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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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A008277 Triangle of Stirling numbers of the second kind, S2(n,k), n >= 1, 1 <= k <= n.

Original entry on oeis.org

1, 1, 1, 1, 3, 1, 1, 7, 6, 1, 1, 15, 25, 10, 1, 1, 31, 90, 65, 15, 1, 1, 63, 301, 350, 140, 21, 1, 1, 127, 966, 1701, 1050, 266, 28, 1, 1, 255, 3025, 7770, 6951, 2646, 462, 36, 1, 1, 511, 9330, 34105, 42525, 22827, 5880, 750, 45, 1, 1, 1023, 28501, 145750, 246730, 179487, 63987, 11880, 1155, 55, 1
Offset: 1

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Author

N. J. A. Sloane

Keywords

Comments

Also known as Stirling set numbers and written {n, k}.
S2(n,k) counts partitions of an n-set into k nonempty subsets.
From Manfred Boergens, Apr 07 2025: (Start)
With regard to the preceding comment:
For disjoint collections of subsets see A256894.
For arbitrary collections of subsets see A163353.
For arbitrary collections of nonempty subsets see A055154. (End)
Triangle S2(n,k), 1 <= k <= n, read by rows, given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, ...] where DELTA is Deléham's operator defined in A084938.
Number of partitions of {1, ..., n+1} into k+1 nonempty subsets of nonconsecutive integers, including the partition 1|2|...|n+1 if n=k. E.g., S2(3,2)=3 since the number of partitions of {1,2,3,4} into three subsets of nonconsecutive integers is 3, i.e., 13|2|4, 14|2|3, 1|24|3. - Augustine O. Munagi, Mar 20 2005
Draw n cards (with replacement) from a deck of k cards. Let prob(n,k) be the probability that each card was drawn at least once. Then prob(n,k) = S2(n,k)*k!/k^n (see A090582). - Rainer Rosenthal, Oct 22 2005
Define f_1(x), f_2(x), ..., such that f_1(x)=e^x and for n = 2, 3, ..., f_{n+1}(x) = (d/dx)(x*f_n(x)). Then f_n(x) = e^x*Sum_{k=1..n} S2(n,k)*x^(k-1). - Milan Janjic, May 30 2008
From Peter Bala, Oct 03 2008: (Start)
For tables of restricted Stirling numbers of the second kind see A143494 - A143496.
S2(n,k) gives the number of 'patterns' of words of length n using k distinct symbols - see [Cooper & Kennedy] for an exact definition of the term 'pattern'. As an example, the words AADCBB and XXEGTT, both of length 6, have the same pattern of letters. The five patterns of words of length 3 are AAA, AAB, ABA, BAA and ABC giving row 3 of this table as (1,3,1).
Equivalently, S2(n,k) gives the number of sequences of positive integers (N_1,...,N_n) of length n, with k distinct entries, such that N_1 = 1 and N_(i+1) <= 1 + max{j = 1..i} N_j for i >= 1 (restricted growth functions). For example, Stirling(4,2) = 7 since the sequences of length 4 having 2 distinct entries that satisfy the conditions are (1,1,1,2), (1,1,2,1), (1,2,1,1), (1,1,2,2), (1,2,2,2), (1,2,2,1) and (1,2,1,2).
(End)
Number of combinations of subsets in the plane. - Mats Granvik, Jan 13 2009
S2(n+1,k+1) is the number of size k collections of pairwise disjoint, nonempty subsets of [n]. For example: S2(4,3)=6 because there are six such collections of subsets of [3] that have cardinality two: {(1)(23)},{(12)(3)}, {(13)(2)}, {(1)(2)}, {(1)(3)}, {(2)(3)}. - Geoffrey Critzer, Apr 06 2009
Consider a set of A000217(n) balls of n colors in which, for each integer k = 1 to n, exactly one color appears in the set a total of k times. (Each ball has exactly one color and is indistinguishable from other balls of the same color.) a(n+1, k+1) equals the number of ways to choose 0 or more balls of each color in such a way that exactly (n-k) colors are chosen at least once, and no two colors are chosen the same positive number of times. - Matthew Vandermast, Nov 22 2010
S2(n,k) is the number of monotonic-labeled forests on n vertices with exactly k rooted trees, each of height one or less. See link "Counting forests with Stirling and Bell numbers" below. - Dennis P. Walsh, Nov 16 2011
If D is the operator d/dx, and E the operator xd/dx, Stirling numbers are given by: E^n = Sum_{k=1..n} S2(n,k) * x^k*D^k. - Hyunwoo Jang, Dec 13 2011
The Stirling polynomials of the second kind (a.k.a. the Bell / Touchard polynomials) are the umbral compositional inverses of the falling factorials (a.k.a. the Pochhammer symbol or Stirling polynomials of the first kind), i.e., binomial(Bell(.,x),n) = x^n/n! (cf. Copeland's 2007 formulas), implying binomial(xD,n) = binomial(Bell(.,:xD:),n) = :xD:^n/n! where D = d/dx and :xD:^n = x^n*D^n. - Tom Copeland, Apr 17 2014
S2(n,k) is the number of ways to nest n matryoshkas (Russian nesting dolls) so that exactly k matryoshkas are not contained in any other matryoshka. - Carlo Sanna, Oct 17 2015
The row polynomials R(n, x) = Sum_{k=1..n} S2(n, k)*x^k appear in the numerator of the e.g.f. of n-th powers, E(n, x) = Sum_{m>=0} m^n*x^m/m!, as E(n, x) = exp(x)*x*R(n, x), for n >= 1. - Wolfdieter Lang, Apr 02 2017
With offsets 0 for n and k this is the Sheffer product matrix A007318*A048993 denoted by (exp(t), (exp(t) - 1)) with e.g.f. exp(t)*exp(x*(exp(t) - 1)). - Wolfdieter Lang, Jun 20 2017
Number of words on k+1 unlabeled letters of length n+1 with no repeated letters. - Thomas Anton, Mar 14 2019
Also coefficients of moments of Poisson distribution about the origin expressed as polynomials in lambda. [Haight] (see also A331155). - N. J. A. Sloane, Jan 14 2020
k!*S2(n,k) is the number of surjections from an n-element set to a k-element set. - Jianing Song, Jun 01 2022

Examples

			The triangle S2(n, k) begins:
\ k    1       2       3        4         5         6         7         8        9
n \   10      11      12       13        14        15       ...
----------------------------------------------------------------------------------
1  |   1
2  |   1       1
3  |   1       3       1
4  |   1       7       6        1
5  |   1      15      25       10         1
6  |   1      31      90       65        15         1
7  |   1      63     301      350       140        21         1
8  |   1     127     966     1701      1050       266        28         1
9  |   1     255    3025     7770      6951      2646       462        36        1
10 |   1     511    9330    34105     42525     22827      5880       750       45
       1
11 |   1    1023   28501   145750    246730    179487     63987     11880     1155
      55       1
12 |   1    2047   86526   611501   1379400   1323652    627396    159027    22275
    1705      66       1
13 |   1    4095  261625  2532530   7508501   9321312   5715424   1899612   359502
   39325    2431      78        1
14 |   1    8191  788970 10391745  40075035  63436373  49329280  20912320  5135130
  752752   66066    3367       91         1
15 |   1   16383 2375101 42355950 210766920 420693273 408741333 216627840 67128490
12662650 1479478  106470     4550       105         1
...
----------------------------------------------------------------------------------
x^4 = 1 x_(1) + 7 x_(2) + 6 x_(3) + 1 x_(4), where x_(k) = P(x,k) = k!*C(x,k). - _Daniel Forgues_, Jan 16 2016

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. 835.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 103ff.
  • B. A. Bondarenko, Generalized Pascal Triangles and Pyramids (in Russian), FAN, Tashkent, 1990, ISBN 5-648-00738-8.
  • G. Boole, Finite Differences, 5th ed. New York, NY: Chelsea, 1970.
  • C. A. Charalambides, Enumerative Combinatorics, Chapman & Hall/CRC, 2002, Theorem 8.11, pp. 298-299.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 310.
  • J. H. Conway and R. K. Guy, The Book of Numbers, Springer, p. 92.
  • F. N. David, M. G. Kendall, and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 223.
  • S.N. Elaydi, An Introduction to Difference Equations, 3rd ed. Springer, 2005.
  • H. H. Goldstine, A History of Numerical Analysis, Springer-Verlag, 1977; Section 2.7.
  • R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 244.
  • Frank Avery Haight, Handbook of the Poisson distribution, John Wiley, 1967. See pages 6,7.
  • A. D. Korshunov, Asymptotic behavior of Stirling numbers of the second kind. (Russian) Metody Diskret. Analiz. No. 39 (1983), 24-41.
  • E. Kuz'min and A. I. Shirshov: On the number e, pp. 111-119, eq.(6), in: Kvant Selecta: Algebra and Analysis, I, ed. S. Tabachnikov, Am.Math.Soc., 1999, p. 116, eq. (11).
  • J. Riordan, An Introduction to Combinatorial Analysis, p. 48.
  • J. Stirling, The Differential Method, London, 1749; see p. 7.

Links

Crossrefs

Cf. A008275 (Stirling numbers of first kind), A048993 (another version of this triangle).
Cf. A000217, A001296, A001297, A001298, A007318, A028246, A039810-A039813, A048994, A087107-A087111, A087127, A094262, A127701.
See also A331155.
Cf. A038207, A086885, A132440, A138378, A173018, A238385.
Cf. A055154, A163353, A256894.
Cf. A000110 (row sums), A102661 (partial row sums).

Programs

  • Haskell
    a008277 n k = a008277_tabl !! (n-1) !! (k-1)
a008277_row n = a008277_tabl !! (n-1)
a008277_tabl = map tail $ a048993_tabl  -- Reinhard Zumkeller, Mar 26 2012
  • J
    n ((] (1 % !)) * +/@((^~ * (] (1 ^ |.)) * (! {:)@]) i.@>:)) k NB. _Stephen Makdisi, Apr 06 2016
  • Magma
    [[StirlingSecond(n,k): k in [1..n]]: n in [1..12]]; // G. C. Greubel, May 22 2019
  • Maple
    seq(seq(combinat[stirling2](n, k), k=1..n), n=1..10); # Zerinvary Lajos, Jun 02 2007
stirling_2 := (n,k) -> (1/k!) * add((-1)^(k-i)*binomial(k,i)*i^n, i=0..k);
  • Mathematica
    Table[StirlingS2[n, k], {n, 11}, {k, n}] // Flatten (* Robert G. Wilson v, May 23 2006 *)
BellMatrix[f_, len_] := With[{t = Array[f, len, 0]}, Table[BellY[n, k, t], {n, 0, len - 1}, {k, 0, len - 1}]];
rows = 12;
B = BellMatrix[1&, rows];
Table[B[[n, k]], {n, 2, rows}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jun 28 2018, after Peter Luschny *)
a[n_, n_] := 1; a[n_, 1] := 1;
a[n_, k_] := a[n, k] = a[n-1, k-1] + k a[n-1, k]; Flatten@
Table[a[n, k], {n, 1, 11}, {k, 1, n}] (* Oliver Seipel, Jun 12 2024 *)
With[{m = 11},
 Flatten@MapIndexed[Take[#, #2[[1]]] &,
   Transpose@
    Table[Range[1, m]! Coefficient[(E^x-1)^k/k! + O[x]^(m+1), x,
Range[1, m]], {k, 1, m}]]] (* Oliver Seipel, Jun 12 2024 *)
  • Maxima
    create_list(stirling2(n+1,k+1),n,0,30,k,0,n); /* Emanuele Munarini, Jun 01 2012 */
  • PARI
    for(n=1,22,for(k=1,n,print1(stirling(n,k,2),", "));print()); \\ Joerg Arndt, Apr 21 2013
  • PARI
    Stirling2(n,k)=sum(i=0,k,(-1)^i*binomial(k,i)*i^n)*(-1)^k/k!  \\ M. F. Hasler, Mar 06 2012
  • Sage
    stirling_number2 # Danny Rorabaugh, Oct 11 2015

Formula

S2(n, k) = k*S2(n-1, k) + S2(n-1, k-1), n > 1. S2(1, k) = 0, k > 1. S2(1, 1) = 1.
E.g.f.: A(x, y) = e^(y*e^x-y). E.g.f. for m-th column: (e^x-1)^m/m!.
S2(n, k) = (1/k!) * Sum_{i=0..k} (-1)^(k-i)*binomial(k, i)*i^n.
Row sums: Bell number A000110(n) = Sum_{k=1..n} S2(n, k), n>0.
S(n, k) = Sum (i_1*i_2*...*i_(n-k)) summed over all (n-k)-combinations {i_1, i_2, ..., i_k} with repetitions of the numbers {1, 2, ..., k}. Also S(n, k) = Sum (1^(r_1)*2^(r_2)*...* k^(r_k)) summed over integers r_j >= 0, for j=1..k, with Sum{j=1..k} r_j = n-k. [Charalambides]. - Wolfdieter Lang, Aug 15 2019.
A019538(n, k) = k! * S2(n, k).
A028248(n, k) = (k-1)! * S2(n, k).
For asymptotics see Hsu (1948), among other sources.
Sum_{n>=0} S2(n, k)*x^n = x^k/((1-x)(1-2x)(1-3x)...(1-kx)).
Let P(n) = the number of integer partitions of n (A000041), p(i) = the number of parts of the i-th partition of n, d(i) = the number of distinct parts of the i-th partition of n, p(j, i) = the j-th part of the i-th partition of n, m(i, j) = multiplicity of the j-th part of the i-th partition of n, and Sum_{i=1..P(n), p(i)=m} = sum running from i=1 to i=P(n) but taking only partitions with p(i)=m parts into account. Then S2(n, m) = Sum_{i=1..P(n), p(i)=m} n!/(Product_{j=1..p(i)} p(i, j)!) * 1/(Product_{j=1..d(i)} m(i, j)!). For example, S2(6, 3) = 90 because n=6 has the following partitions with m=3 parts: (114), (123), (222). Their complexions are: (114): 6!/(1!*1!*4!) * 1/(2!*1!) = 15, (123): 6!/(1!*2!*3!) * 1/(1!*1!*1!) = 60, (222): 6!/(2!*2!*2!) * 1/(3!) = 15. The sum of the complexions is 15+60+15 = 90 = S2(6, 3). - Thomas Wieder, Jun 02 2005
Sum_{k=1..n} k*S2(n,k) = B(n+1)-B(n), where B(q) are the Bell numbers (A000110). - Emeric Deutsch, Nov 01 2006
Recurrence: S2(n+1,k) = Sum_{i=0..n} binomial(n,i)*S2(i,k-1). With the starting conditions S2(n,k) = 1 for n = 0 or k = 1 and S2(n,k) = 0 for k = 0 we have the closely related recurrence S2(n,k) = Sum_{i=k..n} binomial(n-1,i-1)*S2(i-1,k-1). - Thomas Wieder, Jan 27 2007
Representation of Stirling numbers of the second kind S2(n,k), n=1,2,..., k=1,2,...,n, as special values of hypergeometric function of type (n)F(n-1): S2(n,k)= (-1)^(k-1)*hypergeom([ -k+1,2,2,...,2],[1,1,...,1],1)/(k-1)!, i.e., having n parameters in the numerator: one equal to -k+1 and n-1 parameters all equal to 2; and having n-1 parameters in the denominator all equal to 1 and the value of the argument equal to 1. Example: S2(6,k)= seq(evalf((-1)^(k-1)*hypergeom([ -k+1,2,2,2,2,2],[1,1,1,1,1],1)/(k-1)!),k=1..6)=1,31,90,65,15,1. - Karol A. Penson, Mar 28 2007
From Tom Copeland, Oct 10 2007: (Start)
Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..n} E(n,j) * binomial(Bell.(x)+j, n) umbrally where Bell_n(x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m.
For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).
Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..n}E(n,k)). Also (n!*Lag(n, -1, j-n)) is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x=1; n!*Bell_n(1) = n!*Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)).
(Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (End)
n-th row = leftmost column of nonzero terms of A127701^(n-1). Also, (n+1)-th row of the triangle = A127701 * n-th row; deleting the zeros. Example: A127701 * [1, 3, 1, 0, 0, 0, ...] = [1, 7, 6, 1, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007
The row polynomials are given by D^n(e^(x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. Cf. A147315 and A094198. See also A185422. - Peter Bala, Nov 25 2011
Let f(x) = e^(e^x). Then for n >= 1, 1/f(x)*(d/dx)^n(f(x)) = 1/f(x)*(d/dx)^(n-1)(e^x*f(x)) = Sum_{k=1..n} S2(n,k)*e^(k*x). Similar formulas hold for A039755, A105794, A111577, A143494 and A154537. - Peter Bala, Mar 01 2012
S2(n,k) = A048993(n,k), 1 <= k <= n. - Reinhard Zumkeller, Mar 26 2012
O.g.f. for the n-th diagonal is D^n(x), where D is the operator x/(1-x)*d/dx. - Peter Bala, Jul 02 2012
n*i!*S2(n-1,i) = Sum_{j=(i+1)..n} (-1)^(j-i+1)*j!/(j-i)*S2(n,j). - Leonid Bedratyuk, Aug 19 2012
G.f.: (1/Q(0)-1)/(x*y), where Q(k) = 1 - (y+k)*x - (k+1)*y*x^2/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 09 2013
From Tom Copeland, Apr 17 2014: (Start)
Multiply each n-th diagonal of the Pascal lower triangular matrix by x^n and designate the result as A007318(x) = P(x).
With Bell(n,x)=B(n,x) defined above, D = d/dx, and :xD:^n = x^n*D^n, a Dobinski formula gives umbrally f(y)^B(.,x) = e^(-x)*e^(f(y)*x). Then f(y)^B(.,:xD:)g(x) = [f(y)^(xD)]g(x) = e^[-(1-f(y)):xD:]g(x) = g[f(y)x].
In particular, for f(y) = (1+y),
A) (1+y)^B(.,x) = e^(-x)*e^((1+y)*x) = e^(x*y) = e^[log(1+y)B(.,x)],
B) (I+dP)^B(.,x) = e^(x*dP) = P(x) = e^[x*(e^M-I)]= e^[M*B(.,x)] with dP = A132440, M = A238385-I = log(I+dP), and I = identity matrix, and
C) (1+dP)^(xD) = e^(dP:xD:) = P(:xD:) = e^[(e^M-I):xD:] = e^[M*xD] with action e^(dP:xD:)g(x) = g[(I+dP)*x].
D) P(x)^m = P(m*x), which implies (Sum_{k=1..m} a_k)^j = B(j,m*x) where the sum is umbrally evaluated only after exponentiation with (a_k)^q = B(.,x)^q = B(q,x). E.g., (a1+a2+a3)^2=a1^2+a2^2+a3^2+2(a1*a2+a1*a3+a2*a3) = 3*B(2,x)+6*B(1,x)^2 = 9x^2+3x = B(2,3x).
E) P(x)^2 = P(2x) = e^[M*B(.,2x)] = A038207(x), the face vectors of the n-Dim hypercubes.
(End)
As a matrix equivalent of some inversions mentioned above, A008277*A008275 = I, the identity matrix, regarded as lower triangular matrices. - Tom Copeland, Apr 26 2014
O.g.f. for the n-th diagonal of the triangle (n = 0,1,2,...): Sum_{k>=0} k^(k+n)*(x*e^(-x))^k/k!. Cf. the generating functions of the diagonals of A039755. Also cf. A112492. - Peter Bala, Jun 22 2014
Floor(1/(-1 + Sum_{n>=k} 1/S2(n,k))) = A034856(k-1), for k>=2. The fractional portion goes to zero at large k. - Richard R. Forberg, Jan 17 2015
From Daniel Forgues, Jan 16 2016: (Start)
Let x_(n), called a factorial term (Boole, 1970) or a factorial polynomial (Elaydi, 2005: p. 60), denote the falling factorial Product_{k=0..n-1} (x-k). Then, for n >= 1, x_(n) = Sum_{k=1..n} A008275(n,k) * x^k, x^n = Sum_{k=1..n} T(n,k) * x_(k), where A008275(n,k) are Stirling numbers of the first kind.
For n >= 1, the row sums yield the exponential numbers (or Bell numbers): Sum_{k=1..n} T(n,k) = A000110(n), and Sum_{k=1..n} (-1)^(n+k) * T(n,k) = (-1)^n * Sum_{k=1..n} (-1)^k * T(n,k) = (-1)^n * A000587(n), where A000587 are the complementary Bell numbers. (End)
Sum_{k=1..n} k*S2(n,k) = A138378(n). - Alois P. Heinz, Jan 07 2022
O.g.f. for the m-th column: x^m/(Product_{j=1..m} 1-j*x). - Daniel Checa, Aug 25 2022
S2(n,k) ~ (k^n)/k!, for fixed k as n->oo. - Daniel Checa, Nov 08 2022
S2(2n+k, n) ~ (2^(2n+k-1/2) * n^(n+k-1/2)) / (sqrt(Pi*(1-c)) * exp(n) * c^n * (2-c)^(n+k)), where c = -LambertW(-2 * exp(-2)). - Miko Labalan, Dec 21 2024
From Mikhail Kurkov, Mar 05 2025: (Start)
For a general proof of the formulas below via generating functions, see Mathematics Stack Exchange link.
Recursion for the n-th row (independently of other rows): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} (j-2)!*binomial(-k,j)*T(n,k+j-1) for 1 <= k < n with T(n,n) = 1 (see Fedor Petrov link).
Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} binomial(n,j)*T(n-j+1,k)*(-1)^j for 1 <= k < n with T(n,n) = 1. (End)

A000680 a(n) = (2n)!/2^n.

Original entry on oeis.org

1, 1, 6, 90, 2520, 113400, 7484400, 681080400, 81729648000, 12504636144000, 2375880867360000, 548828480360160000, 151476660579404160000, 49229914688306352000000, 18608907752179801056000000, 8094874872198213459360000000, 4015057936610313875842560000000
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

Denominators in the expansion of cos(sqrt(2)*x) = 1 - (sqrt(2)*x)^2/2! + (sqrt(2)*x)^4/4! - (sqrt(2)*x)^6/6! + ... = 1 - x^2 + x^4/6 - x^6/90 + ... By Stirling's formula in A000142: a(n) ~ 2^(n+1) * (n/e)^(2n) * sqrt(Pi*n) - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 20 2001
a(n) is also the constant term in the product: Product_{1<=i, j<=n, i!=j} (1 - x_i/x_j)^2. - Sharon Sela (sharonsela(AT)hotmail.com), Feb 12 2002
a(n) is also the number of lattice paths in the n-dimensional lattice [0..2]^n. - T. D. Noe, Jun 06 2002
Representation as the n-th moment of a positive function on the positive half-axis: a(n) = Integral_{x>=0} (x^n*exp(-sqrt(2*x))/sqrt(2*x)), n=0,1,... - Karol A. Penson, Mar 10 2003
Number of permutations of [2n] with no increasing runs of odd length. Example: a(2) = 6 because we have 1234, 13/24, 14/23, 23/14, 24/13 and 34/12 (runs separated by slashes). - Emeric Deutsch, Aug 29 2004
This is also the number of ways of arranging the elements of n distinct pairs, assuming the order of elements is significant and the pairs are distinguishable. When the pairs are not distinguishable, see A001147 and A132101. For example, there are 6 ways of arranging 2 pairs [1,1], [2,2]: {[1122], [1212], [1221], [2211], [2121], [2112]}. - Ross Drewe, Mar 16 2008
n married couples are seated in a row so that every wife is to the left of her husband. The recurrence a(n+1) = a(n)*((2*n + 1) + binomial(2*n+1, 2)) conditions on whether the (n+1)st couple is seated together or separated by at least one other person. - Geoffrey Critzer, Jun 10 2009
a(n) is the number of functions f:[2n]->[n] such that the preimage of {y} has cardinality 2 for every y in [n]. Note that [k] denotes the set {1,2,...,k} and [0] denotes the empty set. - Dennis P. Walsh, Nov 17 2009
a(n) is also the number of n X 2n (0,1)-matrices with row sum 2 and column sum 1. - Shanzhen Gao, Feb 12 2010
Number of ways that 2n people of different heights can be arranged (for a photograph) in two rows of equal length so that every person in the front row is shorter than the person immediately behind them in the back row.
a(n) is the number of functions f:[n]->[n^2] such that, if floor((f(x))^.5) = floor((f(y))^.5), then x = y. For example, with n = 4, the range of f consists of one element from each of the four sets {1,2,3}, {4,5,6,7,8}, {9,10,11,12,13,14,15}, and {16}. Hence there are 1*3*5*7 = 105 ways to choose the range for f, and there are 4! ways to injectively map {1,2,3,4} to the four elements of the range. Thus there are 105*24 = 2520 such functions. Note also that a(n) = n!*(product of the first n odd numbers). - Dennis P. Walsh, Nov 28 2012
a(n) is also the 2*n th difference of n-powers of A000217 (triangular numbers). For example a(2) is the 4th difference of the squares of triangular numbers. - Enric Reverter i Bigas, Jun 24 2013
a(n) is the multinomial coefficient (2*n) over (2, 2, 2, ..., 2) where there are n 2's in the last parenthesis. It is therefore also the number of words of length 2n obtained with n letters, each letter appearing twice. - Robert FERREOL, Jan 14 2018
Number of ways to put socks and shoes on an n-legged animal, if a sock must be put on before a shoe. - Daniel Bishop, Jan 29 2018

Examples

			For n = 2, a(2) = 6 since there are 6 functions f:[4]->[2] with size 2 preimages for both {1} and {2}. In this case, there are binomial(4, 2) = 6 ways to choose the 2 elements of [4] f maps to {1} and the 2 elements of [4] that f maps to {2}. - _Dennis P. Walsh_, Nov 17 2009

References

  • G. E. Andrews, R. Askey and R. Roy, Special Functions, Cambridge University Press, 1998.
  • H. T. Davis, Tables of the Mathematical Functions. Vols. 1 and 2, 2nd ed., 1963, Vol. 3 (with V. J. Fisher), 1962; Principia Press of Trinity Univ., San Antonio, TX, Vol. 2, p. 283.
  • A. Fletcher, J. C. P. Miller, L. Rosenhead and L. J. Comrie, An Index of Mathematical Tables. Vols. 1 and 2, 2nd ed., Blackwell, Oxford and Addison-Wesley, Reading, MA, 1962, Vol. 1, p. 112.
  • Shanzhen Gao and Kenneth Matheis, Closed formulas and integer sequences arising from the enumeration of (0,1)-matrices with row sum two and some constant column sums. In Proceedings of the Forty-First Southeastern International Conference on Combinatorics, Graph Theory and Computing. Congr. Numer. 202 (2010), 45-53.
  • 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).
  • C. B. Tompkins, Methods of successive restrictions in computational problems involving discrete variables. 1963, Proc. Sympos. Appl. Math., Vol. XV pp. 95-106; Amer. Math. Soc., Providence, R.I.

Links

Crossrefs

Cf. A084939, A084940, A084941, A084942, A084943, A084944, A087127, A001147, A132101.
A diagonal of the triangle in A241171.
Main diagonal of A267479, row sums of A267480.
Row n=2 of A089759.
Column n=2 of A187783.
Even bisection of column k=0 of A097591.

Programs

  • Maple
    A000680 := n->(2*n)!/(2^n);
a[0]:=1:a[1]:=1:for n from 2 to 50 do a[n]:=a[n-1]*(2*n-1)*n od: seq(a[n], n=0..16); # Zerinvary Lajos, Mar 08 2008
seq(product(binomial(2*n-2*k,2),k=0..n-1),n=0..16); # Dennis P. Walsh, Nov 17 2009
  • Mathematica
    Table[Product[Binomial[2 i, 2], {i, 1, n}], {n, 0, 16}]
polygorial[k_, n_] := FullSimplify[ n!/2^n (k -2)^n*Pochhammer[2/(k -2), n]]; Array[ polygorial[6, #] &, 17, 0] (* Robert G. Wilson v, Dec 26 2016 *)
Table[(2n)!/2^n,{n,0,20}] (* Harvey P. Dale, Sep 21 2020 *)
  • PARI
    a(n) = (2*n)! / 2^n

Formula

E.g.f.: 1/(1 - x^2/2) (with interpolating zeros). - Paul Barry, May 26 2003
a(n) = polygorial(n, 6) = (A000142(n)/A000079(n))*A001813(n) = (n!/2^n)*Product_{i=0..n-1} (4*i + 2) = (n!/2^n)*4^n*Pochhammer(1/2, n) = gamma(2*n+1)/2^n. - Daniel Dockery (peritus(AT)gmail.com), Jun 13 2003
a(n) = A087127(n,2*n) = Sum_{i=0..2*n} (-1)^(2*n-i)*binomial(2*n, i)*binomial(i+2, 2)^n. Let T(n,k,j) = ((n - k + j)*(2*n - 2*k + 1))^n*binomial(2*n, 2*k-j+1) then a(n) = Sum{k=0..n} (T(n,k,1) - T(n,k,0)). For example a(12) = A087127(12,24) = Sum_{k=0..12} (T(12,k,1) - T(12,k,0)) = 24!/2^12. - André F. Labossière, Mar 29 2004 [Corrected by Jianing Song, Jan 08 2019]
For even n, a(n) = binomial(2n, n)*(a(n/2))^2. For odd n, a(n) = binomial(2n, n+1)*a((n+1)/2)*a((n-1)/2). For positive n, a(n) = binomial(2n, 2)*a(n-1) with a(0) = 1. - Dennis P. Walsh, Nov 17 2009
a(n) = Product_{i=1..n} binomial(2i, 2).
a(n) = a(n-1)*binomial(2n, 2).
From Peter Bala, Feb 21 2011: (Start)
a(n) = Product_{k = 0..n-1} (T(n) - T(k)), where T(n) = n*(n + 1)/2 is the n-th triangular number.
Compare with n! = Product_{k = 0..n-1} (n - k).
Thus we may view a(n) as a generalized factorial function associated with the triangular numbers A000217. Cf. A010050. The corresponding generalized binomial coefficients a(n)/(a(k)*a(n-k)) are triangle A086645. Also cf. A186432.
a(n) = n*(n + n-1)*(n + n-1 + n-2)*...*(n + n-1 + n-2 + ... + 1).
For example, a(5) = 5*(5+4)*(5+4+3)*(5+4+3+2)*(5+4+3+2+1) = 113400. (End).
G.f.: 1/U(0) where U(k)= x*(2*k - 1)*k + 1 - x*(2*k + 1)*(k + 1)/U(k+1); (continued fraction, Euler's 1st kind, 1-step). - Sergei N. Gladkovskii, Oct 28 2012
a(n) = n!*(product of the first n odd integers). - Dennis P. Walsh, Nov 28 2012
a(0) = 1, a(n) = a(n-1)*T(2*n-1), where T(n) is the n-th triangular number. For example: a(4) = a(3)*T(7) = 90*28 = 2520. - Enric Reverter i Bigas, Jun 24 2013
E.g.f.: 1/(1 - x/(1 - 2*x/(1 - 3*x/(1 - 4*x/(1 - 5*x/(1 - ...)))))), a continued fraction. - Ilya Gutkovskiy, May 10 2017
From Amiram Eldar, Jun 25 2020: (Start)
Sum_{n>=0} 1/a(n) = cosh(sqrt(2)).
Sum_{n>=0} (-1)^n/a(n) = cos(sqrt(2)). (End)
D-finite with recurrence a(n) -n*(2*n-1)*a(n-1)=0. - R. J. Mathar, Jan 28 2022
a(n) = n *A007019(n-1), n>0. - R. J. Mathar, Jan 28 2022

A028246 Triangular array a(n,k) = (1/k)*Sum_{i=0..k} (-1)^(k-i)*binomial(k,i)*i^n; n >= 1, 1 <= k <= n, read by rows.

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 7, 12, 6, 1, 15, 50, 60, 24, 1, 31, 180, 390, 360, 120, 1, 63, 602, 2100, 3360, 2520, 720, 1, 127, 1932, 10206, 25200, 31920, 20160, 5040, 1, 255, 6050, 46620, 166824, 317520, 332640, 181440, 40320, 1, 511, 18660, 204630, 1020600, 2739240, 4233600, 3780000, 1814400, 362880
Offset: 1

Views

Author

N. J. A. Sloane, Doug McKenzie (mckfam4(AT)aol.com)

Keywords

Comments

Let M = n X n matrix with (i,j)-th entry a(n+1-j, n+1-i), e.g., if n = 3, M = [1 1 1; 3 1 0; 2 0 0]. Given a sequence s = [s(0)..s(n-1)], let b = [b(0)..b(n-1)] be its inverse binomial transform and let c = [c(0)..c(n-1)] = M^(-1)*transpose(b). Then s(k) = Sum_{i=0..n-1} b(i)*binomial(k,i) = Sum_{i=0..n-1} c(i)*k^i, k=0..n-1. - Gary W. Adamson, Nov 11 2001
From Gary W. Adamson, Aug 09 2008: (Start)
Julius Worpitzky's 1883 algorithm generates Bernoulli numbers.
By way of example [Wikipedia]:
B0 = 1;
B1 = 1/1 - 1/2;
B2 = 1/1 - 3/2 + 2/3;
B3 = 1/1 - 7/2 + 12/3 - 6/4;
B4 = 1/1 - 15/2 + 50/3 - 60/4 + 24/5;
B5 = 1/1 - 31/2 + 180/3 - 390/4 + 360/5 - 120/6;
B6 = 1/1 - 63/2 + 602/3 - 2100/4 + 3360/5 - 2520/6 + 720/7;
...
Note that in this algorithm, odd n's for the Bernoulli numbers sum to 0, not 1, and the sum for B1 = 1/2 = (1/1 - 1/2). B3 = 0 = (1 - 7/2 + 13/3 - 6/4) = 0. The summation for B4 = -1/30. (End)
Pursuant to Worpitzky's algorithm and given M = A028246 as an infinite lower triangular matrix, M * [1/1, -1/2, 1/3, ...] (i.e., the Harmonic series with alternate signs) = the Bernoulli numbers starting [1/1, 1/2, 1/6, ...]. - Gary W. Adamson, Mar 22 2012
From Tom Copeland, Oct 23 2008: (Start)
G(x,t) = 1/(1 + (1-exp(x*t))/t) = 1 + 1 x + (2 + t)*x^2/2! + (6 + 6t + t^2)*x^3/3! + ... gives row polynomials for A090582, the f-polynomials for the permutohedra (see A019538).
G(x,t-1) = 1 + 1*x + (1 + t)*x^2 / 2! + (1 + 4t + t^2)*x^3 / 3! + ... gives row polynomials for A008292, the h-polynomials for permutohedra.
G[(t+1)x,-1/(t+1)] = 1 + (1+ t) x + (1 + 3t + 2 t^2) x^2 / 2! + ... gives row polynomials for the present triangle. (End)
The Worpitzky triangle seems to be an apt name for this triangle. - Johannes W. Meijer, Jun 18 2009
If Pascal's triangle is written as a lower triangular matrix and multiplied by A028246 written as an upper triangular matrix, the product is a matrix where the (i,j)-th term is (i+1)^j. For example,
1,0,0,0 1,1,1, 1 1,1, 1, 1
1,1,0,0 * 0,1,3, 7 = 1,2, 4, 8
1,2,1,0 0,0,2,12 1,3, 9,27
1,3,3,1 0,0,0, 6 1,4,16,64
So, numbering all three matrices' rows and columns starting at 0, the (i,j) term of the product is (i+1)^j. - Jack A. Cohen (ProfCohen(AT)comcast.net), Aug 03 2010
The Fi1 and Fi2 triangle sums are both given by sequence A000670. For the definition of these triangle sums see A180662. The mirror image of the Worpitzky triangle is A130850. - Johannes W. Meijer, Apr 20 2011
Let S_n(m) = 1^m + 2^m + ... + n^m. Then, for n >= 0, we have the following representation of S_n(m) as a linear combination of the binomial coefficients:
S_n(m) = Sum_{i=1..n+1} a(i+n*(n+1)/2)*C(m,i). E.g., S_2(m) = a(4)*C(m,1) + a(5)*C(m,2) + a(6)*C(m,3) = C(m,1) + 3*C(m,2) + 2*C(m,3). - Vladimir Shevelev, Dec 21 2011
Given the set X = [1..n] and 1 <= k <= n, then a(n,k) is the number of sets T of size k of subset S of X such that S is either empty or else contains 1 and another element of X and such that any two elemements of T are either comparable or disjoint. - Michael Somos, Apr 20 2013
Working with the row and column indexing starting at -1, a(n,k) gives the number of k-dimensional faces in the first barycentric subdivision of the standard n-dimensional simplex (apply Brenti and Welker, Lemma 2.1). For example, the barycentric subdivision of the 2-simplex (a triangle) has 1 empty face, 7 vertices, 12 edges and 6 triangular faces giving row 4 of this triangle as (1,7,12,6). Cf. A053440. - Peter Bala, Jul 14 2014
See A074909 and above g.f.s for associations among this array and the Bernoulli polynomials and their umbral compositional inverses. - Tom Copeland, Nov 14 2014
An e.g.f. G(x,t) = exp[P(.,t)x] = 1/t - 1/[t+(1-t)(1-e^(-xt^2))] = (1-t) * x + (-2t + 3t^2 - t^3) * x^2/2! + (6t^2 - 12t^3 + 7t^4 - t^5) * x^3/3! + ... for the shifted, reverse, signed polynomials with the first element nulled, is generated by the infinitesimal generator g(u,t)d/du = [(1-u*t)(1-(1+u)t)]d/du, i.e., exp[x * g(u,t)d/du] u eval. at u=0 generates the polynomials. See A019538 and the G. Rzadkowski link below for connections to the Bernoulli and Eulerian numbers, a Ricatti differential equation, and a soliton solution to the KdV equation. The inverse in x of this e.g.f. is Ginv(x,t) = (-1/t^2)*log{[1-t(1+x)]/[(1-t)(1-tx)]} = [1/(1-t)]x + [(2t-t^2)/(1-t)^2]x^2/2 + [(3t^2-3t^3+t^4)/(1-t)^3]x^3/3 + [(4t^3-6t^4+4t^5-t^6)/(1-t)^4]x^4/4 + ... . The numerators are signed, shifted A135278 (reversed A074909), and the rational functions are the columns of A074909. Also, dG(x,t)/dx = g(G(x,t),t) (cf. A145271). (Analytic G(x,t) added, and Ginv corrected and expanded on Dec 28 2015.) - Tom Copeland, Nov 21 2014
The operator R = x + (1 + t) + t e^{-D} / [1 + t(1-e^(-D))] = x + (1+t) + t - (t+t^2) D + (t+3t^2+2t^3) D^2/2! - ... contains an e.g.f. of the reverse row polynomials of the present triangle, i.e., A123125 * A007318 (with row and column offset 1 and 1). Umbrally, R^n 1 = q_n(x;t) = (q.(0;t)+x)^n, with q_m(0;t) = (t+1)^(m+1) - t^(m+1), the row polynomials of A074909, and D = d/dx. In other words, R generates the Appell polynomials associated with the base sequence A074909. For example, R 1 = q_1(x;t) = (q.(0;t)+x) = q_1(0;t) + q__0(0;t)x = (1+2t) + x, and R^2 1 = q_2(x;t) = (q.(0;t)+x)^2 = q_2(0:t) + 2q_1(0;t)x + q_0(0;t)x^2 = 1+3t+3t^2 + 2(1+2t)x + x^2. Evaluating the polynomials at x=0 regenerates the base sequence. With a simple sign change in R, R generates the Appell polynomials associated with A248727. - Tom Copeland, Jan 23 2015
For a natural refinement of this array, see A263634. - Tom Copeland, Nov 06 2015
From Wolfdieter Lang, Mar 13 2017: (Start)
The e.g.f. E(n, x) for {S(n, m)}{m>=0} with S(n, m) = Sum{k=1..m} k^n, n >= 0, (with undefined sum put to 0) is exp(x)*R(n+1, x) with the exponential row polynomials R(n, x) = Sum_{k=1..n} a(n, k)*x^k/k!. E.g., e.g.f. for n = 2, A000330: exp(x)*(1*x/1!+3*x^2/2!+2*x^3/3!).
The o.g.f. G(n, x) for {S(n, m)}{m >=0} is then found by Laplace transform to be G(n, 1/p) = p*Sum{k=1..n} a(n+1, k)/(p-1)^(2+k).
Hence G(n, x) = x/(1 - x)^(n+2)*Sum_{k=1..n} A008292(n,k)*x^(k-1).
E.g., n=2: G(2, 1/p) = p*(1/(p-1)^2 + 3/(p-1)^3 + 2/(p-1)^4) = p^2*(1+p)/(p-1)^4; hence G(2, x) = x*(1+x)/(1-x)^4.
This works also backwards: from the o.g.f. to the e.g.f. of {S(n, m)}_{m>=0}. (End)
a(n,k) is the number of k-tuples of pairwise disjoint and nonempty subsets of a set of size n. - Dorian Guyot, May 21 2019
From Rajesh Kumar Mohapatra, Mar 16 2020: (Start)
a(n-1,k) is the number of chains of length k in a partially ordered set formed from subsets of an n-element set ordered by inclusion such that the first term of the chains is either the empty set or an n-element set.
Also, a(n-1,k) is the number of distinct k-level rooted fuzzy subsets of an n-set ordered by set inclusion. (End)
The relations on p. 34 of Hasan (also p. 17 of Franco and Hasan) agree with the relation between A019538 and this entry given in the formula section. - Tom Copeland, May 14 2020
T(n,k) is the size of the Green's L-classes in the D-classes of rank (k-1) in the semigroup of partial transformations on an (n-1)-set. - Geoffrey Critzer, Jan 09 2023
T(n,k) is the number of strongly connected binary relations on [n] that have period k (A367948) and index 1. See Theorem 5.4.25(6) in Ki Hang Kim reference. - Geoffrey Critzer, Dec 07 2023

Examples

			The triangle a(n, k) starts:
n\k 1   2    3     4      5      6      7      8     9
1:  1
2:  1   1
3:  1   3    2
4:  1   7   12     6
5:  1  15   50    60     24
6:  1  31  180   390    360    120
7:  1  63  602  2100   3360   2520    720
8:  1 127 1932 10206  25200  31920  20160   5040
9:  1 255 6050 46620 166824 317520 332640 181440 40320
... [Reformatted by _Wolfdieter Lang_, Mar 26 2015]
-----------------------------------------------------
Row 5 of triangle is {1,15,50,60,24}, which is {1,15,25,10,1} times {0!,1!,2!,3!,4!}.
From _Vladimir Shevelev_, Dec 22 2011: (Start)
Also, for power sums, we have
S_0(n) = C(n,1);
S_1(n) = C(n,1) +    C(n,2);
S_2(n) = C(n,1) +  3*C(n,2) +  2*C(n,3);
S_3(n) = C(n,1) +  7*C(n,2) + 12*C(n,3) +  6*C(n,4);
S_4(n) = C(n,1) + 15*C(n,2) + 50*C(n,3) + 60*C(n,4) + 24*C(n,5); etc.
(End)
For X = [1,2,3], the sets T are {{}}, {{},{1,2}}, {{},{1,3}}, {{},{1,2,3}}, {{},{1,2},{1,2,3}}, {{},{1,3},{1,2,3}} and so a(3,1)=1, a(3,2)=3, a(3,3)=2. - _Michael Somos_, Apr 20 2013

References

  • Ki Hang Kim, Boolean Matrix Theory and Applications, Marcel Dekker, New York and Basel (1982).

Links

Crossrefs

Dropping the column of 1's gives A053440.
Without the k in the denominator (in the definition), we get A019538. See also the Stirling number triangle A008277.
Cf. A087127, A087107, A087108, A087109, A087110, A087111, A084938 A075263.
Row sums give A000629(n-1) for n >= 1.
Cf. A027642, A002445. - Gary W. Adamson, Aug 09 2008
Appears in A161739 (RSEG2 triangle), A161742 and A161743. - Johannes W. Meijer, Jun 18 2009
Binomial transform is A038719. Cf. A131689.
Cf. A007318, A008292, A046802, A074909, A090582, A123125, A130850, A135278, A141618, A145271, A163626, A248727, A263634.
Cf. A119879.
From Rajesh Kumar Mohapatra, Mar 29 2020: (Start)
A000007(n-1) (column k=1), A000225(n-1) (column k=2), A028243(n-1) (column k=3), A028244(n-1) (column k=4), A028245(n-1) (column k=5), for n > 0.
Diagonal gives A000142(n-1), for n >=1.
Next-to-last diagonal gives A001710,
Third, fourth, fifth, sixth, seventh external diagonal respectively give A005460, A005461, A005462, A005463, A005464. (End)

Programs

  • GAP
    Flat(List([1..10], n-> List([1..n], k-> Stirling2(n,k)* Factorial(k-1) ))); # G. C. Greubel, May 30 2019
  • Magma
    [[StirlingSecond(n,k)*Factorial(k-1): k in [1..n]]: n in [1..10]]; // G. C. Greubel, May 30 2019
  • Maple
    a := (n,k) -> add((-1)^(k-i)*binomial(k,i)*i^n, i=0..k)/k;
seq(print(seq(a(n,k),k=1..n)),n=1..10);
T := (n,k) -> add(eulerian1(n,j)*binomial(n-j,n-k), j=0..n):
seq(print(seq(T(n,k),k=0..n)),n=0..9); # Peter Luschny, Jul 12 2013
  • Mathematica
    a[n_, k_] = Sum[(-1)^(k-i) Binomial[k,i]*i^n, {i,0,k}]/k; Flatten[Table[a[n, k], {n, 10}, {k, n}]] (* Jean-François Alcover, May 02 2011 *)
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, n! * polcoeff( (x / log(1 + x + x^2 * O(x^n) ))^(n+1), n-k))}; /* Michael Somos, Oct 02 2002 */
  • PARI
    {T(n,k) = stirling(n,k,2)*(k-1)!}; \\ G. C. Greubel, May 31 2019
  • Python
    # Assuming offset (n, k) = (0, 0).
def T(n, k):
    if k >  n: return 0
    if k == 0: return 1
    return k*T(n - 1, k - 1) + (k + 1)*T(n - 1, k)
for n in range(9):
    print([T(n, k) for k in range(n + 1)])  # Peter Luschny, Apr 26 2022
  • Sage
    def A163626_row(n) :
    x = polygen(ZZ,'x')
    A = []
    for m in range(0, n, 1) :
        A.append((-x)^m)
        for j in range(m, 0, -1):
            A[j - 1] = j * (A[j - 1] - A[j])
    return list(A[0])
for i in (1..7) : print(A163626_row(i))  # Peter Luschny, Jan 25 2012
  • Sage
    [[stirling_number2(n,k)*factorial(k-1) for k in (1..n)] for n in (1..10)] # G. C. Greubel, May 30 2019

Formula

E.g.f.: -log(1-y*(exp(x)-1)). - Vladeta Jovovic, Sep 28 2003
a(n, k) = S2(n, k)*(k-1)! where S2(n, k) is a Stirling number of the second kind (cf. A008277). Also a(n,k) = T(n,k)/k, where T(n, k) = A019538.
Essentially same triangle as triangle [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, 0, 7, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] where DELTA is Deléham's operator defined in A084938, but the notation is different.
Sum of terms in n-th row = A000629(n) - Gary W. Adamson, May 30 2005
The row generating polynomials P(n, t) are given by P(1, t)=t, P(n+1, t) = t(t+1)(d/dt)P(n, t) for n >= 1 (see the Riskin and Beckwith reference). - Emeric Deutsch, Aug 09 2005
From Gottfried Helms, Jul 12 2006: (Start)
Delta-matrix as can be read from H. Hasse's proof of a connection between the zeta-function and Bernoulli numbers (see link below).
Let P = lower triangular matrix with entries P[row,col] = binomial(row,col).
Let J = unit matrix with alternating signs J[r,r]=(-1)^r.
Let N(m) = column matrix with N(m)(r) = (r+1)^m, N(1)--> natural numbers.
Let V = Vandermonde matrix with V[r,c] = (r+1)^c.
V is then also N(0)||N(1)||N(2)||N(3)... (indices r,c always beginning at 0).
Then Delta = P*J * V and B' = N(-1)' * Delta, where B is the column matrix of Bernoulli numbers and ' means transpose, or for the single k-th Bernoulli number B_k with the appropriate column of Delta,
B_k = N(-1)' * Delta[ *,k ] = N(-1)' * P*J * N(k).
Using a single column instead of V and assuming infinite dimension, H. Hasse showed that in x = N(-1) * P*J * N(s), where s can be any complex number and s*zeta(1-s) = x.
His theorem reads: s*zeta(1-s) = Sum_{n>=0..inf} (n+1)^-1*delta(n,s), where delta(n,s) = Sum_{j=0..n} (-1)^j * binomial(n,j) * (j+1)^s.
(End)
a(n,k) = k*a(n-1,k) + (k-1)*a(n-1,k-1) with a(n,1) = 1 and a(n,n) = (n-1)!. - Johannes W. Meijer, Jun 18 2009
Rephrasing the Meijer recurrence above: Let M be the (n+1)X(n+1) bidiagonal matrix with M(r,r) = M(r,r+1) = r, r >= 1, in the two diagonals and the rest zeros. The row a(n+1,.) of the triangle is row 1 of M^n. - Gary W. Adamson, Jun 24 2011
From Tom Copeland, Oct 11 2011: (Start)
With e.g.f.. A(x,t) = G[(t+1)x,-1/(t+1)]-1 (from 2008 comment) = -1 + 1/[1-(1+t)(1-e^(-x))] = (1+t)x + (1+3t+2t^2)x^2/2! + ..., the comp. inverse in x is
B(x,t)= -log(t/(1+t)+1/((1+t)(1+x))) = (1/(1+t))x - ((1+2t)/(1+t)^2)x^2/2 + ((1+3t+3t^2)/(1+t)^3)x^3/3 + .... The numerators are the row polynomials of A074909, and the rational functions are (omitting the initial constants) signed columns of the re-indexed Pascal triangle A007318.
Let h(x,t)= 1/(dB/dx) = (1+x)(1+t(1+x)), then the row polynomial P(n,t) = (1/n!)(h(x,t)*d/dx)^n x, evaluated at x=0, A=exp(x*h(y,t)*d/dy) y, eval. at y=0, and dA/dx = h(A(x,t),t), with P(1,t)=1+t. (Series added Dec 29 2015.)(End)
Let denote the Eulerian numbers A173018(n,k), then T(n,k) = Sum_{j=0..n} *binomial(n-j,n-k). - Peter Luschny, Jul 12 2013
Matrix product A007318 * A131689. The n-th row polynomial R(n,x) = Sum_{k >= 1} k^(n-1)*(x/(1 + x))^k, valid for x in the open interval (-1/2, inf). Cf A038719. R(n,-1/2) = (-1)^(n-1)*(2^n - 1)*Bernoulli(n)/n. - Peter Bala, Jul 14 2014
a(n,k) = A141618(n,k) / C(n,k-1). - Tom Copeland, Oct 25 2014
For the row polynomials, A028246(n,x) = A019538(n-1,x) * (1+x). - Tom Copeland, Dec 28 2015
A248727 = A007318*(reversed A028246) = A007318*A130850 = A007318*A123125*A007318 = A046802*A007318. - Tom Copeland, Nov 14 2016
n-th row polynomial R(n,x) = (1+x) o (1+x) o ... o (1+x) (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See example E11 in the Bala link. - Peter Bala, Jan 12 2018
From Dorian Guyot, May 21 2019: (Start)
Sum_{i=0..k} binomial(k,i) * a(n,i) = (k+1)^n.
Sum_{k=0..n} a(n,k) = 2*A000670(n).
(End)
With all offsets 0, let A_n(x;y) = (y + E.(x))^n, an Appell sequence in y where E.(x)^k = E_k(x) are the Eulerian polynomials of A123125. Then the row polynomials of this entry, A028246, are given by x^n * A_n(1 + 1/x;0). Other specializations of A_n(x;y) give A046802, A090582, A119879, A130850, and A248727. - Tom Copeland, Jan 24 2020
The row generating polynomials R(n,x) = Sum_{i=1..n} a(n,i) * x^i satisfy the recurrence equation R(n+1,x) = R(n,x) + Sum_{k=0..n-1} binomial(n-1,k) * R(k+1,x) * R(n-k,x) for n >= 1 with initial value R(1,x) = x. - Werner Schulte, Jun 17 2021

Extensions

Definition corrected by Li Guo, Dec 16 2006
Typo in link corrected by Johannes W. Meijer, Oct 17 2009
Error in title corrected by Johannes W. Meijer, Sep 24 2010
Edited by M. F. Hasler, Oct 29 2014

A024166 a(n) = Sum_{1 <= i < j <= n} (j-i)^3.

Original entry on oeis.org

0, 1, 10, 46, 146, 371, 812, 1596, 2892, 4917, 7942, 12298, 18382, 26663, 37688, 52088, 70584, 93993, 123234, 159334, 203434, 256795, 320804, 396980, 486980, 592605, 715806, 858690, 1023526, 1212751, 1428976, 1674992, 1953776, 2268497, 2622522, 3019422
Offset: 0

Views

Author

Clark Kimberling

Keywords

Comments

Convolution of the cubes (A000578) with the positive integers a(n)=n+1, where all sequences have offset zero. - Graeme McRae, Jun 06 2006
a(n) gives the n-th antidiagonal sum of the convolution array A212891. - Clark Kimberling, Jun 16 2012
In general, the r-th successive summation of the cubes from 1 to n is (6*n^2 + 6*n*r + r^2 - r)*(n+r)!/((r+3)!*(n-1)!), n>0. Here r = 2. - Gary Detlefs, Mar 01 2013
The inverse binomial transform is (essentially) row n=2 of A087127. - R. J. Mathar, Aug 31 2022

Examples

			4*a(7) = 6384 = (0*1)^2 + (1*2)^2 + (2*3)^2 + (3*4)^2 + (4*5)^2 + (5*6)^2 + (6*7)^2 + (7*8)^2. - _Bruno Berselli_, Feb 05 2014

References

  • Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.

Links

Crossrefs

Cf. A000292, A000332, A000389, A000579, A000580, A024166, A027555, A085438, A085439, A085440, A085441, A085442, A086020, A086021, A086022, A086023, A086024, A086025, A086026, A086027, A086028, A086029, A086030, A087127.
Cf. A000330, A000537 (first diffs), A001286, A003215, A100536, A101094 (partial sums), A101097, A101102, A222716.

Programs

  • Haskell
    a024166 n = sum $ zipWith (*) [n+1,n..0] a000578_list
-- Reinhard Zumkeller, Oct 14 2001
  • Magma
    [n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: n in [0..30]]; // G. C. Greubel, Nov 21 2017
  • Maple
    A024166:=n->n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60: seq(A024166(n), n=0..50); # Wesley Ivan Hurt, Nov 21 2017
  • Mathematica
    Nest[Accumulate,Range[0,40]^3,2] (* Harvey P. Dale, Jan 10 2016 *)
Table[n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, {n,0,30}] (* G. C. Greubel, Nov 21 2017 *)
  • PARI
    a(n)=sum(j=1,n, sum(m=1, j, sum(i=m*(m+1)/2-m+1, m*(m+1)/2, (2*i-1)))) \\ Alexander R. Povolotsky, May 17 2008
  • PARI
    for(n=0,30, print1(n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60, ", ")) \\ G. C. Greubel, Nov 21 2017

Formula

From Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Dec 29 1999: (Start)
a(n) = Sum_{i=0..n} A000537(i), partial sums of A000537.
a(n) = n*(n+1)*(n+2)*(3*n^2 + 6*n + 1)/60. (End)
a(A004772(n)) mod 2 = 0; a(A016813(n)) mod 2 = 1. - Reinhard Zumkeller, Oct 14 2001
a(n) = Sum_{k=0..n} k^3*(n+1-k). - Paul Barry, Sep 14 2003; edited by Jon E. Schoenfield, Dec 29 2014
a(n) = 2*n*(n+1)*(n+2)*((n+1)^2 + 2*n*(n+2))/5!. This sequence could be obtained from the general formula a(n) = n*(n+1)*(n+2)*(n+3)* ...* (n+k) *(n*(n+k) + (k-1)*k/6)/((k+3)!/6) at k=2. - Alexander R. Povolotsky, May 17 2008
O.g.f.: x*(1 + 4*x + x^2)/(-1 + x)^6. - R. J. Mathar, Jun 06 2008
a(n) = (6*n^2 + 12*n + 2)*(n+2)!/(120*(n-1)!), n > 0. - Gary Detlefs, Mar 01 2013
a(n) = A222716(n+1)/10 = A000292(n)*A100536(n+1)/10. - Jonathan Sondow, Mar 04 2013
4*a(n) = Sum_{i=0..n} A000290(i)*A000290(i+1). - Bruno Berselli, Feb 05 2014
a(n) = Sum_{i=1..n} Sum_{j=1..n} i*j*(n - max(i, j) + 1). - Melvin Peralta, May 12 2016
a(n) = n*binomial(n+3, 4) + binomial(n+2, 5). - Tony Foster III, Nov 14 2017
a(n) = Sum_{i=1..n} i*A143037(n,n-i+1). - J. M. Bergot, Aug 30 2022

A086020 a(n) = Sum_(i=1..n) binomial(i+2,3)^2 [ Sequential sums of the tetragonal numbers or "tetras" (pyramidal, square) raised to power 2 (drawn from the 4th diagonal - left or right - of Pascal's Triangle) ].

Original entry on oeis.org

1, 17, 117, 517, 1742, 4878, 11934, 26334, 53559, 101959, 183755, 316251, 523276, 836876, 1299276, 1965132, 2904093, 4203693, 5972593, 8344193, 11480634, 15577210, 20867210, 27627210, 36182835, 46915011, 60266727, 76750327
Offset: 1

Views

Author

André F. Labossière, Jul 17 2003

Keywords

Comments

Kekulé numbers for certain benzenoids (see the Cyvin-Gutman reference, p. 243; expression in (13.26) yields same sequence with offset 0). - Emeric Deutsch, Aug 02 2005
Partial sums of A001249. - R. J. Mathar, Aug 19 2008

Examples

			a(8) = Sum_{i=1..8} binomial(i+2,3)^2 = (20*(8^7) + 210*(8^6) + 854*(8^5) + 1680*(8^4) + 1610*(8^3) + 630*(8^2) + 36*8)/7! = 26334.

Links

Crossrefs

Cf. A000292, A087127, A024166, A085438, A085439, A085440, A085441, A085442, A000332, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030.

Programs

  • Magma
    [n*(n+1)*(n+2)*(n+3)*(2*n+3)*(5*n^2+15*n+1)/2520: n in [1..30]]; // G. C. Greubel, Nov 22 2017
  • Maple
    a:=n->n*(n+1)*(n+2)*(n+3)*(2*n+3)*(5*n^2+15*n+1)/2520: seq(a(n),n=1..31); # Emeric Deutsch
  • Mathematica
    Accumulate[Binomial[Range[30]+2,3]^2]  (* Harvey P. Dale, Mar 24 2011 *)
LinearRecurrence[{8,-28,56,-70,56,-28,8,-1},{1,17,117,517,1742,4878, 11934, 26334},30] (* Harvey P. Dale, Aug 17 2014 *)
  • PARI
    a(n)=n*(n+1)*(n+2)*(n+3)*(2*n+3)*(5*n^2+15*n+1)/2520 \\ Charles R Greathouse IV, May 18 2015

Formula

a(n) = Sum_(i=1..n) binomial(i+2, 3)^2.
a(n) = ( C(n+3, 4)/35 )*( 35 + 84*C(n-1, 1) + 70*C(n-1, 2) + 20*C(n-1, 3) ).
a(n) = n*(n+1)*(n+2)*(n+3)*(2*n+3)(5*n^2 + 15*n + 1)/2520. - Emeric Deutsch, Aug 02 2005
O.g.f: x*(1+x)*(1 + 8*x + x^2)/(1-x)^8. - R. J. Mathar, Aug 19 2008

A085438 a(n) = Sum_{i=1..n} binomial(i+1,2)^3.

Original entry on oeis.org

1, 28, 244, 1244, 4619, 13880, 35832, 82488, 173613, 339988, 627484, 1102036, 1855607, 3013232, 4741232, 7256688, 10838265, 15838476, 22697476, 31958476, 44284867, 60479144, 81503720, 108503720, 142831845, 186075396, 240085548, 307008964, 389321839
Offset: 1

Views

Author

André F. Labossière, Jun 30 2003

Keywords

Examples

			a(10) = (90*(10^7)+630*(10^6)+1638*(10^5)+1890*(10^4)+840*(10^3)-48*(10))/5040 = 339988.

References

  • Elisabeth Busser and Gilles Cohen, Neuro-Logies - "Chercher, jouer, trouver", La Recherche, April 1999, No. 319, page 97.

Links

Crossrefs

Column k=3 of A334781.
Cf. A000292, A087127, A024166, A024166, A085439, A085440, A085441, A085442, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030.

Programs

  • Magma
    [(90*n^7 +630*n^6 +1638*n^5 +1890*n^4+ 840*n^3 -48*n)/ Factorial(7): n in [1..30]]; // G. C. Greubel, Nov 22 2017
  • Mathematica
    Table[(90*n^7 + 630*n^6 + 1638*n^5 + 1890*n^4 + 840*n^3 - 48*n)/7!, {n, 1, 50}] (* G. C. Greubel, Nov 22 2017 *)
  • PARI
    Vec(x*(x^4+20*x^3+48*x^2+20*x+1)/(x-1)^8 + O(x^100)) \\ Colin Barker, May 02 2014
  • PARI
    a(n) = sum(i=1, n, binomial(i+1, 2)^3); \\ Michel Marcus, Nov 22 2017

Formula

a(n) = (90*n^7 +630*n^6 +1638*n^5 +1890*n^4+ 840*n^3 -48*n)/7!.
a(n) = (C(n+2, 3)/35)*(35 +210*C(n-1, 1) +399*C(n-1, 2) +315*C(n-1, 3) +90*C(n-1, 4)).
G.f.: x*(x^4+20*x^3+48*x^2+20*x+1) / (x-1)^8. - Colin Barker, May 02 2014

Extensions

More terms from Colin Barker, May 02 2014
Formula and example edited by Colin Barker, May 02 2014

A085442 a(n) = Sum_{i=1..n} binomial(i+1,2)^7.

Original entry on oeis.org

1, 2188, 282124, 10282124, 181141499, 1982230040, 15475158552, 93839322648, 467508775773, 1989944010148, 7445104711204, 25010673566116, 76686775501847, 217396817767472, 575714897767472, 1436257466526768, 3398894618986905, 7674255436599996, 16612972826599996
Offset: 1

Views

Author

André F. Labossière, Jul 07 2003

Keywords

Links

Crossrefs

Column k=7 of A334781.
Cf. A000292, A087127, A024166, A024166, A085438, A085439, A085440, A085441, A000332, A086020, A086021, A086022, A000389, A086023, A086024, A000579, A086025, A086026, A000580, A086027, A086028, A027555, A086029, A086030.

Programs

  • Magma
    [(1/823680) *n*(n+1)*(n+2)*(429*n^12 +5148*n^11 +24123*n^10 +52470*n^9 +43047*n^8 -8856*n^7 +4109*n^6 +50430*n^5 -18796*n^4 -44472*n^3 +26864*n^2 +8352*n -5568): n in [1..30]]; // G. C. Greubel, Nov 22 2017
  • Mathematica
    Table[Sum[Binomial[k+1,2]^7, {k,1,n}], {n,1,30}] (* G. C. Greubel, Nov 22 2017 *)
LinearRecurrence[{16,-120,560,-1820,4368,-8008,11440,-12870,11440,-8008,4368,-1820,560,-120,16,-1},{1,2188,282124,10282124,181141499,1982230040,15475158552,93839322648,467508775773,1989944010148,7445104711204,25010673566116,76686775501847,217396817767472,575714897767472,1436257466526768},20] (* Harvey P. Dale, May 11 2022 *)
  • PARI
    for(n=1,30, print1(sum(k=1,n, binomial(k+1,2)^7), ", ")) \\ G. C. Greubel, Nov 22 2017

Formula

a(n) = (1/823680) *n*(n+1)*(n+2)*(429*n^12 +5148*n^11 +24123*n^10 +52470*n^9 +43047*n^8 -8856*n^7 +4109*n^6 +50430*n^5 -18796*n^4 -44472*n^3 +26864*n^2 +8352*n -5568). - Vladeta Jovovic, Jul 07 2003
G.f.: x*(x^12 +2172*x^11 +247236*x^10 +6030140*x^9 +49258935*x^8 +163809288*x^7 +242384856*x^6 +163809288*x^5 +49258935*x^4 +6030140*x^3 +247236*x^2 +2172*x+ 1) / (x -1)^16. - Colin Barker, May 02 2014

A086030 a(n) = Sum_{i=1..n} C(i+6,7)^3.

Original entry on oeis.org

1, 513, 47169, 1775169, 37712169, 534505257, 5587534953, 46011772521, 312480135396, 1809674119396, 9165388162788, 41395684407012, 169328324418084, 635173167426084, 2207399512578084, 7167715400927268, 21902130296812161, 63361228916945025, 174437774859945025
Offset: 1

Views

Author

André F. Labossière, Jul 11 2003

Keywords

Examples

			a(3) = Sum_{i=1..3} C(6+i,7)^3 = C(10,8)*(5*C(16,14) + 210*C(15,14) + 1491*C(14,14))/5 = 47169.

Links

Crossrefs

Cf. A087127, A024166, A085438 - A085442, A086020 - A086029.

Programs

  • Magma
    [(1/28165294080000)*n^2*(1 + n)^2*(2 + n)^2*(3 + n)^2*(4 + n)^2*(5 + n)^2*(6 + n)^2*(7 + n)^2*(-3 + 3234*n + 6979*n^2 + 5292*n^3 + 1603*n^4 + 210*n^5 + 10*n^6): n in [1..30]]; // G. C. Greubel, Nov 22 2017
  • Maple
    A086030:=n->add(binomial(i+6,7)^3, i=1..n); seq(A086030(n), n=1..30); # Wesley Ivan Hurt, Dec 22 2013
  • Mathematica
    Table[Sum[Binomial[i + 6, 7]^3, {i, n}], {n, 30}] (* Wesley Ivan Hurt, Dec 22 2013 *)
  • PARI
    a(n) = sum(i=1, n, binomial(i+6, 7)^3); \\ Michel Marcus, Dec 22 2013

Formula

a(n) = C(n+7,8)*(5*C(n+13,14) + 210*C(n+12,14) + 1491*C(n+11,14) + 2828*C(n+10,14) + 1491*C(n+9,14) + 210*C(n+8,14) + 5*C(n+7,14))/5. - Yahia Kahloune, Dec 22 2013
-(n-1)^3*a(n) +(2*n+5)*(n^2+5*n+43)*a(n-1) -(n+6)^3*a(n-2)=0. - R. J. Mathar, Dec 22 2013
G.f.: -x*(x^14 + 490*x^13 + 35623*x^12 + 818300*x^11 + 7917371*x^10 + 37215794*x^9 + 91789005*x^8 + 123519792*x^7 + 91789005*x^6 + 37215794*x^5 + 7917371*x^4 + 818300*x^3 + 35623*x^2 + 490*x + 1)/(x-1)^23. - Vaclav Kotesovec, Dec 23 2013
a(n) = (1/28165294080000)*n^2*(1 + n)^2*(2 + n)^2*(3 + n)^2*(4 + n)^2*(5 + n)^2*(6 + n)^2*(7 + n)^2*(-3 + 3234*n + 6979*n^2 + 5292*n^3 + 1603*n^4 + 210*n^5 + 10*n^6). - G. C. Greubel, Nov 22 2017

A086023 a(n) = Sum_{i=1..n} C(i+3,4)^2.

Original entry on oeis.org

1, 26, 251, 1476, 6376, 22252, 66352, 175252, 420277, 931502, 1933503, 3796728, 7109128, 12773528, 22137128, 37160504, 60634529, 96454754, 149963979, 228375004, 341286880, 501309380, 724811880, 1032814380, 1452040005, 2016150006, 2767184031, 3757230256
Offset: 1

Views

Author

André F. Labossière, Jul 11 2003

Keywords

Links

Crossrefs

Cf. A087127, A024166, A085438, A085439, A085440, A085441, A085442, A086020, A086021, A086022, A086024, A086025, A086026, A086027, A086028, A086029, A086030.

Programs

  • Magma
    [n*(n+1)*(n+2)*(n+3)*(n+4)*(35*n^4 +280*n^3 +685*n^2 +500*n +12 )/181440: n in [1..30]]; // G. C. Greubel, Nov 22 2017
  • Mathematica
    Table[n*(n + 1)*(n + 2)*(n + 3)*(n + 4)*(35*n^4 + 280*n^3 + 685*n^2 + 500*n + 12)/181440, {n, 1, 50}] (* G. C. Greubel, Nov 22 2017 *)
Accumulate[Binomial[Range[30]+3,4]^2] (* or *) LinearRecurrence[{10,-45,120,-210,252,-210,120,-45,10,-1},{1,26,251,1476,6376,22252,66352,175252,420277,931502},30] (* Harvey P. Dale, May 06 2018 *)
  • PARI
    a(n) = sum(i=1, n, binomial(i+3, 4)^2); \\ Michel Marcus, Sep 05 2013
  • PARI
    Vec(x*(x^4+16*x^3+36*x^2+16*x+1)/(x-1)^10 + O(x^100)) \\ Colin Barker, May 02 2014

Formula

a(n) = ( C(n+4,5)/126 )*( 126 +420*C(n-1,1) +540*C(n-1,2) +315*C(n-1,3) +70*C(n-1,4) ).
a(n) = n*(n+1)*(n+2)*(n+3)*(n+4)*(35*n^4 +280*n^3 +685*n^2 +500*n +12 )/181440. - Bruno Berselli, Sep 05 2013
G.f.: x*(x^4+16*x^3+36*x^2+16*x+1) / (x-1)^10. - Colin Barker, May 02 2014

Extensions

More terms from Michel Marcus, Sep 05 2013

A086025 a(n) = Sum_{i=1..n} C(i+4,5)^2.

Original entry on oeis.org

1, 37, 478, 3614, 19490, 82994, 296438, 923702, 2580071, 6588075, 15606084, 34685508, 72976852, 146387476, 281597860, 521971876, 936053677, 1629533233, 2761788434, 4568378450, 7391175350, 11718183750, 18235516650, 27894475050, 41997225075, 62305185111
Offset: 1

Views

Author

André F. Labossière, Jul 11 2003

Keywords

Links

Crossrefs

Cf. A087127, A024166, A085438, A085439, A085440, A085441, A085442, A086020, A086021, A086022, A086023, A086024, A086026, A086027, A086028, A086029, A086030.

Programs

  • Magma
    [n*(2*n+5)*(n+5)*(n+4)*(n+3)*(n+2)*(n+1)*(63*n^4 +630*n^3 +1855*n^2 +1400*n +12)/19958400: n in [1..30]]; // G. C. Greubel, Nov 22 2017
  • Mathematica
    Table[n*(2*n+5)*(n+5)*(n+4)*(n+3)*(n+2)*(n+1)*(63*n^4 +630*n^3 +1855*n^2 +1400*n +12)/19958400, {n,1,30}] (* G. C. Greubel, Nov 22 2017 *)
LinearRecurrence[{12,-66,220,-495,792,-924,792,-495,220,-66,12,-1},{1,37,478,3614,19490,82994,296438,923702,2580071,6588075,15606084,34685508},30] (* Harvey P. Dale, Dec 22 2024 *)
  • PARI
    for(n=1,30, print1(sum(i=1,n, binomial(i+4,5)^2), ", ")) \\ G. C. Greubel, Nov 22 2017

Formula

From R. J. Mathar, Jun 16 2010: (Start)
G.f.: x*(1+x)*(x^4+24*x^3+76*x^2+24*x+1)/(x-1)^12.
a(n) = n*(2*n+5)*(n+5)*(n+4)*(n+3)*(n+2)*(n+1)*(63*n^4 +630*n^3 +1855*n^2 +1400*n +12) / 19958400. (End)

Extensions

More terms from R. J. Mathar, Jun 16 2010
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