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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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

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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

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

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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

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

A099193 a(n) = n*(4*n^6 + 70*n^4 + 196*n^2 + 45)/315.

Original entry on oeis.org

0, 1, 14, 99, 476, 1765, 5418, 14407, 34232, 74313, 149830, 284075, 511380, 880685, 1459810, 2340495, 3644272, 5529233, 8197758, 11905267, 16970060, 23784309, 32826266, 44673751, 60018984, 79684825, 104642486, 136030779, 175176964, 223619261, 283131090, 355747103
Offset: 0

Views

Author

Jonathan Vos Post, Nov 16 2004

Keywords

Comments

Kim asserts that every nonnegative integer can be represented by the sum of no more than 21 of these numbers.
Starting with 1 = binomial transform of [1, 13, 72, 220, 400, 432, 256, 0, 0, 0, ...], where (1, 13, 72, 220, 400, 432, 256) = row 7 of the Chebyshev triangle A081277. Also = row 7 of the array in A142978. - Gary W. Adamson, Jul 19 2008

Crossrefs

Similar sequences: A005900 (m=3), A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099195 (m=8), A099196 (m=9), A099197 (m=10).
Cf. A000332.

Programs

  • Mathematica
    Table[SeriesCoefficient[x (1 + x)^6/(1 - x)^8, {x, 0, n}], {n, 0, 31}] (* Michael De Vlieger, Dec 14 2015 *)
  • PARI
    concat(0, Vec(x*(1+x)^6/(1-x)^8 + O(x^40))) \\ Michel Marcus, Dec 14 2015

Formula

a(n) = n*(4*n^6 + 70*n^4 + 196*n^2 + 45)/315.
G.f.: x*(1+x)^6/(1-x)^8. - R. J. Mathar, Jul 18 2009
a(n) = 14*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018

Extensions

More terms from Michel Marcus, Dec 14 2015

A099196 a(n) = n*(2*n^8 + 84*n^6 + 798*n^4 + 1636*n^2 + 315)/2835.

Original entry on oeis.org

0, 1, 18, 163, 996, 4645, 17718, 57799, 166344, 432073, 1030490, 2286955, 4772780, 9446125, 17852030, 32398735, 56730512, 96220561, 158611106, 254831667, 400030580, 614859189, 927052742, 1373356887, 2001853784, 2874747225, 4071671786, 5693596923, 7867403068, 10751213181
Offset: 0

Views

Author

Jonathan Vos Post, Nov 16 2004

Keywords

Crossrefs

Similar sequences: A005900 (m=3), A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193 (m=7), A099195 (m=8), A099197 (m=10).
Cf. A000332.

Programs

  • PARI
    concat(0, Vec(x*(1+x)^8/(1-x)^10 + O(x^40))) \\ Michel Marcus, Dec 14 2015

Formula

a(n) = n*(2*n^8 + 84*n^6 + 798*n^4 + 1636*n^2 + 315)/2835.
G.f.: x*(1+x)^8/(1-x)^10. [Colin Barker, May 01 2012]
a(n) = 18*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018

Extensions

More terms from Michel Marcus, Dec 14 2015

A006976 Coefficients of Chebyshev T polynomials: a(n) = A053120(n+12, n), n >= 0.

Original entry on oeis.org

1, 13, 98, 560, 2688, 11424, 44352, 160512, 549120, 1793792, 5637632, 17145856, 50692096, 146227200, 412778496, 1143078912, 3111714816, 8341487616, 22052208640, 57567870976, 148562247680, 379364311040, 959384125440
Offset: 0

Views

Author

Keywords

Comments

Binomial transform of A069039. - Paul Barry, Feb 19 2003
If X_1, X_2, ..., X_n are 2-blocks of a (2n+1)-set X then, for n >= 5, a(n-5) is the number of (n+6)-subsets of X intersecting each X_i, (i = 1, 2, ..., n). - Milan Janjic, Nov 18 2007

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 795.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

a(n) = A039991(n+12, 12), A053120.
Partial sums are in A002409.

Programs

  • GAP
    List([0..25], n-> 2^(n-1)*Binomial(n+5,5)*(n+12)/6); # G. C. Greubel, Aug 27 2019
  • Magma
    [2^(n-1)/6*Binomial(n+5,5)*(n+12) : n in [0..25]]; // Brad Clardy, Mar 10 2012
    
  • Maple
    seq(2^(n-1)*binomial(n+5,5)*(n+12)/6, n=0..25); # G. C. Greubel, Aug 27 2019
  • Mathematica
    Table[2^(n-1)*Binomial[n+5,5]*(n+12)/6, {n,0,25}] (* G. C. Greubel, Aug 27 2019 *)
    LinearRecurrence[{14,-84,280,-560,672,-448,128},{1,13,98,560,2688,11424,44352},30] (* Harvey P. Dale, Sep 26 2024 *)
  • PARI
    vector(26, n, 2^(n-2)*binomial(n+4,5)*(n+11)/6) \\ G. C. Greubel, Aug 27 2019
    
  • Sage
    [2^(n-1)*binomial(n+5,5)*(n+12)/6 for n in (0..25)] # G. C. Greubel, Aug 27 2019
    

Formula

G.f.: (1-x)/(1-2*x)^7.
a(n) = 2^n*binomial(n+5, 5)*(n+12)/12. [See a comment in A053120 on subdiagonal sequences. - Wolfdieter Lang, Jan 03 2020]
a(n) = Sum_{k = 0..floor((n+12)/2)} C(n+12,2*k)*C(k,6). - Paul Barry, May 15 2003
E.g.f.: (1/45)*exp(2*x)*(45 + 495*x + 1125*x^2 + 900*x^3 + 300*x^4 + 42*x^5 + 2*x^6). - Stefano Spezia, Jan 03 2020

Extensions

More terms from James Sellers, Aug 21 2000
Name clarified by Wolfdieter Lang, Nov 26 2019

A099195 a(n) = (n^2)*( n^6 + 28*n^4 + 154*n^2 + 132 )/315.

Original entry on oeis.org

0, 1, 16, 129, 704, 2945, 10128, 29953, 78592, 187137, 411280, 845185, 1640640, 3032705, 5373200, 9173505, 15158272, 24331777, 38058768, 58161793, 87037120, 127791489, 184402064, 261902081, 366594816, 506298625, 690625936, 931299201, 1242506944, 1641303169, 2148053520
Offset: 0

Views

Author

Jonathan Vos Post, Nov 16 2004

Keywords

References

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

Crossrefs

Similar sequences: A005900 (m=3), A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193 (m=7), A099196 (m=9), A099197 (m=10).
Cf. A000332.

Programs

  • Mathematica
    LinearRecurrence[{9,-36,84,-126,126,-84,36,-9,1},{0,1,16,129,704,2945,10128,29953,78592},40] (* Harvey P. Dale, Jan 23 2019 *)
  • PARI
    concat(0, Vec(x*(1+x)^7/(1-x)^9 + O(x^40))) \\ Michel Marcus, Dec 14 2015

Formula

a(n) = (n^2)*( n^6 + 28*n^4 + 154*n^2 + 132 )/315.
G.f.: x*(1+x)^7/(1-x)^9. [R. J. Mathar, Jul 18 2009]
a(n) = 16*a(n-1)/(n-1) + a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018

A099197 a(n) = (n^2)*(2*n^8+120*n^6+1806*n^4+7180*n^2+5067)/14175.

Original entry on oeis.org

0, 1, 20, 201, 1360, 7001, 29364, 104881, 329024, 927441, 2390004, 5707449, 12767184, 26986089, 54284244, 104535009, 193664256, 346615329, 601446996, 1014889769, 1669752016, 2684641785, 4226553716, 6526963345, 9902174016, 14778775025, 21725194036, 31490462745
Offset: 0

Views

Author

Jonathan Vos Post, Nov 16 2004

Keywords

Crossrefs

Similar sequences: A005900 (m=3), A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193 (m=7), A099195 (m=8), A099196 (m=9).
Cf. A000332.

Programs

Formula

a(n) = (n^2)*(2*n^8+120*n^6+1806*n^4+7180*n^2+5067)/14175.

A035600 Number of points of L1 norm 6 in cubic lattice Z^n.

Original entry on oeis.org

0, 2, 24, 146, 608, 1970, 5336, 12642, 27008, 53154, 97880, 170610, 284000, 454610, 703640, 1057730, 1549824, 2220098, 3116952, 4298066, 5831520, 7796978, 10286936, 13408034, 17282432, 22049250, 27866072, 34910514, 43381856, 53502738, 65520920, 79711106
Offset: 0

Views

Author

Keywords

Crossrefs

Programs

  • Magma
    [( 4*n^6 +40*n^4 +46*n^2 )/45: n in [0..30]]; // Vincenzo Librandi, Apr 23 2012
  • Maple
    f := proc(d,m) local i; sum( 2^i*binomial(d,i)*binomial(m-1,i-1),i=1..min(d,m)); end; # n=dimension, m=norm
  • Mathematica
    CoefficientList[Series[2*x*(1+x)^5/(1-x)^7,{x,0,33}],x] (* Vincenzo Librandi, Apr 23 2012 *)
  • PARI
    a(n)=(4*n^6+40*n^4+46*n^2)/45 \\ Charles R Greathouse IV, Dec 07 2011
    

Formula

a(n) = (4*n^6 + 40*n^4 + 46*n^2)/45. - Frank Ellermann, Mar 16 2002
G.f.: 2*x*(1+x)^5/(1-x)^7. - Colin Barker, Apr 15 2012
a(n) = 2*A069039(n). - R. J. Mathar, Dec 10 2013

A300624 Figurate numbers based on the 11-dimensional regular convex polytope called the 11-dimensional cross-polytope, or 11-dimensional hyperoctahedron.

Original entry on oeis.org

0, 1, 22, 243, 1804, 10165, 46530, 180775, 614680, 1871145, 5188590, 13286043, 31760676, 71513949, 152784282, 311603535, 609802800, 1150082385, 2098144710, 3714481475, 6399123260, 10753517061, 17664712562, 28418229623, 44847366984, 69528316025, 106032285086
Offset: 0

Views

Author

Keywords

Comments

The 11-dimensional cross-polytope is represented by the Schlaefli symbol {3, 3, 3, 3, 3, 3, 3, 3, 3, 4}. It is the dual of the 11-dimensional hypercube.

Crossrefs

Similar sequences: A005900 (m=3), A014820(n-1) (m=4), A069038 (m=5), A069039 (m=6), A099193 (m=7), A099195 (m=8), A099196 (m=9), A099197 (m=10).

Programs

  • Magma
    [(n*(14175 + 83754*n^2 + 50270*n^4 + 7392*n^6 + 330*n^8 + 4*n^10)) / 155925 : n in [0..40]]; // Wesley Ivan Hurt, Jul 17 2020
  • PARI
    concat(0, Vec(x*(1 + x)^10 / (1 - x)^12 + O(x^40))) \\ Colin Barker, Aug 15 2018
    
  • PARI
    a(n) = (n*(14175 + 83754*n^2 + 50270*n^4 + 7392*n^6 + 330*n^8 + 4*n^10)) / 155925 \\ Colin Barker, Aug 15 2018
    

Formula

a(n) = 11-crosspolytope(n).
From Colin Barker, Aug 15 2018: (Start)
G.f.: x*(1 + x)^10 / (1 - x)^12.
a(n) = (n*(14175 + 83754*n^2 + 50270*n^4 + 7392*n^6 + 330*n^8 + 4*n^10)) / 155925.
(End)
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