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 29 results. Next

A000583 Fourth powers: a(n) = n^4.

Original entry on oeis.org

0, 1, 16, 81, 256, 625, 1296, 2401, 4096, 6561, 10000, 14641, 20736, 28561, 38416, 50625, 65536, 83521, 104976, 130321, 160000, 194481, 234256, 279841, 331776, 390625, 456976, 531441, 614656, 707281, 810000, 923521, 1048576, 1185921
Offset: 0

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Author

Keywords

Comments

Figurate numbers based on 4-dimensional regular convex polytope called the 4-measure polytope, 4-hypercube or tesseract with Schlaefli symbol {4,3,3}. - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004
Totally multiplicative sequence with a(p) = p^4 for prime p. - Jaroslav Krizek, Nov 01 2009
The binomial transform yields A058649. The inverse binomial transforms yields the (finite) 0, 1, 14, 36, 24, the 4th row in A019538 and A131689. - R. J. Mathar, Jan 16 2013
Generate Pythagorean triangles with parameters a and b to get sides of lengths x = b^2-a^2, y = 2*a*b, and z = a^2 + b^2. In particular use a=n-1 and b=n for a triangle with sides (x1,y1,z1) and a=n and b=n+1 for another triangle with sides (x2,y2,z2). Then x1*x2 + y1*y2 + z1*z2 = 8*a(n). - J. M. Bergot, Jul 22 2013
For n > 0, a(n) is the largest integer k such that k^4 + n is a multiple of k + n. Also, for n > 0, a(n) is the largest integer k such that k^2 + n^2 is a multiple of k + n^2. - Derek Orr, Sep 04 2014
Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017
a(n+2)/2 is the area of a trapezoid with vertices at (T(n), T(n+1)), (T(n+1), T(n)), (T(n+1), T(n+2)), and (T(n+2), T(n+1)) with T(n)=A000292(n) for n >= 0. - J. M. Bergot, Feb 16 2018

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 64.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 255; 2nd. ed., p. 269. Worpitzky's identity (6.37).
  • Dov Juzuk, Curiosa 56: An interesting observation, Scripta Mathematica 6 (1939), 218.
  • 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, Page 47.

Crossrefs

Programs

Formula

a(n) = A123865(n)+1 = A002523(n)-1.
Multiplicative with a(p^e) = p^(4e). - David W. Wilson, Aug 01 2001
G.f.: x*(1 + 11*x + 11*x^2 + x^3)/(1 - x)^5. More generally, g.f. for n^m is Euler(m, x)/(1-x)^(m+1), where Euler(m, x) is Eulerian polynomial of degree m (cf. A008292).
Dirichlet generating function: zeta(s-4). - Franklin T. Adams-Watters, Sep 11 2005
E.g.f.: (x + 7*x^2 + 6*x^3 + x^4)*e^x. More generally, the general form for the e.g.f. for n^m is phi_m(x)*e^x, where phi_m is the exponential polynomial of order n. - Franklin T. Adams-Watters, Sep 11 2005
Sum_{k>0} 1/a(k) = Pi^4/90 = A013662. - Jaume Oliver Lafont, Sep 20 2009
a(n) = C(n+3,4) + 11*C(n+2,4) + 11*C(n+1,4) + C(n,4). [Worpitzky's identity for powers of 4. See, e.g., Graham et al., eq. (6.37). - Wolfdieter Lang, Jul 17 2019]
a(n) = n*A177342(n) - Sum_{i=1..n-1} A177342(i) - (n - 1), with n > 1. - Bruno Berselli, May 07 2010
a(n) + a(n+1) + 1 = 2*A002061(n+1)^2. - Charlie Marion, Jun 13 2013
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) + 24. - Ant King, Sep 23 2013
From Amiram Eldar, Jan 20 2021: (Start)
Sum_{n>=1} (-1)^(n+1)/a(n) = 7*Pi^4/720 (A267315).
Product_{n>=2} (1 - 1/a(n)) = sinh(Pi)/(4*Pi). (End)

A000332 Binomial coefficient binomial(n,4) = n*(n-1)*(n-2)*(n-3)/24.

Original entry on oeis.org

0, 0, 0, 0, 1, 5, 15, 35, 70, 126, 210, 330, 495, 715, 1001, 1365, 1820, 2380, 3060, 3876, 4845, 5985, 7315, 8855, 10626, 12650, 14950, 17550, 20475, 23751, 27405, 31465, 35960, 40920, 46376, 52360, 58905, 66045, 73815, 82251, 91390, 101270, 111930, 123410
Offset: 0

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Author

Keywords

Comments

Number of intersection points of diagonals of convex n-gon where no more than two diagonals intersect at any point in the interior.
Also the number of equilateral triangles with vertices in an equilateral triangular array of points with n rows (offset 1), with any orientation. - Ignacio Larrosa Cañestro, Apr 09 2002. [See Les Reid link for proof. - N. J. A. Sloane, Apr 02 2016] [See Peter Kagey link for alternate proof. - Sameer Gauria, Jul 29 2025]
Start from cubane and attach amino acids according to the reaction scheme that describes the reaction between the active sites. See the hyperlink on chemistry. - Robert G. Wilson v, Aug 02 2002
For n>0, a(n) = (-1/8)*(coefficient of x in Zagier's polynomial P_(2n,n)). (Zagier's polynomials are used by PARI/GP for acceleration of alternating or positive series.)
Figurate numbers based on the 4-dimensional regular convex polytope called the regular 4-simplex, pentachoron, 5-cell, pentatope or 4-hypertetrahedron with Schlaefli symbol {3,3,3}. a(n)=((n*(n-1)*(n-2)*(n-3))/4!). - Michael J. Welch (mjw1(AT)ntlworld.com), Apr 01 2004, R. J. Mathar, Jul 07 2009
Maximal number of crossings that can be created by connecting n vertices with straight lines. - Cameron Redsell-Montgomerie (credsell(AT)uoguelph.ca), Jan 30 2007
If X is an n-set and Y a fixed (n-1)-subset of X then a(n) is equal to the number of 4-subsets of X intersecting Y. - Milan Janjic, Aug 15 2007
Product of four consecutive numbers divided by 24. - Artur Jasinski, Dec 02 2007
The only prime in this sequence is 5. - Artur Jasinski, Dec 02 2007
For strings consisting entirely of 0's and 1's, the number of distinct arrangements of four 1's such that 1's are not adjacent. The shortest possible string is 7 characters, of which there is only one solution: 1010101, corresponding to a(5). An eight-character string has 5 solutions, nine has 15, ten has 35 and so on, congruent to A000332. - Gil Broussard, Mar 19 2008
For a(n)>0, a(n) is pentagonal if and only if 3 does not divide n. All terms belong to the generalized pentagonal sequence (A001318). Cf. A000326, A145919, A145920. - Matthew Vandermast, Oct 28 2008
Nonzero terms = row sums of triangle A158824. - Gary W. Adamson, Mar 28 2009
Except for the 4 initial 0's, is equivalent to the partial sums of the tetrahedral numbers A000292. - Jeremy Cahill (jcahill(AT)inbox.com), Apr 15 2009
If the first 3 zeros are disregarded, that is, if one looks at binomial(n+3, 4) with n>=0, then it becomes a 'Matryoshka doll' sequence with alpha=0: seq(add(add(add(i,i=alpha..k),k=alpha..n),n=alpha..m),m=alpha..50). - Peter Luschny, Jul 14 2009
For n>=1, a(n) is the number of n-digit numbers the binary expansion of which contains two runs of 0's. - Vladimir Shevelev, Jul 30 2010
For n>0, a(n) is the number of crossing set partitions of {1,2,..,n} into n-2 blocks. - Peter Luschny, Apr 29 2011
The Kn3, Ca3 and Gi3 triangle sums of A139600 are related to the sequence given above, e.g., Gi3(n) = 2*A000332(n+3) - A000332(n+2) + 7*A000332(n+1). For the definitions of these triangle sums, see A180662. - Johannes W. Meijer, Apr 29 2011
For n > 3, a(n) is the hyper-Wiener index of the path graph on n-2 vertices. - Emeric Deutsch, Feb 15 2012
Except for the four initial zeros, number of all possible tetrahedra of any size, having the same orientation as the original regular tetrahedron, formed when intersecting the latter by planes parallel to its sides and dividing its edges into n equal parts. - V.J. Pohjola, Aug 31 2012
a(n+3) is the number of different ways to color the faces (or the vertices) of a regular tetrahedron with n colors if we count mirror images as the same.
a(n) = fallfac(n,4)/4! is also the number of independent components of an antisymmetric tensor of rank 4 and dimension n >= 1. Here fallfac is the falling factorial. - Wolfdieter Lang, Dec 10 2015
Does not satisfy Benford's law [Ross, 2012] - N. J. A. Sloane, Feb 12 2017
Number of chiral pairs of colorings of the vertices (or faces) of a regular tetrahedron with n available colors. Chiral colorings come in pairs, each the reflection of the other. - Robert A. Russell, Jan 22 2020
From Mircea Dan Rus, Aug 26 2020: (Start)
a(n+3) is the number of lattice rectangles (squares included) in a staircase of order n; this is obtained by stacking n rows of consecutive unit lattice squares, aligned either to the left or to the right, which consist of 1, 2, 3, ..., n squares and which are stacked either in the increasing or in the decreasing order of their lengths. Below, there is a staircase or order 4 which contains a(7) = 35 rectangles. [See the Teofil Bogdan and Mircea Dan Rus link, problem 3, under A004320]
_
||
|||_
|||_|_
|||_|_|
(End)
a(n+4) is the number of strings of length n on an ordered alphabet of 5 letters where the characters in the word are in nondecreasing order. E.g., number of length-2 words is 15: aa,ab,ac,ad,ae,bb,bc,bd,be,cc,cd,ce,dd,de,ee. - Jim Nastos, Jan 18 2021
From Tom Copeland, Jun 07 2021: (Start)
Aside from the zeros, this is the fifth diagonal of the Pascal matrix A007318, the only nonvanishing diagonal (fifth) of the matrix representation IM = (A132440)^4/4! of the differential operator D^4/4!, when acting on the row vector of coefficients of an o.g.f., or power series.
M = e^{IM} is the matrix of coefficients of the Appell sequence p_n(x) = e^{D^4/4!} x^n = e^{b. D} x^n = (b. + x)^n = Sum_{k=0..n} binomial(n,k) b_n x^{n-k}, where the (b.)^n = b_n have the e.g.f. e^{b.t} = e^{t^4/4!}, which is that for A025036 aerated with triple zeros, the first column of M.
See A099174 and A000292 for analogous relationships for the third and fourth diagonals of the Pascal matrix. (End)
For integer m and positive integer r >= 3, the polynomial a(n) + a(n + m) + a(n + 2*m) + ... + a(n + r*m) in n has its zeros on the vertical line Re(n) = (3 - r*m)/2 in the complex plane. - Peter Bala, Jun 02 2024

Examples

			a(5) = 5 from the five independent components of an antisymmetric tensor A of rank 4 and dimension 5, namely A(1,2,3,4), A(1,2,3,5), A(1,2,4,5), A(1,3,4,5) and A(2,3,4,5). See the Dec 10 2015 comment. - _Wolfdieter Lang_, Dec 10 2015
		

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. 828.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 196.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 74, Problem 8.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See p. 70.
  • L. E. Dickson, History of the Theory of Numbers. Carnegie Institute Public. 256, Washington, DC, Vol. 1, 1919; Vol. 2, 1920; Vol. 3, 1923, see vol. 2, p. 7.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 294.
  • J. C. P. Miller, editor, Table of Binomial Coefficients. Royal Society Mathematical Tables, Vol. 3, Cambridge Univ. Press, 1954.
  • 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).
  • Charles W. Trigg, Mathematical Quickies, New York: Dover Publications, Inc., 1985, p. 53, #191.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 127.

Crossrefs

binomial(n, k): A161680 (k = 2), A000389 (k = 5), A000579 (k = 6), A000580 (k = 7), A000581 (k = 8), A000582 (k = 9).
Cf. A000217, A000292, A007318 (column k = 4).
Cf. A158824.
Cf. A006008 (Number of ways to color the faces (or vertices) of a regular tetrahedron with n colors when mirror images are counted as two).
Cf. A104712 (third column, k=4).
See A269747 for a 3-D analog.
Cf. A006008 (oriented), A006003 (achiral) tetrahedron colorings.
Row 3 of A325000, col. 4 of A007318.

Programs

  • GAP
    A000332 := List([1..10^2], n -> Binomial(n, 4)); # Muniru A Asiru, Oct 16 2017
    
  • Magma
    [Binomial(n,4): n in [0..50]]; // Vincenzo Librandi, Nov 23 2014
    
  • Maple
    A000332 := n->binomial(n,4); [seq(binomial(n,4), n=0..100)];
  • Mathematica
    Table[ Binomial[n, 4], {n, 0, 45} ] (* corrected by Harvey P. Dale, Aug 22 2011 *)
    Table[(n-4)(n-3)(n-2)(n-1)/24, {n, 100}] (* Artur Jasinski, Dec 02 2007 *)
    LinearRecurrence[{5,-10,10,-5,1}, {0,0,0,0,1}, 45] (* Harvey P. Dale, Aug 22 2011 *)
    CoefficientList[Series[x^4 / (1 - x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Nov 23 2014 *)
  • PARI
    a(n)=binomial(n,4);
    
  • Python
    # Starts at a(3), i.e. computes n*(n+1)*(n+2)*(n+3)/24
    # which is more in line with A000217 and A000292.
    def A000332():
        x, y, z, u = 1, 1, 1, 1
        yield 0
        while True:
            yield x
            x, y, z, u = x + y + z + u + 1, y + z + u + 1, z + u + 1, u + 1
    a = A000332(); print([next(a) for i in range(41)]) # Peter Luschny, Aug 03 2019
    
  • Python
    print([n*(n-1)*(n-2)*(n-3)//24 for n in range(50)])
    # Gennady Eremin, Feb 06 2022

Formula

a(n) = n*(n-1)*(n-2)*(n-3)/24.
G.f.: x^4/(1-x)^5. - Simon Plouffe in his 1992 dissertation
a(n) = n*a(n-1)/(n-4). - Benoit Cloitre, Apr 26 2003, R. J. Mathar, Jul 07 2009
a(n) = Sum_{k=1..n-3} Sum_{i=1..k} i*(i+1)/2. - Benoit Cloitre, Jun 15 2003
Convolution of natural numbers {1, 2, 3, 4, ...} and A000217, the triangular numbers {1, 3, 6, 10, ...}. - Jon Perry, Jun 25 2003
a(n) = A110555(n+1,4). - Reinhard Zumkeller, Jul 27 2005
a(n+1) = ((n^5-(n-1)^5) - (n^3-(n-1)^3))/24 - (n^5-(n-1)^5-1)/30; a(n) = A006322(n-2)-A006325(n-1). - Xavier Acloque, Oct 20 2003; R. J. Mathar, Jul 07 2009
a(4*n+2) = Pyr(n+4, 4*n+2) where the polygonal pyramidal numbers are defined for integers A>2 and B>=0 by Pyr(A, B) = B-th A-gonal pyramid number = ((A-2)*B^3 + 3*B^2 - (A-5)*B)/6; For all positive integers i and the pentagonal number function P(x) = x*(3*x-1)/2: a(3*i-2) = P(P(i)) and a(3*i-1) = P(P(i) + i); 1 + 24*a(n) = (n^2 + 3*n + 1)^2. - Jonathan Vos Post, Nov 15 2004
First differences of A000389(n). - Alexander Adamchuk, Dec 19 2004
For n > 3, the sum of the first n-2 tetrahedral numbers (A000292). - Martin Steven McCormick (mathseq(AT)wazer.net), Apr 06 2005 [Corrected by Doug Bell, Jun 25 2017]
Starting (1, 5, 15, 35, ...), = binomial transform of [1, 4, 6, 4, 1, 0, 0, 0, ...]. - Gary W. Adamson, Dec 28 2007
Sum_{n>=4} 1/a(n) = 4/3, from the Taylor expansion of (1-x)^3*log(1-x) in the limit x->1. - R. J. Mathar, Jan 27 2009
A034263(n) = (n+1)*a(n+4) - Sum_{i=0..n+3} a(i). Also A132458(n) = a(n)^2 - a(n-1)^2 for n>0. - Bruno Berselli, Dec 29 2010
a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5); a(0)=0, a(1)=0, a(2)=0, a(3)=0, a(4)=1. - Harvey P. Dale, Aug 22 2011
a(n) = (binomial(n-1,2)^2 - binomial(n-1,2))/6. - Gary Detlefs, Nov 20 2011
a(n) = Sum_{k=1..n-2} Sum_{i=1..k} i*(n-k-2). - Wesley Ivan Hurt, Sep 25 2013
a(n) = (A000217(A000217(n-2) - 1))/3 = ((((n-2)^2 + (n-2))/2)^2 - (((n-2)^2 + (n-2))/2))/(2*3). - Raphie Frank, Jan 16 2014
Sum_{n>=0} a(n)/n! = e/24. Sum_{n>=3} a(n)/(n-3)! = 73*e/24. See A067764 regarding the second ratio. - Richard R. Forberg, Dec 26 2013
Sum_{n>=4} (-1)^(n+1)/a(n) = 32*log(2) - 64/3 = A242023 = 0.847376444589... . - Richard R. Forberg, Aug 11 2014
4/(Sum_{n>=m} 1/a(n)) = A027480(m-3), for m>=4. - Richard R. Forberg, Aug 12 2014
E.g.f.: x^4*exp(x)/24. - Robert Israel, Nov 23 2014
a(n+3) = C(n,1) + 3*C(n,2) + 3*C(n,3) + C(n,4). Each term indicates the number of ways to use n colors to color a tetrahedron with exactly 1, 2, 3, or 4 colors.
a(n) = A080852(1,n-4). - R. J. Mathar, Jul 28 2016
From Gary W. Adamson, Feb 06 2017: (Start)
G.f.: Starting (1, 5, 14, ...), x/(1-x)^5 can be written
as (x * r(x) * r(x^2) * r(x^4) * r(x^8) * ...) where r(x) = (1+x)^5;
as (x * r(x) * r(x^3) * r(x^9) * r(x^27) * ...) where r(x) = (1+x+x^2)^5;
as (x * r(x) * r(x^4) * r(x^16) * r(x^64) * ...) where r(x) = (1+x+x^2+x^3)^5;
... (as a conjectured infinite set). (End)
From Robert A. Russell, Jan 22 2020: (Start)
a(n) = A006008(n) - a(n+3) = (A006008(n) - A006003(n)) / 2 = a(n+3) - A006003(n).
a(n+3) = A006008(n) - a(n) = (A006008(n) + A006003(n)) / 2 = a(n) + A006003(n).
a(n) = A007318(n,4).
a(n+3) = A325000(3,n). (End)
Product_{n>=5} (1 - 1/a(n)) = cosh(sqrt(15)*Pi/2)/(100*Pi). - Amiram Eldar, Jan 21 2021

Extensions

Some formulas that referred to another offset corrected by R. J. Mathar, Jul 07 2009

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

A001845 Centered octahedral numbers (crystal ball sequence for cubic lattice).

Original entry on oeis.org

1, 7, 25, 63, 129, 231, 377, 575, 833, 1159, 1561, 2047, 2625, 3303, 4089, 4991, 6017, 7175, 8473, 9919, 11521, 13287, 15225, 17343, 19649, 22151, 24857, 27775, 30913, 34279, 37881, 41727, 45825, 50183, 54809, 59711, 64897, 70375, 76153, 82239
Offset: 0

Views

Author

Keywords

Comments

Number of points in simple cubic lattice at most n steps from origin.
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 6-subsets of X intersecting each Y_i (i=1,2,3). - Milan Janjic, Aug 26 2007
Equals binomial transform of [1, 6, 12, 8, 0, 0, 0, ...] where (1, 6, 12, 8) = row 3 of the Chebyshev triangle A013609. - Gary W. Adamson, Jul 19 2008
Let A be the Hessenberg matrix of order n, defined by: A[1,j]=1, A[i,i]:=2,(i>1), A[i,i-1]=-1, and A[i,j]=0 otherwise. Then, for n >= 4, a(n-2) = -coeff(charpoly(A,x),x^(n-3)). - Milan Janjic, Jan 26 2010
a(n) = A005408(n) * A097080(n-1) / 3. - Reinhard Zumkeller, Dec 15 2013
a(n) = D(3,n) where D are the Delannoy numbers (A008288). As such, a(n) gives the number of grid paths from (0,0) to (3,n) using steps that move one unit north, east, or northeast. - David Eppstein, Sep 07 2014
The first comment above can be re-expressed and generalized as follows: a(n) is the number of points in Z^3 that are L1 (Manhattan) distance <= n from any given point. Equivalently, due to a symmetry that is easier to see in the Delannoy numbers array (A008288), as a special case of Dmitry Zaitsev's Dec 10 2015 comment on A008288, a(n) is the number of points in Z^n that are L1 (Manhattan) distance <= 3 from any given point. - Shel Kaphan, Jan 02 2023

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 81.
  • 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

Sums of 2 consecutive terms give A008412.
(1/12)*t*(2*n^3 - 3*n^2 + n) + 2*n - 1 for t = 2, 4, 6, ... gives A049480, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.
Partial sums of A005899.
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/column 3 of A008288.

Programs

Formula

G.f.: (1+x)^3 /(1-x)^4. [conjectured (correctly) by Simon Plouffe in his 1992 dissertation]
a(n) = (2*n+1)*(2*n^2 + 2*n + 3)/3.
First differences of A014820(n). - Alexander Adamchuk, May 23 2006
a(n) = a(n-1) + 4*n^2 + 2, a(0)=1. - Vincenzo Librandi, Mar 27 2011
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4), with a(0)=1, a(1)=7, a(2)=25, a(3)=63. - Harvey P. Dale, Jun 05 2013
a(n) = Sum_{k=0..min(3,n)} 2^k * binomial(3,k) * binomial(n,k). See Bump et al. - Tom Copeland, Sep 05 2014
From Luciano Ancora, Jan 08 2015: (Start)
a(n) = 2 * A000330(n) + A000330(n+1) + A000330(n-1).
a(n) = A005900(n) + A005900(n+1).
a(n) = A005900(n) + A000330(n) + A000330(n+1).
a(n) = A000330(n-1) + A000330(n) + A005900(n+1). (End)
a(n) = A002412(n+1) + A016061(n-1) for n > 0. - Bruce J. Nicholson, Nov 12 2017
E.g.f.: exp(x)*(3 + 18*x + 18*x^2 + 4*x^3)/3. - Stefano Spezia, Mar 14 2024
Sum_{n >= 1} (-1)^(n+1)/(n*a(n-1)*a(n)) = 5/6 - log(2) = (1 - 1/2 + 1/3) - log(2). - Peter Bala, Mar 21 2024

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

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

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
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