>=1} (-1)^(n+1)/a(n) = 3*Pi/2 - 9/2. (End)
a(n) = A081277(3, n-1) = (1+2*n)*binomial(n+1, n-2)*2^2/(n-1) for n > 0. - Mathias Zechmeister, Jul 26 2022
E.g.f.: 2*exp(x)*x*(6 + 9*x + 2*x^2)/3. - Stefano Spezia, Jul 31 2022

Extensions

Minor errors corrected and edited by Johannes W. Meijer, Feb 04 2010
Title modified by Charles R Greathouse IV at the suggestion of J. M. Bergot, Apr 05 2014

A014820 a(n) = (1/3)*(n^2 + 2*n + 3)*(n+1)^2.

Original entry on oeis.org

1, 8, 33, 96, 225, 456, 833, 1408, 2241, 3400, 4961, 7008, 9633, 12936, 17025, 22016, 28033, 35208, 43681, 53600, 65121, 78408, 93633, 110976, 130625, 152776, 177633, 205408, 236321, 270600, 308481
Offset: 0

Views

Author

Keywords

Comments

a(n) is the number of 4 X 4 pandiagonal magic squares with sum 2n. - Sharon Sela (sharonsela(AT)hotmail.com), May 10 2002
Figurate numbers based on the 4-dimensional regular convex polytope called the 16-cell, hexadecachoron, 4-cross polytope or 4-hyperoctahedron with Schlaefli symbol {3,3,4}. a(n)=(n^2*(n^2+2))/3 if the offset were 1. - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004, R. J. Mathar, Jul 18 2009
If X is an n-set and Y_i (i=1,2,3) mutually disjoint 2-subsets of X then a(n-6) is equal to the number of 7-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Aug 26 2007
Equals binomial transform of [1, 7, 18, 20, 8, 0, 0, 0, ...], where (1, 7, 18, 20, 8) = row 4 of the Chebyshev triangle A081277. Also = row 4 of the array in A142978. - Gary W. Adamson, Jul 19 2008

References

  • T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.

Crossrefs

Programs

  • GAP
    List([0..40], n -> (n+1)^2*((n+1)^2 +2)/3); # G. C. Greubel, Feb 10 2019
  • Magma
    [(1/3)*(n^2+2*n+3)*(n+1)^2: n in [0..40]]; // Vincenzo Librandi, May 22 2011
    
  • Maple
    al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n),r=0..d-1); end; [seq(be(4,n),n=0..100)];
  • Mathematica
    LinearRecurrence[{5, -10, 10, -5, 1}, {1, 8, 33, 96, 225}, 31] (* Jean-François Alcover, Jan 17 2018 *)
  • PARI
    a(n)=(n+1)^2*(n^2+2*n+3)/3 \\ Charles R Greathouse IV, Apr 17 2012
    
  • R
    a <- c(1, 8, 33, 96,225)
    for(n in (length(a)+1):30) a[n] <- 5*a[n-1]-10*a[n-2]+10*a[n-3]-5*a[n-4]+a[n-5]
    a # Yosu Yurramendi, Sep 03 2013
    
  • Sage
    [((n+1)^2+2)*(n+1)^2/3 for n in range(40)] # G. C. Greubel, Feb 10 2019
    

Formula

Or, a(n-1) = n^2*(n^2+2)/3. - Corrected by R. J. Mathar, Jul 18 2009
From Vladeta Jovovic, Apr 03 2002: (Start)
G.f.: (1+x)^3/(1-x)^5.
Recurrence: a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5). (End)
a(n-1) = C(n+3,4) + 3 C(n+2,4) + 3 C(n+1,4) + C(n,4).
Sum_{n>=0} 1/((1/3*(n^2 + 2*n + 3))*(n+1)^2) = (1/4)*Pi^2 - 3*sqrt(2)*Pi*coth(Pi*sqrt(2))*(1/8) + 3/8 = 1.1758589... - Stephen Crowley, Jul 14 2009
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5), with n > 4, a(0)=1, a(1)=8, a(2)=33, a(3)=96, a(4)=225. - Yosu Yurramendi, Sep 03 2013
From Bruce J. Nicholson, Jan 23 2019: (Start)
Sum_{i=0..n} a(i) = A061927(n+1).
a(n) = 4*A002415(n+1) + A000290(n+1) = A039623(n+1) + A002415(n+1). (End)
E.g.f.: (3 + 21*x + 27*x^2 + 10*x^3 + x^4)*exp(x)/3. - G. C. Greubel, Feb 10 2019
Sum_{n >= 0} (-1)^n/(a(n)*a(n+1)) = 17/3 - 8*log(2) = 1/(8 + 2/(8 + 6/(8 + ... + n*(n-1)/(8 + ...)))). See A142983. - Peter Bala, Mar 06 2024

Extensions

Formula index corrected by R. J. Mathar, Jul 18 2009

A028297 Coefficients of Chebyshev polynomials of the first kind: triangle of coefficients in expansion of cos(n*x) in descending powers of cos(x).

Original entry on oeis.org

1, 1, 2, -1, 4, -3, 8, -8, 1, 16, -20, 5, 32, -48, 18, -1, 64, -112, 56, -7, 128, -256, 160, -32, 1, 256, -576, 432, -120, 9, 512, -1280, 1120, -400, 50, -1, 1024, -2816, 2816, -1232, 220, -11, 2048, -6144, 6912, -3584, 840, -72, 1, 4096, -13312, 16640, -9984
Offset: 0

Views

Author

Keywords

Comments

Rows are of lengths 1, 1, 2, 2, 3, 3, ... (A008619).
This triangle is generated from A118800 by shifting down columns to allow for (1, 1, 2, 2, 3, 3, ...) terms in each row. - Gary W. Adamson, Dec 16 2007
Unsigned triangle = A034839 * A007318. - Gary W. Adamson, Nov 28 2008
Triangle, with zeros omitted, given by (1, 1, 0, 0, 0, 0, 0, 0, 0, ...) DELTA (0, -1, 1, 0, 0, 0, 0, 0, 0, ...) where DELTA is the operator defined in A084938. - Philippe Deléham, Dec 16 2011
From Wolfdieter Lang, Aug 02 2014: (Start)
This irregular triangle is the row reversed version of the Chebyshev T-triangle A053120 given by A039991 with vanishing odd-indexed columns removed.
If zeros are appended in each row n >= 1, in order to obtain a regular triangle (see the Philippe Deléham comment, g.f. and example) this becomes the Riordan triangle (1-x)/(1-2*x), -x^2/(1-2*x). See also the unsigned version A201701 of this regular triangle.
(End)
Apparently, unsigned diagonals of this array are rows of A200139. - Tom Copeland, Oct 11 2014
It appears that the coefficients are generated by the following: Let SM_k = Sum( d_(t_1, t_2)* eM_1^t_1 * eM_2^t_2) summed over all length 2 integer partitions of k, i.e., 1*t_1 + 2*t_2 = k, where SM_k are the averaged k-th power sum symmetric polynomials in 2 data (i.e., SM_k = S_k/2 where S_k are the k-th power sum symmetric polynomials, and where eM_k are the averaged k-th elementary symmetric polynomials, eM_k = e_k/binomial(2,k) with e_k being the k-th elementary symmetric polynomials. The data d_(t_1, t_2) form an irregular triangle, with one row for each k value starting with k=1. Thus this procedure and associated OEIS sequences A287768, A288199, A288207, A288211, A288245, A288188 are generalizations of Chebyshev polynomials of the first kind. - Gregory Gerard Wojnar, Jul 01 2017

Examples

			Letting c = cos x, we have: cos 0x = 1, cos 1x = 1c; cos 2x = 2c^2-1; cos 3x = 4c^3-3c, cos 4x = 8c^4-8c^2+1, etc.
T4 = 8x^4 - 8x^2 + 1 = 8, -8, +1 = 2^(3) - (4)(2) + [2^(-1)](4)/2.
From _Wolfdieter Lang_, Aug 02 2014: (Start)
The irregular triangle T(n,k) begins:
n\k     1      2     3      4     5     6   7   8 ....
0:      1
1:      1
2:      2     -1
3:      4     -3
4:      8     -8     1
5:     16    -20     5
6:     32    -48    18     -1
7:     64   -112    56     -7
8:    128   -256   160    -32     1
9:    256   -576   432   -120     9
10:   512  -1280  1120   -400    50    -1
11:  1024  -2816  2816  -1232   220   -11
12:  2048  -6144  6912  -3584   840   -72   1
13:  4096 -13312 16640  -9984  2912  -364  13
14:  8192 -28672 39424 -26880  9408 -1568  98  -1
15: 16384 -61440 92160 -70400 28800 -6048 560 -15
...
T(4,x) = 8*x^4 -8*x^2 + 1*x^0, T(5,x) = 16*x^5 - 20*x^3 + 5*x^1, with Chebyshev's T-polynomials (A053120). (End)
From _Philippe Deléham_, Dec 16 2011: (Start)
The triangle (1,1,0,0,0,0,...) DELTA (0,-1,1,0,0,0,0,...) includes zeros and begins:
   1;
   1,   0;
   2,  -1,  0;
   4,  -3,  0,  0;
   8,  -8,  1,  0, 0;
  16, -20,  5,  0, 0, 0;
  32, -48, 18, -1, 0, 0, 0; (End)
		

References

  • I. S. Gradshteyn and I. M. Ryzhik, Tables of Integrals, Series and Products, 5th ed., Section 1.335, p. 35.
  • S. Selby, editor, CRC Basic Mathematical Tables, CRC Press, 1970, p. 106. [From Rick L. Shepherd, Jul 06 2010]

Crossrefs

Cf. A028298.
Reflection of A008310, the main entry. With zeros: A039991.
Cf. A053120 (row reversed table including zeros).
Cf. A001333 (row sums 1), A001333 (alternating row sums). - Wolfdieter Lang, Aug 02 2014

Programs

  • Maple
    b:= proc(n) b(n):= `if`(n<2, 1, expand(2*b(n-1)-x*b(n-2))) end:
    T:= n-> (p-> seq(coeff(p, x, i), i=0..degree(p)))(b(n)):
    seq(T(n), n=0..15);  # Alois P. Heinz, Sep 04 2019
  • Mathematica
    t[n_] := (Cos[n x] // TrigExpand) /. Sin[x]^m_ /; EvenQ[m] -> (1 - Cos[x]^2)^(m/2) // Expand; Flatten[Table[ r = Reverse @ CoefficientList[t[n], Cos[x]]; If[OddQ[Length[r]], AppendTo[r,0]]; Partition[r,2][[All, 1]],{n, 0, 13}] ][[1 ;; 53]] (* Jean-François Alcover, May 06 2011 *)
    Tpoly[n_] := HypergeometricPFQ[{(1 - n)/2, -n/2}, {1/2}, 1 - x];
    Table[CoefficientList[Tpoly[n], x], {n, 0, 12}] // Flatten (* Peter Luschny, Feb 03 2021 *)

Formula

cos(n*x) = 2 * cos((n-1)*x) * cos(x) - cos((n-2)*x) (from CRC's Multiple-angle relations). - Rick L. Shepherd, Jul 06 2010
G.f.: (1-x) / (1-2x+y*x^2). - Philippe Deléham, Dec 16 2011
Sum_{k=0..n} T(n,k)*x^k = A011782(n), A000012(n), A146559(n), A087455(n), A138230(n), A006495(n), A138229(n) for x = 0, 1, 2, 3, 4, 5, 6, respectively. - Philippe Deléham, Dec 16 2011
T(n,k) = [x^k] hypergeom([1/2 - n/2, -n/2], [1/2], 1 - x). - Peter Luschny, Feb 03 2021
T(n,k) = (-1)^k * 2^(n-1-2*k) * A034807(n,k). - Hoang Xuan Thanh, Jun 21 2025

Extensions

More terms from David W. Wilson
Row length sequence and link to Abramowitz-Stegun added by Wolfdieter Lang, Aug 02 2014

A142978 Table of figurate numbers for the n-dimensional cross polytopes.

Original entry on oeis.org

1, 1, 2, 1, 4, 3, 1, 6, 9, 4, 1, 8, 19, 16, 5, 1, 10, 33, 44, 25, 6, 1, 12, 51, 96, 85, 36, 7, 1, 14, 73, 180, 225, 146, 49, 8, 1, 16, 99, 304, 501, 456, 231, 64, 9, 1, 18, 129, 476, 985, 1182, 833, 344, 81, 10
Offset: 1

Views

Author

Peter Bala, Jul 15 2008

Keywords

Comments

The n-th row entries for this array are the regular polytope numbers for the n-dimensional cross polytope as defined by [Kim]. The rows are the partial sums of the rows of the square array of Delannoy numbers A008288.
The odd numbered rows of this array form A142977. For a triangular version of this table see A104698. Cf. also A101603.
The n-th row of the array is the binomial transform of n-th row of triangle A081277, followed by zeros. Example: row 4 (1, 6, 19, 44, 85, ...) = binomial transform of row 3 of A081277: (1, 5, 8, 4, 0, 0, 0, ...). - Gary W. Adamson, Jul 17 2008
The main diagonal of the array T(n,k) is A047781 Sum_{k=0..n-1} binomial(n-1,k)*binomial(n+k,k). Also a(n) = T(n,n), array T as in A049600. The link from A099193 to J. V. Post, Table of polytope numbers, Sorted, Through 1,000,000, includes all n-D Hyperoctahedron (n-Cross Polytope) Numbers through 10-Cross(20) = 1669752016. - Jonathan Vos Post, Jul 16 2008

Examples

			The square array A(n, k) begins:
  n\k| 1   2    3     4     5       6
  ---+-------------------------------
   1 | 1   2    3     4      5      6    A000027
   2 | 1   4    9    16     25     36    A000290
   3 | 1   6   19    44     85    146    A005900
   4 | 1   8   33    96    225    456    A014820
   5 | 1  10   51   180    501   1182    A069038
   6 | 1  12   73   304    985   2668    A069039
   7 | 1  14   99   476   1765   5418    A099193
		

Crossrefs

Cf. A008288 (Delannoy numbers), A005900 (row 3), A014820 (row 4), A069038 (row 5), A069039 (row 6), A099193 (row 7), A099195 (row 8), A099196 (row 9), A099197 (row 10), A101603, A104698 (triangle version), A142977, A142983.

Programs

  • Haskell
    a142978 n k = a142978_tabl !! (n-1) !! (k-1)
    a142978_row n = a142978_tabl !! (n-1)
    a142978_tabl = map reverse a104698_tabl
    -- Reinhard Zumkeller, Jul 17 2015
  • Maple
    with(combinat): T:=(n,k) -> add(binomial(n-1,i)*binomial(k+i,n),i = 0..n-1); for n from 1 to 10 do seq(T(n,k),k = 1..10) end do; # Program restored by Peter Bala, Oct 02 2008
    A := (n, k) -> k*hypergeom([1 - n, 1 - k], [2], 2):
    seq(print(seq(simplify(A(n, k)), k = 1..9)), n=1..7); # Peter Luschny, Mar 23 2023
  • Mathematica
    t[n_, k_] := Sum[ Binomial[n-1, i]*Binomial[k+i, n], {i, 0, n-1}]; Table[t[n-k, k], {n, 1, 11}, {k, 1, n-1}] // Flatten (* Jean-François Alcover, Mar 06 2013 *)

Formula

T(n,k) = Sum_{i = 0..n-1} C(n-1,i)*C(k+i,n).
Reciprocity law: n*T(n,k) = k*T(k,n).
Recurrence relation: T(n,1) = 1, T(1,k) = k, T(n,k) = T(n,k-1) + T(n-1,k-1) + T(n-1,k), n,k > 1.
O.g.f. row n: x*(1 + x)^(n-1)/(1 - x)^(n+1).
O.g.f. for array: Sum_{n >= 1, k >= 1} T(n, k)*x^k*y^n = x*y/((1 - x)*(1 - x - y - x*y)).
The n-th row entries are the values [p_n(k)], k >= 1, of the polynomial function p_n(x) = Sum_{k = 1..n} 2^(k-1)*C(n-1,k-1)*C(x,k). The first few values are p_1(x) = x, p_2(x) = x^2, p_3(x) = (2*x^3 + x)/3 and p_4(x) = (x^4 + 2*x^2)/3.
The polynomial p_n(x) is the unique polynomial solution of the difference equation x*( f(x+1) - f(x-1) ) = 2*n*f(x), normalized so that f(1) = 1.
The o.g.f. for the p_n(x) is 1/2*((1 + t)/(1 - t))^x = 1/2 + x*t + x^2*t^2 + (2*x^3 + x)/3*t^3 + .... Thus p_n(x) is, apart from a constant factor, the Meixner polynomial of the first kind M_n(x;b,c) at b = 0, c = -1, also known as a Mittag-Leffler polynomial.
The entries in the n-th row appear in the series acceleration formula for the constant log(2): Sum_{k >= 1} (-1)^(k+1)/(T(n,k)*T(n,k+1)) = 1 + (-1)^(n+1) * (2*n)*(log(2) - (1 - 1/2 + 1/3 - ... + (-1)^(n+1)/n)). For example, n = 3 gives log(2) = 4/6 + (1/6)*(1/(1*6) - 1/(6*19) + 1/(19*44) - 1/(44*85) + ...). See A142983 for further details.
From Peter Bala, Oct 02 2008: (Start)
The odd-indexed columns of this array form the array A142992 of crystal ball sequences for lattices of type C_n.
Conjectural congruences for main diagonal entries: Put A(n) = T(n,n). Calculation suggests the following congruences: for prime p > 3 and m, r >= 1, A(m*p^r) == A(m*p^(r-1)) (mod p^(3*r));
Sum_{k = 0..p-1} A(k)^2 == 0 (mod p) if p is a prime of the form 8*n+1 or 8*n+7;
Sum_{k = 0..p-1} A(k)^2 == -1 (mod p) if p is a prime of the form 8*n+3 or 8*n+5.
(End)
From Peter Bala, Sep 27 2021: (Start)
T(n,k) = (1/2)*Sum_{i = 0..k} binomial(k,i)*binomial(n+k-1-i,k-1).
T(n,k) = (1/2)*[x^n] ((1+x)/(1-x))^k = (1/2)*(k/n)*[x^k] ((1+x)/(1-x))^n.
n*T(n,k) = 2*k*T(n-1,k) + (n - 2)*T(n-2,k). (End)
A(n,k) = k*hypergeom([1 - n, 1 - k], [2], 2). - Peter Luschny, Mar 23 2023
T(n,k) = 2*(Sum_{j=1..k-1} T(n-1,j)) + T(n-1,k) for n > 1. - Robert FERREOL, Jun 25 2024

A069038 Expansion of g.f. x*(1+x)^4/(1-x)^6.

Original entry on oeis.org

0, 1, 10, 51, 180, 501, 1182, 2471, 4712, 8361, 14002, 22363, 34332, 50973, 73542, 103503, 142544, 192593, 255834, 334723, 432004, 550725, 694254, 866295, 1070904, 1312505, 1595906, 1926315, 2309356, 2751085, 3258006, 3837087, 4495776, 5242017, 6084266
Offset: 0

Views

Author

Vladeta Jovovic, Apr 03 2002

Keywords

Comments

Hyun Kwang Kim asserts that every nonnegative integer can be represented by the sum of no more than 14 of these numbers. - Jonathan Vos Post, Nov 16 2004
If Y_i (i=1,2,3,4) are 2-blocks of a (n+4)-set X then a(n-4) is the number of 9-subsets of X intersecting each Y_i (i=1,2,3,4). - Milan Janjic, Oct 28 2007
Starting with 1 = binomial transform of [1, 9, 32, 56, 48, 16, 0, 0, 0, ...], where (1, 9, 32, 56, 48, 16) = row 5 of the Chebyshev triangle A081277. Also = row 5 of the array in A142978. - Gary W. Adamson, Jul 19 2008
Starting with the term 1 this is the self-convolution of A001844(n). - Anton Zakharov, Sep 02 2016

References

  • H. S. M. Coxeter, Regular Polytopes, New York: Dover Publications, 1973.

Crossrefs

Programs

  • Magma
    [n*(2*n^4 + 10*n^2 + 3)/15: n in [0..40]]; // Vincenzo Librandi, May 22 2011
    
  • Maple
    al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n), r=0..d-1); end; [seq(be(5,n),n=0..100)];
  • Mathematica
    CoefficientList[Series[x (1 + x)^4/(1 - x)^6, {x, 0, 32}], x] (* Michael De Vlieger, Sep 02 2016 *)
    LinearRecurrence[{6,-15,20,-15,6,-1},{0,1,10,51,180,501},40] (* Harvey P. Dale, Jun 19 2021 *)
  • PARI
    concat(0, Vec(x*(1+x)^4/(1-x)^6 + O(x^99))) \\ Altug Alkan, Sep 02 2016
    
  • PARI
    {a(n) = n * (2*n^4 + 10*n^2 + 3) / 15}; /* Michael Somos, Jun 17 2018 */

Formula

Recurrence: a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6).
a(n) = n*(2*n^4 + 10*n^2 + 3)/15. - Jonathan Vos Post, Nov 16 2004
a(n) = C(n+4,5) + 4*C(n+3,5) + 6*C(n+2,5) + 4*C(n+1,5) + C(n,5).
Sum_{n>=1} 1/((1/15)*n*(2*n^4 + 10*n^2 + 3)*n!) = hypergeom([1, 1, 1+i*sqrt(10-2*sqrt(19))*(1/2), 1-i*sqrt(10-2*sqrt(19))*(1/2), 1+i*sqrt(10+2*sqrt(19))*(1/2), 1-i*sqrt(10+2*sqrt(19))*(1/2)], [2, 2, 2+i*sqrt(10-2*sqrt(19))*(1/2), 2-i*sqrt(10-2*sqrt(19))*(1/2), 2+i*sqrt(10+2*sqrt(19))*(1/2), 2-i*sqrt(10+2*sqrt(19))*(1/2)], 1) = 1.05351734968093116819345664995829700099916... - Stephen Crowley, Jul 14 2009
a(n) = a(n-1) + A014820(n-1) + A014820(n-2). - Bruce J. Nicholson, Apr 18 2018
a(n) = 10*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018
Euler transform of length 2 sequence [10, -4]. - Michael Somos, Jun 19 2018
Sum_{k >= 1} (-1)^k/(a(k)*a(k+1)) = 10*log(2) - 41/6 = 1/(10 + 2/(10 + 6/(10 + ... + n*(n-1)/(10 + ...)))). See A142983. Cf. A005900 and A014820. - Peter Bala, Mar 08 2024
E.g.f.: exp(x)*x*(15 + 60*x + 60*x^2 + 20*x^3 + 2*x^4)/15. - Stefano Spezia, Mar 10 2024

A069039 Expansion of g.f. x*(1+x)^5/(1-x)^7.

Original entry on oeis.org

0, 1, 12, 73, 304, 985, 2668, 6321, 13504, 26577, 48940, 85305, 142000, 227305, 351820, 528865, 774912, 1110049, 1558476, 2149033, 2915760, 3898489, 5143468, 6704017, 8641216, 11024625, 13933036, 17455257, 21690928, 26751369, 32760460, 39855553, 48188416, 57926209
Offset: 0

Views

Author

Vladeta Jovovic, Apr 03 2002

Keywords

Comments

Figurate numbers based on the 6-dimensional regular convex polytope called the 6-dimensional cross-polytope, or 6-dimensional hyperoctahedron, which is represented by the Schlaefli symbol {3, 3, 3, 3, 4}. It is the dual of the 6-dimensional hypercube. Kim asserts that every nonnegative integer can be represented by the sum of no more than 19 of these 6-crosspolytope numbers. - Jonathan Vos Post, Nov 16 2004
Starting with 1 = binomial transform of [1, 11, 50, 120, 160, 112, 32, 0, 0, 0, ...] where (1, 11, 50, 120, 160, 112, 32) = row 6 of the Chebyshev triangle A081277. Also = row 6 of the array in A142978. - Gary W. Adamson, Jul 19 2008

References

  • H. S. M. Coxeter, Regular Polytopes, New York: Dover, 1973.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 240.
  • Jonathan Vos Post, "4-Dimensional Jonathan numbers: polytope numbers and Centered polytope numbers of Higher Than 3 Dimensions", Draft 1.5 of 9 a.m., 12 March 2004, circulated by e-mail.

Crossrefs

Similar sequence: A005900 (m=3), A014820(n-1) (m=4), A069038 (m=5), A099193 (m=7), A099195 (m=8), A099196 (m=9), A099197 (m=10).
Cf. A000332.

Programs

  • Maple
    al:=proc(s,n) binomial(n+s-1,s); end; be:=proc(d,n) local r; add( (-1)^r*binomial(d-1,r)*2^(d-1-r)*al(d-r,n), r=0..d-1); end; [seq(be(6,n),n=0..100)];
  • Mathematica
    a[n_] := n^2*(2*n^4 + 20*n^2 + 23)/45; Table[a[n], {n, 0, 30}] (* Jean-François Alcover, Jan 29 2014 *)
    CoefficientList[Series[x (1+x)^5/(1-x)^7,{x,0,40}],x] (* or *) LinearRecurrence[ {7,-21,35,-35,21,-7,1},{0,1,12,73,304,985,2668},40] (* Harvey P. Dale, Aug 05 2018 *)
  • PARI
    x='x+O('x^100); concat(0, Vec(x*(1+x)^5/(1-x)^7)) \\ Altug Alkan, Dec 14 2015

Formula

Recurrence: a(n) = 7*a(n-1) - 21*a(n-2) + 35*a(n-3) - 35*a(n-4) + 21*a(n-5) - 7*a(n-6) + a(n-7).
a(n) = (n^2)*(2*n^4 + 20*n^2 + 23 )/45. - Jonathan Vos Post, Nov 16 2004
From Stephen Crowley, Jul 14 2009: (Start)
Sum_{n >= 1} 1/a(n) = -5*(Sum(_alpha*(77*_alpha^2+655)*Psi(1-_alpha), _alpha = RootOf(2*_Z^4+20*_Z^2+23)))*(1/3174)+15*Pi^2*(1/46)=1.10203455013915915542552577192042916250524...
Sum_{n>=1} 1/(a(n)*n!) = hypergeom([1, 1, 1, 1-a, 1+b, 1-b, 1+a], [2, 2, 2, 2+b, 2-b, 2+a, 2-a], 1) = 1.04409584723862654376639417281585634150689... where a = (i/2)*sqrt(20+6*sqrt(6)), b = (i/2)*sqrt(20-6*sqrt(6)), and i = sqrt(-1). (End)
a(n) = 12*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018
E.g.f.: exp(x)*x*(45 + 225*x + 300*x^2 + 150*x^3 + 30*x^4 + 2*x^5)/45. - Stefano Spezia, Mar 10 2024
Showing 1-10 of 19 results. Next