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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A000330 Square pyramidal numbers: a(n) = 0^2 + 1^2 + 2^2 + ... + n^2 = n*(n+1)*(2*n+1)/6.

Original entry on oeis.org

0, 1, 5, 14, 30, 55, 91, 140, 204, 285, 385, 506, 650, 819, 1015, 1240, 1496, 1785, 2109, 2470, 2870, 3311, 3795, 4324, 4900, 5525, 6201, 6930, 7714, 8555, 9455, 10416, 11440, 12529, 13685, 14910, 16206, 17575, 19019, 20540, 22140, 23821, 25585, 27434, 29370
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

The sequence contains exactly one square greater than 1, namely 4900 (according to Gardner). - Jud McCranie, Mar 19 2001, Mar 22 2007 [This is a result from Watson. - Charles R Greathouse IV, Jun 21 2013] [See A351830 for further related comments and references.]
Number of rhombi in an n X n rhombus. - Matti De Craene (Matti.DeCraene(AT)rug.ac.be), May 14 2000
Number of acute triangles made from the vertices of a regular n-polygon when n is odd (cf. A007290). - Sen-Peng Eu, Apr 05 2001
Gives number of squares with sides parallel to the axes formed from an n X n square. In a 1 X 1 square, one is formed. In a 2 X 2 square, five squares are formed. In a 3 X 3 square, 14 squares are formed and so on. - Kristie Smith (kristie10spud(AT)hotmail.com), Apr 16 2002; edited by Eric W. Weisstein, Mar 05 2025
a(n-1) = B_3(n)/3, where B_3(x) = x(x-1)(x-1/2) is the third Bernoulli polynomial. - Michael Somos, Mar 13 2004
Number of permutations avoiding 13-2 that contain the pattern 32-1 exactly once.
Since 3*r = (r+1) + r + (r-1) = T(r+1) - T(r-2), where T(r) = r-th triangular number r*(r+1)/2, we have 3*r^2 = r*(T(r+1) - T(r-2)) = f(r+1) - f(r-1) ... (i), where f(r) = (r-1)*T(r) = (r+1)*T(r-1). Summing over n, the right hand side of relation (i) telescopes to f(n+1) + f(n) = T(n)*((n+2) + (n-1)), whence the result Sum_{r=1..n} r^2 = n*(n+1)*(2*n+1)/6 immediately follows. - Lekraj Beedassy, Aug 06 2004
Also as a(n) = (1/6)*(2*n^3 + 3*n^2 + n), n > 0: structured trigonal diamond numbers (vertex structure 5) (cf. A006003 = alternate vertex; A000447 = structured diamonds; A100145 for more on structured numbers). - James A. Record (james.record(AT)gmail.com), Nov 07 2004
Number of triples of integers from {1, 2, ..., n} whose last component is greater than or equal to the others.
Kekulé numbers for certain benzenoids. - Emeric Deutsch, Jun 12 2005
Sum of the first n positive squares. - Cino Hilliard, Jun 18 2007
Maximal number of cubes of side 1 in a right pyramid with a square base of side n and height n. - Pasquale CUTOLO (p.cutolo(AT)inwind.it), Jul 09 2007
If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-3) is the number of 4-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007
We also have the identity 1 + (1+4) + (1+4+9) + ... + (1+4+9+16+ ... + n^2) = n(n+1)(n+2)(n+(n+1)+(n+2))/36; ... and in general the k-fold nested sum of squares can be expressed as n(n+1)...(n+k)(n+(n+1)+...+(n+k))/((k+2)!(k+1)/2). - Alexander R. Povolotsky, Nov 21 2007
The terms of this sequence are coefficients of the Engel expansion of the following converging sum: 1/(1^2) + (1/1^2)*(1/(1^2+2^2)) + (1/1^2)*(1/(1^2+2^2))*(1/(1^2+2^2+3^2)) + ... - Alexander R. Povolotsky, Dec 10 2007
Convolution of A000290 with A000012. - Sergio Falcon, Feb 05 2008
Hankel transform of binomial(2*n-3, n-1) is -a(n). - Paul Barry, Feb 12 2008
Starting (1, 5, 14, 30, ...) = binomial transform of [1, 4, 5, 2, 0, 0, 0, ...]. - Gary W. Adamson, Jun 13 2008
Starting (1,5,14,30,...) = second partial sums of binomial transform of [1,2,0,0,0,...]. a(n) = Sum_{i=0..n} binomial(n+2,i+2)*b(i), where b(i)=1,2,0,0,0,... - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009
Convolution of A001477 with A005408: a(n) = Sum_{k=0..n} (2*k+1)*(n-k). - Reinhard Zumkeller, Mar 07 2009
Sequence of the absolute values of the z^1 coefficients of the polynomials in the GF1 denominators of A156921. See A157702 for background information. - Johannes W. Meijer, Mar 07 2009
The sequence is related to A000217 by a(n) = n*A000217(n) - Sum_{i=0..n-1} A000217(i) and this is the case d = 1 in the identity n^2*(d*n-d+2)/2 - Sum_{i=0..n-1} i*(d*i-d+2)/2 = n*(n+1)(2*d*n-2*d+3)/6, or also the case d = 0 in n^2*(n+2*d+1)/2 - Sum_{i=0..n-1} i*(i+2*d+1)/2 = n*(n+1)*(2*n+3*d+1)/6. - Bruno Berselli, Apr 21 2010, Apr 03 2012
a(n)/n = k^2 (k = integer) for n = 337; a(337) = 12814425, a(n)/n = 38025, k = 195, i.e., the number k = 195 is the quadratic mean (root mean square) of the first 337 positive integers. There are other such numbers -- see A084231 and A084232. - Jaroslav Krizek, May 23 2010
Also the number of moves to solve the "alternate coins game": given 2n+1 coins (n+1 Black, n White) set alternately in a row (BWBW...BWB) translate (not rotate) a pair of adjacent coins at a time (1 B and 1 W) so that at the end the arrangement shall be BBBBB..BW...WWWWW (Blacks separated by Whites). Isolated coins cannot be moved. - Carmine Suriano, Sep 10 2010
From J. M. Bergot, Aug 23 2011: (Start)
Using four consecutive numbers n, n+1, n+2, and n+3 take all possible pairs (n, n+1), (n, n+2), (n, n+3), (n+1, n+2), (n+1, n+3), (n+2, n+3) to create unreduced Pythagorean triangles. The sum of all six areas is 60*a(n+1).
Using three consecutive odd numbers j, k, m, (j+k+m)^3 - (j^3 + k^3 + m^3) equals 576*a(n) = 24^2*a(n) where n = (j+1)/2. (End)
From Ant King, Oct 17 2012: (Start)
For n > 0, the digital roots of this sequence A010888(a(n)) form the purely periodic 27-cycle {1, 5, 5, 3, 1, 1, 5, 6, 6, 7, 2, 2, 9, 7, 7, 2, 3, 3, 4, 8, 8, 6, 4, 4, 8, 9, 9}.
For n > 0, the units' digits of this sequence A010879(a(n)) form the purely periodic 20-cycle {1, 5, 4, 0, 5, 1, 0, 4, 5, 5, 6, 0, 9, 5, 0, 6, 5, 9, 0, 0}. (End)
Length of the Pisano period of this sequence mod n, n>=1: 1, 4, 9, 8, 5, 36, 7, 16, 27, 20, 11, 72, 13, 28, 45, 32, 17, 108, 19, 40, ... . - R. J. Mathar, Oct 17 2012
Sum of entries of n X n square matrix with elements min(i,j). - Enrique Pérez Herrero, Jan 16 2013
The number of intersections of diagonals in the interior of regular n-gon for odd n > 1 divided by n is a square pyramidal number; that is, A006561(2*n+1)/(2*n+1) = A000330(n-1) = (1/6)*n*(n-1)*(2*n-1). - Martin Renner, Mar 06 2013
For n > 1, a(n)/(2n+1) = A024702(m), for n such that 2n+1 = prime, which results in 2n+1 = A000040(m). For example, for n = 8, 2n+1 = 17 = A000040(7), a(8) = 204, 204/17 = 12 = A024702(7). - Richard R. Forberg, Aug 20 2013
A formula for the r-th successive summation of k^2, for k = 1 to n, is (2*n+r)*(n+r)!/((r+2)!*(n-1)!) (H. W. Gould). - Gary Detlefs, Jan 02 2014
The n-th square pyramidal number = the n-th triangular dipyramidal number (Johnson 12), which is the sum of the n-th + (n-1)-st tetrahedral numbers. E.g., the 3rd tetrahedral number is 10 = 1+3+6, the 2nd is 4 = 1+3. In triangular "dipyramidal form" these numbers can be written as 1+3+6+3+1 = 14. For "square pyramidal form", rebracket as 1+(1+3)+(3+6) = 14. - John F. Richardson, Mar 27 2014
Beukers and Top prove that no square pyramidal number > 1 equals a tetrahedral number A000292. - Jonathan Sondow, Jun 21 2014
Odd numbered entries are related to dissections of polygons through A100157. - Tom Copeland, Oct 05 2014
From Bui Quang Tuan, Apr 03 2015: (Start)
We construct a number triangle from the integers 1, 2, 3, ..., n as follows. The first column contains 2*n-1 integers 1. The second column contains 2*n-3 integers 2, ... The last column contains only one integer n. The sum of all the numbers in the triangle is a(n).
Here is an example with n = 5:
1
1 2
1 2 3
1 2 3 4
1 2 3 4 5
1 2 3 4
1 2 3
1 2
1
(End)
The Catalan number series A000108(n+3), offset 0, gives Hankel transform revealing the square pyramidal numbers starting at 5, A000330(n+2), offset 0 (empirical observation). - Tony Foster III, Sep 05 2016; see Dougherty et al. link p. 2. - Andrey Zabolotskiy, Oct 13 2016
Number of floating point additions in the factorization of an (n+1) X (n+1) real matrix by Gaussian elimination as e.g. implemented in LINPACK subroutines sgefa.f or dgefa.f. The number of multiplications is given by A007290. - Hugo Pfoertner, Mar 28 2018
The Jacobi polynomial P(n-1,-n+2,2,3) or equivalently the sum of dot products of vectors from the first n rows of Pascal's triangle (A007318) with the up-diagonal Chebyshev T coefficient vector (1,3,2,0,...) (A053120) or down-diagonal vector (1,-7,32,-120,400,...) (A001794). a(5) = 1 + (1,1).(1,3) + (1,2,1).(1,3,2) + (1,3,3,1).(1,3,2,0) + (1,4,6,4,1).(1,3,2,0,0) = (1 + (1,1).(1,-7) + (1,2,1).(1,-7,32) + (1,3,3,1).(1,-7,32,-120) + (1,4,6,4,1).(1,-7,32,-120,400))*(-1)^(n-1) = 55. - Richard Turk, Jul 03 2018
Coefficients in the terminating series identity 1 - 5*n/(n + 4) + 14*n*(n - 1)/((n + 4)*(n + 5)) - 30*n*(n - 1)*(n - 2)/((n + 4)*(n + 5)*(n + 6)) + ... = 0 for n = 1,2,3,.... Cf. A002415 and A108674. - Peter Bala, Feb 12 2019
n divides a(n) iff n == +- 1 (mod 6) (see A007310). (See De Koninck reference.) Examples: a(11) = 506 = 11 * 46, and a(13) = 819 = 13 * 63. - Bernard Schott, Jan 10 2020
For n > 0, a(n) is the number of ternary words of length n+2 having 3 letters equal to 2 and 0 only occurring as the last letter. For example, for n=2, the length 4 words are 2221,2212,2122,1222,2220. - Milan Janjic, Jan 28 2020
Conjecture: Every integer can be represented as a sum of three generalized square pyramidal numbers. A related conjecture is given in A336205 corresponding to pentagonal case. A stronger version of these conjectures is that every integer can be expressed as a sum of three generalized r-gonal pyramidal numbers for all r >= 3. In here "generalized" means negative indices are included. - Altug Alkan, Jul 30 2020
The natural number y is a term if and only if y = a(floor((3 * y)^(1/3))). - Robert Israel, Dec 04 2024
Also the number of directed bishop moves on an n X n chessboard, where two moves are considered the same if one can be obtained from the other by a rotation of the board. Reflections are ignored. Equivalently, number of directed bishop moves on an n X n chessboard, where two moves are considered the same if one can be obtained from the other by an axial reflection of the board (horizontal or vertical). Rotations and diagonal reflections are ignored. - Hilko Koning, Aug 22 2025

Examples

			G.f. = x + 5*x^2 + 14*x^3 + 30*x^4 + 55*x^5 + 91*x^6 + 140*x^7 + 204*x^8 + ...

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. 813.
  • A. H. Beiler, Recreations in the Theory of Numbers, Dover Publications, NY, 1964, p. 194.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, id. 215,223.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 122, see #19 (3(1)), I(n); p. 155.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 47-49.
  • 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.
  • S. J. Cyvin and I. Gutman, Kekulé structures in benzenoid hydrocarbons, Lecture Notes in Chemistry, No. 46, Springer, New York, 1988 (p.165).
  • J. M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 310, pp. 46-196, Ellipses, Paris, 2004.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 93.
  • 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. 2.
  • M. Gardner, Fractal Music, Hypercards and More, Freeman, NY, 1991, p. 293.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.6 Figurate Numbers, p. 293.
  • M. Holt, Math puzzles and games, Walker Publishing Company, 1977, p. 2 and p. 89.
  • Simon Singh, The Simpsons and Their Mathematical Secrets. London: Bloomsbury Publishing PLC (2013): 188.
  • 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).
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 126.

Links

Crossrefs

Cf. A000217, A000292, A000537, A005408, A006003, A006331, A033994, A033999, A046092, A050409, A050446, A050447, A060493, A100157, A132124, A132112, A156921, A157702, A258708, A351105, A351830.
Sums of 2 consecutive terms give A005900.
Column 0 of triangle A094414.
Column 1 of triangle A008955.
Right side of triangle A082652.
Row 2 of array A103438.
Partial sums of A000290.
Cf. similar sequences listed in A237616 and A254142.
Cf. |A084930(n, 1)|.
Cf. A253903 (characteristic function).
Cf. A034705 (differences of any two terms).

Programs

  • GAP
    List([0..30], n-> n*(n+1)*(2*n+1)/6); # G. C. Greubel, Dec 31 2019
  • Haskell
    a000330 n = n * (n + 1) * (2 * n + 1) `div` 6
a000330_list = scanl1 (+) a000290_list
-- Reinhard Zumkeller, Nov 11 2012, Feb 03 2012
  • Magma
    [n*(n+1)*(2*n+1)/6: n in [0..50]]; // Wesley Ivan Hurt, Jun 28 2014
  • Magma
    [0] cat [((2*n+3)*Binomial(n+2,2))/3: n in [0..40]]; // Vincenzo Librandi, Jul 30 2014
  • Maple
    A000330 := n -> n*(n+1)*(2*n+1)/6;
a := n->(1/6)*n*(n+1)*(2*n+1): seq(a(n),n=0..53); # Emeric Deutsch
with(combstruct): ZL:=[st, {st=Prod(left, right), left=Set(U, card=r), right=Set(U, card=r), U=Sequence(Z, card>=1)}, unlabeled]: subs(r=1, stack): seq(count(subs(r=2, ZL), size=m*2), m=1..45) ; # Zerinvary Lajos, Jan 02 2008
nmax := 44; for n from 0 to nmax do fz(n) := product( (1-(2*m-1)*z)^(n+1-m) , m=1..n); c(n) := abs(coeff(fz(n),z,1)); end do: a := n-> c(n): seq(a(n), n=0..nmax); # Johannes W. Meijer, Mar 07 2009
  • Mathematica
    Table[Binomial[w+2, 3] + Binomial[w+1, 3], {w, 0, 30}]
CoefficientList[Series[x(1+x)/(1-x)^4, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 30 2014 *)
Accumulate[Range[0,50]^2] (* Harvey P. Dale, Sep 25 2014 *)
  • Maxima
    A000330(n):=binomial(n+2,3)+binomial(n+1,3)$
makelist(A000330(n),n,0,20); /* Martin Ettl, Nov 12 2012 */
  • PARI
    {a(n) = n * (n+1) * (2*n+1) / 6};
  • PARI
    upto(n) = [x*(x+1)*(2*x+1)/6 | x<-[0..n]] \\ Cino Hilliard, Jun 18 2007, edited by M. F. Hasler, Jan 02 2024
  • Python
    a=lambda n: (n*(n+1)*(2*n+1))//6 # Indranil Ghosh, Jan 04 2017
  • Sage
    [n*(n+1)*(2*n+1)/6 for n in (0..30)] # G. C. Greubel, Dec 31 2019

Formula

G.f.: x*(1+x)/(1-x)^4. - Simon Plouffe (in his 1992 dissertation: generating function for sequence starting at a(1))
E.g.f.: (x + 3*x^2/2 + x^3/3)*exp(x).
a(n) = n*(n+1)*(2*n+1)/6 = binomial(n+2, 3) + binomial(n+1, 3).
2*a(n) = A006331(n). - N. J. A. Sloane, Dec 11 1999
Can be extended to Z with a(n) = -a(-1-n) for all n in Z.
a(n) = A002492(n)/4. - Paul Barry, Jul 19 2003
a(n) = (((n+1)^4 - n^4) - ((n+1)^2 - n^2))/12. - Xavier Acloque, Oct 16 2003
From Alexander Adamchuk, Oct 26 2004: (Start)
a(n) = sqrt(A271535(n)).
a(n) = (Sum_{k=1..n} Sum_{j=1..n} Sum_{i=1..n} (i*j*k)^2)^(1/3). (End)
a(n) = Sum_{i=1..n} i*(2*n-2*i+1); sum of squares gives 1 + (1+3) + (1+3+5) + ... - Jon Perry, Dec 08 2004
a(n+1) = A000217(n+1) + 2*A000292(n). - Creighton Dement, Mar 10 2005
Sum_{n>=1} 1/a(n) = 6*(3-4*log(2)); Sum_{n>=1} (-1)^(n+1)*1/a(n) = 6*(Pi-3). - Philippe Deléham, May 31 2005
Sum of two consecutive tetrahedral (or pyramidal) numbers a(n) = A000292(n-1) + A000292(n). - Alexander Adamchuk, May 17 2006
Euler transform of length-2 sequence [ 5, -1 ]. - Michael Somos, Sep 04 2006
a(n) = a(n-1) + n^2. - Rolf Pleisch, Jul 22 2007
a(n) = A132121(n,0). - Reinhard Zumkeller, Aug 12 2007
a(n) = binomial(n, 2) + 2*binomial(n, 3). - Borislav St. Borisov (b.st.borisov(AT)abv.bg), Mar 05 2009, corrected by M. F. Hasler, Jan 02 2024
a(n) = A168559(n) + 1 for n > 0. - Reinhard Zumkeller, Feb 03 2012
a(n) = Sum_{i=1..n} J_2(i)*floor(n/i), where J_2 is A007434. - Enrique Pérez Herrero, Feb 26 2012
a(n) = s(n+1, n)^2 - 2*s(n+1, n-1), where s(n, k) are Stirling numbers of the first kind, A048994. - Mircea Merca, Apr 03 2012
a(n) = A001477(n) + A000217(n) + A007290(n+2) + 1. - J. M. Bergot, May 31 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) + 2. - Ant King, Oct 17 2012
a(n) = Sum_{i = 1..n} Sum_{j = 1..n} min(i,j). - Enrique Pérez Herrero, Jan 15 2013
a(n) = A000217(n) + A007290(n+1). - Ivan N. Ianakiev, May 10 2013
a(n) = (A047486(n+2)^3 - A047486(n+2))/24. - Richard R. Forberg, Dec 25 2013
a(n) = Sum_{i=0..n-1} (n-i)*(2*i+1), with a(0) = 0. After 0, row sums of the triangle in A101447. - Bruno Berselli, Feb 10 2014
a(n) = n + 1 + Sum_{i=1..n+1} (i^2 - 2i). - Wesley Ivan Hurt, Feb 25 2014
a(n) = A000578(n+1) - A002412(n+1). - Wesley Ivan Hurt, Jun 28 2014
a(n) = Sum_{i = 1..n} Sum_{j = i..n} max(i,j). - Enrique Pérez Herrero, Dec 03 2014
a(n) = A055112(n)/6, see Singh (2013). - Alonso del Arte, Feb 20 2015
For n >= 2, a(n) = A028347(n+1) + A101986(n-2). - Bui Quang Tuan, Apr 03 2015
For n > 0: a(n) = A258708(n+3,n-1). - Reinhard Zumkeller, Jun 23 2015
a(n) = A175254(n) + A072481(n), n >= 1. - Omar E. Pol, Aug 12 2015
a(n) = A000332(n+3) - A000332(n+1). - Antal Pinter, Dec 27 2015
Dirichlet g.f.: zeta(s-3)/3 + zeta(s-2)/2 + zeta(s-1)/6. - Ilya Gutkovskiy, Jun 26 2016
a(n) = A080851(2,n-1). - R. J. Mathar, Jul 28 2016
a(n) = (A005408(n) * A046092(n))/12 = (2*n+1)*(2*n*(n+1))/12. - Bruce J. Nicholson, May 18 2017
12*a(n) = (n+1)*A001105(n) + n*A001105(n+1). - Bruno Berselli, Jul 03 2017
a(n) = binomial(n-1, 1) + binomial(n-1, 2) + binomial(n, 3) + binomial(n+1, 2) + binomial(n+1, 3). - Tony Foster III, Aug 24 2018
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Nathan Fox, Dec 04 2019
Let T(n) = A000217(n), the n-th triangular number. Then a(n) = (T(n)+1)^2 + (T(n)+2)^2 + ... + (T(n)+n)^2 - (n+2)*T(n)^2. - Charlie Marion, Dec 31 2019
a(n) = 2*n - 1 - a(n-2) + 2*a(n-1). - Boštjan Gec, Nov 09 2023
a(n) = 2/(2*n)! * Sum_{j = 1..n} (-1)^(n+j) * j^(2*n+2) * binomial(2*n, n-j). Cf. A060493. - Peter Bala, Mar 31 2025

Extensions

Partially edited by Joerg Arndt, Mar 11 2010

A052502 Number of permutations sigma of [3n] without fixed points such that sigma^3 = Id.

Original entry on oeis.org

1, 2, 40, 2240, 246400, 44844800, 12197785600, 4635158528000, 2345390215168000, 1524503639859200000, 1237896955565670400000, 1227993779921145036800000, 1461312598106162593792000000, 2054605512937264606871552000000
Offset: 0

Views

Author

encyclopedia(AT)pommard.inria.fr, Jan 25 2000

Keywords

Comments

For n >= 1 a(n) is the size of the conjugacy class in the symmetric group S_(3n) consisting of permutations whose cycle decomposition is a product of n disjoint 3-cycles.

References

  • F. W. J. Olver, Asymptotics and special functions, Academic Press, 1974, pages 336-344.

Links

Crossrefs

Cf. A000142. Row sums of triangle A060063.
First column of array A091752 (also negative of second column).
Equals row sums of A157702. - Johannes W. Meijer, Mar 07 2009
Karol A. Penson suggested that the row sums of A060063 coincide with this entry.
Cf. A001147, A052504, A060706, A261317, A261381.
Trisection of column k=3 of A261430.

Programs

  • GAP
    List([0..20], n-> Factorial(3*n)/(3^n*Factorial(n))) # G. C. Greubel, May 14 2019
  • Magma
    [Factorial(3*n)/(3^n*Factorial(n)): n in [0..20]]; // G. C. Greubel, May 14 2019
  • Maple
    spec := [S,{S=Set(Union(Cycle(Z,card=3)))},labeled]: seq(combstruct[count](spec,size=n), n=0..20);
  • Mathematica
    Table[(3*n)!/(3^n*n!), {n, 0, 20}] (* G. C. Greubel, May 14 2019 *)
  • PARI
    {a(n) = (3*n)!/(3^n*n!)}; \\ G. C. Greubel, May 14 2019
  • Sage
    [factorial(3*n)/(3^n*factorial(n)) for n in (0..20)] # G. C. Greubel, May 14 2019

Formula

From Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 21 2001: (Start)
a(n) = (3*n)!/(3^n * n!).
a(n) ~ sqrt(3) * 9^n * (n/e)^(2n). (End)
E.g.f.: (every third coefficient of) exp(x^3/3).
G.f.: hypergeometric3F0([1/3, 2/3, 1], [], 9*x).
D-finite with recurrence a(n) = (3*n-1)*(3*n-2)*a(n-1) for n >= 1, with a(0) = 1.
Write the generating function for this sequence in the form A(x) = Sum_{n >= 0} a(n)* x^(2*n+1)/(2*n+1)!. The g.f. A(x) satisfies A'(x)*( 1 - A(x)^2) = 1. Robert Israel remarks that consequently A(x) is a root of z^3 - 3*z + 3*x with A(0) = 0. Cf. A001147, A052504 and A060706. - Peter Bala, Jan 02 2015
From Peter Bala, Feb 27 2024: (Start)
u(n) := a(n+1) satisfies the second-order recurrence u(n) = 18*n*u(n-1) + (3*n - 1)^2*(3*n - 2)^2*u(n-2) with u(0) = 2 and u(1) = 40.
A second solution to the recurrence is given by v(n) := u(n)*Sum_{k = 0..n} (-1)^k/((3*k + 1)*(3*k + 2)) with v(0) = 1 and v(1) = 18.
This leads to the continued fraction expansion (2/3)*log(2) = Sum_{k = 0..n} (-1)^k/((3*k + 1)*(3*k + 2)) = Limit_{n -> oo} v(n)/u(n) = 1/(2 + (1*2)^2/(18 + (4*5)^2/(2*18 + (7*8)^2/(3*18 + (10*11)^2/(4*18 + ... ))))). (End)
From Gabriel B. Apolinario, Jul 30 2024: (Start)
a(n) = 3 * Integral_{t=0..oo} Ai(t)*t^(3*n) dt, where Ai(t) is the Airy function.
a(n) = Integral_{t=-oo..oo} Ai(t)*t^(3*n) dt. (End)

Extensions

Edited by Wolfdieter Lang, Feb 13 2004
Title improved by Geoffrey Critzer, Aug 14 2015

A156921 FP1 polynomials related to the generating functions of the right hand columns of the A156920 triangle.

Original entry on oeis.org

1, 1, 1, 1, -6, 1, 7, -79, 119, 126, -270, 1, 28, -515, 1654, 8689, -65864, 142371, -82242, -99090, 113400, 1, 86, -2255, 5784, 300930, -3904584, 20663714, -41517272, -80232259, 657717054
Offset: 0

Views

Author

Johannes W. Meijer, Feb 20 2009

Keywords

Comments

The FP1 polynomials appear in the numerators of the GF1 o.g.f.s. of the right hand columns of A156920. The FP1 can be calculated with the formula for the RHC sequence, see A156920, and the formula for the general structure of the generating function GF1, see below.
An appropriate name for the FP1 polynomials seems to be the flower polynomials of the first kind because the zero patterns of these polynomials look like flowers. The zero patterns of the FP2, see A156925, and the FP1 resemble each other closely.
A Maple program that generates for a right hand column with a certain RHCnr its GF1 and FP1 can be found below. RHCnr stands for right hand column number and starts from 1.

Examples

			The first few rows of the "triangle" of the coefficients of the FP1 polynomials.
In the columns the coefficients of the powers of z^m, m=0,1,2,... , appear.
  [1]
  [1]
  [1, 1, -6]
  [1, 7, -79, 119, 126, -270]
  [1, 28, -515, 1654, 8689, -65864, 142371, -82242, -99090, 113400]
Matrix of the coefficients of the FP1 polynomials. The coefficients in the columns of this matrix are the powers of z^m, m=0,1,2,.. .
  [1, 0 ,0, 0, 0, 0, 0, 0, 0, 0]
  [1, 0 ,0, 0, 0, 0, 0, 0, 0, 0]
  [1, 1, -6, 0 ,0, 0, 0, 0, 0, 0]
  [1, 7, -79, 119, 126, -270, 0, 0, 0, 0]
  [1, 28, -515, 1654, 8689, -65864, 142371, -82242, -99090, 113400]
The first few FP1 polynomials are:
  FP1(z; RHCnr=1) = 1
  FP1(z; RHCnr=2) = 1
  FP1(z; RHCnr =3) = 1+z-6*z^2
Some GF1(z;RHCnr) are:
  GF1(z;RHCnr= 3) = (1+z-6*z^2)/((1-5*z)*(1-3*z)^2*(1-z)^3)
  GF1(z;RHCnr= 4) = (1+7*z-79*z^2+119*z^3+126*z^4-270*z^5)/((1-7*z)*(1-5*z)^2*(1-3*z)^3*(1-z)^4)

Crossrefs

Cf. A156920, A156925, A156927, A156933.
For the first few GF1's see A000340, A156922, A156923, A156924.
The number of FP1 terms follow the triangular numbers A000217, with quite surprisingly one exception here a(0)=1.
Abs(Row sums (n)) = A098695(n).
For the polynomials in the denominators of the GF1(z;RHCnr) see A157702.

Programs

  • Maple
    RHCnr:=4: if RHCnr=1 then RHCmax :=1; else RHCmax:=(RHCnr-1)*(RHCnr)/2 end if: RHCend:=RHCnr+RHCmax: for k from RHCnr to RHCend do for n from 0 to k do S2[k,n]:=sum((-1)^(n+i)*binomial(n,i)*i^k/n!,i=0..n) end do: G(k,x):= sum(S2[k,p]*((2*p)!/p!) *x^p/(1-4*x)^(p+1),p=0..k)/(((-1)^(k+1)*2*x)/(-1+4*x)^(k+1)): fx:=simplify(G(k,x)): nmax:=degree(fx); RHC[k-RHCnr+1]:= coeff(fx,x,k-RHCnr)/2^(k-RHCnr) end do: a:=n-> RHC[n]: seq(a(n), n=1..RHCend-RHCnr+1); for nx from 0 to RHCmax do num:=sort(sum(A[t]*z^t, t=0..RHCmax)); nom:=Product((1-(2*u-1)*z)^(RHCnr-u+1),u=1..RHCnr): RHCa:= series(num/nom,z,nx+1); y:=coeff(RHCa,z,nx)-A[nx]; x:=RHC[nx+1]; A[nx]:=x-y; end do: FP1[RHCnr]:=sort(num,z, ascending); GenFun[RHCnr] :=FP1[RHCnr]/product((1-(2*m-1)*z)^(RHCnr-m+1),m=1..RHCnr);

Formula

G.f.: GF1(z;RHCnr) := FP1(z;RHCnr)/product((1-(2*m-1)*z)^(RHCnr+1-m),m=1..RHCnr)
Row sums (n) = (-1)^(1+(n+1)*(n+2)/2)*A098695(n).

A157704 G.f.s of the z^p coefficients of the polynomials in the GF3 denominators of A156927.

Original entry on oeis.org

1, 1, 5, 32, 186, 132, 10, 56, 2814, 17834, 27324, 11364, 1078, 10, 48, 17988, 494720, 3324209, 7526484, 6382271, 2004296, 203799, 4580, 5, 16, 72210, 7108338, 146595355, 1025458635, 2957655028, 3828236468
Offset: 0

Views

Author

Johannes W. Meijer, Mar 07 2009

Keywords

Comments

The formula for the PDGF3(z;n) polynomials in the GF3 denominators of A156927 can be found below.
The general structure of the GFKT3(z;p) that generate the z^p coefficients of the PDGF3(z; n) polynomials can also be found below. The KT3(z;p) polynomials in the numerators of the GFKT3(z; p) have a nice symmetrical structure.
The sequence of the number of terms of the first few KT3(z;p) polynomials is 1, 2, 4, 7, 10, 13, 14, 17, 20, 23, 26, 29, 32, 34, 36, 39, 42. The differences of this sequence and that of the number of terms of the KT4(z;p), see A157705, follow a simple pattern.
A Maple algorithm that generates relevant GFKT3(z;p) information can be found below.

Examples

			Some PDGF3 (z;n) are:
  PDGF3(z;n=3) = (1-z)*(1-2*z)^4*(1-3*z)^7*(1-4*z)^10
  PDGF3(z;n=4) = (1-z)*(1-2*z)^4*(1-3*z)^7*(1-4*z)^10*(1-5*z)^13
The first few GFKT3's are:
  GFKT3(z;p=0) = 1/(1-z)
  GFKT3(z;p=1) = -(5*z+1)/(1-z)^4
  GFKT3(z;p=2) = z*(32+186*z+132*z^2+10*z^3)/(1-z)^7
Some KT3(z,p) polynomials are:
  KT3(z;p=2) = 32+186*z+132*z^2+10*z^3
  KT3(z;p=3) = 56+2814*z+17834*z^2+27324*z^3+11364*z^4+1078*z^5+10*z^6

Crossrefs

Originator sequence A156927.
See A002414 for the z^1 coefficients and A157707 for the z^2 coefficients divided by 2.
Row sums equal A064350 and those of A157705.
Cf. A157702, A157703, A157705.

Programs

  • Maple
    p:=2; fn:=sum((-1)^(n1+1)*binomial(3*p+1,n1) *a(n-n1),n1=1..3*p+1): fk:=rsolve(a(n) = fn,a(k)): for n2 from 0 to 3*p+1 do fz(n2):=product((1-(k+1)*z)^(1+3*k), k=0..n2): a(n2):= coeff(fz(n2),z,p); end do: b:=n-> a(n): seq(b(n), n=0..3*p+1); a(n)=fn; a(k)=sort(simplify(fk)); GFKT3(p):=sum((fk)*z^k, k=0..infinity); q3:=ldegree((numer(GFKT3(p)))): KT3(p):=sort((-1)^(p)*simplify((GFKT3(p)*(1-z)^(3*p+1))/z^q3),z, ascending);

Formula

PDGF3(z;n) = Product_{k=0..n} (1-(k+1)*z)^(1+3*k) with n = 1, 2, 3, ...
GFKT3(z;p) = (-1)^(p)*(z^q3)*KT3(z, p)/(1-z)^(3*p+1) with p = 0, 1, 2, ...
The recurrence relation for the z^p coefficients a(n) is a(n) = Sum_{k=1..3*p+1} (-1)^(k+1)*binomial(3*p + 1, k)*a(n-k) with p = 0, 1, 2, ... .

A157705 G.f.s of the z^p coefficients of the polynomials in the GF4 denominators of A156933.

Original entry on oeis.org

1, 1, 3, 2, 18, 128, 171, 42, 1, 22, 1348, 11738, 26293, 17693, 3271, 115, 13, 6122, 228986, 2070813, 6324083, 7397855, 3361536, 544247, 24590, 155, 3, 17248, 2413434, 67035224, 612026240, 2274148882
Offset: 0

Views

Author

Johannes W. Meijer, Mar 07 2009

Keywords

Comments

The formula for the PDGF4(z;n) polynomials in the GF4 denominators of A156933 can be found below.
The general structure of the GFKT4(z;p) that generate the z^p coefficients of the PDGF4(z;n) polynomials can also be found below. The KT4(z;p) polynomials in the numerators of the GFKT4(z;p) have a nice symmetrical structure.
The sequence of the number of terms of the first few KT4(z;p) polynomials is 1, 3, 5, 7, 10, 13, 15, 18, 20, 23, 26, 29, 32, 34, 37, 40, 42. The differences of this sequence and that of the number of terms of the KT3(z;p), see A157704, follow a simple pattern.
A Maple algorithm that generates relevant GFKT4(z;p) information can be found below.

Examples

			Some PDGF4 (z;n) are:
  PDGF4(z; n=3) = (1-7*z)*(1-5*z)^4*(1-3*z)^7*(1-z)^10
  PDGF4(z; n=4) = (1-9*z)*(1-7*z)^4*(1-5*z)^7*(1-3*z)^10*(1-z)^13
The first few GFKT4's are:
  GFKT4(z;p=0) = 1/(1-z)
  GFKT4(z;p=1) = -(1+3*z+2*z^2)/(1-z)^4
  GFKT4(z;p=2) = z*(18+128*z+171*z^2+42*z^3+z^4)/(1-z)^7
Some KT4(z,p) polynomials are:
  KT4(z;p=2) = 18+128*z+171*z^2+42*z^3+z^4
  KT4(z;p=3) = 22+1348*z+11738*z^2+26293*z^3+17693*z^4+3271*z^5+115*z^6

Crossrefs

Originator sequence A156933.
See A081436 for the z^1 coefficients and A157708 for the z^2 coefficients.
Row sums equal A064350 and those of A157704.
Cf. A157702, A157703, A157704.

Programs

  • Maple
    p:=2; fn:=sum((-1)^(n1+1)*binomial(3*p+1,n1) *a(n-n1),n1=1..3*p+1): fk:=rsolve(a(n) = fn,a(k)): for n2 from 0 to 3*p+1 do fz(n2):=product((1-(2*n2+1-(2*k))*z)^(3*k+1), k=0..n2): a(n2):= coeff(fz(n2),z,p): end do: b:=n-> a(n): seq(b(n), n=0..3*p+1); a(n)=fn; a(k)= sort (simplify(fk)); GFKT4(p):=sum((fk)*z^k,k=0..infinity); q4:=ldegree((numer (GFKT4(p)))): KT4(p):=sort((-1)^(p)*simplify((GFKT4(p)*(1-z)^(3*p+1))/z^q4),z, ascending);

Formula

PDGF4(z;n) = Product_{k=0..n} (1-(2*n+1-2*k)*z)^(3*k+1) with n = 1, 2, 3, ...
GFKT4(z;p) = (-1)^(p)*(z^q4)*KT4(z, p)/(1-z)^(3*p+1) with p = 0, 1, 2, ...
The recurrence relation for the z^p coefficients a(n) is a(n) = Sum_{k=1..3*p+1} (-1)^(k+1)*binomial(3*p + 1, k)*a(n-k) with p = 0, 1, 2, ... .

A157703 G.f.s of the z^p coefficients of the polynomials in the GF2 denominators of A156925.

Original entry on oeis.org

1, 1, 5, 5, 2, 62, 152, 62, 2, 91, 1652, 5957, 5957, 1652, 91, 52, 5240, 77630, 342188, 551180, 342188, 77630, 5240, 52, 12, 8549, 424921, 5629615, 28123559, 61108544, 61108544, 28123559, 5629615, 424921, 8549, 12
Offset: 0

Views

Author

Johannes W. Meijer, Mar 07 2009

Keywords

Comments

The formula for the PDGF2(z;n) polynomials in the GF2 denominators of A156925 can be found below.
The general structure of the GFKT2(z;p) that generate the z^p coefficients of the PDGF2(z; n) polynomials can also be found below. The KT2(z;p) polynomials in the numerators of the GFKT2(z;p) have a nice symmetrical structure.
The sequence of the number of terms of the first few KT2(z;p) polynomials is: 1, 1, 2, 5, 6, 9, 12, 13, 16, 19, 22, 23, 26. The first differences follow a simple pattern. The positions of the 1's follow the Lazy Caterer's sequence A000124 with one exception, here a(0) = 0.
A Maple algorithm that generates relevant GFKT2(z;p) information can be found below.

Examples

			Some PDGF2 (z;n) are:
  PDGF2(z;n=3) = (1-z)^3*(1-2*z)^2*(1-3*z)
  PDGF2(z;n=4) = (1-z)^4*(1-2*z)^3*(1-3*z)^2*(1-4*z)
The first few GFKT2's are:
  GFKT2(z;p=0) = 1/(1-z)
  GFKT2(z;p=1) = -z/(z-1)^4
  GFKT2(z;p=2) = z^2*(5+5*z)/(1-z)^7
Some KT2(z,p) polynomials are:
  KT2(z;p=2) = 5+5*z
  KT2(z;p=3) = 2+62*z+152*z^2+62*z^3+2*z^4
  KT2(z;p=4) = 91+1652*z+5957*z^2+5957*z^3+1652*z^4+91*z^5

Crossrefs

Originator sequence A156925.
See A000292 for the z^1 coefficients and A040977 for the z^2 coefficients divided by 5.
Row sums equal A025035.
Cf. A157702, A157704, A157705.

Programs

  • Maple
    p:=2; fn:=sum((-1)^(n1+1)*binomial(3*p+1,n1) *a(n-n1),n1=1..3*p+1): fk:=rsolve(a(n) = fn,a(k)): for n2 from 0 to 3*p+1 do fz(n2):=product((1-m*z)^(n2+1-m),m=1..n2): a(n2):= coeff(fz(n2),z,p): end do: b:=n-> a(n): seq(b(n), n=0..3*p+1); a(n)=fn; a(k)=sort(simplify(fk)); GFKT2(p):=sum((fk)*z^k,k=0..infinity); q2:=ldegree((numer(GFKT2(p)))): KT2(p):=sort((-1)^p*simplify((GFKT2(p)*(1-z)^(3*p+1))/z^q2),z, ascending);

Formula

PDGF2(z;n) = Product_{m=1..n} (1-m*z)^(n+1-m) with n = 1, 2, 3, ...
GFKT2(z;p) = (-1)^(p)*(z^q2)*KT2(z, p)/(1-z)^(3*p+1) with p = 0, 1, 2, ...
The recurrence relation for the z^p coefficients a(n) is a(n) = Sum_{k=1..3*p+1} (-1)^(k+1)*binomial(3*p + 1, k)*a(n-k) with p = 0, 1, 2, ... .

A157706 The z^2 coefficients of the polynomials in the GF1 denominators of A156921.

Original entry on oeis.org

7, 75, 385, 1365, 3850, 9282, 19950, 39270, 72105, 125125, 207207, 329875, 507780, 759220, 1106700, 1577532, 2204475, 3026415, 4089085, 5445825, 7158382, 9297750, 11945050, 15192450, 19144125, 23917257
Offset: 2

Views

Author

Johannes W. Meijer, Mar 07 2009

Keywords

Comments

See A157702 for background information.

Crossrefs

Cf. A156921, A157702.

Programs

  • Maple
    nmax:=27; for n from 0 to nmax do fz(n):= product( (1-(2*m-1)*z)^(n+1-m) , m=1..n); c(n):= coeff(fz(n),z,2); end do: a:=n-> c(n): seq(a(n), n=2..nmax);

Formula

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) = 1/18*n^6+1/6*n^5+1/72*n^4-1/4*n^3-5/72*n^2+1/12*n
G.f.: (7+26*z+7*z^2)/(1-z)^7
Showing 1-7 of 7 results.

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