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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A135218 A007318^(-5) * A001286, 5th inverse binomial transform of the Lah numbers.

Original entry on oeis.org

1, 1, 1, 25, 25, 745, 3145, 40945, 353425, 4200625, 49949425, 666656425, 9492440425, 145513569625, 2378864325625, 41355134658625, 761412592638625, 14800232282598625, 302834714220150625
Offset: 0

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Author

Gary W. Adamson, Nov 23 2007

Keywords

Comments

(Note this uses the incorrect offset 0 for A001286. - R. J. Mathar, Apr 04 2012)

Examples

			a(3) = (1, -2, 1) dot (1, 6, 36) = (1 - 12 + 36), where A001286 = (1, 6, 36, 240, 1800, ...).
		

Crossrefs

Cf. A001286.

A000290 The squares: a(n) = n^2.

Original entry on oeis.org

0, 1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401, 2500
Offset: 0

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Author

Keywords

Comments

To test if a number is a square, see Cohen, p. 40. - N. J. A. Sloane, Jun 19 2011
Zero followed by partial sums of A005408 (odd numbers). - Jeremy Gardiner, Aug 13 2002
Begin with n, add the next number, subtract the previous number and so on ending with subtracting a 1: a(n) = n + (n+1) - (n-1) + (n+2) - (n-2) + (n+3) - (n-3) + ... + (2n-1) - 1 = n^2. - Amarnath Murthy, Mar 24 2004
Sum of two consecutive triangular numbers A000217. - Lekraj Beedassy, May 14 2004
Numbers with an odd number of divisors: {d(n^2) = A048691(n); for the first occurrence of 2n + 1 divisors, see A071571(n)}. - Lekraj Beedassy, Jun 30 2004
See also A000037.
First sequence ever computed by electronic computer, on EDSAC, May 06 1949 (see Renwick link). - Russ Cox, Apr 20 2006
Numbers k such that the imaginary quadratic field Q(sqrt(-k)) has four units. - Marc LeBrun, Apr 12 2006
For n > 0: number of divisors of (n-1)th power of any squarefree semiprime: a(n) = A000005(A006881(k)^(n-1)); a(n) = A000005(A000400(n-1)) = A000005(A011557(n-1)) = A000005(A001023(n-1)) = A000005(A001024(n-1)). - Reinhard Zumkeller, Mar 04 2007
If a 2-set Y and an (n-2)-set Z are disjoint subsets of an n-set X then a(n-2) is the number of 3-subsets of X intersecting both Y and Z. - Milan Janjic, Sep 19 2007
Numbers a such that a^1/2 + b^1/2 = c^1/2 and a^2 + b = c. - Cino Hilliard, Feb 07 2008 (this comment needs clarification, Joerg Arndt, Sep 12 2013)
Numbers k such that the geometric mean of the divisors of k is an integer. - Ctibor O. Zizka, Jun 26 2008
Equals row sums of triangle A143470. Example: 36 = sum of row 6 terms: (23 + 7 + 3 + 1 + 1 + 1). - Gary W. Adamson, Aug 17 2008
Equals row sums of triangles A143595 and A056944. - Gary W. Adamson, Aug 26 2008
Number of divisors of 6^(n-1) for n > 0. - J. Lowell, Aug 30 2008
Denominators of Lyman spectrum of hydrogen atom. Numerators are A005563. A000290-A005563 = A000012. - Paul Curtz, Nov 06 2008
a(n) is the number of all partitions of the sum 2^2 + 2^2 + ... + 2^2, (n-1) times, into powers of 2. - Valentin Bakoev, Mar 03 2009
a(n) is the maximal number of squares that can be 'on' in an n X n board so that all the squares turn 'off' after applying the operation: in any 2 X 2 sub-board, a square turns from 'on' to 'off' if the other three are off. - Srikanth K S, Jun 25 2009
Zero together with the numbers k such that 2 is the number of perfect partitions of k. - Juri-Stepan Gerasimov, Sep 26 2009
Totally multiplicative sequence with a(p) = p^2 for prime p. - Jaroslav Krizek, Nov 01 2009
Satisfies A(x)/A(x^2), A(x) = A173277: (1, 4, 13, 32, 74, ...). - Gary W. Adamson, Feb 14 2010
Positive members are the integers with an odd number of odd divisors and an even number of even divisors. See also A120349, A120359, A181792, A181793, A181795. - Matthew Vandermast, Nov 14 2010
Besides the first term, this sequence is the denominator of Pi^2/6 = 1 + 1/4 + 1/9 + 1/16 + 1/25 + 1/36 + ... . - Mohammad K. Azarian, Nov 01 2011
Partial sums give A000330. - Omar E. Pol, Jan 12 2013
Drmota, Mauduit, and Rivat proved that the Thue-Morse sequence along the squares is normal; see A228039. - Jonathan Sondow, Sep 03 2013
a(n) can be decomposed into the sum of the four numbers [binomial(n, 1) + binomial(n, 2) + binomial(n-1, 1) + binomial(n-1, 2)] which form a "square" in Pascal's Triangle A007318, or the sum of the two numbers [binomial(n, 2) + binomial(n+1, 2)], or the difference of the two numbers [binomial(n+2, 3) - binomial(n, 3)]. - John Molokach, Sep 26 2013
In terms of triangular tiling, the number of equilateral triangles with side length 1 inside an equilateral triangle with side length n. - K. G. Stier, Oct 30 2013
Number of positive roots in the root systems of type B_n and C_n (when n > 1). - Tom Edgar, Nov 05 2013
Squares of squares (fourth powers) are also called biquadratic numbers: A000583. - M. F. Hasler, Dec 29 2013
For n > 0, a(n) is the largest integer k such that k^2 + n is a multiple of k + n. More generally, for m > 0 and n > 0, the largest integer k such that k^(2*m) + n is a multiple of k + n is given by k = n^(2*m). - Derek Orr, Sep 03 2014
For n > 0, a(n) is the number of compositions of n + 5 into n parts avoiding the part 2. - Milan Janjic, Jan 07 2016
a(n), for n >= 3, is also the number of all connected subtrees of a cycle graph, having n vertices. - Viktar Karatchenia, Mar 02 2016
On every sequence of natural continuous numbers with an even number of elements, the summatory of the second half of the sequence minus the summatory of the first half of the sequence is always a square. Example: Sequence from 61 to 70 has an even number of elements (10). Then 61 + 62 + 63 + 64 + 65 = 315; 66 + 67 + 68 + 69 + 70 = 340; 340 - 315 = 25. (n/2)^2 for n = number of elements. - César Aguilera, Jun 20 2016
On every sequence of natural continuous numbers from n^2 to (n+1)^2, the sum of the differences of pairs of elements of the two halves in every combination possible is always (n+1)^2. - César Aguilera, Jun 24 2016
Suppose two circles with radius 1 are tangent to each other as well as to a line not passing through the point of tangency. Create a third circle tangent to both circles as well as the line. If this process is continued, a(n) for n > 0 is the reciprocals of the radii of the circles, beginning with the largest circle. - Melvin Peralta, Aug 18 2016
Does not satisfy Benford's law [Ross, 2012]. - N. J. A. Sloane, Feb 08 2017
Numerators of the solution to the generalization of the Feynman triangle problem, with an offset of 2. If each vertex of a triangle is joined to the point (1/p) along the opposite side (measured say clockwise), then the area of the inner triangle formed by these lines is equal to (p - 2)^2/(p^2 - p + 1) times the area of the original triangle, p > 2. For example, when p = 3, the ratio of the areas is 1/7. The denominators of the ratio of the areas is given by A002061. [Cook & Wood, 2004] - Joe Marasco, Feb 20 2017
Equals row sums of triangle A004737, n >= 1. - Martin Michael Musatov, Nov 07 2017
Right-hand side of the binomial coefficient identity Sum_{k = 0..n} (-1)^(n+k+1)*binomial(n,k)*binomial(n + k,k)*(n - k) = n^2. - Peter Bala, Jan 12 2022
Conjecture: For n>0, min{k such that there exist subsets A,B of {0,1,2,...,a(n)-1} such that |A|=|B|=k and A+B contains {0,1,2,...,a(n)-1}} = n. - Michael Chu, Mar 09 2022
Number of 3-permutations of n elements avoiding the patterns 132, 213, 321. See Bonichon and Sun. - Michel Marcus, Aug 20 2022
Number of intercalates in cyclic Latin squares of order 2n (cyclic Latin squares of odd order do not have intercalates). - Eduard I. Vatutin, Feb 15 2024
a(n) is the number of ternary strings of length n with at most one 0, exactly one 1, and no restriction on the number of 2's. For example, a(3)=9, consisting of the 6 permutations of the string 102 and the 3 permutations of the string 122. - Enrique Navarrete, Mar 12 2025

Examples

			For n = 8, a(8) = 8 * 15 - (1 + 3 + 5 + 7 + 9 + 11 + 13) - 7 = 8 * 15 - 49 - 7 = 64. - _Bruno Berselli_, May 04 2010
G.f. = x + 4*x^2 + 9*x^3 + 16*x^4 + 25*x^5 + 36*x^6 + 49*x^7 + 64*x^8 + 81*x^9 + ...
a(4) = 16. For n = 4 vertices, the cycle graph C4 is A-B-C-D-A. The subtrees are: 4 singles: A, B, C, D; 4 pairs: A-B, BC, C-D, A-D; 4 triples: A-B-C, B-C-D, C-D-A, D-A-B; 4 quads: A-B-C-D, B-C-D-A, C-D-A-B, D-A-B-C; 4 + 4 + 4 + 4 = 16. - _Viktar Karatchenia_, Mar 02 2016
		

References

  • G. L. Alexanderson et al., The William Lowell Putnam Mathematical Competition, Problems and Solutions: 1965-1984, "December 1967 Problem B4(a)", pp. 8(157) MAA Washington DC 1985.
  • T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976, page 2.
  • Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966. See Chapter XV, pp. 135-167.
  • R. P. Burn & A. Chetwynd, A Cascade Of Numbers, "The prison door problem" Problem 4 pp. 5-7; 79-80 Arnold London 1996.
  • H. Cohen, A Course in Computational Algebraic Number Theory, Springer, 1996, p. 40.
  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 31, 36, 38, 63.
  • E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), p. 6.
  • M. Gardner, Time Travel and Other Mathematical Bewilderments, Chapter 6 pp. 71-2, W. H. Freeman NY 1988.
  • Granino A. Korn and Theresa M. Korn, Mathematical Handbook for Scientists and Engineers, McGraw-Hill Book Company, New York (1968), p. 982.
  • Jan Gullberg, Mathematics from the Birth of Numbers, W. W. Norton & Co., NY & London, 1997, §8.1 Terminology and §8.6 Figurate Numbers, pp. 264, 290-291.
  • Alfred S. Posamentier, The Art of Problem Solving, Section 2.4 "The Long Cell Block" pp. 10-1; 12; 156-7 Corwin Press Thousand Oaks CA 1996.
  • Alfred S. Posamentier, Math Charmers, Tantalizing Tidbits for the Mind, Prometheus Books, NY, 2003, pages 35, 52-53, 129-132, 244.
  • Michel Rigo, Formal Languages, Automata and Numeration Systems, 2 vols., Wiley, 2014. Mentions this sequence - see "List of Sequences" in Vol. 2.
  • 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).
  • J. K. Strayer, Elementary Number Theory, Exercise Set 3.3 Problems 32, 33, p. 88, PWS Publishing Co. Boston MA 1996.
  • C. W. Trigg, Mathematical Quickies, "The Lucky Prisoners" Problem 141 pp. 40, 141, Dover NY 1985.
  • R. Vakil, A Mathematical Mosaic, "The Painted Lockers" pp. 127;134 Brendan Kelly Burlington Ontario 1996.
  • David Wells, The Penguin Dictionary of Curious and Interesting Numbers. Penguin Books, NY, 1986, Revised edition 1987. See p. 123.

Crossrefs

Cf. A092205, A128200, A005408, A128201, A002522, A005563, A008865, A059100, A143051, A143470, A143595, A056944, A001157 (inverse Möbius transform), A001788 (binomial transform), A228039, A001105, A004159, A159918, A173277, A095794, A162395, A186646 (Pisano periods), A028338 (2nd diagonal).
A row or column of A132191.
This sequence is related to partitions of 2^n into powers of 2, as it is shown in A002577. So A002577 connects the squares and A000447. - Valentin Bakoev, Mar 03 2009
Boustrophedon transforms: A000697, A000745.
Cf. A342819.
Cf. A013661.

Programs

Formula

G.f.: x*(1 + x) / (1 - x)^3.
E.g.f.: exp(x)*(x + x^2).
Dirichlet g.f.: zeta(s-2).
a(n) = a(-n).
Multiplicative with a(p^e) = p^(2*e). - David W. Wilson, Aug 01 2001
Sum of all matrix elements M(i, j) = 2*i/(i+j) (i, j = 1..n). a(n) = Sum_{i = 1..n} Sum_{j = 1..n} 2*i/(i + j). - Alexander Adamchuk, Oct 24 2004
a(0) = 0, a(1) = 1, a(n) = 2*a(n-1) - a(n-2) + 2. - Miklos Kristof, Mar 09 2005
From Pierre CAMI, Oct 22 2006: (Start)
a(n) is the sum of the odd numbers from 1 to 2*n - 1.
a(0) = 0, a(1) = 1, then a(n) = a(n-1) + 2*n - 1. (End)
For n > 0: a(n) = A130064(n)*A130065(n). - Reinhard Zumkeller, May 05 2007
a(n) = Sum_{k = 1..n} A002024(n, k). - Reinhard Zumkeller, Jun 24 2007
Left edge of the triangle in A132111: a(n) = A132111(n, 0). - Reinhard Zumkeller, Aug 10 2007
Binomial transform of [1, 3, 2, 0, 0, 0, ...]. - Gary W. Adamson, Nov 21 2007
a(n) = binomial(n+1, 2) + binomial(n, 2).
This sequence could be derived from the following general formula (cf. A001286, A000330): n*(n+1)*...*(n+k)*(n + (n+1) + ... + (n+k))/((k+2)!*(k+1)/2) at k = 0. Indeed, using the formula for the sum of the arithmetic progression (n + (n+1) + ... + (n+k)) = (2*n + k)*(k + 1)/2 the general formula could be rewritten as: n*(n+1)*...*(n+k)*(2*n+k)/(k+2)! so for k = 0 above general formula degenerates to n*(2*n + 0)/(0 + 2) = n^2. - Alexander R. Povolotsky, May 18 2008
From a(4) recurrence formula a(n+3) = 3*a(n+2) - 3*a(n+1) + a(n) and a(1) = 1, a(2) = 4, a(3) = 9. - Artur Jasinski, Oct 21 2008
The recurrence a(n+3) = 3*a(n+2) - 3*a(n+1) + a(n) is satisfied by all k-gonal sequences from a(3), with a(0) = 0, a(1) = 1, a(2) = k. - Jaume Oliver Lafont, Nov 18 2008
a(n) = floor(n*(n+1)*(Sum_{i = 1..n} 1/(n*(n+1)))). - Ctibor O. Zizka, Mar 07 2009
Product_{i >= 2} 1 - 2/a(i) = -sin(A063448)/A063448. - R. J. Mathar, Mar 12 2009
a(n) = A002378(n-1) + n. - Jaroslav Krizek, Jun 14 2009
a(n) = n*A005408(n-1) - (Sum_{i = 1..n-2} A005408(i)) - (n-1) = n*A005408(n-1) - a(n-1) - (n-1). - Bruno Berselli, May 04 2010
a(n) == 1 (mod n+1). - Bruno Berselli, Jun 03 2010
a(n) = a(n-1) + a(n-2) - a(n-3) + 4, n > 2. - Gary Detlefs, Sep 07 2010
a(n+1) = Integral_{x >= 0} exp(-x)/( (Pn(x)*exp(-x)*Ei(x) - Qn(x))^2 +(Pi*exp(-x)*Pn(x))^2 ), with Pn the Laguerre polynomial of order n and Qn the secondary Laguerre polynomial defined by Qn(x) = Integral_{t >= 0} (Pn(x) - Pn(t))*exp(-t)/(x-t). - Groux Roland, Dec 08 2010
Euler transform of length-2 sequence [4, -1]. - Michael Somos, Feb 12 2011
A162395(n) = -(-1)^n * a(n). - Michael Somos, Mar 19 2011
a(n) = A004201(A000217(n)); A007606(a(n)) = A000384(n); A007607(a(n)) = A001105(n). - Reinhard Zumkeller, Feb 12 2011
Sum_{n >= 1} 1/a(n)^k = (2*Pi)^k*B_k/(2*k!) = zeta(2*k) with Bernoulli numbers B_k = -1, 1/6, 1/30, 1/42, ... for k >= 0. See A019673, A195055/10 etc. [Jolley eq 319].
Sum_{n>=1} (-1)^(n+1)/a(n)^k = 2^(k-1)*Pi^k*(1-1/2^(k-1))*B_k/k! [Jolley eq 320] with B_k as above.
A007968(a(n)) = 0. - Reinhard Zumkeller, Jun 18 2011
A071974(a(n)) = n; A071975(a(n)) = 1. - Reinhard Zumkeller, Jul 10 2011
a(n) = A199332(2*n - 1, n). - Reinhard Zumkeller, Nov 23 2011
For n >= 1, a(n) = Sum_{d|n} phi(d)*psi(d), where phi is A000010 and psi is A001615. - Enrique Pérez Herrero, Feb 29 2012
a(n) = A000217(n^2) - A000217(n^2 - 1), for n > 0. - Ivan N. Ianakiev, May 30 2012
a(n) = (A000217(n) + A000326(n))/2. - Omar E. Pol, Jan 11 2013
a(n) = A162610(n, n) = A209297(n, n) for n > 0. - Reinhard Zumkeller, Jan 19 2013
a(A000217(n)) = Sum_{i = 1..n} Sum_{j = 1..n} i*j, for n > 0. - Ivan N. Ianakiev, Apr 20 2013
a(n) = A133280(A000217(n)). - Ivan N. Ianakiev, Aug 13 2013
a(2*a(n)+2*n+1) = a(2*a(n)+2*n) + a(2*n+1). - Vladimir Shevelev, Jan 24 2014
a(n+1) = Sum_{t1+2*t2+...+n*tn = n} (-1)^(n+t1+t2+...+tn)*multinomial(t1+t2 +...+tn,t1,t2,...,tn)*4^(t1)*7^(t2)*8^(t3+...+tn). - Mircea Merca, Feb 27 2014
a(n) = floor(1/(1-cos(1/n)))/2 = floor(1/(1-n*sin(1/n)))/6, n > 0. - Clark Kimberling, Oct 08 2014
a(n) = ceiling(Sum_{k >= 1} log(k)/k^(1+1/n)) = -Zeta'[1+1/n]. Thus any exponent greater than 1 applied to k yields convergence. The fractional portion declines from A073002 = 0.93754... at n = 1 and converges slowly to 0.9271841545163232... for large n. - Richard R. Forberg, Dec 24 2014
a(n) = Sum_{j = 1..n} Sum_{i = 1..n} ceiling((i + j - n + 1)/3). - Wesley Ivan Hurt, Mar 12 2015
a(n) = Product_{j = 1..n-1} 2 - 2*cos(2*j*Pi/n). - Michel Marcus, Jul 24 2015
From Ilya Gutkovskiy, Jun 21 2016: (Start)
Product_{n >= 1} (1 + 1/a(n)) = sinh(Pi)/Pi = A156648.
Sum_{n >= 0} 1/a(n!) = BesselI(0, 2) = A070910. (End)
a(n) = A028338(n, n-1), n >= 1 (second diagonal). - Wolfdieter Lang, Jul 21 2017
For n >= 1, a(n) = Sum_{d|n} sigma_2(d)*mu(n/d) = Sum_{d|n} A001157(d)*A008683(n/d). - Ridouane Oudra, Apr 15 2021
a(n) = Sum_{i = 1..2*n-1} ceiling(n - i/2). - Stefano Spezia, Apr 16 2021
From Richard L. Ollerton, May 09 2021: (Start) For n >= 1,
a(n) = Sum_{k=1..n} psi(n/gcd(n,k)).
a(n) = Sum_{k=1..n} psi(gcd(n,k))*phi(gcd(n,k))/phi(n/gcd(n,k)).
a(n) = Sum_{k=1..n} sigma_2(n/gcd(n,k))*mu(gcd(n,k))/phi(n/gcd(n,k)).
a(n) = Sum_{k=1..n} sigma_2(gcd(n,k))*mu(n/gcd(n,k))/phi(n/gcd(n,k)). (End)
a(n) = (A005449(n) + A000326(n))/3. - Klaus Purath, May 13 2021
Let T(n) = A000217(n), then a(T(n)) + a(T(n+1)) = T(a(n+1)). - Charlie Marion, Jun 27 2022
a(n) = Sum_{k=1..n} sigma_1(k) + Sum_{i=1..n} (n mod i). - Vadim Kataev, Dec 07 2022
a(n^2) + a(n^2+1) + ... + a(n^2+n) + 4*A000537(n) = a(n^2+n+1) + ... + a(n^2+2n). In general, if P(k,n) = the n-th k-gonal number, then P(2k,n^2) + P(2k,n^2+1) + ... + P(2k,n^2+n) + 4*(k-1)*A000537(n) = P(2k,n^2+n+1) + ... + P(2k,n^2+2n). - Charlie Marion, Apr 26 2024
Sum_{n>=1} 1/a(n) = A013661. - Alois P. Heinz, Oct 19 2024
a(n) = 1 + 3^3*((n-1)/(n+1))^2 + 5^3*((n-1)*(n-2)/((n+1)*(n+2)))^2 + 7^3*((n-1)*(n-2)*(n-3)/((n+1)*(n+2)*(n+3)))^2 + ... for n >= 1. - Peter Bala, Dec 09 2024

Extensions

Incorrect comment and example removed by Joerg Arndt, Mar 11 2010

A001710 Order of alternating group A_n, or number of even permutations of n letters.

Original entry on oeis.org

1, 1, 1, 3, 12, 60, 360, 2520, 20160, 181440, 1814400, 19958400, 239500800, 3113510400, 43589145600, 653837184000, 10461394944000, 177843714048000, 3201186852864000, 60822550204416000, 1216451004088320000, 25545471085854720000, 562000363888803840000
Offset: 0

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Keywords

Comments

For n >= 3, a(n-1) is also the number of ways that a 3-cycle in the symmetric group S_n can be written as a product of 2 long cycles (of length n). - Ahmed Fares (ahmedfares(AT)my-deja.com), Aug 14 2001
a(n) is the number of Hamiltonian circuit masks for an n X n adjacency matrix of an undirected graph. - Chad Brewbaker, Jan 31 2003
a(n-1) is the number of necklaces one can make with n distinct beads: n! bead permutations, divide by two to represent flipping the necklace over, divide by n to represent rotating the necklace. Related to Stirling numbers of the first kind, Stirling cycles. - Chad Brewbaker, Jan 31 2003
Number of increasing runs in all permutations of [n-1] (n>=2). Example: a(4)=12 because we have 12 increasing runs in all the permutations of [3] (shown in parentheses): (123), (13)(2), (3)(12), (2)(13), (23)(1), (3)(2)(1). - Emeric Deutsch, Aug 28 2004
Minimum permanent over all n X n (0,1)-matrices with exactly n/2 zeros. - Simone Severini, Oct 15 2004
The number of permutations of 1..n that have 2 following 1 for n >= 1 is 0, 1, 3, 12, 60, 360, 2520, 20160, ... . - Jon Perry, Sep 20 2008
Starting (1, 3, 12, 60, ...) = binomial transform of A000153: (1, 2, 7, 32, 181, ...). - Gary W. Adamson, Dec 25 2008
First column of A092582. - Mats Granvik, Feb 08 2009
The asymptotic expansion of the higher order exponential integral E(x,m=1,n=3) ~ exp(-x)/x*(1 - 3/x + 12/x^2 - 60/x^3 + 360/x^4 - 2520/x^5 + 20160/x^6 - 81440/x^7 + ...) leads to the sequence given above. See A163931 and A130534 for more information. - Johannes W. Meijer, Oct 20 2009
For n>1: a(n) = A173333(n,2). - Reinhard Zumkeller, Feb 19 2010
Starting (1, 3, 12, 60, ...) = eigensequence of triangle A002260, (a triangle with k terms of (1,2,3,...) in each row given k=1,2,3,...). Example: a(6) = 360, generated from (1, 2, 3, 4, 5) dot (1, 1, 3, 12, 60) = (1 + 2 + 9 + 48 + 300). - Gary W. Adamson, Aug 02 2010
For n>=2: a(n) is the number of connected 2-regular labeled graphs on (n+1) nodes (Cf. A001205). - Geoffrey Critzer, Feb 16 2011.
The Fi1 and Fi2 triangle sums of A094638 are given by the terms of this sequence (n>=1). For the definition of these triangle sums see A180662. - Johannes W. Meijer, Apr 20 2011
Also [1, 1] together with the row sums of triangle A162608. - Omar E. Pol, Mar 09 2012
a(n-1) is, for n>=2, also the number of necklaces with n beads (only C_n symmetry, no turnover) with n-1 distinct colors and signature c[.]^2 c[.]^(n-2). This means that two beads have the same color, and for n=2 the second factor is omitted. Say, cyclic(c[1]c[1]c[2]c[3]..c[n-1]), in short 1123...(n-1), taken cyclically. E.g., n=2: 11, n=3: 112, n=4: 1123, 1132, 1213, n=5: 11234, 11243, 11324, 11342, 11423, 11432, 12134, 12143, 13124, 13142, 14123, 14132. See the next-to-last entry in line n>=2 of the representative necklace partition array A212359. - Wolfdieter Lang, Jun 26 2012
For m >= 3, a(m-1) is the number of distinct Hamiltonian circuits in a complete simple graph with m vertices. See also A001286. - Stanislav Sykora, May 10 2014
In factorial base (A007623) these numbers have a simple pattern: 1, 1, 1, 11, 200, 2200, 30000, 330000, 4000000, 44000000, 500000000, 5500000000, 60000000000, 660000000000, 7000000000000, 77000000000000, 800000000000000, 8800000000000000, 90000000000000000, 990000000000000000, etc. See also the formula based on this observation, given below. - Antti Karttunen, Dec 19 2015
Also (by definition) the independence number of the n-transposition graph. - Eric W. Weisstein, May 21 2017
Number of permutations of n letters containing an even number of even cycles. - Michael Somos, Jul 11 2018
Equivalent to Brewbaker's and Sykora's comments, a(n - 1) is the number of undirected cycles covering n labeled vertices, hence the logarithmic transform of A002135. - Gus Wiseman, Oct 20 2018
For n >= 2 and a set of n distinct leaf labels, a(n) is the number of binary, rooted, leaf-labeled tree topologies that have a caterpillar shape (column k=1 of A306364). - Noah A Rosenberg, Feb 11 2019
Also the clique covering number of the n-Bruhat graph. - Eric W. Weisstein, Apr 19 2019
a(n) is the number of lattices of the form [s,w] in the weak order on S_n, for a fixed simple reflection s. - Bridget Tenner, Jan 16 2020
For n > 3, a(n) = p_1^e_1*...*p_m^e_m, where p_1 = 2 and e_m = 1. There exists p_1^x where x <= e_1 such that p_1^x*p_m^e_m is a primitive Zumkeller number (A180332) and p_1^e_1*p_m^e_m is a Zumkeller number (A083207). Therefore, for n > 3, a(n) = p_1^e_1*p_m^e_m*r, where r is relatively prime to p_1*p_m, is also a Zumkeller number. - Ivan N. Ianakiev, Mar 11 2020
For n>1, a(n) is the number of permutations of [n] that have 1 and 2 as cycle-mates, that is, 1 and 2 are contained in the same cycle of a cyclic representation of permutations of [n]. For example, a(4) counts the 12 permutations with 1 and 2 as cycle-mates, namely, (1 2 3 4), (1 2 4 3), (1 3 2 4), (1 3 4 2), (1 4 2 3), (1 4 3 2), (1 2 3) (4), (1 3 2) (4), (1 2 4 )(3), (1 4 2)(3), (1 2)(3 4), and (1 2)(3)(4). Since a(n+2)=row sums of A162608, our result readily follows. - Dennis P. Walsh, May 28 2020

Examples

			G.f. = 1 + x + x^2 + 3*x^3 + 12*x^4 + 60*x^5 + 360*x^6 + 2520*x^7 + ...
		

References

  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, pp. 87-8, 20. (a), c_n^e(t=1).
  • 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

a(n+1)= A046089(n, 1), n >= 1 (first column of triangle), A161739 (q(n) sequence).
Bisections are A002674 and A085990 (essentially).
Row 3 of A265609 (essentially).
Row sums of A307429.

Programs

  • Magma
    [1] cat [Order(AlternatingGroup(n)): n in [1..20]]; // Arkadiusz Wesolowski, May 17 2014
    
  • Maple
    seq(mul(k, k=3..n), n=0..20); # Zerinvary Lajos, Sep 14 2007
  • Mathematica
    a[n_]:= If[n > 2, n!/2, 1]; Array[a, 21, 0]
    a[n_]:= If[n<3, 1, n*a[n-1]]; Array[a, 21, 0]; (* Robert G. Wilson v, Apr 16 2011 *)
    a[ n_]:= If[n<0, 0, n! SeriesCoefficient[(2-x^2)/(2-2x), {x, 0, n}]]; (* Michael Somos, May 22 2014 *)
    a[ n_]:= If[n<0, 0, n! SeriesCoefficient[1 +Sinh[-Log[1-x]], {x, 0, n}]]; (* Michael Somos, May 22 2014 *)
    Numerator[Range[0, 20]!/2] (* Eric W. Weisstein, May 21 2017 *)
    Table[GroupOrder[AlternatingGroup[n]], {n, 0, 20}] (* Eric W. Weisstein, May 21 2017 *)
  • PARI
    {a(n) = if( n<2, n>=0, n!/2)};
    
  • PARI
    a(n)=polcoeff(1+x*sum(m=0,n,m^m*x^m/(1+m*x+x*O(x^n))^m),n) \\ Paul D. Hanna
    
  • PARI
    A001710=n->n!\2+(n<2) \\ M. F. Hasler, Dec 01 2013
    
  • Python
    from math import factorial
    def A001710(n): return factorial(n)>>1 if n > 1 else 1 # Chai Wah Wu, Feb 14 2023
    
  • SageMath
    def A001710(n): return (factorial(n) +int(n<2))//2
    [A001710(n) for n in range(31)] # G. C. Greubel, Sep 28 2024
  • Scheme
    ;; Using memoization-macro definec for which an implementation can be found in http://oeis.org/wiki/Memoization
    (definec (A001710 n) (cond ((<= n 2) 1) (else (* n (A001710 (- n 1))))))
    ;; Antti Karttunen, Dec 19 2015
    

Formula

a(n) = numerator(n!/2) and A141044(n) = denominator(n!/2).
D-finite with recurrence: a(0) = a(1) = a(2) = 1; a(n) = n*a(n-1) for n>2. - Chad Brewbaker, Jan 31 2003 [Corrected by N. J. A. Sloane, Jul 25 2008]
a(0) = 0, a(1) = 1; a(n) = Sum_{k=1..n-1} k*a(k). - Amarnath Murthy, Oct 29 2002
Stirling transform of a(n+1) = [1, 3, 12, 160, ...] is A083410(n) = [1, 4, 22, 154, ...]. - Michael Somos, Mar 04 2004
First Eulerian transform of A000027. See A000142 for definition of FET. - Ross La Haye, Feb 14 2005
From Paul Barry, Apr 18 2005: (Start)
a(n) = 0^n + Sum_{k=0..n} (-1)^(n-k-1)*T(n-1, k)*cos(Pi*(n-k-1)/2)^2.
T(n,k) = abs(A008276(n, k)). (End)
E.g.f.: (2 - x^2)/(2 - 2*x).
E.g.f. of a(n+2), n>=0, is 1/(1-x)^3.
E.g.f.: 1 + sinh(log(1/(1-x))). - Geoffrey Critzer, Dec 12 2010
a(n+1) = (-1)^n * A136656(n,1), n>=1.
a(n) = n!/2 for n>=2 (proof from the e.g.f). - Wolfdieter Lang, Apr 30 2010
a(n) = (n-2)! * t(n-1), n>1, where t(n) is the n-th triangular number (A000217). - Gary Detlefs, May 21 2010
a(n) = ( A000254(n) - 2* A001711(n-3) )/3, n>2. - Gary Detlefs, May 24 2010
O.g.f.: 1 + x*Sum_{n>=0} n^n*x^n/(1 + n*x)^n. - Paul D. Hanna, Sep 13 2011
a(n) = if n < 2 then 1, otherwise Pochhammer(n,n)/binomial(2*n,n). - Peter Luschny, Nov 07 2011
a(n) = Sum_{k=0..floor(n/2)} s(n,n-2*k) where s(n,k) are Stirling number of the first kind, A048994. - Mircea Merca, Apr 07 2012
a(n-1), n>=3, is M_1([2,1^(n-2)])/n = (n-1)!/2, with the M_1 multinomial numbers for the given n-1 part partition of n. See the second to last entry in line n>=3 of A036038, and the above necklace comment by W. Lang. - Wolfdieter Lang, Jun 26 2012
G.f.: A(x) = 1 + x + x^2/(G(0)-2*x) where G(k) = 1 - (k+1)*x/(1 - x*(k+3)/G(k+1)); (continued fraction). - Sergei N. Gladkovskii, Dec 26 2012.
G.f.: 1 + x + (Q(0)-1)*x^2/(2*(sqrt(x)+x)), where Q(k) = 1 + (k+2)*sqrt(x)/(1 - sqrt(x)/(sqrt(x) + 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 15 2013
G.f.: 1 + x + (x*Q(x)-x^2)/(2*(sqrt(x)+x)), where Q(x) = Sum_{n>=0} (n+1)!*x^n*sqrt(x)*(sqrt(x) + x*(n+2)). - Sergei N. Gladkovskii, May 15 2013
G.f.: 1 + x/2 + (Q(0)-1)*x/(2*(sqrt(x)+x)), where Q(k) = 1 + (k+1)*sqrt(x)/(1 - sqrt(x)/(sqrt(x) + 1/Q(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 15 2013
G.f.: 1 + x + x^2*G(0)/2, where G(k) = 1 + 1/(1 - x/(x + 1/(k+3)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
G.f.: 1+x + x^2*W(0), where W(k) = 1 - x*(k+3)/( x*(k+3) - 1/(1 - x*(k+1)/( x*(k+1) - 1/W(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Aug 26 2013
From Antti Karttunen, Dec 19 2015: (Start)
a(0)=a(1)=1; after which, for even n: a(n) = (n/2) * (n-1)!, and for odd n: a(n) = (n-1)/2 * ((n-1)! + (n-2)!). [The formula was empirically found after viewing these numbers in factorial base, A007623, and is easily proved by considering formulas from Lang (Apr 30 2010) and Detlefs (May 21 2010) shown above.]
For n >= 1, a(2*n+1) = a(2*n) + A153880(a(2*n)). [Follows from above.] (End)
Inverse Stirling transform of a(n) is (-1)^(n-1)*A009566(n). - Anton Zakharov, Aug 07 2016
a(n) ~ sqrt(Pi/2)*n^(n+1/2)/exp(n). - Ilya Gutkovskiy, Aug 07 2016
a(n) = A006595(n-1)*n/A000124(n) for n>=2. - Anton Zakharov, Aug 23 2016
a(n) = A001563(n-1) - A001286(n-1) for n>=2. - Anton Zakharov, Sep 23 2016
From Peter Bala, May 24 2017: (Start)
The o.g.f. A(x) satisfies the Riccati equation x^2*A'(x) + (x - 1)*A(x) + 1 - x^2 = 0.
G.f.: A(x) = 1 + x + x^2/(1 - 3*x/(1 - x/(1 - 4*x/(1 - 2*x/(1 - 5*x/(1 - 3*x/(1 - ... - (n + 2)*x/(1 - n*x/(1 - ... ))))))))) (apply Stokes, 1982).
A(x) = 1 + x + x^2/(1 - 2*x - x/(1 - 3*x/(1 - 2*x/(1 - 4*x/(1 - 3*x/(1 - 5*x/(1 - ... - n*x/(1 - (n+2)*x/(1 - ... ))))))))). (End)
H(x) = (1 - (1 + x)^(-2)) / 2 = x - 3*x^2/2! + 12*x^3/3! - ..., an e.g.f. for the signed sequence here (n!/2!), ignoring the first two terms, is the compositional inverse of G(x) = (1 - 2*x)^(-1/2) - 1 = x + 3*x^2/2! + 15*x^3/3! + ..., an e.g.f. for A001147. Cf. A094638. H(x) is the e.g.f. for the sequence (-1)^m * m!/2 for m = 2,3,4,... . Cf. A001715 for n!/3! and A001720 for n!/4!. Cf. columns of A094587, A173333, and A213936 and rows of A138533. - Tom Copeland, Dec 27 2019
From Amiram Eldar, Jan 08 2023: (Start)
Sum_{n>=0} 1/a(n) = 2*(e-1).
Sum_{n>=0} (-1)^n/a(n) = 2/e. (End)

Extensions

More terms from Larry Reeves (larryr(AT)acm.org), Aug 20 2001
Further terms from Simone Severini, Oct 15 2004

A019538 Triangle of numbers T(n,k) = k!*Stirling2(n,k) read by rows (n >= 1, 1 <= k <= n).

Original entry on oeis.org

1, 1, 2, 1, 6, 6, 1, 14, 36, 24, 1, 30, 150, 240, 120, 1, 62, 540, 1560, 1800, 720, 1, 126, 1806, 8400, 16800, 15120, 5040, 1, 254, 5796, 40824, 126000, 191520, 141120, 40320, 1, 510, 18150, 186480, 834120, 1905120, 2328480, 1451520, 362880, 1, 1022, 55980, 818520, 5103000, 16435440, 29635200, 30240000, 16329600, 3628800
Offset: 1

Views

Author

N. J. A. Sloane and Manfred Goebel (goebel(AT)informatik.uni-tuebingen.de), Dec 11 1996

Keywords

Comments

Number of ways n labeled objects can be distributed into k nonempty parcels. Also number of special terms in n variables with maximal degree k.
In older terminology these are called differences of 0. - Michael Somos, Oct 08 2003
Number of surjections (onto functions) from an n-element set to a k-element set.
Also coefficients (in ascending order) of so-called ordered Bell polynomials.
(k-1)!*Stirling2(n,k-1) is the number of chain topologies on an n-set having k open sets [Stephen].
Number of set compositions (ordered set partitions) of n items into k parts. Number of k dimensional 'faces' of the n dimensional permutohedron (see Simion, p. 162). - Mitch Harris, Jan 16 2007
Correction of comment before: Number of (n-k)-dimensional 'faces' of the permutohedron of order n (an (n-1)-dimensional polytope). - Tilman Piesk, Oct 29 2014
This array is related to the reciprocal of an e.g.f. as sketched in A133314. For example, the coefficient of the fourth-order term in the Taylor series expansion of 1/(a(0) + a(1) x + a(2) x^2/2! + a(3) x^3/3! + ...) is a(0)^(-5) * {24 a(1)^4 - 36 a(1)^2 a(2) a(0) + [8 a(1) a(3) + 6 a(2)^2] a(0)^2 - a(4) a(0)^3}. The unsigned coefficients characterize the P3 permutohedron depicted on page 10 in the Loday link with 24 vertices (0-D faces), 36 edges (1-D faces), 6 squares (2-D faces), 8 hexagons (2-D faces) and 1 3-D permutohedron. Summing coefficients over like dimensions gives A019538 and A090582. Compare to A133437 for the associahedron. - Tom Copeland, Sep 29 2008, Oct 07 2008
Further to the comments of Tom Copeland above, the permutohedron of type A_3 can be taken as the truncated octahedron. Its dual is the tetrakis hexahedron, a simplicial polyhedron, with f-vector (1,14,36,24) giving the fourth row of this triangle. See the Wikipedia entry and [Fomin and Reading p. 21]. The corresponding h-vectors of permutohedra of type A give the rows of the triangle of Eulerian numbers A008292. See A145901 and A145902 for the array of f-vectors for type B and type D permutohedra respectively. - Peter Bala, Oct 26 2008
Subtriangle of triangle in A131689. - Philippe Deléham, Nov 03 2008
Since T(n,k) counts surjective functions and surjective functions are "consistent", T(n,k) satisfies a binomial identity, namely, T(n,x+y) = Sum_{j=0..n} C(n,j)*T(j,x)*T(n-j,y). For definition of consistent functions and a generalized binomial identity, see "Toy stories and combinatorial identities" in the link section below. - Dennis P. Walsh, Feb 24 2012
T(n,k) is the number of labeled forests on n+k vertices satisfying the following two conditions: (i) each forest consists of exactly k rooted trees with roots labeled 1, 2, ..., k; (ii) every root has at least one child vertex. - Dennis P. Walsh, Feb 24 2012
The triangle is the inverse binomial transform of triangle A028246, deleting the left column and shifting up one row. - Gary W. Adamson, Mar 05 2012
See A074909 for associations among this array and the Bernoulli polynomials and their umbral compositional inverses. - Tom Copeland, Nov 14 2014
E.g.f. for the shifted signed polynomials is G(x,t) = (e^t-1)/[1+(1+x)(e^t-1)] = 1-(1+x)(e^t-1) + (1+x)^2(e^t-1)^2 - ... (see also A008292 and A074909), which has the infinitesimal generator g(x,u)d/du = [(1-x*u)(1-(1+x)u)]d/du, i.e., exp[t*g(x,u)d/du]u eval. at u=0 gives G(x,t), and dG(x,t)/dt = g(x,G(x,t)). The compositional inverse is log((1-xt)/(1-(1+x)t)). G(x,t) is a generating series associated to the generalized Hirzebruch genera. See the G. Rzadowski link for the relation of the derivatives of g(x,u) to solutions of the Riccatt differential equation, soliton solns. to the KdV equation, and the Eulerian and Bernoulli numbers. In addition A145271 connects products of derivatives of g(x,u) and the refined Eulerian numbers to the inverse of G(x,t), which gives the normalized, reverse face polynomials of the simplices (A135278, divided by n+1). See A028246 for the generator g(x,u)d/dx. - Tom Copeland, Nov 21 2014
For connections to toric varieties and Eulerian polynomials, see the Dolgachev and Lunts and the Stembridge links. - Tom Copeland, Dec 31 2015
See A008279 for a relation between the e.g.f.s enumerating the faces of permutahedra (this entry) and stellahedra. - Tom Copeland, Nov 14 2016
T(n, k) appears in a Worpitzky identity relating monomials to binomials: x^n = Sum_{k=1..n} T(n, k)*binomial(x,k), n >= 1. See eq. (11.) of the Worpitzky link on p. 209. The relation to the Eulerian numbers is given there in eqs. (14.) and (15.). See the formula below relating to A008292. See also Graham et al. eq. (6.10) (relating monomials to falling factorials) on p. 248 (2nd ed. p. 262). The Worpitzky identity given in the Graham et al. reference as eq. (6.37) (2nd ed. p. 269) is eq. (5.), p. 207, of Worpitzky. - Wolfdieter Lang, Mar 10 2017
T(n, m) is also the number of minimum clique coverings and minimum matchings in the complete bipartite graph K_{m,n}. - Eric W. Weisstein, Apr 26 2017
From the Hasan and Franco and Hasan papers: The m-permutohedra for m=1,2,3,4 are the line segment, hexagon, truncated octahedron and omnitruncated 5-cell. The first three are well-known from the study of elliptic models, brane tilings and brane brick models. The m+1 torus can be tiled by a single (m+2)-permutohedron. Relations to toric Calabi-Yau Kahler manifolds are also discussed. - Tom Copeland, May 14 2020
From Manfred Boergens, Jul 25 2021: (Start)
Number of n X k binary matrices with row sums = 1 and no zero columns. These matrices are a subset of the matrices defining A183109.
The distribution into parcels in the leading comment can be regarded as a covering of [n] by tuples (A_1,...,A_k) in P([n])^k with nonempty and disjoint A_j, with P(.) denoting the power set (corrected for clarity by Manfred Boergens, May 26 2024). For the non-disjoint case see A183109 and A218695.
For tuples with "nonempty" dropped see A089072. For tuples with "nonempty and disjoint" dropped see A092477 and A329943 (amendment by Manfred Boergens, Jun 24 2024). (End)

Examples

			The triangle T(n, k) begins:
  n\k 1    2     3      4       5        6        7        8        9      10
  1:  1
  2:  1    2
  3:  1    6     6
  4:  1   14    36     24
  5:  1   30   150    240     120
  6:  1   62   540   1560    1800      720
  7:  1  126  1806   8400   16800    15120     5040
  8:  1  254  5796  40824  126000   191520   141120    40320
  9:  1  510 18150 186480  834120  1905120  2328480  1451520   362880
  10: 1 1022 55980 818520 5103000 16435440 29635200 30240000 16329600 3628800
  ... Reformatted and extended - _Wolfdieter Lang_, Oct 04 2014
---------------------------------------------------------------------------
T(4,1) = 1: {1234}. T(4,2) = 14: {1}{234} (4 ways), {12}{34} (6 ways), {123}{4} (4 ways). T(4,3) = 36: {12}{3}{4} (12 ways), {1}{23}{4} (12 ways), {1}{2}{34} (12 ways). T(4,4) = 1: {1}{2}{3}{4} (1 way).
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 89, ex. 1; also p. 210.
  • Miklos Bona, Combinatorics of Permutations, Chapman and Hall,2004, p.12.
  • G. Boole, A Treatise On The Calculus of Finite Differences, Dover Publications, 1960, p. 20.
  • 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. 212.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, 1989, p. 155. Also eqs.(6.10) and (6.37).
  • Kiran S. Kedlaya and Andrew V. Sutherland, Computing L -Series of Hyperelliptic Curves in Algorithmic Number Theory Lecture Notes in Computer Science Volume 5011/2008.
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015, Section 5.6.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 33.
  • J. F. Steffensen, Interpolation, 2nd ed., Chelsea, NY, 1950, see p. 54.
  • A. H. Voigt, Theorie der Zahlenreihen und der Reihengleichungen, Goschen, Leipzig, 1911, p. 31.
  • E. Whittaker and G. Robinson, The Calculus of Observations, Blackie, London, 4th ed., 1949; p. 7.

Crossrefs

Row sums give A000670. Maximal terms in rows give A002869. Central terms T(2k-1,k) give A233734.
Diagonal is n! (A000142). 2nd diagonal is A001286. 3rd diagonal is A037960.
Reflected version of A090582. A371568 is another version.
See also the two closely related triangles: A008277(n, k) = T(n, k)/k! (Stirling numbers of second kind) and A028246(n, k) = T(n, k)/k.
Cf. A033282 'faces' of the associahedron.
Cf. A008292, A047969, A145901, A145902. - Peter Bala, Oct 26 2008
Visible in the 3-D array in A249042.
See also A000182.

Programs

  • Haskell
    a019538 n k = a019538_tabl !! (n-1) !! (k-1)
    a019538_row n = a019538_tabl !! (n-1)
    a019538_tabl = iterate f [1] where
       f xs = zipWith (*) [1..] $ zipWith (+) ([0] ++ xs) (xs ++ [0])
    -- Reinhard Zumkeller, Dec 15 2013
    
  • Maple
    with(combinat): A019538 := (n,k)->k!*stirling2(n,k);
  • Mathematica
    Table[k! StirlingS2[n, k], {n, 9}, {k, n}] // Flatten
  • PARI
    {T(n, k) = if( k<0 || k>n, 0, sum(i=0, k, (-1)^i * binomial(k, i) * (k-i)^n))}; /* Michael Somos, Oct 08 2003 */
    
  • Sage
    def T(n, k): return factorial(k)*stirling_number2(n,k) # Danny Rorabaugh, Oct 10 2015

Formula

T(n, k) = k*(T(n-1, k-1)+T(n-1, k)) with T(0, 0) = 1 [or T(1, 1) = 1]. - Henry Bottomley, Mar 02 2001
E.g.f.: (y*(exp(x)-1) - exp(x))/(y*(exp(x)-1) - 1). - Vladeta Jovovic, Jan 30 2003
Equals [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, ...] DELTA [1, 1, 2, 2, 3, 3, 4, 4, 5, 5, ...] where DELTA is Deléham's operator defined in A084938.
T(n, k) = Sum_{j=0..k} (-1)^(k-j)*j^n*binomial(k, j). - Mario Catalani (mario.catalani(AT)unito.it), Nov 28 2003. See Graham et al., eq. (6.19), p. 251. For a proof see Bert Seghers, Jun 29 2013.
Sum_{k=0..n} T(n, k)(-1)^(n-k) = 1, Sum_{k=0..n} T(n, k)(-1)^k = (-1)^n. - Mario Catalani (mario.catalani(AT)unito.it), Dec 11 2003
O.g.f. for n-th row: polylog(-n, x/(1+x))/(x+x^2). - Vladeta Jovovic, Jan 30 2005
E.g.f.: 1 / (1 + t*(1-exp(x))). - Tom Copeland, Oct 13 2008
From Peter Bala, Oct 26 2008: (Start)
O.g.f. as a continued fraction: 1/(1 - x*t/(1 - (x + 1)*t/(1 - 2*x*t/(1 - 2*(x + 1)*t/(1 - ...))))) = 1 + x*t + (x + 2*x^2)*t^2 + (x + 6*x^2 + 6*x^3)*t^3 + ... .
The row polynomials R(n,x), which begin R(1,x) = x, R(2,x) = x + 2*x^2, R(3,x) = x + 6*x^2 + 6*x^3, satisfy the recurrence x*d/dx ((x + 1)*R(n,x)) = R(n+1,x). It follows that the zeros of R(n,x) are real and negative (apply Corollary 1.2 of [Liu and Wang]).
Since this is the triangle of f-vectors of the (simplicial complexes dual to the) type A permutohedra, whose h-vectors form the Eulerian number triangle A008292, the coefficients of the polynomial (x-1)^n*R(n,1/(x-1)) give the n-th row of A008292. For example, from row 3 we have x^2 + 6*x + 6 = 1 + 4*y + y^2, where y = x + 1, producing [1,4,1] as the third row of A008292. The matrix product A008292 * A007318 gives the mirror image of this triangle (see A090582).
For n,k >= 0, T(n+1,k+1) = Sum_{j=0..k} (-1)^(k-j)*binomial(k,j)*[(j+1)^(n+1) - j^(n+1)]. The matrix product of Pascal's triangle A007318 with the current array gives (essentially) A047969. This triangle is also related to triangle A047969 by means of the S-transform of [Hetyei], a linear transformation of polynomials whose value on the basis monomials x^k is given by S(x^k) = binomial(x,k). The S-transform of the shifted n-th row polynomial Q(n,x) := R(n,x)/x is S(Q(n,x)) = (x+1)^n - x^n. For example, from row 3 we obtain S(1 + 6*x + 6*x^2) = 1 + 6*x + 6*x*(x-1)/2 = 1 + 3*x + 3*x^2 = (x+1)^3 - x^3. For fixed k, the values S(Q(n,k)) give the nonzero entries in column (k-1) of the triangle A047969 (the Hilbert transform of the Eulerian numbers). (End)
E.g.f.: (exp(x)-1)^k = sum T(n,k)x^n/n!. - Vladimir Kruchinin, Aug 10 2010
T(n,k) = Sum_{i=1..k} A(n,i)*Binomial(n-i,k-i) where A(n,i) is the number of n-permutations that have i ascending runs, A008292.
From Tom Copeland, Oct 11 2011: (Start)
With e.g.f. A(x,t) = -1 + 1/(1+t*(1-exp(x))), the comp. inverse in x is B(x,t) = log(((1+t)/t) - 1/(t(1+x))).
With h(x,t) = 1/(dB/dx)= (1+x)((1+t)(1+x)-1), the row polynomial P(n,t) is given by (h(x,t)*d/dx)^n x, eval. 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(0,t)=0.
(A factor of -1/n! was removed by Copeland on Aug 25 2016.) (End)
The term linear in x of [x*h(d/dx,t)]^n 1 gives the n-th row polynomial. (See A134685.) - Tom Copeland, Nov 07 2011
Row polynomials are given by D^n(1/(1-x*t)) evaluated at x = 0, where D is the operator (1+x)*d/dx. - Peter Bala, Nov 25 2011
T(n,x+y) = Sum_{j=0..n} binomial(n,j)*T(j,x)*T(n-j,y). - Dennis P. Walsh, Feb 24 2012
Let P be a Rota-Baxter operator of weight 1 satisfying the identity P(x)*P(y) = P(P(x)*y) + P(x*P(y)) + P(x*y). Then P(1)^2 = P(1) + 2*P^2(1). More generally, Guo shows that P(1)^n = Sum_{k=1..n} T(n,k)*P^k(1). - Peter Bala, Jun 08 2012
Sum_{i=1..n} (-1)^i*T(n,i)/i = 0, for n > 1. - Leonid Bedratyuk, Aug 09 2012
T(n, k) = Sum_{j=0..k} (-1)^j*binomial(k, j)*(k-j)^n. [M. Catalani's re-indexed formula from Nov 28 2003] Proof: count the surjections of [n] onto [k] with the inclusion-exclusion principle, as an alternating sum of the number of functions from [n] to [k-j]. - Bert Seghers, Jun 29 2013
n-th row polynomial = 1/(1 + x)*( Sum_{k>=0} k^n*(x/(1 + x))^k ), valid for x in the open interval (-1/2, inf). See Tanny link. Cf. A145901. - Peter Bala, Jul 22 2014
T(n,k) = k * A141618(n,k-1) / binomial(n,k-1). - Tom Copeland, Oct 25 2014
Sum_{n>=0} n^k*a^n = Sum_{i=1..k} (a / (1 - a))^i * T(k, i)/(1-a) for |a| < 1. - David A. Corneth, Mar 09 2015
From Peter Bala, May 26 2015: (Start)
The row polynomials R(n,x) satisfy (1 + x)*R(n,x) = (-1)^n*x*R(n,-(1 + x)).
For a fixed integer k, the expansion of the function A(k,z) := exp( Sum_{n >= 1} R(n,k)*z^n/n ) has integer coefficients and satisfies the functional equation A(k,z)^(k + 1) = BINOMIAL(A(k,z))^k, where BINOMIAL(F(z))= 1/(1 - z)*F(z/(1 - z)) denotes the binomial transform of the o.g.f. F(z). Cf. A145901. For cases see A084784 (k = 1), A090352 (k = 2), A090355 (k = 3), A090357 (k = 4), A090362 (k = 5) and A084785 (k = -2 with z -> -z).
A(k,z)^(k + 1) = A(-(k + 1),-z)^k and hence BINOMIAL(A(k,z)) = A(-(k + 1),-z). (End)
From Tom Copeland, Oct 19 2016: (Start)
Let a(1) = 1 + x + B(1) = x + 1/2 and a(n) = B(n) = (B.)^n, where B(n) are the Bernoulli numbers defined by e^(B.t) = t / (e^t-1), then t / e^(a.t) = t / [(x + 1) * t + exp(B.t)] = (e^t - 1) /[ 1 + (x + 1) (e^t - 1)] = exp(p.(x)t), where (p.(x))^n = p_n(x) are the shifted, signed row polynomials of this array: p_0(x) = 0, p_1(x) = 1, p_2(x) = -(1 + 2 x), p_3(x) = 1 + 6 x + 6 x^2, ... and p_n(x) = n * b(n-1), where b(n) are the partition polynomials of A133314 evaluated with these a(n).
Sum_{n > 0} R(n,-1/2) x^n/n! = 2 * tanh(x/2), where R(n,x) = Sum_{k = 1..n} T(n,k) x^(k-1) are the shifted row polynomials of this entry, so R(n,-1/2) = 4 * (2^(n+1)-1) B(n+1)/(n+1). (Cf. A000182.)
(End)
Also the Bernoulli numbers are given by B(n) = Sum_{k =1..n} (-1)^k T(n,k) / (k+1). - Tom Copeland, Nov 06 2016
G.f. for column k: k! x^k / Product_{i=1..k} (1-i*x). - Robert A. Russell, Sep 25 2018
a(j) <= A183109(j). - Manfred Boergens, Jul 25 2021

A008297 Triangle of Lah numbers.

Original entry on oeis.org

-1, 2, 1, -6, -6, -1, 24, 36, 12, 1, -120, -240, -120, -20, -1, 720, 1800, 1200, 300, 30, 1, -5040, -15120, -12600, -4200, -630, -42, -1, 40320, 141120, 141120, 58800, 11760, 1176, 56, 1, -362880, -1451520, -1693440, -846720, -211680, -28224, -2016, -72, -1, 3628800, 16329600, 21772800, 12700800
Offset: 1

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Author

Keywords

Comments

|a(n,k)| = number of partitions of {1..n} into k lists, where a list means an ordered subset.
Let N be a Poisson random variable with parameter (mean) lambda, and Y_1,Y_2,... independent exponential(theta) variables, independent of N, so that their density is given by (1/theta)*exp(-x/theta), x > 0. Set S=Sum_{i=1..N} Y_i. Then E(S^n), i.e., the n-th moment of S, is given by (theta^n) * L_n(lambda), n >= 0, where L_n(y) is the Lah polynomial Sum_{k=0..n} |a(n,k)| * y^k. - Shai Covo (green355(AT)netvision.net.il), Feb 09 2010
For y = lambda > 0, formula 2) for the Lah polynomial L_n(y) dated Feb 02 2010 can be restated as follows: L_n(lambda) is the n-th ascending factorial moment of the Poisson distribution with parameter (mean) lambda. - Shai Covo (green355(AT)netvision.net.il), Feb 10 2010
See A111596 for an expression of the row polynomials in terms of an umbral composition of the Bell polynomials and relation to an inverse Mellin transform and a generalized Dobinski formula. - Tom Copeland, Nov 21 2011
Also the Bell transform of the sequence (-1)^(n+1)*(n+1)! without column 0. For the definition of the Bell transform see A264428. - Peter Luschny, Jan 28 2016
Named after the Slovenian mathematician and actuary Ivo Lah (1896-1979). - Amiram Eldar, Jun 13 2021

Examples

			|a(2,1)| = 2: (12), (21); |a(2,2)| = 1: (1)(2). |a(4,1)| = 24: (1234) (24 ways); |a(4,2)| = 36: (123)(4) (6*4 ways), (12)(34) (3*4 ways); |a(4,3)| = 12: (12)(3)(4) (6*2 ways); |a(4,4)| = 1: (1)(2)(3)(4) (1 way).
Triangle:
    -1;
     2,    1;
    -6,   -6,   -1;
    24,   36,   12,   1;
  -120, -240, -120, -20, -1; ...
		

References

  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 156.
  • Shai Covo, The moments of a compound Poisson process with exponential or centered normal jumps, J. Probab. Stat. Sci., Vol. 7, No. 1 (2009), pp. 91-100.
  • Theodore S. Motzkin, Sorting numbers for cylinders and other classification numbers, in Combinatorics, Proc. Symp. Pure Math. 19, AMS, 1971, pp. 167-176; the sequence called {!}^{n+}. For a link to this paper see A000262.
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 44.
  • S. Gill Williamson, Combinatorics for Computer Science, Computer Science Press, 1985; see p. 176.

Crossrefs

Same as A066667 and A105278 except for signs. Other variants: A111596 (differently signed triangle and (0,0)-based), A271703 (unsigned and (0,0)-based), A089231.
A293125 (row sums) and A000262 (row sums of unsigned triangle).
Columns 1-6 (unsigned): A000142, A001286, A001754, A001755, A001777, A001778.
A002868 gives maximal element (in magnitude) in each row.
A248045 (central terms, negated). A130561 is a natural refinement.

Programs

  • Haskell
    a008297 n k = a008297_tabl !! (n-1) !! (k-1)
    a008297_row n = a008297_tabl !! (n-1)
    a008297_tabl = [-1] : f [-1] 2 where
       f xs i = ys : f ys (i + 1) where
         ys = map negate $
              zipWith (+) ([0] ++ xs) (zipWith (*) [i, i + 1 ..] (xs ++ [0]))
    -- Reinhard Zumkeller, Sep 30 2014
    
  • Maple
    A008297 := (n,m) -> (-1)^n*n!*binomial(n-1,m-1)/m!;
  • Mathematica
    a[n_, m_] := (-1)^n*n!*Binomial[n-1, m-1]/m!; Table[a[n, m], {n, 1, 10}, {m, 1, n}] // Flatten (* Jean-François Alcover, Dec 12 2012, after Maple *)
    T[n_, n_] := (-1)^n; T[n_, k_]/;0Oliver Seipel, Dec 06 2024 *)
  • PARI
    T(n, m) = (-1)^n*n!*binomial(n-1, m-1)/m!
    for(n=1,9, for(m=1,n, print1(T(n,m)", "))) \\ Charles R Greathouse IV, Mar 09 2016
    
  • Perl
    use bigint; use ntheory ":all"; my @L; for my $n (1..9) { push @L, map { stirling($n,$,3)*(-1)**$n } 1..$n; } say join(", ",@L); # _Dana Jacobsen, Mar 16 2017
  • Sage
    def A008297_triangle(dim): # computes unsigned T(n, k).
        M = matrix(ZZ,dim,dim)
        for n in (0..dim-1): M[n,n] = 1
        for n in (1..dim-1):
            for k in (0..n-1):
                M[n,k] = M[n-1,k-1]+(2+2*k)*M[n-1,k]+((k+1)*(k+2))*M[n-1,k+1]
        return M
    A008297_triangle(9) # Peter Luschny, Sep 19 2012
    

Formula

a(n, m) = (-1)^n*n!*A007318(n-1, m-1)/m!, n >= m >= 1.
a(n+1, m) = (n+m)*a(n, m)+a(n, m-1), a(n, 0) := 0; a(n, m) := 0, n < m; a(1, 1)=1.
a(n, m) = ((-1)^(n-m+1))*L(1, n-1, m-1) where L(1, n, m) is the triangle of coefficients of the generalized Laguerre polynomials n!*L(n, a=1, x). These polynomials appear in the radial l=0 eigen-functions for discrete energy levels of the H-atom.
|a(n, m)| = Sum_{k=m..n} |A008275(n, k)|*A008277(k, m), where A008275 = Stirling numbers of first kind, A008277 = Stirling numbers of second kind. - Wolfdieter Lang
If L_n(y) = Sum_{k=0..n} |a(n, k)|*y^k (a Lah polynomial) then the e.g.f. for L_n(y) is exp(x*y/(1-x)). - Vladeta Jovovic, Jan 06 2001
E.g.f. for the k-th column (unsigned): x^k/(1-x)^k/k!. - Vladeta Jovovic, Dec 03 2002
a(n, k) = (n-k+1)!*N(n, k) where N(n, k) is the Narayana triangle A001263. - Philippe Deléham, Jul 20 2003
From Shai Covo (green355(AT)netvision.net.il), Feb 02 2010: (Start)
We have the following expressions for the Lah polynomial L_n(y) = Sum_{k=0..n} |a(n, k)|*y^k -- exact generalizations of results in A000262 for A000262(n) = L_n(1):
1) L_n(y) = y*exp(-y)*n!*M(n+1,2,y), n >= 1, where M (=1F1) is the confluent hypergeometric function of the first kind;
2) L_n(y) = exp(-y)* Sum_{m>=0} y^m*[m]^n/m!, n>=0, where [m]^n = m*(m+1)*...*(m+n-1) is the rising factorial;
3) L_n(y) = (2n-2+y)L_{n-1}(y)-(n-1)(n-2)L_{n-2}(y), n>=2;
4) L_n(y) = y*(n-1)!*Sum_{k=1..n} (L_{n-k}(y) k!)/((n-k)! (k-1)!), n>=1. (End)
The row polynomials are given by D^n(exp(-x*t)) evaluated at x = 0, where D is the operator (1-x)^2*d/dx. Cf. A008277 and A035342. - Peter Bala, Nov 25 2011
n!C(-xD,n) = Lah(n,:xD:) where C(m,n) is the binomial coefficient, xD= x d/dx, (:xD:)^k = x^k D^k, and Lah(n,x) are the row polynomials of this entry. E.g., 2!C(-xD,2)= 2 xD + x^2 D^2. - Tom Copeland, Nov 03 2012
From Tom Copeland, Sep 25 2016: (Start)
The Stirling polynomials of the second kind A048993 (A008277), i.e., the Bell-Touchard-exponential polynomials B_n[x], are umbral compositional inverses of the Stirling polynomials of the first kind signed A008275 (A130534), i.e., the falling factorials, (x)_n = n! binomial(x,n); that is, umbrally B_n[(x).] = x^n = (B.[x])_n.
An operational definition of the Bell polynomials is (xD_x)^n = B_n[:xD:], where, by definition, (:xD_x:)^n = x^n D_x^n, so (B.[:xD_x:])_n = (xD_x)_n = :xD_x:^n = x^n (D_x)^n.
Let y = 1/x, then D_x = -y^2 D_y; xD_x = -yD_y; and P_n(:yD_y:) = (-yD_y)_n = (-1)^n (1/y)^n (y^2 D_y)^n, the row polynomials of this entry in operational form, e.g., P_3(:yD_y:) = (-yD_y)_3 = (-yD_y) (yD_y-1) (yD_y-2) = (-1)^3 (1/y)^3 (y^2 D_y)^3 = -( 6 :yD_y: + 6 :yD_y:^2 + :yD_y:^3 ) = - ( 6 y D_y + 6 y^2 (D_y)^2 + y^3 (D_y)^3).
Therefore, P_n(y) = e^(-y) P_n(:yD_y:) e^y = e^(-y) (-1/y)^n (y^2 D_y)^n e^y = e^(-1/x) x^n (D_x)^n e^(1/x) = P_n(1/x) and P_n(x) = e^(-1/x) x^n (D_x)^n e^(1/x) = e^(-1/x) (:x D_x:)^n e^(1/x). (Cf. also A094638.) (End)
T(n,k) = Sum_{j=k..n} (-1)^j*A008296(n,j)*A360177(j,k). - Mélika Tebni, Feb 02 2023

A131689 Triangle of numbers T(n,k) = k!*Stirling2(n,k) = A000142(k)*A048993(n,k) read by rows, T(n, k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 2, 0, 1, 6, 6, 0, 1, 14, 36, 24, 0, 1, 30, 150, 240, 120, 0, 1, 62, 540, 1560, 1800, 720, 0, 1, 126, 1806, 8400, 16800, 15120, 5040, 0, 1, 254, 5796, 40824, 126000, 191520, 141120, 40320, 0, 1, 510, 18150, 186480, 834120, 1905120, 2328480, 1451520, 362880
Offset: 0

Views

Author

Philippe Deléham, Sep 14 2007

Keywords

Comments

Triangle T(n,k), 0 <= k <= n, read by rows given by [0,1,0,2,0,3,0,4,0,5,0,6,0,7,0,...] DELTA [1,1,2,2,3,3,4,4,5,5,6,6,...] where DELTA is the operator defined in A084938; another version of A019538.
See also A019538: version with n > 0 and k > 0. - Philippe Deléham, Nov 03 2008
From Peter Bala, Jul 21 2014: (Start)
T(n,k) gives the number of (k-1)-dimensional faces in the interior of the first barycentric subdivision of the standard (n-1)-dimensional simplex. For example, the barycentric subdivision of the 1-simplex is o--o--o, with 1 interior vertex and 2 interior edges, giving T(2,1) = 1 and T(2,2) = 2.
This triangle is used when calculating the face vectors of the barycentric subdivision of a simplicial complex. Let S be an n-dimensional simplicial complex and write f_k for the number of k-dimensional faces of S, with the usual convention that f_(-1) = 1, so that F := (f_(-1), f_0, f_1,...,f_n) is the f-vector of S. If M(n) denotes the square matrix formed from the first n+1 rows and n+1 columns of the present triangle, then the vector F*M(n) is the f-vector of the first barycentric subdivision of the simplicial complex S (Brenti and Welker, Lemma 2.1). For example, the rows of Pascal's triangle A007318 (but with row and column indexing starting at -1) are the f-vectors for the standard n-simplexes. It follows that A007318*A131689, which equals A028246, is the array of f-vectors of the first barycentric subdivision of standard n-simplexes. (End)
This triangle T(n, k) appears in the o.g.f. G(n, x) = Sum_{m>=0} S(n, m)*x^m with S(n, m) = Sum_{j=0..m} j^n for n >= 1 as G(n, x) = Sum_{k=1..n} (x^k/(1 - x)^(k+2))*T(n, k). See also the Eulerian triangle A008292 with a Mar 31 2017 comment for a rewritten form. For the e.g.f. see A028246 with a Mar 13 2017 comment. - Wolfdieter Lang, Mar 31 2017
T(n,k) = the number of alignments of length k of n strings each of length 1. See Slowinski. An example is given below. Cf. A122193 (alignments of strings of length 2) and A299041 (alignments of strings of length 3). - Peter Bala, Feb 04 2018
The row polynomials R(n,x) are the Fubini polynomials. - Emanuele Munarini, Dec 05 2020
From Gus Wiseman, Feb 18 2022: (Start)
Also the number of patterns of length n with k distinct parts (or with maximum part k), where we define a pattern to be a finite sequence covering an initial interval of positive integers. For example, row n = 3 counts the following patterns:
(1,1,1) (1,2,2) (1,2,3)
(2,1,2) (1,3,2)
(2,2,1) (2,1,3)
(1,1,2) (2,3,1)
(1,2,1) (3,1,2)
(2,1,1) (3,2,1)
(End)
Regard A048994 as a lower-triangular matrix and divide each term A048994(n,k) by n!, then this is the matrix inverse. Because Sum_{k=0..n} (A048994(n,k) * x^n / n!) = A007318(x,n), Sum_{k=0..n} (A131689(n,k) * A007318(x,k)) = x^n. - Natalia L. Skirrow, Mar 23 2023
T(n,k) is the number of ordered partitions of [n] into k blocks. - Alois P. Heinz, Feb 21 2025

Examples

			The triangle T(n,k) begins:
  n\k 0 1    2     3      4       5        6        7        8        9      10 ...
  0:  1
  1:  0 1
  2:  0 1    2
  3:  0 1    6     6
  4:  0 1   14    36     24
  5:  0 1   30   150    240     120
  6:  0 1   62   540   1560    1800      720
  7:  0 1  126  1806   8400   16800    15120     5040
  8:  0 1  254  5796  40824  126000   191520   141120    40320
  9:  0 1  510 18150 186480  834120  1905120  2328480  1451520   362880
  10: 0 1 1022 55980 818520 5103000 16435440 29635200 30240000 16329600 3628800
  ... reformatted and extended. - _Wolfdieter Lang_, Mar 31 2017
From _Peter Bala_, Feb 04 2018: (Start)
T(4,2) = 14 alignments of length 2 of 4 strings of length 1. Examples include
  (i) A -    (ii) A -    (iii) A -
      B -         B -          - B
      C -         - C          - C
      - D         - D          - D
There are C(4,1) = 4 alignments of type (i) with a single gap character - in column 1, C(4,2) = 6 alignments of type (ii) with two gap characters in column 1 and C(4,3) = 4 alignments of type (iii) with three gap characters in column 1, giving a total of 4 + 6 + 4 = 14 alignments. (End)
		

Crossrefs

Case m=1 of the polynomials defined in A278073.
Cf. A000142 (diagonal), A000670 (row sums), A000012 (alternating row sums), A210029 (central terms).
Cf. A008292, A028246 (o.g.f. and e.g.f. of sums of powers).
A version for partitions is A116608, or by maximum A008284.
A version for compositions is A235998, or by maximum A048004.
Classes of patterns:
- A000142 = strict
- A005649 = anti-run, complement A069321
- A019536 = necklace
- A032011 = distinct multiplicities
- A060223 = Lyndon
- A226316 = (1,2,3)-avoiding, weakly A052709, complement A335515
- A296975 = aperiodic
- A345194 = alternating, up/down A350354, complement A350252
- A349058 = weakly alternating
- A351200 = distinct runs
- A351292 = distinct run-lengths

Programs

  • Julia
    function T(n, k)
        if k < 0 || k > n return 0 end
        if n == 0 && k == 0 return 1 end
        k*(T(n-1, k-1) + T(n-1, k))
    end
    for n in 0:7
        println([T(n, k) for k in 0:n])
    end
    # Peter Luschny, Mar 26 2020
    
  • Maple
    A131689 := (n,k) -> Stirling2(n,k)*k!: # Peter Luschny, Sep 17 2011
    # Alternatively:
    A131689_row := proc(n) 1/(1-t*(exp(x)-1)); expand(series(%,x,n+1)); n!*coeff(%,x,n); PolynomialTools:-CoefficientList(%,t) end:
    for n from 0 to 9 do A131689_row(n) od; # Peter Luschny, Jan 23 2017
  • Mathematica
    t[n_, k_] := k!*StirlingS2[n, k]; Table[t[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Feb 25 2014 *)
    T[n_, k_] := If[n <= 0 || k <= 0, Boole[n == 0 && k == 0], Sum[(-1)^(i + k) Binomial[k, i] i^(n + k), {i, 0, k}]]; (* Michael Somos, Jul 08 2018 *)
  • PARI
    {T(n, k) = if( n<0, 0, sum(i=0, k, (-1)^(k + i) * binomial(k, i) * i^n))};
    /* Michael Somos, Jul 08 2018 */
    
  • SageMath
    @cached_function
    def F(n): # Fubini polynomial
        R. = PolynomialRing(ZZ)
        if n == 0: return R(1)
        return R(sum(binomial(n, k)*F(n - k)*x for k in (1..n)))
    for n in (0..9): print(F(n).list()) # Peter Luschny, May 21 2021

Formula

T(n,k) = k*(T(n-1,k-1) + T(n-1,k)) with T(0,0)=1. Sum_{k=0..n} T(n,k)*x^k = (-1)^n*A000629(n), A033999(n), A000007(n), A000670(n), A004123(n+1), A032033(n), A094417(n), A094418(n), A094419(n) for x = -2, -1, 0, 1, 2, 3, 4, 5, 6 respectively. [corrected by Philippe Deléham, Feb 11 2013]
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000142(n), A000670(n), A122704(n) for x=-1, 0, 1, 2 respectively. - Philippe Deléham, Oct 09 2007
Sum_{k=0..n} (-1)^k*T(n,k)/(k+1) = Bernoulli numbers A027641(n)/A027642(n). - Peter Luschny, Sep 17 2011
G.f.: F(x,t) = 1 + x*t + (x+x^2)*t^2/2! + (x+6*x^2+6*x^3)*t^3/3! + ... = Sum_{n>=0} R(n,x)*t^n/n!.
The row polynomials R(n,x) satisfy the recursion R(n+1,x) = (x+x^2)*R'(n,x) + x*R(n,x) where ' indicates differentiation with respect to x. - Philippe Deléham, Feb 11 2013
T(n,k) = [t^k] (n! [x^n] (1/(1-t*(exp(x)-1)))). - Peter Luschny, Jan 23 2017
The n-th row polynomial has the form x o x o ... o x (n factors), where o denotes the black diamond multiplication operator of Dukes and White. See also Bala, Example E8. - Peter Bala, Jan 08 2018

A133314 Coefficients of list partition transform: reciprocal of an exponential generating function (e.g.f.).

Original entry on oeis.org

1, -1, -1, 2, -1, 6, -6, -1, 8, 6, -36, 24, -1, 10, 20, -60, -90, 240, -120, -1, 12, 30, -90, 20, -360, 480, -90, 1080, -1800, 720, -1, 14, 42, -126, 70, -630, 840, -420, -630, 5040, -4200, 2520, -12600, 15120, -5040, -1, 16, 56, -168, 112, -1008, 1344, 70
Offset: 0

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Author

Tom Copeland, Oct 18 2007, Oct 29 2007, Nov 16 2007

Keywords

Comments

The list partition transform of a sequence a(n) for which a(0)=1 is illustrated by:
b_0 = 1
b_1 = -a_1
b_2 = -a_2 + 2 a_1^2
b_3 = -a_3 + 6 a_2 a_1 - 6 a_1^3
b_4 = -a_4 + 8 a_3 a_1 + 6 a_2^2 - 36 a_2 a_1^2 + 24 a_1^4
... .
The unsigned coefficients are A049019 with a leading 1. The sign is dependent on the partition as evident from inspection (replace a_n's by -1).
Expressed umbrally, i.e., with the umbral operation (a.)^n := a_n,
exp(a.x) exp(b.x) = exp[(a.+b.)x] = 1; i.e., (a.+b.)^n = 1 for n=0 and 0 for all other values of n.
Expressed recursively,
b_0 = 1, b_n = -Sum_{j=1..n} binomial(n,j) a_j b_{n-j}; which is conditionally self-inverse, i.e., the roles of a_k and b_k may be reversed with a_0 = b_0 = 1.
Expressed in matrix form, b_n form the first column of B = matrix inverse of A .
A = Pascal matrix diagonally multiplied by a_n, i.e., A_{n,k} = binomial(n,k)* a_{n-k}.
Some examples of reciprocal pairs of sequences under these operations are:
1) A084358 and -A000262 with the first term set to 1.
2) (1,-1,0,0,...) and (0!,1!,2!,3!,...) with the unsigned associated matrices A128229 and A094587.
3) (1,-1,-1,-1,...) and A000670.
5) (1,-2,-2,0,0,0,...) and (0! c_1,1! c_2,2! c_3,3! c_4,...) where c_n = A000129(n) with the associated matrices A110327 and A110330.
6) (1,-2,2,0,0,0,...) and (1!,2!,3!,4!,...).
7) Sequences of rising and signed lowering factorials form reciprocal pairs where a_n = (-1)^n m!/(m-n)! and b_n = (m-1+n)!/(m-1)! for m=0,1,2,... .
Denote the action of the list partition transform on the sequence a. or an invertible matrix M by LPT(a.) = b. or LPT(M)= M^(-1).
If the matrix equation M = exp(T) also holds, then exp[a.*T]*exp[b.*T] = exp[(a.+b.)*T] = I, the identity matrix, because (a.+b.)^n = delta_n, the Kronecker delta with delta_n = 1 and delta_n = 0 otherwise, i.e., (0)^n = delta_n.
Therefore, [exp(a.*T)]^(-1) = exp[b.*T] = exp[LPT(a.)*T] = LPT[exp(a.*T)].
The fundamental Pascal (A007318), unsigned Lah (A105278) and associated Laguerre matrices can be generated by exponentiation of special infinitesimal matrices (see A132440, A132710 and A132681) such that finding LPT(a.) amounts to multiplying the k'th diagonal of the fundamental matrices by a_k for every diagonal followed by matrix inversion and then extraction of the b_n factors from the first column (simplest for the Pascal formulas above).
Conversely, the inverses of matrices formed by diagonally multiplying the three fundamental matrices by a_k are given by diagonally multiplying the fundamental matrices by b_k.
If LPT(M) is defined differently as application of the top formula to a_n = M^n, then b_n = (-M)^n and the formalism could even be applied to more general sequences of matrices M., providing the reciprocal of exp[t*M.].
The group of fundamental lower triangular matrices M = exp(T) such that LPT[exp(a.*T)] = exp[LPT(a.)*T] = [exp[a.*T]]^(-1) are obtained by infinitesimal generator matrices of the form T =
0;
t(0), 0;
0, t(1), 0;
0, 0, t(2), 0;
0, 0, 0, t(3), 0;
... .
T^m has trivially vanishing terms except along the m'th subdiagonal, which is a sequence of generalized factorials:
[ t(0)*t(1)...t(m-2)*t(m-1), t(1)*t(2)...t(m-1)*t(m), t(2)*t(3)...t(m)*t(m+1), ... ].
Therefore the principal submatrices of T (given by setting t(j) = 0 for j > n-1) are nilpotent with at least [Tsub_n]^(n+1) = 0.
The general group of matrices GM[a.] = exp[a.*T] can also be obtained through diagonal multiplication of M = exp(T) by the sequence a_n, as in the Pascal matrix example above and their inverses by diagonal multiplication by b. = LPT(a.).
Weighted-mappings interpretation for the top partition equation:
Given n pre-nodes (Pre) and k post-nodes (Post), each Pre is connected to only one Post and each Post has at least one Pre connected to it (surjections or onto functions/maps). Weight each Post by -a_m where m is the number of connections to the Post.
Weight each map by the product of the Post weights and multiply by the number of maps that share the same connectivity. Sum over the possible mappings for n Pre. The result is b_n.
E.g., b_3 = [ 3 Pre to 1 Post ] + [ 3 Pre to 2 Post ] + [ 3 Pre to 3 Post ]
= [1 map with 1 Post with 3 connections] + [ 6 maps with 1 Post with 2 connections and 1 Post with 1 connection] + [6 maps with 3 Post with 1 connection each]
= -a_3 + 6 * [-a_2*(-a_1)] + 6 * [-a_1*(-a_1)*(-a_1)].
See A263633 for the complementary formulation for the reciprocal of o.g.f.s rather than e.g.f.s and computations of these partition polynomials as Gram determinants. - Tom Copeland, Dec 04 2016
The coefficients of the partition polynomials enumerate the faces of the convex, bounded polytopes called permutohedra, and the absolute value of the sum of the coefficients gives the Euler characteristic of unity for each polytope; i.e., the absolute value of the sum of each row of the array is unity. In addition, the signs of the faces alternate with dimension, and the coefficients of faces with the same dimension for each polytope have the same sign. - Tom Copeland, Nov 13 2019
With the fundamental matrix chosen to be the lower triangular Pascal matrix M, the matrix MA whose n-th diagonals are multiplied by a_n (i.e., MA_{i,j} = PM_{i,j} * a_{i-j}) gives a matrix representation of the e.g.f. associated to the Appell polynomial sequence defined by e^{a.t}e^{xt}= e^{(a.+x)t} = e^{A.(x)t} where umbrally (A.(x))^n = A_n(x) = (a. + x)^n = sum_{k=0..n} binomial(n,k) a_k x^{n-k} are the associated Appell polynomials. Left multiplication of the column vector (1,x,x^2,..) by MA gives the Appell polynomial sequence, and multiplication of the two e.g.f.s e^{a.t} and e^{b.t} corresponds to multiplication of their respective matrix representations MA and MB. Forming the reciprocal of an e.g.f. corresponds to taking the matrix inverse of its matrix representation as noted above. A263634 gives an associated modified Pascal matrix representation of the raising operator for the Appell sequence. - Tom Copeland, Nov 13 2019
The diagonal of MA consists of all ones. Let MAN be the truncated square submatrix of MA containing the coefficients of the first N Appell polynomials A_k=(a.+x)^k = Sum(j=0 to k) MAN(k,j) x^j. Then by the Cayley-Hamilton theorem (I-MAN)^N = 0; therefore, MAN^(-1) = Sum(k=1 to N) binomial(N,k) (-MAN)^{k-1} = MBN, the inverse of MAN, containing the coefficients of the first N rows of the Appell polynomials B_k(x) = (b. + x)^k = Sum(j=0 to k) MBN(k,j) x^j, which are the umbral compositional inverses of the Appell row polynomials A_k(x) of MAN; that is, A_k(B.(x)) = x^k = B_k(A.(x)), where, e.g., (A.(x))^k = A_k(x). - Tom Copeland, May 13 2020
The use of the term 'list partition transform' resulted from one of my first uses of these partition polynomials in relating A000262 to A084358 with their simple e.g.f.s. Other appropriate names would be the permutohedra polynomials since they are refined Euler characteristics of the permutohedra or the reciprocal polynomials since they give the multiplicative inverses of e.g.f.s with a constant of 1. - Tom Copeland, Oct 09 2022

Examples

			Table starts:
[0] [ 1]
[1] [-1]
[2] [-1,  2]
[3] [-1,  6, -6]
[4] [-1,  8,  6, -36,  24]
[5] [-1, 10, 20, -60, -90,  240, -120]
[6] [-1, 12, 30, -90,  20, -360,  480, -90, 1080, -1800, 720]
		

Crossrefs

Programs

  • Mathematica
    b[0] = 1; b[n_] := b[n] = -Sum[Binomial[n, j]*a[j]*b[n-j], {j, 1, n}];
    row[0] = {1}; row[n_] := Coefficient[b[n], #]& /@ (Times @@ (a /@ #)&) /@ IntegerPartitions[n];
    Table[row[n], {n, 0, 8}] // Flatten (* Jean-François Alcover, Apr 23 2014 *)
  • Sage
    def A133314_row(n): return [(-1)^len(s)*factorial(len(s))*SetPartitions(sum(s), s).cardinality() for s in Partitions(n)]
    for n in (0..10): print(A133314_row(n)) # Peter Luschny, Sep 18 2015

Formula

b_{n-1} = (1/n)(d/da(1))p_n[a_1, a_2, ..., a_n] where p_n are the row partition polynomials of the cumulant generator A127671. - Tom Copeland, Oct 13 2012
(E.g.f. of matrix B) = (e.g.f. of b)·exp(xt) = exp(b.t)·exp(xt) = exp(xt)/exp(a.t) = (e.g.f. of A^(-1)) and (e.g.f. of matrix A) = exp(a.t)·exp(xt) = exp(xt)/exp(b.t) = (e.g.f. of B^(-1)), where the umbral evaluation of exp(b.t) = Sum{n >= 0} (b.t)^n / n! = Sum_{n >= 0} b_n t^n / n! is understood in the denominator. These e.g.f.s define Appell sequences of polynomials. - Tom Copeland, Mar 22 2014
Sum of the n-th row is (-1)^n. - Peter Luschny, Sep 18 2015
The unsigned coefficients for the partitions a_2*a_1^n for n >= 0 are the Lah numbers A001286. - Tom Copeland, Aug 06 2016
G.f.: 1 / (1 + Sum_{n > 0} a_n x^n/n!) = 1 / exp(a.x). - Tom Copeland, Oct 18 2016
Let a_1 = 1 + x + B_1 = x + 1/2 and a_n = B_n = (B.)^n, where B_n are the Bernoulli numbers defined by e^(B.t) = t / (e^t-1), then t / e^(a.t) = t / [(x + 1) * t + exp(B.t)] = (e^t - 1) /[ 1 + (x + 1) (e^t - 1)] = exp(p.(x)t), where (p.(x))^n = p_n(x) are the shifted signed polynomials of A019538: p_0(x) = 0, p_1(x) = 1, p_2(x) = -(1 + 2 x), p_3(x) = 1 + 6 x + 6 x^2, ... , p_n(x) = n * b_{n-1}. - Tom Copeland, Oct 18 2016
With a_n = 1/(n+1), b_n = B_n, the Bernoulli numbers. - Tom Copeland, Nov 08 2016
Indeterminate substitutions as illustrated in A356145 lead to [E] = [L][P] = [P][E]^(-1)[P] = [P][RT] and [E]^(-1) = [P][L] = [P][E][P] = [RT][P], where [E] contains the refined Eulerian partition polynomials of A145271; [E]^(-1), A356145, the inverse set to [E]; [P], the permutohedra polynomials of this entry; [L], the classic Lagrange inversion polynomials of A134685; and [RT], the reciprocal tangent polynomials of A356144. Since [L]^2 = [P]^2 = [RT]^2 = [I], the substitutional identity, [L] = [E][P] = [P][E]^(-1) = [RT][P], [RT] = [E]^(-1)[P] = [P][L][P] = [P][E], and [P] = [L][E] = [E][RT] = [E]^(-1)[L] = [RT][E]^(-1). - Tom Copeland, Oct 05 2022

Extensions

More terms from Jean-François Alcover, Apr 23 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

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Author

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.

Crossrefs

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

A030298 List of permutations of 1,2,3,...,n for n=1,2,3,..., in lexicographic order.

Original entry on oeis.org

1, 1, 2, 2, 1, 1, 2, 3, 1, 3, 2, 2, 1, 3, 2, 3, 1, 3, 1, 2, 3, 2, 1, 1, 2, 3, 4, 1, 2, 4, 3, 1, 3, 2, 4, 1, 3, 4, 2, 1, 4, 2, 3, 1, 4, 3, 2, 2, 1, 3, 4, 2, 1, 4, 3, 2, 3, 1, 4, 2, 3, 4, 1, 2, 4, 1, 3, 2, 4, 3, 1, 3, 1, 2, 4, 3, 1, 4, 2, 3, 2, 1, 4, 3, 2, 4, 1, 3, 4, 1, 2, 3, 4, 2, 1, 4, 1, 2, 3, 4, 1, 3, 2, 4, 2
Offset: 1

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Author

Keywords

Comments

Contains every finite sequence of distinct numbers, infinitely many times.

Examples

			The permutations can be written as
  1,
  12, 21,
  123, 132, 213, 231, 312, 321, etc.
Write them in order and insert commas.
		

Crossrefs

A030299 gives the initial portion of these same permutations as decimally encoded numbers.
Cf. A001563 (row lengths), A001286 (row sums).
Cf. A030496 for another ordering.
The same information is essentially given in A055089, but in more compact way, by skipping those permutations which start with a fixed element (cf. A220696).
A220660(n) tells the zero-based rank r of the n-th permutation in this sequence, among all finite permutations of the same size.
A220663(n) tells the zero-based position (from the left) of that a(n) in that permutation of rank r.
A084557(n) tells that the n-th term a(n) belongs to the a(n):th lexicographically ordered permutation from the start (its "global rank").
A220660(A084557(n)) tells the "local rank" of the permutation (amongst the permutations of the same size) to which the n-th term a(n) belongs.
(A130664(n),A084555(n)) = (1,1),(2,3),(4,5),(6,8),(9,11),(12,14),... gives the starting and ending offsets of the n-th permutation in this list.

Programs

  • Haskell
    import Data.List (permutations, sort)
    a030298 n k = a030298_tabf !! (n-1) (k-1)
    a030298_row = concat . sort . permutations . enumFromTo 1
    a030298_tabf = map a030298_row [1..]
    -- Reinhard Zumkeller, Mar 29 2012
    (MIT/GNU Scheme, with Antti Karttunen's intseq-library):
    ;; Note that in Scheme, vector indexing is zero-based.
    ;; Requires also A055089permvec from A055089.
    (define (A030298 n) (vector-ref (A030298permvec (A084556 (A084557 n)) (A220660 (A084557 n))) (A220663 n)))
    (define (A030298permvec size rank) (vector-reverse (vector1invert (A055089permvec size rank))))
    (define (vector1invert vec) (make-initialized-vector (vector-length vec) (lambda (i) (1+ (- (vector-length vec) (vector-ref vec i))))))
    (define (vector-reverse vec) (make-initialized-vector (vector-length vec) (lambda (i) (vector-ref vec (- (vector-length vec) i 1)))))
    
  • Mathematica
    f[n_] := Permutations[Range@ n, {n}]; Array[f, 4] // Flatten (* Robert G. Wilson v, Dec 18 2012 *)
  • Python
    from itertools import permutations, count, chain, islice
    def A030298_gen(): # generator of terms
        return chain.from_iterable(p for l in count(2) for p in permutations(range(1,l)))
    A030298_list = list(islice(A030298_gen(),30)) # Chai Wah Wu, Mar 21 2022

Formula

Start with 1, then 12 and 21, then the 6 permutations of 123 in lexical order: 123, 132, 213, 231, 312, 321 and so on.

Extensions

Entry revised by N. J. A. Sloane, Feb 02 2006
Keyword tabf added by Reinhard Zumkeller, Mar 29 2012

A111596 The matrix inverse of the unsigned Lah numbers A271703.

Original entry on oeis.org

1, 0, 1, 0, -2, 1, 0, 6, -6, 1, 0, -24, 36, -12, 1, 0, 120, -240, 120, -20, 1, 0, -720, 1800, -1200, 300, -30, 1, 0, 5040, -15120, 12600, -4200, 630, -42, 1, 0, -40320, 141120, -141120, 58800, -11760, 1176, -56, 1, 0, 362880, -1451520, 1693440, -846720, 211680, -28224, 2016, -72, 1
Offset: 0

Views

Author

Wolfdieter Lang, Aug 23 2005

Keywords

Comments

Also the associated Sheffer triangle to Sheffer triangle A111595.
Coefficients of Laguerre polynomials (-1)^n * n! * L(n,-1,x), which equals (-1)^n * Lag(n,x,-1) below. Lag(n,Lag(.,x,-1),-1) = x^n evaluated umbrally, i.e., with (Lag(.,x,-1))^k = Lag(k,x,-1). - Tom Copeland, Apr 26 2014
Without row n=0 and column m=0 this is, up to signs, the Lah triangle A008297.
The unsigned column sequences are (with leading zeros): A000142, A001286, A001754, A001755, A001777, A001778, A111597-A111600 for m=1..10.
The row polynomials p(n,x) := Sum_{m=0..n} a(n,m)*x^m, together with the row polynomials s(n,x) of A111595 satisfy the exponential (or binomial) convolution identity s(n,x+y) = Sum_{k=0..n} binomial(n,k)*s(k,x)*p(n-k,y), n>=0.
Exponential Riordan array [1,x/(1+x)]. Inverse of the exponential Riordan array [1,x/(1-x)], which is the unsigned version of A111596. - Paul Barry, Apr 12 2007
For the unsigned subtriangle without column number m=0 and row number n=0, see A105278.
Unsigned triangle also matrix product |S1|*S2 of Stirling number matrices.
The unsigned row polynomials are Lag(n,-x,-1), the associated Laguerre polynomials of order -1 with negated argument. See Gradshteyn and Ryzhik, Abramowitz and Stegun and Rota (Finite Operator Calculus) for extensive formulas. - Tom Copeland, Nov 17 2007, Sep 09 2008
An infinitesimal matrix generator for unsigned A111596 is given by A132792. - Tom Copeland, Nov 22 2007
From the formalism of A132792 and A133314 for n > k, unsigned A111596(n,k) = a(k) * a(k+1)...a(n-1) / (n-k)! = a generalized factorial, where a(n) = A002378(n) = n-th term of first subdiagonal of unsigned A111596. Hence Deutsch's remark in A002378 provides an interpretation of A111596(n,k) in terms of combinations of certain circular binary words. - Tom Copeland, Nov 22 2007
Given T(n,k)= A111596(n,k) and matrices A and B with A(n,k) = T(n,k)*a(n-k) and B(n,k) = T(n,k)*b(n-k), then A*B = C where C(n,k) = T(n,k)*[a(.)+b(.)]^(n-k), umbrally. - Tom Copeland, Aug 27 2008
Operationally, the unsigned row polynomials may be expressed as p_n(:xD:) = x*:Dx:^n*x^{-1}=x*D^nx^n*x^{-1}= n!*binomial(xD+n-1,n) = (-1)^n n! binomial(-xD,n) = n!L(n,-1,-:xD:), where, by definition, :AB:^n = A^nB^n for any two operators A and B, D = d/dx, and L(n,-1,x) is the Laguerre polynomial of order -1. A similarity transformation of the operators :Dx:^n generates the higher order Laguerre polynomials, which can also be expressed in terms of rising or falling factorials or Kummer's confluent hypergeometric functions (cf. the Mathoverflow post). - Tom Copeland, Sep 21 2019

Examples

			Binomial convolution of row polynomials: p(3,x) = 6*x-6*x^2+x^3; p(2,x) = -2*x+x^2, p(1,x) = x, p(0,x) = 1,
together with those from A111595: s(3,x) = 9*x-6*x^2+x^3; s(2,x) = 1-2*x+x^2, s(1,x) = x, s(0,x) = 1; therefore
9*(x+y)-6*(x+y)^2+(x+y)^3 = s(3,x+y) = 1*s(0,x)*p(3,y) + 3*s(1,x)*p(2,y) + 3*s(2,x)*p(1,y) +1*s(3,x)*p(0,y) = (6*y-6*y^2+y^3) + 3*x*(-2*y+y^2) + 3*(1-2*x+x^2)*y + 9*x-6*x^2+x^3.
From _Wolfdieter Lang_, Apr 28 2014: (Start)
The triangle a(n,m) begins:
n\m  0     1       2       3      4     5   6  7
0:   1
1:   0     1
2:   0    -2       1
3:   0     6      -6       1
4:   0   -24      36     -12      1
5:   0   120    -240     120    -20     1
6:   0  -720    1800   -1200    300   -30   1
7:   0  5040  -15120   12600  -4200   630 -42  1
...
For more rows see the link.
(End)
		

Crossrefs

Row sums: A111884. Unsigned row sums: A000262.
A002868 gives maximal element (in magnitude) in each row.
Cf. A130561 for a natural refinement.
Cf. A264428, A264429, A271703 (unsigned).
Cf. A008297, A089231, A105278 (variants).

Programs

  • Maple
    # The function BellMatrix is defined in A264428.
    BellMatrix(n -> `if`(n::odd, -(n+1)!, (n+1)!), 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    a[0, 0] = 1; a[n_, m_] := ((-1)^(n-m))*(n!/m!)*Binomial[n-1, m-1]; Table[a[n, m], {n, 0, 10}, {m, 0, n}] // Flatten (* Jean-François Alcover, Jul 05 2013 *)
    T[ n_, k_] := (-1)^n n! Coefficient[ LaguerreL[ n, -1, x], x, k]; (* Michael Somos, Dec 15 2014 *)
    rows = 9;
    t = Table[(-1)^(n+1) n!, {n, 1, rows}];
    T[n_, k_] := BellY[n, k, t];
    Table[T[n, k], {n, 0, rows}, {k, 0, n}]  // Flatten (* Jean-François Alcover, Jun 22 2018, after Peter Luschny *)
  • PARI
    {T(n, k) = if( n<1 || k<1, n==0 && k==0, (-1)^n * n! * polcoeff( sum(k=1, n, binomial( n-1, k-1) * (-x)^k / k!), k))}; /* Michael Somos, Dec 15 2014 */
  • Sage
    lah_number = lambda n, k: factorial(n-k)*binomial(n,n-k)*binomial(n-1,n-k)
    A111596_row = lambda n: [(-1)^(n-k)*lah_number(n, k) for k in (0..n)]
    for n in range(10): print(A111596_row(n)) # Peter Luschny, Oct 05 2014
    
  • Sage
    # uses[inverse_bell_transform from A264429]
    def A111596_matrix(dim):
        fact = [factorial(n) for n in (1..dim)]
        return inverse_bell_transform(dim, fact)
    A111596_matrix(10) # Peter Luschny, Dec 20 2015
    

Formula

E.g.f. m-th column: ((x/(1+x))^m)/m!, m>=0.
E.g.f. for row polynomials p(n, x) is exp(x*y/(1+y)).
a(n, m) = ((-1)^(n-m))*|A008297(n, m)| = ((-1)^(n-m))*(n!/m!)*binomial(n-1, m-1), n>=m>=1; a(0, 0)=1; else 0.
a(n, m) = -(n-1+m)*a(n-1, m) + a(n-1, m-1), n>=m>=0, a(n, -1):=0, a(0, 0)=1; a(n, m)=0 if n
|a(n,m)| = Sum_{k=m..n} |S1(n,k)|*S2(k,m), n>=0. S2(n,m):=A048993. S1(n,m):=A048994. - Wolfdieter Lang, May 04 2007
From Tom Copeland, Nov 21 2011: (Start)
For this Lah triangle, the n-th row polynomial is given umbrally by
(-1)^n n! binomial(-Bell.(-x),n), where Bell_n(-x)= exp(x)(xd/dx)^n exp(-x), the n-th Bell / Touchard / exponential polynomial with neg. arg., (cf. A008277). E.g., 2! binomial(-Bell.(-x),2) = -Bell.(-x)*(-Bell.(-x)-1) = Bell_2(-x)+Bell_1(-x) = -2x+x^2.
A Dobinski relation is (-1)^n n! binomial(-Bell.(-x),n)= (-1)^n n! e^x Sum_{j>=0} (-1)^j binomial(-j,n)x^j/j!= n! e^x Sum_{j>=0} (-1)^j binomial(j-1+n,n)x^j/j!. See the Copeland link for the relation to inverse Mellin transform. (End)
The n-th row polynomial is (-1/x)^n e^x (x^2*D_x)^n e^(-x). - Tom Copeland, Oct 29 2012
Let f(.,x)^n = f(n,x) = x!/(x-n)!, the falling factorial,and r(.,x)^n = r(n,x) = (x-1+n)!/(x-1)!, the rising factorial, then the Lah polynomials, Lah(n,t)= n!*Sum{k=1..n} binomial(n-1,k-1)(-t)^k/k! (extra sign factor on odd rows), give the transform Lah(n,-f(.,x))= r(n,x), and Lah(n,r(.,x))= (-1)^n * f(n,x). - Tom Copeland, Oct 04 2014
|T(n,k)| = Sum_{j=0..2*(n-k)} A254881(n-k,j)*k^j/(n-k)!. Note that A254883 is constructed analogously from A254882. - Peter Luschny, Feb 10 2015
The T(n,k) are the inverse Bell transform of [1!,2!,3!,...] and |T(n,k)| are the Bell transform of [1!,2!,3!,...]. See A264428 for the definition of the Bell transform and A264429 for the definition of the inverse Bell transform. - Peter Luschny, Dec 20 2015
Dividing each n-th diagonal by n!, where the main diagonal is n=1, generates a shifted, signed Narayana matrix A001263. - Tom Copeland, Sep 23 2020

Extensions

New name using a comment from Wolfdieter Lang by Peter Luschny, May 10 2021
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