cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 218 results. Next

A077134 Duplicate of A001710.

Original entry on oeis.org

1, 1, 1, 3, 12, 60, 360, 2520, 20160, 181440, 1814400, 19958400, 239500800
Offset: 0

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Keywords

A375300 Triangle T(n,k) read by rows that has the A001710(n)-th lexicographic permutation of [1,...,n] as row n.

Original entry on oeis.org

1, 1, 2, 2, 1, 3, 2, 4, 3, 1, 3, 2, 5, 4, 1, 3, 6, 5, 4, 2, 1, 4, 3, 7, 6, 5, 2, 1, 4, 8, 7, 6, 5, 3, 2, 1, 5, 4, 9, 8, 7, 6, 3, 2, 1, 5, 10, 9, 8, 7, 6, 4, 3, 2, 1, 6, 5, 11, 10, 9, 8, 7, 4, 3, 2, 1, 6, 12, 11, 10, 9, 8, 7, 5, 4, 3, 2, 1, 7, 6, 13, 12, 11, 10, 9, 8, 5, 4, 3, 2, 1
Offset: 1

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Author

Hugo Pfoertner, Aug 24 2024

Keywords

Examples

			The triangle begins
  1,
  1, 2,
  2, 1, 3,
  2, 4, 3, 1,
  3, 2, 5, 4, 1,
  3, 6, 5, 4, 2, 1,
  4, 3, 7, 6, 5, 2, 1,
  4, 8, 7, 6, 5, 3, 2, 1,
  5, 4, 9, 8, 7, 6, 3, 2, 1,
  5, 10, 9, 8, 7, 6, 4, 3, 2, 1,
  ...
		

Crossrefs

Programs

  • PARI
    a375300_row(n) = if (n<2, [1], my(m=n!\2); forperm(n, p, m--; if (m==0, return(Vec(p)))))

A002260 Triangle read by rows: T(n,k) = k for n >= 1, k = 1..n.

Original entry on oeis.org

1, 1, 2, 1, 2, 3, 1, 2, 3, 4, 1, 2, 3, 4, 5, 1, 2, 3, 4, 5, 6, 1, 2, 3, 4, 5, 6, 7, 1, 2, 3, 4, 5, 6, 7, 8, 1, 2, 3, 4, 5, 6, 7, 8, 9, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14
Offset: 1

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Author

Angele Hamel (amh(AT)maths.soton.ac.uk)

Keywords

Comments

Old name: integers 1 to k followed by integers 1 to k+1 etc. (a fractal sequence).
Start counting again and again.
This is a "doubly fractal sequence" - see the Franklin T. Adams-Watters link.
The PARI functions t1, t2 can be used to read a square array T(n,k) (n >= 1, k >= 1) by antidiagonals downwards: n -> T(t1(n), t2(n)). - Michael Somos, Aug 23 2002
Reading this sequence as the antidiagonals of a rectangular array, row n is (n,n,n,...); this is the weight array (Cf. A144112) of the array A127779 (rectangular). - Clark Kimberling, Sep 16 2008
The upper trim of an arbitrary fractal sequence s is s, but the lower trim of s, although a fractal sequence, need not be s itself. However, the lower trim of A002260 is A002260. (The upper trim of s is what remains after the first occurrence of each term is deleted; the lower trim of s is what remains after all 0's are deleted from the sequence s-1.) - Clark Kimberling, Nov 02 2009
Eigensequence of the triangle = A001710 starting (1, 3, 12, 60, 360, ...). - Gary W. Adamson, Aug 02 2010
The triangle sums, see A180662 for their definitions, link this triangle of natural numbers with twenty-three different sequences, see the crossrefs. The mirror image of this triangle is A004736. - Johannes W. Meijer, Sep 22 2010
A002260 is the self-fission of the polynomial sequence (q(n,x)), where q(n,x) = x^n + x^(n-1) + ... + x + 1. See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
Sequence B is called a reluctant sequence of sequence A, if B is triangle array read by rows: row number k coincides with first k elements of the sequence A. Sequence A002260 is reluctant sequence of sequence 1,2,3,... (A000027). - Boris Putievskiy, Dec 12 2012
This is the maximal sequence of positive integers, such that once an integer k has occurred, the number of k's always exceeds the number of (k+1)'s for the remainder of the sequence, with the first occurrence of the integers being in order. - Franklin T. Adams-Watters, Oct 23 2013
A002260 are the k antidiagonal numerators of rationals in Cantor's proof of 1-to-1 correspondence between rationals and naturals; the denominators are k-numerator+1. - Adriano Caroli, Mar 24 2015
T(n,k) gives the distance to the largest triangular number < n. - Ctibor O. Zizka, Apr 09 2020

Examples

			First six rows:
  1
  1   2
  1   2   3
  1   2   3   4
  1   2   3   4   5
  1   2   3   4   5   6
		

References

  • Clark Kimberling, "Fractal sequences and interspersions," Ars Combinatoria 45 (1997) 157-168. (Introduces upper trimming, lower trimming, and signature sequences.)
  • M. Myers, Smarandache Crescendo Subsequences, R. H. Wilde, An Anthology in Memoriam, Bristol Banner Books, Bristol, 1998, p. 19.
  • F. Smarandache, Sequences of Numbers Involved in Unsolved Problems, Hexis, Phoenix, 2006.

Crossrefs

Cf. A140756 (alternating signs).
Triangle sums (see the comments): A000217 (Row1, Kn11); A004526 (Row2); A000096 (Kn12); A055998 (Kn13); A055999 (Kn14); A056000 (Kn15); A056115 (Kn16); A056119 (Kn17); A056121 (Kn18); A056126 (Kn19); A051942 (Kn110); A101859 (Kn111); A132754 (Kn112); A132755 (Kn113); A132756 (Kn114); A132757 (Kn115); A132758 (Kn116); A002620 (Kn21); A000290 (Kn3); A001840 (Ca2); A000326 (Ca3); A001972 (Gi2); A000384 (Gi3).
Cf. A108872.

Programs

  • Haskell
    a002260 n k = k
    a002260_row n = [1..n]
    a002260_tabl = iterate (\row -> map (+ 1) (0 : row)) [1]
    -- Reinhard Zumkeller, Aug 04 2014, Jul 03 2012
    
  • Maple
    at:=0; for n from 1 to 150 do for i from 1 to n do at:=at+1; lprint(at,i); od: od: # N. J. A. Sloane, Nov 01 2006
    seq(seq(i,i=1..k),k=1..13); # Peter Luschny, Jul 06 2009
  • Mathematica
    FoldList[{#1, #2} &, 1, Range[2, 13]] // Flatten (* Robert G. Wilson v, May 10 2011 *)
    Flatten[Table[Range[n],{n,20}]] (* Harvey P. Dale, Jun 20 2013 *)
  • Maxima
    T(n,k):=sum((i+k)*binomial(i+k-1,i)*binomial(k,n-i-k+1)*(-1)^(n-i-k+1),i,max(0,n+1-2*k),n-k+1); /* Vladimir Kruchinin, Oct 18 2013 */
    
  • PARI
    t1(n)=n-binomial(floor(1/2+sqrt(2*n)),2) /* this sequence */
    
  • PARI
    A002260(n)=n-binomial((sqrtint(8*n)+1)\2,2) \\ M. F. Hasler, Mar 10 2014
    
  • Python
    from math import isqrt, comb
    def A002260(n): return n-comb((m:=isqrt(k:=n<<1))+(k>m*(m+1)),2) # Chai Wah Wu, Nov 08 2024

Formula

a(n) = 1 + A002262(n).
n-th term is n - m*(m+1)/2 + 1, where m = floor((sqrt(8*n+1) - 1) / 2).
The above formula is for offset 0; for offset 1, use a(n) = n-m*(m+1)/2 where m = floor((-1+sqrt(8*n-7))/2). - Clark Kimberling, Jun 14 2011
a(k * (k + 1) / 2 + i) = i for k >= 0 and 0 < i <= k + 1. - Reinhard Zumkeller, Aug 14 2001
a(n) = (2*n + round(sqrt(2*n)) - round(sqrt(2*n))^2)/2. - Brian Tenneson, Oct 11 2003
a(n) = n - binomial(floor((1+sqrt(8*n))/2), 2). - Paul Barry, May 25 2004
T(n,k) = A001511(A118413(n,k)); T(n,k) = A003602(A118416(n,k)). - Reinhard Zumkeller, Apr 27 2006
a(A000217(n)) = A000217(n) - A000217(n-1), a(A000217(n-1) + 1) = 1, a(A000217(n) - 1) = A000217(n) - A000217(n-1) - 1. - Alexander R. Povolotsky, May 28 2008
a(A169581(n)) = A038566(n). - Reinhard Zumkeller, Dec 02 2009
T(n,k) = Sum_{i=1..k} i*binomial(k,i)*binomial(n-k,n-i) (regarded as triangle, see the example). - Mircea Merca, Apr 11 2012
T(n,k) = Sum_{i=max(0,n+1-2*k)..n-k+1} (i+k)*binomial(i+k-1,i)*binomial(k,n-i-k+1)*(-1)^(n-i-k+1). - Vladimir Kruchinin, Oct 18 2013
G.f.: x*y / ((1 - x) * (1 - x*y)^2) = Sum_{n,k>0} T(n,k) * x^n * y^k. - Michael Somos, Sep 17 2014
a(n) = n - S(n) where S(n) = sum of distinct terms in {a(1), a(2), ..., a(n-1)}. - David James Sycamore, Mar 10 2025

Extensions

More terms from Reinhard Zumkeller, Apr 27 2006
Incorrect program removed by Franklin T. Adams-Watters, Mar 19 2010
New name from Omar E. Pol, Jul 15 2012

A000254 Unsigned Stirling numbers of first kind, s(n+1,2): a(n+1) = (n+1)*a(n) + n!.

Original entry on oeis.org

0, 1, 3, 11, 50, 274, 1764, 13068, 109584, 1026576, 10628640, 120543840, 1486442880, 19802759040, 283465647360, 4339163001600, 70734282393600, 1223405590579200, 22376988058521600, 431565146817638400, 8752948036761600000, 186244810780170240000
Offset: 0

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Author

Keywords

Comments

Number of permutations of n+1 elements with exactly two cycles.
Number of cycles in all permutations of [n]. Example: a(3) = 11 because the permutations (1)(2)(3), (1)(23), (12)(3), (13)(2), (132), (123) have 11 cycles altogether. - Emeric Deutsch, Aug 12 2004
Row sums of A094310: In the symmetric group S_n, each permutation factors into k independent cycles; a(n) = sum k over S_n. - Harley Flanders (harley(AT)umich.edu), Jun 28 2004
The sum of the top levels of the last column over all deco polyominoes of height n. A deco polyomino is a directed column-convex polyomino in which the height, measured along the diagonal, is attained only in the last column. Example: a(2)=3 because the deco polyominoes of height 2 are the vertical and horizontal dominoes, the levels of their last columns being 2 and 1, respectively. - Emeric Deutsch, Aug 12 2006
a(n) is divisible by n for all composite n >= 6. a(2*n) is divisible by 2*n + 1. - Leroy Quet, May 20 2007
For n >= 2 the determinant of the n-1 X n-1 matrix M(i,j) = i + 2 for i = j and 1 otherwise (i,j = 1..n-1). E.g., for n = 3 the determinant of [(3, 1), (1, 4)]. See 53rd Putnam Examination, 1992, Problem B5. - Franz Vrabec, Jan 13 2008, Mar 26 2008
The numerator of the fraction when we sum (without simplification) the terms in the harmonic sequence. (1 + 1/2 = 2/2 + 1/2 = 3/2; 3/2 + 1/3 = 9/6 + 2/6 = 11/6; 11/6 + 1/4 = 44/24 + 6/24 = 50/24;...). The denominator of this fraction is n!*A000142. - Eric Desbiaux, Jan 07 2009
The asymptotic expansion of the higher order exponential integral E(x,m=2,n=1) ~ exp(-x)/x^2*(1 - 3/x + 11/x^2 - 50/x^3 + 274/x^4 - 1764/x^5 + 13068/x^6 - ...) leads to the sequence given above. See A163931 and A028421 for more information. - Johannes W. Meijer, Oct 20 2009
a(n) is the number of permutations of [n+1] containing exactly 2 cycles. Example: a(2) = 3 because the permutations (1)(23), (12)(3), (13)(2) are the only permutations of [3] with exactly 2 cycles. - Tom Woodward (twoodward(AT)macalester.edu), Nov 12 2009
It appears that, with the exception of n= 4, a(n) mod n = 0 if n is composite and = n-1 if n is prime. - Gary Detlefs, Sep 11 2010
a(n) is a multiple of A025527(n). - Charles R Greathouse IV, Oct 16 2012
Numerator of harmonic number H(n) = Sum_{i=1..n} 1/i when not reduced. See A001008 (Wolstenholme numbers) for the reduced numerators. - Rahul Jha, Feb 18 2015
The Stirling transform of this sequence is A222058(n) (Harmonic-geometric numbers). - Anton Zakharov, Aug 07 2016
a(n) is the (n-1)-st elementary symmetric function of the first n numbers. - Anton Zakharov, Nov 02 2016
The n-th iterated integral of log(x) is x^n * (n! * log(x) - a(n))/(n!)^2 + a polynomial of degree n-1 with arbitrary coefficients. This can be proven using the recurrence relation a(n) = (n-1)! + n*a(n-1). - Mohsen Maesumi, Oct 31 2018
Primes p such that p^3 | a(p-1) are the Wolstenholme primes A088164. - Amiram Eldar and Thomas Ordowski, Aug 08 2019
Total number of left-to-right maxima (or minima) in all permutations of [n]. a(3) = 11 = 3+2+2+2+1+1: (1)(2)(3), (1)(3)2, (2)1(3), (2)(3)1, (3)12, (3)21. - Alois P. Heinz, Aug 01 2020

Examples

			(1-x)^-1 * (-log(1-x)) = x + 3/2*x^2 + 11/6*x^3 + 25/12*x^4 + ...
G.f. = x + x^2 + 5*x^3 + 14*x^4 + 94*x^5 + 444*x^6 + 3828*x^7 + 25584*x^8 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, 1964 (and various reprintings), p. 833.
  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, identities 186-190.
  • N. Bleistein and R. A. Handelsman, Asymptotic Expansions of Integrals, Dover Publications, 1986, see page 2. MR0863284 (89d:41049)
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 217.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 226.
  • Shanzhen Gao, Permutations with Restricted Structure (in preparation).
  • K. Javorszky, Natural Orders: De Ordinibus Naturalibus, 2016, ISBN 978-3-99057-139-2.
  • 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).

Crossrefs

Programs

  • Magma
    a:=[]; for n in [1..22] do a:=a cat [Abs(StirlingFirst(n,2))]; end for; a; // Marius A. Burtea, Jan 01 2020
  • Maple
    A000254 := proc(n) option remember; if n<=1 then n else n*A000254(n-1)+(n-1)!; fi; end: seq(A000254(n),n=0..21);
    a := n -> add(n!/k, k=1..n): seq(a(n), n=0..21); # Zerinvary Lajos, Jan 22 2008
  • Mathematica
    Table[ (PolyGamma[ m ]+EulerGamma) (m-1)!, {m, 1, 24} ] (* Wouter Meeussen *)
    Table[ n!*HarmonicNumber[n], {n, 0, 19}] (* Robert G. Wilson v, May 21 2005 *)
    Table[Sum[1/i,{i,1,n}]/Product[1/i,{i,1,n}],{n,1,30}] (* Alexander Adamchuk, Jul 11 2006 *)
    Abs[StirlingS1[Range[20],2]] (* Harvey P. Dale, Aug 16 2011 *)
    Table[Gamma'[n + 1] /. EulerGamma -> 0, {n, 0, 30}] (* Li Han, Feb 14 2024*)
  • Maxima
    a(n):=(-1)^(n+1)/2*(n+1)*sum(k*bern(k-1)*stirling1(n,k),k,1,n); /* Vladimir Kruchinin, Nov 20 2016 */
    
  • MuPAD
    A000254 := proc(n) begin n*A000254(n-1)+fact(n-1) end_proc: A000254(1) := 1:
    
  • PARI
    {a(n) = if( n<0, 0, (n+1)! / 2 * sum( k=1, n, 1 / k / (n+1-k)))} /* Michael Somos, Feb 05 2004 */
    
  • Sage
    [stirling_number1(i, 2) for i in range(1, 22)]  # Zerinvary Lajos, Jun 27 2008
    

Formula

Let P(n,X) = (X+1)*(X+2)*(X+3)*...*(X+n); then a(n) is the coefficient of X; or a(n) = P'(n,0). - Benoit Cloitre, May 09 2002
Sum_{k > 0} a(k) * x^k/ k!^2 = exp(x) *(Sum_{k>0} (-1)^(k+1) * x^k / (k * k!)). - Michael Somos, Mar 24 2004; corrected by Warren D. Smith, Feb 12 2006
a(n) is the coefficient of x^(n+2) in (-log(1-x))^2, multiplied by (n+2)!/2.
a(n) = n! * Sum_{i=1..n} 1/i = n! * H(n), where H(n) = A001008(n)/A002805(n) is the n-th harmonic number.
a(n) ~ 2^(1/2)*Pi^(1/2)*log(n)*n^(1/2)*e^-n*n^n. - Joe Keane (jgk(AT)jgk.org), Jun 06 2002
E.g.f.: log(1 - x) / (x-1). (= (log(1 - x))^2 / 2 if offset 1). - Michael Somos, Feb 05 2004
D-finite with recurrence: a(n) = a(n-1) * (2*n - 1) - a(n-2) * (n - 1)^2, if n > 1. - Michael Somos, Mar 24 2004
a(n) = A081358(n)+A092691(n). - Emeric Deutsch, Aug 12 2004
a(n) = n!*Sum_{k=1..n} (-1)^(k+1)*binomial(n, k)/k. - Vladeta Jovovic, Jan 29 2005
p^2 divides a(p-1) for prime p > 3. a(n) = (Sum_{i=1..n} 1/i) / Product_{i=1..n} 1/i. - Alexander Adamchuk, Jul 11 2006
a(n) = 3* A001710(n) + 2* A001711(n-3) for n > 2; e.g., 11 = 3*3 + 2*1, 50 = 3*12 + 2*7, 274 = 3*60 + 2*47, ... - Gary Detlefs, May 24 2010
a(n) = A138772(n+1) - A159324(n). - Gary Detlefs, Jul 05 2010
a(n) = A121633(n) + A002672(n). - Gary Detlefs, Jul 18 2010
a(n+1) = Sum_{i=1..floor((n-1)/2)} n!/((n-i)*i) + Sum_{i=ceiling(n/2)..floor(n/2)} n!/(2*(n-i)*i). - Shanzhen Gao, Sep 14 2010
From Gary Detlefs, Sep 11 2010: (Start)
a(n) = (a(n-1)*(n^2 - 2*n + 1) + (n + 1)!)/(n - 1) for n > 2.
It appears that, with the exception of n = 2, (a(n+1)^2 - a(n)^2) mod n^2 = 0 if n is composite and 4*n if n is prime.
It appears that, with the exception of n = 2, (a(n+1)^3 - a(n)^2) mod n = 0 if n is composite and n - 2 if n is prime.
It appears that, with the exception of n = 2, (a(n)^2 + a(n+1)^2) mod n = 0 if n is composite and = 2 if n is prime. (End)
a(n) = Integral_{x=0..oo} (x^n - n!)*log(x)*exp(-x) dx. - Groux Roland, Mar 28 2011
a(n) = 3*n!/2 + 2*(n-2)!*Sum_{k=0..n-3} binomial(k+2,2)/(n-2-k) for n >= 2. - Gary Detlefs, Sep 02 2011
a(n)/(n-1)! = ml(n) = n*ml(n-1)/(n-1) + 1 for n > 1, where ml(n) is the average number of random draws from an n-set with replacement until the total set has been observed. G.f. of ml: x*(1 - log(1 - x))/(1 - x)^2. - Paul Weisenhorn, Nov 18 2011
a(n) = det(|S(i+2, j+1)|, 1 <= i,j <= n-2), where S(n,k) are Stirling numbers of the second kind. - Mircea Merca, Apr 06 2013
E.g.f.: x/(1 - x)*E(0)/2, where E(k) = 2 + E(k+1)*x*(k + 1)/(k + 2). - Sergei N. Gladkovskii, Jun 01 2013 [Edited by Michael Somos, Nov 28 2013]
0 = a(n) * (a(n+4) - 6*a(n+3) + 7*a(n+2) - a(n+1)) - a(n+1) * (4*a(n+3) - 6*a(n+2) + a(n+1)) + 3*a(n+2)^2 unless n=0. - Michael Somos, Nov 28 2013
For a simple way to calculate the sequence, multiply n! by the integral from 0 to 1 of (1 - x^n)/(1 - x) dx. - Rahul Jha, Feb 18 2015
From Ilya Gutkovskiy, Aug 07 2016: (Start)
Inverse binomial transform of A073596.
a(n) ~ sqrt(2*Pi*n) * n^n * (log(n) + gamma)/exp(n), where gamma is the Euler-Mascheroni constant A001620. (End)
a(n) = ((-1)^(n+1)/2*(n+1))*Sum_{k=1..n} k*Bernoulli(k-1)*Stirling1(n,k). - Vladimir Kruchinin, Nov 20 2016
a(n) = (n)! * (digamma(n+1) + gamma), where gamma is the Euler-Mascheroni constant A001620. - Pedro Caceres, Mar 10 2018
From Andy Nicol, Oct 21 2021: (Start)
Gamma'(x) = a(x-1) - (x-1)!*gamma, where Gamma'(x) is the derivative of the gamma function at positive integers and gamma is the Euler-Mascheroni constant. E.g.:
Gamma'(1) = -gamma, Gamma'(2) = 1-gamma, Gamma'(3) = 3-2*gamma,
Gamma'(22) = 186244810780170240000 - 51090942171709440000*gamma. (End)
From Peter Bala, Feb 03 2022: (Start)
The following are all conjectural:
E.g.f.: for nonzero m, (1/m)*Sum_{n >= 1} (-1)^(n+1)*(1/n)*binomial(m*n,n)* x^n/(1 - x)^(m*n+1) = x + 3*x^2/2! + 11*x^3/3! + 50*x^4/4! + ....
For nonzero m, a(n) = (1/m)*n!*Sum_{k = 1..n} (-1)^(k+1)*(1/k)*binomial(m*k,k)* binomial(n+(m-1)*k,n-k).
a(n)^2 = (1/2)*n!^2*Sum_{k = 1..n} (-1)^(k+1)*(1/k^2)*binomial(n,k)* binomial(n+k,k). (End)
From Mélika Tebni, Jun 20 2022: (Start)
a(n) = -Sum_{k=0..n} k!*A021009(n, k+1).
a(n) = Sum_{k=0..n} k!*A094587(n, k+1). (End)
a(n) = n! * 1/(1 - 1^2/(3 - 2^2/(5 - 3^2/(7 - ... - (n - 1)^2/((2*n - 1)))))). - Peter Bala, Mar 16 2024

A132393 Triangle of unsigned Stirling numbers of the first kind (see A048994), read by rows, T(n,k) for 0 <= k <= n.

Original entry on oeis.org

1, 0, 1, 0, 1, 1, 0, 2, 3, 1, 0, 6, 11, 6, 1, 0, 24, 50, 35, 10, 1, 0, 120, 274, 225, 85, 15, 1, 0, 720, 1764, 1624, 735, 175, 21, 1, 0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, 0, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 1, 0, 362880, 1026576, 1172700, 723680, 269325, 63273, 9450, 870, 45, 1
Offset: 0

Views

Author

Philippe Deléham, Nov 10 2007, Oct 15 2008, Oct 17 2008

Keywords

Comments

Another name: Triangle of signless Stirling numbers of the first kind.
Triangle T(n,k), 0<=k<=n, read by rows given by [0,1,1,2,2,3,3,4,4,5,5,...] DELTA [1,0,1,0,1,0,1,0,1,...] where DELTA is the operator defined in A084938.
A094645*A007318 as infinite lower triangular matrices.
Row sums are the factorial numbers. - Roger L. Bagula, Apr 18 2008
Exponential Riordan array [1/(1-x), log(1/(1-x))]. - Ralf Stephan, Feb 07 2014
Also the Bell transform of the factorial numbers (A000142). For the definition of the Bell transform see A264428 and for cross-references A265606. - Peter Luschny, Dec 31 2015
This is the lower triagonal Sheffer matrix of the associated or Jabotinsky type |S1| = (1, -log(1-x)) (see the W. Lang link under A006232 for the notation and references). This implies the e.g.f.s given below. |S1| is the transition matrix from the monomial basis {x^n} to the rising factorial basis {risefac(x,n)}, n >= 0. - Wolfdieter Lang, Feb 21 2017
T(n, k), for n >= k >= 1, is also the total volume of the n-k dimensional cell (polytope) built from the n-k orthogonal vectors of pairwise different lengths chosen from the set {1, 2, ..., n-1}. See the elementary symmetric function formula for T(n, k) and an example below. - Wolfdieter Lang, May 28 2017
From Wolfdieter Lang, Jul 20 2017: (Start)
The compositional inverse w.r.t. x of y = y(t;x) = x*(1 - t(-log(1-x)/x)) = x + t*log(1-x) is x = x(t;y) = ED(y,t) := Sum_{d>=0} D(d,t)*y^(d+1)/(d+1)!, the e.g.f. of the o.g.f.s D(d,t) = Sum_{m>=0} T(d+m, m)*t^m of the diagonal sequences of the present triangle. See the P. Bala link for a proof (there d = n-1, n >= 1, is the label for the diagonals).
This inversion gives D(d,t) = P(d, t)/(1-t)^(2*d+1), with the numerator polynomials P(d, t) = Sum_{m=0..d} A288874(d, m)*t^m. See an example below. See also the P. Bala formula in A112007. (End)
For n > 0, T(n,k) is the number of permutations of the integers from 1 to n which have k visible digits when viewed from a specific end, in the sense that a higher value hides a lower one in a subsequent position. - Ian Duff, Jul 12 2019

Examples

			Triangle T(n,k) begins:
  1;
  0,    1;
  0,    1,     1;
  0,    2,     3,     1;
  0,    6,    11,     6,    1;
  0,   24,    50,    35,   10,    1;
  0,  120,   274,   225,   85,   15,   1;
  0,  720,  1764,  1624,  735,  175,  21,  1;
  0, 5040, 13068, 13132, 6769, 1960, 322, 28, 1;
  ...
---------------------------------------------------
Production matrix is
  0, 1
  0, 1, 1
  0, 1, 2,  1
  0, 1, 3,  3,  1
  0, 1, 4,  6,  4,  1
  0, 1, 5, 10, 10,  5,  1
  0, 1, 6, 15, 20, 15,  6, 1
  0, 1, 7, 21, 35, 35, 21, 7, 1
  ...
From _Wolfdieter Lang_, May 09 2017: (Start)
Three term recurrence: 50 = T(5, 2) = 1*6 + (5-1)*11 = 50.
Recurrence from the Sheffer a-sequence [1, 1/2, 1/6, 0, ...]: 50 = T(5, 2) = (5/2)*(binomial(1, 1)*1*6 + binomial(2, 1)*(1/2)*11 + binomial(3, 1)*(1/6)*6 + 0) = 50. The vanishing z-sequence produces the k=0 column from T(0, 0) = 1. (End)
Elementary symmetric function T(4, 2) = sigma^{(3)}_2 = 1*2 + 1*3 + 2*3 = 11. Here the cells (polytopes) are 3 rectangles with total area 11. - _Wolfdieter Lang_, May 28 2017
O.g.f.s of diagonals: d=2 (third diagonal) [0, 6, 50, ...] has D(2,t) = P(2, t)/(1-t)^5, with P(2, t) = 2 + t, the n = 2 row of A288874. - _Wolfdieter Lang_, Jul 20 2017
Boas-Buck recurrence for column k = 2 and n = 5: T(5, 2) = (5!*2/3)*((3/8)*T(2,2)/2! + (5/12)*T(3,2)/3! + (1/2)*T(4,2)/4!) = (5!*2/3)*(3/16 + (5/12)*3/3! + (1/2)*11/4!) = 50. The beta sequence begins: {1/2, 5/12, 3/8, ...}. - _Wolfdieter Lang_, Aug 11 2017
		

References

  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, pages 31, 187, 441, 996.
  • R. L. Graham, D. E. Knuth, and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd. ed., Table 259, p. 259.
  • Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), pp. 149-150

Crossrefs

Essentially a duplicate of A048994. Cf. A008275, A008277, A112007, A130534, A288874, A354795.

Programs

  • Haskell
    a132393 n k = a132393_tabl !! n !! k
    a132393_row n = a132393_tabl !! n
    a132393_tabl = map (map abs) a048994_tabl
    -- Reinhard Zumkeller, Nov 06 2013
    
  • Maple
    a132393_row := proc(n) local k; seq(coeff(expand(pochhammer (x,n)),x,k),k=0..n) end: # Peter Luschny, Nov 28 2010
  • Mathematica
    p[t_] = 1/(1 - t)^x; Table[ ExpandAll[(n!)SeriesCoefficient[ Series[p[t], {t, 0, 30}], n]], {n, 0, 10}]; a = Table[(n!)* CoefficientList[SeriesCoefficient[ Series[p[t], {t, 0, 30}], n], x], {n, 0, 10}]; Flatten[a] (* Roger L. Bagula, Apr 18 2008 *)
    Flatten[Table[Abs[StirlingS1[n,i]],{n,0,10},{i,0,n}]] (* Harvey P. Dale, Feb 04 2014 *)
  • Maxima
    create_list(abs(stirling1(n,k)),n,0,12,k,0,n); /* Emanuele Munarini, Mar 11 2011 */
    
  • PARI
    column(n,k) = my(v1, v2); v1 = vector(n-1, i, 0); v2 = vector(n, i, 0); v2[1] = 1; for(i=1, n-1, v1[i] = (i+k)*(i+k-1)/2*v2[i]; for(j=1, i-1, v1[j] *= (i-j)*(i+k)/(i-j+2)); v2[i+1] = vecsum(v1)/i); v2 \\ generates n first elements of the k-th column starting from the first nonzero element. - Mikhail Kurkov, Mar 05 2025

Formula

T(n,k) = T(n-1,k-1)+(n-1)*T(n-1,k), n,k>=1; T(n,0)=T(0,k); T(0,0)=1.
Sum_{k=0..n} T(n,k)*x^(n-k) = A000012(n), A000142(n), A001147(n), A007559(n), A007696(n), A008548(n), A008542(n), A045754(n), A045755(n) for x = 0, 1, 2, 3, 4, 5, 6, 7, 8 respectively. - Philippe Deléham, Nov 13 2007
Expand 1/(1-t)^x = Sum_{n>=0}p(x,n)*t^n/n!; then the coefficients of the p(x,n) produce the triangle. - Roger L. Bagula, Apr 18 2008
Sum_{k=0..n} T(n,k)*2^k*x^(n-k) = A000142(n+1), A000165(n), A008544(n), A001813(n), A047055(n), A047657(n), A084947(n), A084948(n), A084949(n) for x = 1, 2, 3, 4, 5, 6, 7, 8, 9 respectively. - Philippe Deléham, Sep 18 2008
a(n) = Sum_{k=0..n} T(n,k)*3^k*x^(n-k) = A001710(n+2), A001147(n+1), A032031(n), A008545(n), A047056(n), A011781(n), A144739(n), A144756(n), A144758(n) for x=1,2,3,4,5,6,7,8,9,respectively. - Philippe Deléham, Sep 20 2008
Sum_{k=0..n} T(n,k)*4^k*x^(n-k) = A001715(n+3), A002866(n+1), A007559(n+1), A047053(n), A008546(n), A049308(n), A144827(n), A144828(n), A144829(n) for x=1,2,3,4,5,6,7,8,9 respectively. - Philippe Deléham, Sep 21 2008
Sum_{k=0..n} x^k*T(n,k) = x*(1+x)*(2+x)*...*(n-1+x), n>=1. - Philippe Deléham, Oct 17 2008
From Wolfdieter Lang, Feb 21 2017: (Start)
E.g.f. k-th column: (-log(1 - x))^k, k >= 0.
E.g.f. triangle (see the Apr 18 2008 Baluga comment): exp(-x*log(1-z)).
E.g.f. a-sequence: x/(1 - exp(-x)). See A164555/A027642. The e.g.f. for the z-sequence is 0. (End)
From Wolfdieter Lang, May 28 2017: (Start)
The row polynomials R(n, x) = Sum_{k=0..n} T(n, k)*x^k, for n >= 0, are R(n, x) = risefac(x,n-1) := Product_{j=0..n-1} x+j, with the empty product for n=0 put to 1. See the Feb 21 2017 comment above. This implies:
T(n, k) = sigma^{(n-1)}_(n-k), for n >= k >= 1, with the elementary symmetric functions sigma^{(n-1)}_m of degree m in the n-1 symbols 1, 2, ..., n-1, with binomial(n-1, m) terms. See an example below.(End)
Boas-Buck type recurrence for column sequence k: T(n, k) = (n!*k/(n - k)) * Sum_{p=k..n-1} beta(n-1-p)*T(p, k)/p!, for n > k >= 0, with input T(k, k) = 1, and beta(k) = A002208(k+1)/A002209(k+1). See a comment and references in A286718. - Wolfdieter Lang, Aug 11 2017
T(n,k) = Sum_{j=k..n} j^(j-k)*binomial(j-1, k-1)*A354795(n,j) for n > 0. - Mélika Tebni, Mar 02 2023
n-th row polynomial: n!*Sum_{k = 0..2*n} (-1)^k*binomial(-x, k)*binomial(-x, 2*n-k) = n!*Sum_{k = 0..2*n} (-1)^k*binomial(1-x, k)*binomial(-x, 2*n-k). - Peter Bala, Mar 31 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} binomial(-k,j)*T(n,k+j-1)*(-1)^j for 1 <= k < n with T(n,n) = 1.
Recursion for the k-th column (independently of other columns): T(n,k) = 1/(n-k)*Sum_{j=2..n-k+1} (j-2)!*binomial(n,j)*T(n-j+1,k) for 1 <= k < n with T(n,n) = 1 (see Fedor Petrov link). (End)

A001286 Lah numbers: a(n) = (n-1)*n!/2.

Original entry on oeis.org

1, 6, 36, 240, 1800, 15120, 141120, 1451520, 16329600, 199584000, 2634508800, 37362124800, 566658892800, 9153720576000, 156920924160000, 2845499424768000, 54420176498688000, 1094805903679488000, 23112569077678080000, 510909421717094400000
Offset: 2

Views

Author

Keywords

Comments

Number of surjections from {1,...,n} to {1,...,n-1}. - Benoit Cloitre, Dec 05 2003
First Eulerian transform of 0,1,2,3,4,... . - Ross La Haye, Mar 05 2005
With offset 0 : determinant of the n X n matrix m(i,j)=(i+j+1)!/i!/j!. - Benoit Cloitre, Apr 11 2005
These numbers arise when expressing n(n+1)(n+2)...(n+k)[n+(n+1)+(n+2)+...+(n+k)] as sums of squares: n(n+1)[n+(n+1)] = 6(1+4+9+16+ ... + n^2), n(n+1)(n+2)(n+(n+1)+(n+2)) = 36(1+(1+4)+(1+4+9)+...+(1+4+9+16+ ... + n^2)), n(n+1)(n+2)(n+3)(n+(n+1)+(n+2)+(n+3)) = 240(...), ... . - Alexander R. Povolotsky, Oct 16 2006
a(n) is the number of edges in the Hasse diagram for the weak Bruhat order on the symmetric group S_n. For permutations p,q in S_n, q covers p in the weak Bruhat order if p,q differ by an adjacent transposition and q has one more inversion than p. Thus 23514 covers 23154 due to the transposition that interchanges the third and fourth entries. Cf. A002538 for the strong Bruhat order. - David Callan, Nov 29 2007
a(n) is also the number of excedances in all permutations of {1,2,...,n} (an excedance of a permutation p is a value j such p(j)>j). Proof: j is exceeded (n-1)! times by each of the numbers j+1, j+2, ..., n; now, Sum_{j=1..n} (n-j)(n-1)! = n!(n-1)/2. Example: a(3)=6 because the number of excedances of the permutations 123, 132, 312, 213, 231, 321 are 0, 1, 1, 1, 2, 1, respectively. - Emeric Deutsch, Dec 15 2008
(-1)^(n+1)*a(n) is the determinant of the n X n matrix whose (i,j)-th element is 0 for i = j, is j-1 for j>i, and j for j < i. - Michel Lagneau, May 04 2010
Row sums of the triangle in A030298. - Reinhard Zumkeller, Mar 29 2012
a(n) is the total number of ascents (descents) over all n-permutations. a(n) = Sum_{k=1..n} A008292(n,k)*k. - Geoffrey Critzer, Jan 06 2013
For m>=4, a(m-2) is the number of Hamiltonian cycles in a simple graph with m vertices which is complete, except for one edge. Proof: think of distinct round-table seatings of m persons such that persons "1" and "2" may not be neighbors; the count is (m-3)(m-2)!/2. See also A001710. - Stanislav Sykora, Jun 17 2014
Popularity of left (right) children in treeshelves. Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link. Popularity is the sum of a certain statistic (number of left children, in this case) over all objects of size n. See A278677, A278678 or A278679 for more definitions and examples. See A008292 for the distribution of the left (right) children in treeshelves. - Sergey Kirgizov, Dec 24 2016

Examples

			G.f. = x^2 + 6*x^3 + 36*x^4 + 240*x^5 + 1800*x^6 + 15120*x^7 + 141120*x^8 + ...
a(10) = (1+2+3+4+5+6+7+8+9)*(1*2*3*4*5*6*7*8*9) = 16329600. - _Reinhard Zumkeller_, May 15 2010
		

References

  • A. T. Benjamin and J. J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A. 2003, p. 90, ex. 4.
  • Louis Comtet, Advanced Combinatorics, Reidel, 1974, p. 156.
  • A. P. Prudnikov, Yu. A. Brychkov and O.I. Marichev, "Integrals and Series", Volume 1: "Elementary Functions", Chapter 4: "Finite Sums", New York, Gordon and Breach Science Publishers, 1986-1992.
  • John Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 44.
  • 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

A002868 is an essentially identical sequence.
Column 2 of |A008297|.
Third column (m=2) of triangle |A111596(n, m)|: matrix product of |S1|.S2 Stirling number matrices.
Cf. also A000110, A000111.

Programs

Formula

a(n) = Sum_{i=0..n-1} (-1)^(n-i-1) * i^n * binomial(n-1,i). - Yong Kong (ykong(AT)curagen.com), Dec 26 2000 [corrected by Amiram Eldar, May 02 2022]
E.g.f.: x^2/[2(1-x)^2]. - Ralf Stephan, Apr 02 2004
a(n+1) = (-1)^(n+1)*det(M_n) where M_n is the n X n matrix M_(i,j)=max(i*(i+1)/2,j*(j+1)/2). - Benoit Cloitre, Apr 03 2004
Row sums of table A051683. - Alford Arnold, Sep 29 2006
5th binomial transform of A135218: (1, 1, 1, 25, 25, 745, 3145, ...). - Gary W. Adamson, Nov 23 2007
If we define f(n,i,x) = Sum_{k=i..n} Sum_{j=i..k} binomial(k,j)*Stirling1(n,k)*Stirling2(j,i)*x^(k-j) then a(n)=(-1)^n*f(n,2,-2), (n>=2). - Milan Janjic, Mar 01 2009
a(n) = A000217(n-1)*A000142(n-1). - Reinhard Zumkeller, May 15 2010
a(n) = (n+1)!*Sum_{k=1..n-1} 1/(k^2+3*k+2). - Gary Detlefs, Sep 14 2011
Sum_{n>=2} 1/a(n) = 2*(2 - exp(1) - gamma + Ei(1)) = 1.19924064599..., where gamma = A001620 and Ei(1) = A091725. - Ilya Gutkovskiy, Nov 24 2016
a(n+1) = a(n)*n*(n+1)/(n-1). - Chai Wah Wu, Apr 11 2018
Sum_{n>=2} (-1)^n/a(n) = 2*(gamma - Ei(-1)) - 2/e, where e = A001113 and Ei(-1) = -A099285. - Amiram Eldar, May 02 2022

A130534 Triangle T(n,k), 0 <= k <= n, read by rows, giving coefficients of the polynomial (x+1)(x+2)...(x+n), expanded in increasing powers of x. T(n,k) is also the unsigned Stirling number |s(n+1, k+1)|, denoting the number of permutations on n+1 elements that contain exactly k+1 cycles.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 11, 6, 1, 24, 50, 35, 10, 1, 120, 274, 225, 85, 15, 1, 720, 1764, 1624, 735, 175, 21, 1, 5040, 13068, 13132, 6769, 1960, 322, 28, 1, 40320, 109584, 118124, 67284, 22449, 4536, 546, 36, 1, 362880, 1026576, 1172700, 723680, 269325, 63273, 9450, 870, 45, 1
Offset: 0

Views

Author

Philippe Deléham, Aug 09 2007

Keywords

Comments

This triangle is an unsigned version of the triangle of Stirling numbers of the first kind, A008275, which is the main entry for these numbers. - N. J. A. Sloane, Jan 25 2011
Or, triangle T(n,k), 0 <= k <= n, read by rows given by [1,1,2,2,3,3,4,4,5,5,6,6,...] DELTA [1,0,1,0,1,0,1,0,1,0,1,0,...] where DELTA is the operator defined in A084938.
Reversal of A094638.
Equals A132393*A007318, as infinite lower triangular matrices. - Philippe Deléham, Nov 13 2007
From Johannes W. Meijer, Oct 07 2009: (Start)
The higher order exponential integrals E(x,m,n) are defined in A163931. The asymptotic expansion of the exponential integrals E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + n*(n+1)/x^2 - n*(n+1)*(n+2)/x^3 + ...), see Abramowitz and Stegun. This formula follows from the general formula for the asymptotic expansion, see A163932. We rewrite E(x,m=1,n) ~ (exp(-x)/x)*(1 - n/x + (n^2+n)/x^2 - (2*n+3*n^2+n^3)/x^3 + (6*n+11*n^2+6*n^3+n^4)/x^3 - ...) and observe that the T(n,m) are the polynomials coefficients in the denominators. Looking at the a(n,m) formula of A028421, A163932 and A163934, and shifting the offset given above to 1, we can write T(n-1,m-1) = a(n,m) = (-1)^(n+m)*Stirling1(n,m), see the Maple program.
The asymptotic expansion leads for values of n from one to eleven to known sequences, see the cross-references. With these sequences one can form the triangles A008279 (right-hand columns) and A094587 (left-hand columns).
See A163936 for information about the o.g.f.s. of the right-hand columns of this triangle.
(End)
The number of elements greater than i to the left of i in a permutation gives the i-th element of the inversion vector. (Skiena-Pemmaraju 2003, p. 69.) T(n,k) is the number of n-permutations that have exactly k 0's in their inversion vector. See evidence in Mathematica code below. - Geoffrey Critzer, May 07 2010
T(n,k) counts the rooted trees with k+1 trunks in forests of "naturally grown" rooted trees with n+2 nodes. This corresponds to sums of coefficients of iterated derivatives representing vectors, Lie derivatives, or infinitesimal generators for flow fields and formal group laws. Cf. links in A139605. - Tom Copeland, Mar 23 2014
A refinement is A036039. - Tom Copeland, Mar 30 2014
From Tom Copeland, Apr 05 2014: (Start)
With initial n=1 and row polynomials of T as p(n,x)=x(x+1)...(x+n-1), the powers of x correspond to the number of trunks of the rooted trees of the "naturally-grown" forest referred to above. With each trunk allowed m colors, p(n,m) gives the number of such non-plane colored trees for the forest with each tree having n+1 vertices.
p(2,m) = m + m^2 = A002378(m) = 2*A000217(m) = 2*(first subdiag of |A238363|).
p(3,m) = 2m + 3m^2 + m^3 = A007531(m+2) = 3*A007290(m+2) = 3*(second subdiag A238363).
p(4,m) = 6m + 11m^2 + 6m^3 + m^4 = A052762(m+3) = 4*A033487(m) = 4*(third subdiag).
From the Joni et al. link, p(n,m) also represents the disposition of n distinguishable flags on m distinguishable flagpoles.
The chromatic polynomial for the complete graph K_n is the falling factorial, which encodes the colorings of the n vertices of K_n and gives a shifted version of p(n,m).
E.g.f. for the row polynomials: (1-y)^(-x).
(End)
A relation to derivatives of the determinant |V(n)| of the n X n Vandermonde matrix V(n) in the indeterminates c(1) thru c(n):
|V(n)| = Product_{1<=jTom Copeland, Apr 10 2014
From Peter Bala, Jul 21 2014: (Start)
Let M denote the lower unit triangular array A094587 and for k = 0,1,2,... define M(k) to be the lower unit triangular block array
/I_k 0\
\ 0 M/
having the k X k identity matrix I_k as the upper left block; in particular, M(0) = M. Then the present triangle equals the infinite matrix product M(0)*M(1)*M(2)*... (which is clearly well defined). See the Example section. (End)
For the relation of this rising factorial to the moments of Viennot's Laguerre stories, see the Hetyei link, p. 4. - Tom Copeland, Oct 01 2015
Can also be seen as the Bell transform of n! without column 0 (and shifted enumeration). For the definition of the Bell transform see A264428. - Peter Luschny, Jan 27 2016

Examples

			Triangle  T(n,k) begins:
n\k         0        1        2       3       4      5      6     7    8  9 10
n=0:        1
n=1:        1        1
n=2:        2        3        1
n=3:        6       11        6       1
n=4:       24       50       35      10       1
n=5:      120      274      225      85      15      1
n=6:      720     1764     1624     735     175     21      1
n=7:     5040    13068    13132    6769    1960    322     28     1
n=8:    40320   109584   118124   67284   22449   4536    546    36    1
n=9:   362880  1026576  1172700  723680  269325  63273   9450   870   45  1
n=10: 3628800 10628640 12753576 8409500 3416930 902055 157773 18150 1320 55  1
[Reformatted and extended by _Wolfdieter Lang_, Feb 05 2013]
T(3,2) = 6 because there are 6 permutations of {1,2,3,4} that have exactly 2 0's in their inversion vector: {1, 2, 4, 3}, {1, 3, 2, 4}, {1, 3, 4, 2}, {2, 1, 3, 4},{2, 3, 1, 4}, {2, 3, 4, 1}. The respective inversion vectors are {0, 0, 1}, {0, 1, 0}, {0, 2, 0}, {1, 0, 0}, {2, 0, 0}, {3, 0, 0}. - _Geoffrey Critzer_, May 07 2010
T(3,1)=11 since there are exactly 11 permutations of {1,2,3,4} with exactly 2 cycles, namely, (1)(234), (1)(243), (2)(134), (2)(143), (3)(124), (3)(142), (4)(123), (4)(143), (12)(34), (13)(24), and (14)(23). - _Dennis P. Walsh_, Jan 25 2011
From _Peter Bala_, Jul 21 2014: (Start)
With the arrays M(k) as defined in the Comments section, the infinite product M(0*)M(1)*M(2)*... begins
  / 1          \/1        \/1        \      / 1           \
  | 1  1       ||0 1      ||0 1      |      | 1  1        |
  | 2  2  1    ||0 1 1    ||0 0 1    |... = | 2  3  1     |
  | 6  6  3 1  ||0 2 2 1  ||0 0 1 1  |      | 6 11  6  1  |
  |24 24 12 4 1||0 6 6 3 1||0 0 2 2 1|      |24 50 35 10 1|
  |...         ||...      ||...      |      |...          |
(End)
		

References

  • John H. Conway and Richard K. Guy, The Book of Numbers, New York: Springer-Verlag, 1996. See pp. 93-94.
  • Sriram Pemmaraju and Steven Skiena, Computational Discrete Mathematics, Cambridge University Press, 2003, pp. 69-71. [Geoffrey Critzer, May 07 2010]

Crossrefs

See A008275, which is the main entry for these numbers; A094638 (reversed rows).
From Johannes W. Meijer, Oct 07 2009: (Start)
Row sums equal A000142.
The asymptotic expansions lead to A000142 (n=1), A000142(n=2; minus a(0)), A001710 (n=3), A001715 (n=4), A001720 (n=5), A001725 (n=6), A001730 (n=7), A049388 (n=8), A049389 (n=9), A049398 (n=10), A051431 (n=11), A008279 and A094587.
Cf. A163931 (E(x,m,n)), A028421 (m=2), A163932 (m=3), A163934 (m=4), A163936.
(End)
Cf. A136662.

Programs

  • Haskell
    a130534 n k = a130534_tabl !! n !! k
    a130534_row n = a130534_tabl !! n
    a130534_tabl = map (map abs) a008275_tabl
    -- Reinhard Zumkeller, Mar 18 2013
  • Maple
    with(combinat): A130534 := proc(n,m): (-1)^(n+m)*stirling1(n+1,m+1) end proc: seq(seq(A130534(n,m), m=0..n), n=0..10); # Johannes W. Meijer, Oct 07 2009, revised Sep 11 2012
    # The function BellMatrix is defined in A264428.
    # Adds (1,0,0,0, ..) as column 0 (and shifts the enumeration).
    BellMatrix(n -> n!, 9); # Peter Luschny, Jan 27 2016
  • Mathematica
    Table[Table[ Length[Select[Map[ToInversionVector, Permutations[m]], Count[ #, 0] == n &]], {n, 0, m - 1}], {m, 0, 8}] // Grid (* Geoffrey Critzer, May 07 2010 *)
    rows = 10;
    t = Range[0, rows]!;
    T[n_, k_] := BellY[n, k, t];
    Table[T[n, k], {n, 1, rows}, {k, 1, n}] // Flatten (* Jean-François Alcover, Jun 22 2018, after Peter Luschny *)

Formula

T(0,0) = 1, T(n,k) = 0 if k > n or if n < 0, T(n,k) = T(n-1,k-1) + n*T(n-1,k). T(n,0) = n! = A000142(n). T(2*n,n) = A129505(n+1). Sum_{k=0..n} T(n,k) = (n+1)! = A000142(n+1). Sum_{k=0..n} T(n,k)^2 = A047796(n+1). T(n,k) = |Stirling1(n+1,k+1)|, see A008275. (x+1)(x+2)...(x+n) = Sum_{k=0..n} T(n,k)*x^k. [Corrected by Arie Bos, Jul 11 2008]
Sum_{k=0..n} T(n,k)*x^k = A000007(n), A000142(n), A000142(n+1), A001710(n+2), A001715(n+3), A001720(n+4), A001725(n+5), A001730(n+6), A049388(n), A049389(n), A049398(n), A051431(n) for x = -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, respectively. - Philippe Deléham, Nov 13 2007
For k=1..n, let A={a_1,a_2,...,a_k} denote a size-k subset of {1,2,...,n}. Then T(n,n-k) = Sum(Product_{i=1..k} a_i) where the sum is over all subsets A. For example, T(4,1)=50 since 1*2*3 + 1*2*4 + 1*3*4 + 2*3*4 = 50. - Dennis P. Walsh, Jan 25 2011
The preceding formula means T(n,k) = sigma_{n-k}(1,2,3,..,n) with the (n-k)-th elementary symmetric function sigma with the indeterminates chosen as 1,2,...,n. See the Oct 24 2011 comment in A094638 with sigma called there a. - Wolfdieter Lang, Feb 06 2013
From Gary W. Adamson, Jul 08 2011: (Start)
n-th row of the triangle = top row of M^n, where M is the production matrix:
1, 1;
1, 2, 1;
1, 3, 3, 1;
1, 4, 6, 4, 1;
... (End)
Exponential Riordan array [1/(1 - x), log(1/(1 - x))]. Recurrence: T(n+1,k+1) = Sum_{i=0..n-k} (n + 1)!/(n + 1 - i)!*T(n-i,k). - Peter Bala, Jul 21 2014

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

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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).

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.
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. 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
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

A094638 Triangle read by rows: T(n,k) = |s(n,n+1-k)|, where s(n,k) are the signed Stirling numbers of the first kind A008276 (1 <= k <= n; in other words, the unsigned Stirling numbers of the first kind in reverse order).

Original entry on oeis.org

1, 1, 1, 1, 3, 2, 1, 6, 11, 6, 1, 10, 35, 50, 24, 1, 15, 85, 225, 274, 120, 1, 21, 175, 735, 1624, 1764, 720, 1, 28, 322, 1960, 6769, 13132, 13068, 5040, 1, 36, 546, 4536, 22449, 67284, 118124, 109584, 40320, 1, 45, 870, 9450, 63273, 269325, 723680, 1172700, 1026576, 362880
Offset: 1

Views

Author

André F. Labossière, May 17 2004

Keywords

Comments

Triangle of coefficients of the polynomial (x+1)(x+2)...(x+n), expanded in decreasing powers of x. - T. D. Noe, Feb 22 2008
Row n also gives the number of permutation of 1..n with complexity 0,1,...,n-1. See the comments in A008275. - N. J. A. Sloane, Feb 08 2019
T(n,k) is the number of deco polyominoes of height n and having k columns. A deco polyomino is a directed column-convex polyomino in which the height, measured along the diagonal, is attained only in the last column. Example: T(2,1)=1 and T(2,2)=1 because the deco polyominoes of height 2 are the vertical and horizontal dominoes, having, respectively, 1 and 2 columns. - Emeric Deutsch, Aug 14 2006
Sum_{k=1..n} k*T(n,k) = A121586. - Emeric Deutsch, Aug 14 2006
Let the triangle U(n,k), 0 <= k <= n, read by rows, be given by [1,0,1,0,1,0,1,0,1,0,1,...] DELTA [1,1,2,2,3,3,4,4,5,5,6,...] where DELTA is the operator defined in A084938; then T(n,k) = U(n-1,k-1). - Philippe Deléham, Jan 06 2007
From Tom Copeland, Dec 15 2007: (Start)
Consider c(t) = column vector(1, t, t^2, t^3, t^4, t^5, ...).
Starting at 1 and sampling every integer to the right, we obtain (1,2,3,4,5,...). And T * c(1) = (1, 1*2, 1*2*3, 1*2*3*4,...), giving n! for n > 0. Call this sequence the right factorial (n+)!.
Starting at 1 and sampling every integer to the left, we obtain (1,0,-1,-2,-3,-4,-5,...). And T * c(-1) = (1, 1*0, 1*0*-1, 1*0*-1*-2,...) = (1, 0, 0, 0, ...), the left factorial (n-)!.
Sampling every other integer to the right, we obtain (1,3,5,7,9,...). T * c(2) = (1, 1*3, 1*3*5, ...) = (1,3,15,105,945,...), giving A001147 for n > 0, the right double factorial, (n+)!!.
Sampling every other integer to the left, we obtain (1,-1,-3,-5,-7,...). T * c(-2) = (1, 1*-1, 1*-1*-3, 1*-1*-3*-5,...) = (1,-1,3,-15,105,-945,...) = signed A001147, the left double factorial, (n-)!!.
Sampling every 3 steps to the right, we obtain (1,4,7,10,...). T * c(3) = (1, 1*4, 1*4*7,...) = (1,4,28,280,...), giving A007559 for n > 0, the right triple factorial, (n+)!!!.
Sampling every 3 steps to the left, we obtain (1,-2,-5,-8,-11,...), giving T * c(-3) = (1, 1*-2, 1*-2*-5, 1*-2*-5*-8,...) = (1,-2,10,-80,880,...) = signed A008544, the left triple factorial, (n-)!!!.
The list partition transform A133314 of [1,T * c(t)] gives [1,T * c(-t)] with all odd terms negated; e.g., LPT[1,T*c(2)] = (1,-1,-1,-3,-15,-105,-945,...) = (1,-A001147). And e.g.f. for [1,T * c(t)] = (1-xt)^(-1/t).
The above results hold for t any real or complex number. (End)
Let R_n(x) be the real and I_n(x) the imaginary part of Product_{k=0..n} (x + I*k). Then, for n=1,2,..., we have R_n(x) = Sum_{k=0..floor((n+1)/2)}(-1)^k*Stirling1(n+1,n+1-2*k)*x^(n+1-2*k), I_n(x) = Sum_{k=0..floor(n/2)}(-1)^(k+1)*Stirling1(n+1,n-2*k)*x^(n-2*k). - Milan Janjic, May 11 2008
T(n,k) is also the number of permutations of n with "reflection length" k (i.e., obtained from 12..n by k not necessarily adjacent transpositions). For example, when n=3, 132, 213, 321 are obtained by one transposition, while 231 and 312 require two transpositions. - Kyle Petersen, Oct 15 2008
From Tom Copeland, Nov 02 2010: (Start)
[x^(y+1) D]^n = x^(n*y) [T(n,1)(xD)^n + T(n,2)y (xD)^(n-1) + ... + T(n,n)y^(n-1)(xD)], with D the derivative w.r.t. x.
E.g., [x^(y+1) D]^4 = x^(4*y) [(xD)^4 + 6 y(xD)^3 + 11 y^2(xD)^2 + 6 y^3(xD)].
(xD)^m can be further expanded in terms of the Stirling numbers of the second kind and operators of the form x^j D^j. (End)
With offset 0, 0 <= k <= n: T(n,k) is the sum of products of each size k subset of {1,2,...,n}. For example, T(3,2) = 11 because there are three subsets of size two: {1,2},{1,3},{2,3}. 1*2 + 1*3 + 2*3 = 11. - Geoffrey Critzer, Feb 04 2011
The Kn11, Fi1 and Fi2 triangle sums link this triangle with two sequences, see the crossrefs. For the definitions of these triangle sums see A180662. The mirror image of this triangle is A130534. - Johannes W. Meijer, Apr 20 2011
T(n+1,k+1) is the elementary symmetric function a_k(1,2,...,n), n >= 0, k >= 0, (a_0(0):=1). See the T. D. Noe and Geoffrey Critzer comments given above. For a proof see the Stanley reference, p. 19, Second Proof. - Wolfdieter Lang, Oct 24 2011
Let g(t) = 1/d(log(P(j+1,-t)))/dt (see Tom Copeland's 2007 formulas). The Mellin transform (t to s) of t*Dirac[g(t)] gives Sum_{n=1..j} n^(-s), which as j tends to infinity gives the Riemann zeta function for Re(s) > 1. Dirac(x) is the Dirac delta function. The complex contour integral along a circle of radius 1 centered at z=1 of z^s/g(z) gives the same result. - Tom Copeland, Dec 02 2011
Rows are coefficients of the polynomial expansions of the Pochhammer symbol, or rising factorial, Pch(n,x) = (x+n-1)!/(x-1)!. Expansion of Pch(n,xD) = Pch(n,Bell(.,:xD:)) in a polynomial with terms :xD:^k=x^k*D^k gives the Lah numbers A008297. Bell(n,x) are the unsigned Bell polynomials or Stirling polynomials of the second kind A008277. - Tom Copeland, Mar 01 2014
From Tom Copeland, Dec 09 2016: (Start)
The Betti numbers, or dimension, of the pure braid group cohomology. See pp. 12 and 13 of the Hyde and Lagarias link.
Row polynomials and their products appear in presentation of the Jack symmetric functions of R. Stanley. See Copeland link on the Witt differential generator.
(End)
From Tom Copeland, Dec 16 2019: (Start)
The e.g.f. given by Copeland in the formula section appears in a combinatorial Dyson-Schwinger equation of quantum field theory in Yeats in Thm. 2 on p. 62 related to a Hopf algebra of rooted trees. See also the Green function on p. 70.
Per comments above, this array contains the coefficients in the expansion in polynomials of the Euler, or state number, operator xD of the rising factorials Pch(n,xD) = (xD+n-1)!/(xD-1)! = x [:Dx:^n/n!]x^{-1} = L_n^{-1}(-:xD:), where :Dx:^n = D^n x^n and :xD:^n = x^n D^n. The polynomials L_n^{-1} are the Laguerre polynomials of order -1, i.e., normalized Lah polynomials.
The Witt differential operators L_n = x^(n+1) D and the row e.g.f.s appear in Hopf and dual Hopf algebra relations presented by Foissy. The Witt operators satisfy L_n L_k - L_k L_n = (k-n) L_(n+k), as for the dual Hopf algebra. (End)

Examples

			Triangle starts:
  1;
  1,  1;
  1,  3,  2;
  1,  6, 11,  6;
  1, 10, 35, 50, 24;
...
		

References

  • M. Miyata and J. W. Son, On the complexity of permutations and the metric space of bijections, Tensor, 60 (1998), No. 1, 109-116 (MR1768839).
  • Jerome Spanier and Keith B. Oldham, "Atlas of Functions", Hemisphere Publishing Corp., 1987, chapter 18, equations 18:4:2 - 18:4:8 at page 151.
  • R. P. Stanley, Enumerative Combinatorics, Vol. 1, Cambridge University Press, 1997.

Crossrefs

A008276 gives the (signed) Stirling numbers of the first kind.
Cf. A000108, A014137, A001246, A033536, A000984, A094639, A006134, A082894, A002897, A079727, A000217 (2nd column), A000914 (3rd column), A001303 (4th column), A000915 (5th column), A053567 (6th column), A000142 (row sums).
Triangle sums (see the comments): A124380 (Kn11), A001710 (Fi1, Fi2). - Johannes W. Meijer, Apr 20 2011

Programs

  • GAP
    Flat(List([1..10], n-> List([1..n], k-> Stirling1(n,n-k+1) ))); # G. C. Greubel, Dec 29 2019
  • Haskell
    a094638 n k = a094638_tabl !! (n-1) !! (k-1)
    a094638_row n = a094638_tabl !! (n-1)
    a094638_tabl = map reverse a130534_tabl
    -- Reinhard Zumkeller, Aug 01 2014
    
  • Magma
    [(-1)^(k+1)*StirlingFirst(n,n-k+1): k in [1..n], n in [1..10]]; // G. C. Greubel, Dec 29 2019
    
  • Maple
    T:=(n,k)->abs(Stirling1(n,n+1-k)): for n from 1 to 10 do seq(T(n,k),k=1..n) od; # yields sequence in triangular form. # Emeric Deutsch, Aug 14 2006
  • Mathematica
    Table[CoefficientList[Series[Product[1 + i x, {i,n}], {x,0,20}], x], {n,0,6}] (* Geoffrey Critzer, Feb 04 2011 *)
    Table[Abs@StirlingS1[n, n-k+1], {n, 10}, {k, n}]//Flatten (* Michael De Vlieger, Aug 29 2015 *)
  • Maxima
    create_list(abs(stirling1(n+1,n-k+1)),n,0,10,k,0,n); /* Emanuele Munarini, Jun 01 2012 */
    
  • PARI
    {T(n,k)=if(n<1 || k>n,0,(n-1)!*polcoeff(polcoeff(x*y/(1 - x*y+x*O(x^n))^(1 + 1/y),n,x),k,y))} /* Paul D. Hanna, Jul 21 2011 */
    
  • Sage
    [[stirling_number1(n, n-k+1) for k in (1..n)] for n in (1..10)] # G. C. Greubel, Dec 29 2019
    

Formula

With P(n,t) = Sum_{k=0..n-1} T(n,k+1) * t^k = 1*(1+t)*(1+2t)...(1+(n-1)*t) and P(0,t)=1, exp[P(.,t)*x] = (1-tx)^(-1/t). T(n,k+1) = (1/k!) (D_t)^k (D_x)^n [ (1-tx)^(-1/t) - 1 ] evaluated at t=x=0. (1-tx)^(-1/t) - 1 is the e.g.f. for a plane m-ary tree when t=m-1. See Bergeron et al. in "Varieties of Increasing Trees". - Tom Copeland, Dec 09 2007
First comment and formula above rephrased as o.g.f. for row n: Product_{i=0...n} (1+i*x). - Geoffrey Critzer, Feb 04 2011
n-th row polynomials with alternate signs are the characteristic polynomials of the (n-1)x(n-1) matrices with 1's in the superdiagonal, (1,2,3,...) in the main diagonal, and the rest zeros. For example, the characteristic polynomial of [1,1,0; 0,2,1; 0,0,3] is x^3 - 6*x^2 + 11*x - 6. - Gary W. Adamson, Jun 28 2011
E.g.f.: A(x,y) = x*y/(1 - x*y)^(1 + 1/y) = Sum_{n>=1, k=1..n} T(n,k)*x^n*y^k/(n-1)!. - Paul D. Hanna, Jul 21 2011
With F(x,t) = (1-t*x)^(-1/t) - 1 an e.g.f. for the row polynomials P(n,t) of A094638 with P(0,t)=0, G(x,t)= [1-(1+x)^(-t)]/t is the comp. inverse in x. Consequently, with H(x,t) = 1/(dG(x,t)/dx) = (1+x)^(t+1),
P(n,t) = [(H(x,t)*d/dx)^n] x evaluated at x=0; i.e.,
F(x,t) = exp[x*P(.,t)] = exp[x*H(u,t)*d/du] u, evaluated at u = 0.
Also, dF(x,t)/dx = H(F(x,t),t). - Tom Copeland, Sep 20 2011
T(n,k) = |A008276(n,k)|. - R. J. Mathar, May 19 2016
The row polynomials of this entry are the reversed row polynomials of A143491 multiplied by (1+x). E.g., (1+x)(1 + 5x + 6x^2) = (1 + 6x + 11x^2 + 6x^3). - Tom Copeland, Dec 11 2016
Regarding the row e.g.f.s in Copeland's 2007 formulas, e.g.f.s for A001710, A001715, and A001720 give the compositional inverses of the e.g.f. here for t = 2, 3, and 4 respectively. - Tom Copeland, Dec 28 2019

Extensions

Edited by Emeric Deutsch, Aug 14 2006

A001715 a(n) = n!/6.

Original entry on oeis.org

1, 4, 20, 120, 840, 6720, 60480, 604800, 6652800, 79833600, 1037836800, 14529715200, 217945728000, 3487131648000, 59281238016000, 1067062284288000, 20274183401472000, 405483668029440000, 8515157028618240000, 187333454629601280000, 4308669456480829440000
Offset: 3

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Author

Keywords

Comments

The numbers (4, 20, 120, 840, 6720, ...) arise from the divisor values in the general formula a(n) = n*(n+1)*(n+2)*(n+3)* ... *(n+k)*(n*(n+k) + (k-1)*k/6)/((k+3)!/6) (which covers the following sequences: A000578, A000537, A024166, A101094, A101097, A101102). - Alexander R. Povolotsky, May 17 2008
a(n) is also the number of decreasing 3-cycles in the decomposition of permutations as product of disjoint cycles, a(3)=1, a(4)=4, a(5)=20. - Wenjin Woan, Dec 21 2008
Equals eigensequence of triangle A130128 reflected. - Gary W. Adamson, Dec 23 2008
a(n) is the number of n-permutations having 1, 2, and 3 in three distinct cycles. - Geoffrey Critzer, Apr 26 2009
From Johannes W. Meijer, Oct 20 2009: (Start)
The asymptotic expansion of the higher order exponential integral E(x,m=1,n=4) ~ exp(-x)/x*(1 - 4/x + 20/x^2 - 120/x^3 + 840/x^4 - 6720/x^5 + 60480/x^6 - 604800/x^7 + ...) leads to the sequence given above. See A163931 and A130534 for more information.
(End)

References

  • 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

Programs

Formula

a(n) = A049352(n-2, 1) (first column of triangle).
E.g.f. if offset 0: 1/(1-x)^4.
a(n) = A173333(n,3). - Reinhard Zumkeller, Feb 19 2010
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x/(x + 1/(k+4)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
G.f.: W(0), where W(k) = 1 - x*(k+4)/( x*(k+4) - 1/(1 - x*(k+1)/( x*(k+1) - 1/W(k+1) ))); (continued fraction). - Sergei N. Gladkovskii, Aug 26 2013
a(n) = A245334(n,n-3) / 4. - Reinhard Zumkeller, Aug 31 2014
From Peter Bala, May 22 2017: (Start)
The o.g.f. A(x) satisfies the Riccati equation x^2*A'(x) + (4*x - 1)*A(x) + 1 = 0.
G.f. as an S-fraction: A(x) = 1/(1 - 4*x/(1 - x/(1 - 5*x/(1 - 2*x/(1 - 6*x/(1 - 3*x/(1 - ... - (n + 3)*x/(1 - n*x/(1 - ... ))))))))) (apply Stokes, 1982).
A(x) = 1/(1 - 3*x - x/(1 - 4*x/(1 - 2*x/(1 - 5*x/(1 - 3*x/(1 - 6*x/(1 - ... - n*x/(1 - (n+3)*x/(1 - ... ))))))))). (End)
H(x) = (1 - (1 + x)^(-3)) / 3 = x - 4 x^2/2! + 20 x^3/3! - ... is an e.g.f. of the signed sequence (n!/4!), which is the compositional inverse of G(x) = (1 - 3*x)^(-1/3) - 1, an e.g.f. for A007559. Cf. A094638, A001710 (for n!/2!), and A001720 (for n!/4!). Cf. columns of A094587, A173333, and A213936 and rows of A138533.- Tom Copeland, Dec 27 2019
E.g.f.: x^3 / (3! * (1 - x)). - Ilya Gutkovskiy, Jul 09 2021
From Amiram Eldar, Jan 15 2023: (Start)
Sum_{n>=3} 1/a(n) = 6*e - 15.
Sum_{n>=3} (-1)^(n+1)/a(n) = 3 - 6/e. (End)

Extensions

More terms from Harvey P. Dale, Aug 12 2012
Showing 1-10 of 218 results. Next