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 62 results. Next

A112008 Fourth diagonal of second-order Eulerian triangle A008517. Fourth column (m=3) of triangle A112007.

Original entry on oeis.org

1, 52, 1452, 32120, 644020, 12440064, 238904904, 4642163952, 92199790224, 1883079661824, 39689578055808, 865023253219584, 19515249341231616, 455924361142656000, 11030149104146035200, 276260563641659673600
Offset: 0

Views

Author

Wolfdieter Lang, Sep 12 2005

Keywords

Examples

			1452= a(2) = 9*52 + 3*328.
		

Crossrefs

Cf. A002539 (3rd diagonal of A008517; third column of A112007).
Contribution from Johannes W. Meijer, Oct 16 2009: (Start)
Equals fourth left hand column of A163936.
(End)

Formula

a(n)=A112007(n+3, 3), n>=0.
a(n)= (n+7)*a(n-1) + (n+1)*A002539(n+1), n>=1, a(0)=1.
Contribution from Johannes W. Meijer, Oct 16 2009: (Start)
a(n) = sum((-1)^(n+k)*binomial(2*n+9,k)*stirling1(n-k+8,4-k), k=0..3)
(End)

A112485 Fifth diagonal of second-order Eulerian triangle A008517. Fifth column (m=4) of triangle A112007.

Original entry on oeis.org

1, 114, 5610, 195800, 5765500, 155357384, 4002695088, 101180433024, 2549865473424, 64728375139872, 1666424486271456, 43708768764064128, 1171582385481357696, 32157753536587053312, 905080567903692754176
Offset: 0

Views

Author

Wolfdieter Lang, Sep 12 2005

Keywords

Examples

			5610= a(2) = 11*114 + 3*1452.
		

Crossrefs

Cf. A112008 (fourth diagonal of A008517 and fourth column of A112007).
Contribution from Johannes W. Meijer, Oct 16 2009: (Start)
Equals fifth left hand column of A163936.
(End)

Formula

a(n)=A112007(n+4, 4), n>=0.
a(n)= (n+9)*a(n-1) + (n+1)*A112008(n), n>=1, a(0)=1.
Contribution from Johannes W. Meijer, Oct 16 2009: (Start)
a(n) = sum((-1)^(n+k+1)*binomial(2*n+11,k)*stirling1(n-k+10,5-k),k=0..4)
(End)

A112692 Coefficient array of numerator polynomials of o.g.f.s (rising powers) for the columns of triangle A008517 (second-order Eulerian numbers).

Original entry on oeis.org

1, 3, -1, -6, 6, -9, -70, 163, -42, -72, 30, -123, -1110, 8440, -18244, 2423, 43036, -53172, 11232, 8640, 90, -792, -7425, 137760, -771911, 1624514, 2262109, -21114844, 51074797, -54783526, 6214788, 45596664, -40513824, 7309440, 3110400, 630, -10278, -86841, 3685605, -41159454
Offset: 0

Views

Author

Wolfdieter Lang, Oct 14 2005

Keywords

Comments

The sequence of row lengths is A000217 (triangular numbers): [1, 3, 6, 10, 15, 21,..].
The o.g.f. of the k-th column sequence of triangle A008517(n,k), n>=k>=1, is (2^floor(k/2))*(x^k)*p(k,x)/product((1-j*x)^(k+1-j),j=1..k), k>=2, with the row polynomials p(k,x):= sum(a(k-2,m)*x^m,m=0..(k*(k-1)/2)-1).

Examples

			Rows: [1]; [3,-1,-6]; [6,-9,-70,163,-42,-72];...
The k=3, offset 3, column sequence [6,58,328,..] of A008517 has o.g.f. 2*(x^3)*(3-x-6*x^2)/product((1-j*x)^(4-j),j=1..3).
		

Crossrefs

Row sums A112693. Unsigned row sums A112694.

A156141 Triangle T(n,k) = A008517(n,k+1)+A008517(n,n+1-k) read by rows.

Original entry on oeis.org

2, 1, 1, 1, 4, 1, 1, 14, 14, 1, 1, 46, 116, 46, 1, 1, 172, 772, 772, 172, 1, 1, 834, 5160, 8800, 5160, 834, 1, 1, 5280, 39594, 90260, 90260, 39594, 5280, 1, 1, 40814, 361086, 981104, 1288040, 981104, 361086, 40814, 1, 1, 363884, 3801180, 12088796
Offset: 0

Views

Author

Roger L. Bagula, Feb 04 2009

Keywords

Comments

Row sums are 2*A001147(n).

Examples

			{2},
{1, 1},
{1, 4, 1},
{1, 14, 14, 1},
{1, 46, 116, 46, 1},
{1, 172, 772, 772, 172, 1},
{1, 834, 5160, 8800, 5160, 834, 1},
{1, 5280, 39594, 90260, 90260, 39594, 5280, 1},
{1, 40814, 361086, 981104, 1288040, 981104, 361086, 40814, 1},
{1, 363884, 3801180, 12088796, 18205564, 18205564, 12088796, 3801180, 363884, 1},
{1, 3630826, 44557888, 167513072, 283669904, 310714768, 283669904, 167513072, 44557888, 3630826, 1}
		

Crossrefs

Programs

  • Mathematica
    e[n_, 0] := 1;
    e[n_, k_] := 0 /; k >= n;
    e[n_, k_] := (k + 1)e[n - 1, k] + (2n - k - 1)e[n - 1, k - 1];
    Table[Table[e[n, k] + e[n, n - k], {k, 0, n}], {n, 0, 10}];
    Flatten[%]

Formula

T(n,0)=T(n,n) =1, n>=1.
T(n,k)= A008517(n,k+1)+A008517(n,n+1-k), n>1, 0
T(n,k)= T(n,n-k) .

Extensions

Definition simplified by the Associate Editors of the OEIS Sep 17 2009

A219120 Triangle, read by rows, where T(n,k) is defined for n>=1, k=1..2*n-1, by a formula analogous to the second-order Eulerian triangle A008517.

Original entry on oeis.org

1, 1, 1, -1, 1, 5, -2, -2, 1, 1, 15, 13, -19, 3, 3, -1, 1, 37, 128, -26, -74, 46, -4, -4, 1, 1, 83, 679, 755, -654, -68, 230, -90, 5, 5, -1, 1, 177, 2866, 9048, 2091, -5741, 1856, 498, -545, 155, -6, -6, 1, 1, 367, 10721, 67541, 98069, -24675, -35027, 22717, -3773, -1673, 1099, -245, 7, 7, -1, 1
Offset: 1

Author

Paul D. Hanna, Nov 12 2012

Keywords

Comments

Compare to the o.g.f. of row n, E2(x,n), in the second-order Eulerian triangle A008517:
E2(x,n) = (1-x)^(2*n+1) * Sum_{k>=0} k^n * k^k * exp(-k*x) * x^k/k!.

Examples

			Triangle begins:
1;
1, 1, -1;
1, 5, -2, -2, 1;
1, 15, 13, -19, 3, 3, -1;
1, 37, 128, -26, -74, 46, -4, -4, 1;
1, 83, 679, 755, -654, -68, 230, -90, 5, 5, -1;
1, 177, 2866, 9048, 2091, -5741, 1856, 498, -545, 155, -6, -6, 1;
1, 367, 10721, 67541, 98069, -24675, -35027, 22717, -3773, -1673, 1099, -245, 7, 7, -1;
1, 749, 37300, 409170, 1290116, 863168, -590008, -108832, 182806, -65858, 5824, 4228, -1988, 364, -8, -8, 1; ...
		

Crossrefs

Programs

  • PARI
    {T(n,k)=polcoeff((1-x)^(2*n-1)*sum(j=0,2*n,(j^n)*(j+1)^(j-1)*x^j/j!*exp(-(j+1)*x +O(x^k))),k)}
    for(n=1,10,for(k=1,2*n-1,print1(T(n,k),", "));print(""))

Formula

O.g.f. of row n, R(x,n), is given by:
R(x,n) = (1-x)^(2*n-1) * Sum_{k>=0} k^n *(k+1)^(k-1) * exp(-(k+1)*x) * x^k/k!.
Row sums = A001147, which is the odd double factorials.
Column 1 = A050488(n-1), where A050488(n) = 3*(2^n-1) - 2*n.
Central terms of rows = A219121.

A112498 Third column of second-order Eulerian triangle A008517 divided by 2.

Original entry on oeis.org

3, 29, 164, 726, 2805, 9975, 33630, 109424, 347519, 1085313, 3349848, 10253994, 31203945, 94561643, 285716018, 861472836, 2593592883, 7800176565, 23441423340, 70410252350, 211411111133, 634610819679, 1904620987014
Offset: 3

Author

Wolfdieter Lang, Oct 14 2005

Keywords

Comments

See A004301 for the doubled sequence.

Crossrefs

Cf. A000295 (one half of second column).

Formula

a(n)=A008517(n, 3)/2.
G.f.: x^3*(3-x-6*x^2)/(((1-x)^3)*((1-2*x)^2)*(1-3*x)). See the comment on column g.f.s under A008517.
a(n) = 3*a(n-1) + (2*n-3)*(2^(n-1)-n), n>3, with a(3)=3.
a(n)= 3*A112502(n-3) - A112502(n-4) - 6*A112502(n-5), n>=5.

A156529 Triangle, T(n, k) = A008517(n+1, k+1)*A008517(n+1, n-k+1), read by rows.

Original entry on oeis.org

1, 2, 2, 6, 64, 6, 24, 1276, 1276, 24, 120, 23088, 107584, 23088, 120, 720, 422712, 6388800, 6388800, 422712, 720, 5040, 8156160, 326165400, 1031694400, 326165400, 8156160, 5040, 40320, 168521184, 15666814800, 126099116000, 126099116000, 15666814800, 168521184, 40320
Offset: 0

Author

Roger L. Bagula, Feb 09 2009

Keywords

Examples

			Triangle begins as:
     1;
     2,       2;
     6,      64,         6;
    24,    1276,      1276,         24;
   120,   23088,    107584,      23088,       120;
   720,  422712,   6388800,    6388800,    422712,     720;
  5040, 8156160, 326165400, 1031694400, 326165400, 8156160, 5040;
		

Crossrefs

Cf. A008517.

Programs

  • Magma
    A008517:= func< n,k | (&+[ (-1)^(n+j)*Binomial(2*n+1, j)*StiringFirst(2*n-k-j+1, n-k-j+1) : j in [0..n-k]]) >;
    A156529:= func< n,k | A008517(n+1,k+1)*A008517(n+1,n-k+1) >;
    [A156529(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Dec 30 2021
    
  • Mathematica
    f[n_, k_]:= f[n, k]= If[k<0 || k>n, 0, If[k==0, 1, (k+1)*f[n-1, k] + (2*n-k+1)*f[n-1, k-1] ]]; (* f = A008517 *)
    T[n_, k_]:= f[n+1, k+1]*f[n+1, n-k+1];
    Table[T[n,k], {n,0,12}, {k,0,n}]//Flatten (* modified by G. C. Greubel, Dec 30 2021 *)
  • Sage
    @CachedFunction
    def A008517(n,k): return sum( (-1)^(n+j)*binomial(2*n+1, j)*stirling_number1(2*n-k-j+1, n-k-j+1)  for j in (0..n-k) )
    def A156529(n,k): return A008517(n+1, k+1)*A008517(n+1, n-k+1)
    flatten([[A156529(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Dec 30 2021

Formula

T(n, k) = A008517(n+1, k+1)*A008517(n+1, n-k+1).
From G. C. Greubel, Dec 30 2021: (Start)
T(n, n-k) = T(n, k).
T(n, 0) = n!. (End)

Extensions

Edited by G. C. Greubel, Dec 30 2021

A112499 Fourth column of second-order Eulerian triangle A008517 divided by 4.

Original entry on oeis.org

6, 111, 1100, 8030, 48950, 265625, 1331540, 6310976, 28719094, 126814819, 547457452, 2323131730, 9729382150, 40335953245, 165915269268, 678306115284, 2759909133030, 11187839886855, 45220188014220, 182359367356230, 734088513869846, 2950950104332001, 11849511321016340, 47540932962238760
Offset: 4

Author

Wolfdieter Lang, Oct 14 2005

Keywords

Crossrefs

Formula

G.f.: x^4*(6-9*x-70*x^2+163*x^3-42*x^4-72*x^5)/(Product_{j=1..4} (1-j*x)^(5-j)). See the comment on column g.f.s under A008517.
a(n) = 4*a(n-1) + (n-2)*A112498(n-1), n>4, with a(4)=6.
a(n+4) = 6*b(n)-9*b(n-1)-70*b(n-2)+163*b(n-3)-42*b(n-4)-72*b(n-5), n>=0, with b(n):=A112503(n).

Extensions

More terms from Michel Marcus, Jul 07 2025

A001147 Double factorial of odd numbers: a(n) = (2*n-1)!! = 1*3*5*...*(2*n-1).

Original entry on oeis.org

1, 1, 3, 15, 105, 945, 10395, 135135, 2027025, 34459425, 654729075, 13749310575, 316234143225, 7905853580625, 213458046676875, 6190283353629375, 191898783962510625, 6332659870762850625, 221643095476699771875, 8200794532637891559375, 319830986772877770815625
Offset: 0

Keywords

Comments

The solution to Schröder's third problem.
Number of fixed-point-free involutions in symmetric group S_{2n} (cf. A000085).
a(n-2) is the number of full Steiner topologies on n points with n-2 Steiner points. [corrected by Lyle Ramshaw, Jul 20 2022]
a(n) is also the number of perfect matchings in the complete graph K(2n). - Ola Veshta (olaveshta(AT)my-deja.com), Mar 25 2001
Number of ways to choose n disjoint pairs of items from 2*n items. - Ron Zeno (rzeno(AT)hotmail.com), Feb 06 2002
Number of ways to choose n-1 disjoint pairs of items from 2*n-1 items (one item remains unpaired). - Bartosz Zoltak, Oct 16 2012
For n >= 1 a(n) is the number of permutations in the symmetric group S_(2n) whose cycle decomposition is a product of n disjoint transpositions. - Ahmed Fares (ahmedfares(AT)my-deja.com), Apr 21 2001
a(n) is the number of distinct products of n+1 variables with commutative, nonassociative multiplication. - Andrew Walters (awalters3(AT)yahoo.com), Jan 17 2004. For example, a(3)=15 because the product of the four variables w, x, y and z can be constructed in exactly 15 ways, assuming commutativity but not associativity: 1. w(x(yz)) 2. w(y(xz)) 3. w(z(xy)) 4. x(w(yz)) 5. x(y(wz)) 6. x(z(wy)) 7. y(w(xz)) 8. y(x(wz)) 9. y(z(wx)) 10. z(w(xy)) 11. z(x(wy)) 12. z(y(wx)) 13. (wx)(yz) 14. (wy)(xz) 15. (wz)(xy).
a(n) = E(X^(2n)), where X is a standard normal random variable (i.e., X is normal with mean = 0, variance = 1). So for instance a(3) = E(X^6) = 15, etc. See Abramowitz and Stegun or Hoel, Port and Stone. - Jerome Coleman, Apr 06 2004
Second Eulerian transform of 1,1,1,1,1,1,... The second Eulerian transform transforms a sequence s to a sequence t by the formula t(n) = Sum_{k=0..n} E(n,k)s(k), where E(n,k) is a second-order Eulerian number (A008517). - Ross La Haye, Feb 13 2005
Integral representation as n-th moment of a positive function on the positive axis: a(n) = Integral_{x=0..oo} x^n*exp(-x/2)/sqrt(2*Pi*x) dx, n >= 0. - Karol A. Penson, Oct 10 2005
a(n) is the number of binary total partitions of n+1 (each non-singleton block must be partitioned into exactly two blocks) or, equivalently, the number of unordered full binary trees with n+1 labeled leaves (Stanley, ex 5.2.6). - Mitch Harris, Aug 01 2006
a(n) is the Pfaffian of the skew-symmetric 2n X 2n matrix whose (i,j) entry is i for iDavid Callan, Sep 25 2006
a(n) is the number of increasing ordered rooted trees on n+1 vertices where "increasing" means the vertices are labeled 0,1,2,...,n so that each path from the root has increasing labels. Increasing unordered rooted trees are counted by the factorial numbers A000142. - David Callan, Oct 26 2006
Number of perfect multi Skolem-type sequences of order n. - Emeric Deutsch, Nov 24 2006
a(n) = total weight of all Dyck n-paths (A000108) when each path is weighted with the product of the heights of the terminal points of its upsteps. For example with n=3, the 5 Dyck 3-paths UUUDDD, UUDUDD, UUDDUD, UDUUDD, UDUDUD have weights 1*2*3=6, 1*2*2=4, 1*2*1=2, 1*1*2=2, 1*1*1=1 respectively and 6+4+2+2+1=15. Counting weights by height of last upstep yields A102625. - David Callan, Dec 29 2006
a(n) is the number of increasing ternary trees on n vertices. Increasing binary trees are counted by ordinary factorials (A000142) and increasing quaternary trees by triple factorials (A007559). - David Callan, Mar 30 2007
From Tom Copeland, Nov 13 2007, clarified in first and extended in second paragraph, Jun 12 2021: (Start)
a(n) has the e.g.f. (1-2x)^(-1/2) = 1 + x + 3*x^2/2! + ..., whose reciprocal is (1-2x)^(1/2) = 1 - x - x^2/2! - 3*x^3/3! - ... = b(0) - b(1)*x - b(2)*x^2/2! - ... with b(0) = 1 and b(n+1) = -a(n) otherwise. By the formalism of A133314, Sum_{k=0..n} binomial(n,k)*b(k)*a(n-k) = 0^n where 0^0 := 1. In this sense, the sequence a(n) is essentially self-inverse. See A132382 for an extension of this result. See A094638 for interpretations.
This sequence aerated has the e.g.f. e^(t^2/2) = 1 + t^2/2! + 3*t^4/4! + ... = c(0) + c(1)*t + c(2)*t^2/2! + ... and the reciprocal e^(-t^2/2); therefore, Sum_{k=0..n} cos(Pi k/2)*binomial(n,k)*c(k)*c(n-k) = 0^n; i.e., the aerated sequence is essentially self-inverse. Consequently, Sum_{k=0..n} (-1)^k*binomial(2n,2k)*a(k)*a(n-k) = 0^n. (End)
From Ross Drewe, Mar 16 2008: (Start)
This is also the number of ways of arranging the elements of n distinct pairs, assuming the order of elements is significant but the pairs are not distinguishable, i.e., arrangements which are the same after permutations of the labels are equivalent.
If this sequence and A000680 are denoted by a(n) and b(n) respectively, then a(n) = b(n)/n! where n! = the number of ways of permuting the pair labels.
For example, there are 90 ways of arranging the elements of 3 pairs [1 1], [2 2], [3 3] when the pairs are distinguishable: A = { [112233], [112323], ..., [332211] }.
By applying the 6 relabeling permutations to A, we can partition A into 90/6 = 15 subsets: B = { {[112233], [113322], [221133], [223311], [331122], [332211]}, {[112323], [113232], [221313], [223131], [331212], [332121]}, ....}
Each subset or equivalence class in B represents a unique pattern of pair relationships. For example, subset B1 above represents {3 disjoint pairs} and subset B2 represents {1 disjoint pair + 2 interleaved pairs}, with the order being significant (contrast A132101). (End)
A139541(n) = a(n) * a(2*n). - Reinhard Zumkeller, Apr 25 2008
a(n+1) = Sum_{j=0..n} A074060(n,j) * 2^j. - Tom Copeland, Sep 01 2008
From Emeric Deutsch, Jun 05 2009: (Start)
a(n) is the number of adjacent transpositions in all fixed-point-free involutions of {1,2,...,2n}. Example: a(2)=3 because in 2143=(12)(34), 3412=(13)(24), and 4321=(14)(23) we have 2 + 0 + 1 adjacent transpositions.
a(n) = Sum_{k>=0} k*A079267(n,k).
(End)
Hankel transform is A137592. - Paul Barry, Sep 18 2009
(1, 3, 15, 105, ...) = INVERT transform of A000698 starting (1, 2, 10, 74, ...). - Gary W. Adamson, Oct 21 2009
a(n) = (-1)^(n+1)*H(2*n,0), where H(n,x) is the probabilists' Hermite polynomial. The generating function for the probabilists' Hermite polynomials is as follows: exp(x*t-t^2/2) = Sum_{i>=0} H(i,x)*t^i/i!. - Leonid Bedratyuk, Oct 31 2009
The Hankel transform of a(n+1) is A168467. - Paul Barry, Dec 04 2009
Partial products of odd numbers. - Juri-Stepan Gerasimov, Oct 17 2010
See A094638 for connections to differential operators. - Tom Copeland, Sep 20 2011
a(n) is the number of subsets of {1,...,n^2} that contain exactly k elements from {1,...,k^2} for k=1,...,n. For example, a(3)=15 since there are 15 subsets of {1,2,...,9} that satisfy the conditions, namely, {1,2,5}, {1,2,6}, {1,2,7}, {1,2,8}, {1,2,9}, {1,3,5}, {1,3,6}, {1,3,7}, {1,3,8}, {1,3,9}, {1,4,5}, {1,4,6}, {1,4,7}, {1,4,8}, and {1,4,9}. - Dennis P. Walsh, Dec 02 2011
a(n) is the leading coefficient of the Bessel polynomial y_n(x) (cf. A001498). - Leonid Bedratyuk, Jun 01 2012
For n>0: a(n) is also the determinant of the symmetric n X n matrix M defined by M(i,j) = min(i,j)^2 for 1 <= i,j <= n. - Enrique Pérez Herrero, Jan 14 2013
a(n) is also the numerator of the mean value from 0 to Pi/2 of sin(x)^(2n). - Jean-François Alcover, Jun 13 2013
a(n) is the size of the Brauer monoid on 2n points (see A227545). - James Mitchell, Jul 28 2013
For n>1: a(n) is the numerator of M(n)/M(1) where the numbers M(i) have the property that M(n+1)/M(n) ~ n-1/2 (for example, large Kendell-Mann numbers, see A000140 or A181609, as n --> infinity). - Mikhail Gaichenkov, Jan 14 2014
a(n) = the number of upper-triangular matrix representations required for the symbolic representation of a first order central moment of the multivariate normal distribution of dimension 2(n-1), i.e., E[X_1*X_2...*X_(2n-2)|mu=0, Sigma]. See vignette for symmoments R package on CRAN and Phillips reference below. - Kem Phillips, Aug 10 2014
For n>1: a(n) is the number of Feynman diagrams of order 2n (number of internal vertices) for the vacuum polarization with one charged loop only, in quantum electrodynamics. - Robert Coquereaux, Sep 15 2014
Aerated with intervening zeros (1,0,1,0,3,...) = a(n) (cf. A123023), the e.g.f. is e^(t^2/2), so this is the base for the Appell sequence A099174 with e.g.f. e^(t^2/2) e^(x*t) = exp(P(.,x),t) = unsigned A066325(x,t), the probabilist's (or normalized) Hermite polynomials. P(n,x) = (a. + x)^n with (a.)^n = a_n and comprise the umbral compositional inverses for A066325(x,t) = exp(UP(.,x),t), i.e., UP(n,P(.,t)) = x^n = P(n,UP(.,t)), where UP(n,t) are the polynomials of A066325 and, e.g., (P(.,t))^n = P(n,t). - Tom Copeland, Nov 15 2014
a(n) = the number of relaxed compacted binary trees of right height at most one of size n. A relaxed compacted binary tree of size n is a directed acyclic graph consisting of a binary tree with n internal nodes, one leaf, and n pointers. It is constructed from a binary tree of size n, where the first leaf in a post-order traversal is kept and all other leaves are replaced by pointers. These links may point to any node that has already been visited by the post-order traversal. The right height is the maximal number of right-edges (or right children) on all paths from the root to any leaf after deleting all pointers. The number of unbounded relaxed compacted binary trees of size n is A082161(n). See the Genitrini et al. link. - Michael Wallner, Jun 20 2017
Also the number of distinct adjacency matrices in the n-ladder rung graph. - Eric W. Weisstein, Jul 22 2017
From Christopher J. Smyth, Jan 26 2018: (Start)
a(n) = the number of essentially different ways of writing a probability distribution taking n+1 values as a sum of products of binary probability distributions. See comment of Mitch Harris above. This is because each such way corresponds to a full binary tree with n+1 leaves, with the leaves labeled by the values. (This comment is due to Niko Brummer.)
Also the number of binary trees with root labeled by an (n+1)-set S, its n+1 leaves by the singleton subsets of S, and other nodes labeled by subsets T of S so that the two daughter nodes of the node labeled by T are labeled by the two parts of a 2-partition of T. This also follows from Mitch Harris' comment above, since the leaf labels determine the labels of the other vertices of the tree.
(End)
a(n) is the n-th moment of the chi-squared distribution with one degree of freedom (equivalent to Coleman's Apr 06 2004 comment). - Bryan R. Gillespie, Mar 07 2021
Let b(n) = 0 for n odd and b(2k) = a(k); i.e., let the sequence b(n) be an aerated version of this entry. After expanding the differential operator (x + D)^n and normal ordering the resulting terms, the integer coefficient of the term x^k D^m is n! b(n-k-m) / [(n-k-m)! k! m!] with 0 <= k,m <= n and (k+m) <= n. E.g., (x+D)^2 = x^2 + 2xD + D^2 + 1 with D = d/dx. The result generalizes to the raising (R) and lowering (L) operators of any Sheffer polynomial sequence by replacing x by R and D by L and follows from the disentangling relation e^{t(L+R)} = e^{t^2/2} e^{tR} e^{tL}. Consequently, these are also the coefficients of the reordered 2^n permutations of the binary symbols L and R under the condition LR = RL + 1. E.g., (L+R)^2 = LL + LR + RL + RR = LL + 2RL + RR + 1. (Cf. A344678.) - Tom Copeland, May 25 2021
From Tom Copeland, Jun 14 2021: (Start)
Lando and Zvonkin present several scenarios in which the double factorials occur in their role of enumerating perfect matchings (pairings) and as the nonzero moments of the Gaussian e^(x^2/2).
Speyer and Sturmfels (p. 6) state that the number of facets of the abstract simplicial complex known as the tropical Grassmannian G'''(2,n), the space of phylogenetic T_n trees (see A134991), or Whitehouse complex is a shifted double factorial.
These are also the unsigned coefficients of the x[2]^m terms in the partition polynomials of A134685 for compositional inversion of e.g.f.s, a refinement of A134991.
a(n)*2^n = A001813(n) and A001813(n)/(n+1)! = A000108(n), the Catalan numbers, the unsigned coefficients of the x[2]^m terms in the partition polynomials A133437 for compositional inversion of o.g.f.s, a refinement of A033282, A126216, and A086810. Then the double factorials inherit a multitude of analytic and combinatoric interpretations from those of the Catalan numbers, associahedra, and the noncrossing partitions of A134264 with the Catalan numbers as unsigned-row sums. (End)
Connections among the Catalan numbers A000108, the odd double factorials, values of the Riemann zeta function and its derivative for integer arguments, and series expansions of the reduced action for the simple harmonic oscillator and the arc length of the spiral of Archimedes are given in the MathOverflow post on the Riemann zeta function. - Tom Copeland, Oct 02 2021
b(n) = a(n) / (n! 2^n) = Sum_{k = 0..n} (-1)^n binomial(n,k) (-1)^k a(k) / (k! 2^k) = (1-b.)^n, umbrally; i.e., the normalized double factorial a(n) is self-inverse under the binomial transform. This can be proved by applying the Euler binomial transformation for o.g.f.s Sum_{n >= 0} (1-b.)^n x^n = (1/(1-x)) Sum_{n >= 0} b_n (x / (x-1))^n to the o.g.f. (1-x)^{-1/2} = Sum_{n >= 0} b_n x^n. Other proofs are suggested by the discussion in Watson on pages 104-5 of transformations of the Bessel functions of the first kind with b(n) = (-1)^n binomial(-1/2,n) = binomial(n-1/2,n) = (2n)! / (n! 2^n)^2. - Tom Copeland, Dec 10 2022

Examples

			a(3) = 1*3*5 = 15.
From _Joerg Arndt_, Sep 10 2013: (Start)
There are a(3)=15 involutions of 6 elements without fixed points:
  #:    permutation           transpositions
  01:  [ 1 0 3 2 5 4 ]      (0, 1) (2, 3) (4, 5)
  02:  [ 1 0 4 5 2 3 ]      (0, 1) (2, 4) (3, 5)
  03:  [ 1 0 5 4 3 2 ]      (0, 1) (2, 5) (3, 4)
  04:  [ 2 3 0 1 5 4 ]      (0, 2) (1, 3) (4, 5)
  05:  [ 2 4 0 5 1 3 ]      (0, 2) (1, 4) (3, 5)
  06:  [ 2 5 0 4 3 1 ]      (0, 2) (1, 5) (3, 4)
  07:  [ 3 2 1 0 5 4 ]      (0, 3) (1, 2) (4, 5)
  08:  [ 3 4 5 0 1 2 ]      (0, 3) (1, 4) (2, 5)
  09:  [ 3 5 4 0 2 1 ]      (0, 3) (1, 5) (2, 4)
  10:  [ 4 2 1 5 0 3 ]      (0, 4) (1, 2) (3, 5)
  11:  [ 4 3 5 1 0 2 ]      (0, 4) (1, 3) (2, 5)
  12:  [ 4 5 3 2 0 1 ]      (0, 4) (1, 5) (2, 3)
  13:  [ 5 2 1 4 3 0 ]      (0, 5) (1, 2) (3, 4)
  14:  [ 5 3 4 1 2 0 ]      (0, 5) (1, 3) (2, 4)
  15:  [ 5 4 3 2 1 0 ]      (0, 5) (1, 4) (2, 3)
(End)
G.f. = 1 + x + 3*x^2 + 15*x^3 + 105*x^4 + 945*x^5 + 10395*x^6 + 135135*x^7 + ...
		

References

  • M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards Applied Math. Series 55, Tenth Printing, 1972, (26.2.28).
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 317.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 228, #19.
  • Hoel, Port and Stone, Introduction to Probability Theory, Section 7.3.
  • F. K. Hwang, D. S. Richards and P. Winter, The Steiner Tree Problem, North-Holland, 1992, see p. 14.
  • C. Itzykson and J.-B. Zuber, Quantum Field Theory, McGraw-Hill, 1980, pages 466-467.
  • 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).
  • R. P. Stanley, Enumerative Combinatorics, Cambridge, Vol. 2, 1999; see Example 5.2.6 and also p. 178.
  • R. Vein and P. Dale, Determinants and Their Applications in Mathematical Physics, Springer-Verlag, New York, 1999, p. 73.
  • G. Watson, The Theory of Bessel Functions, Cambridge Univ. Press, 1922.

Crossrefs

Cf. A086677; A055142 (for this sequence, |a(n+1)| + 1 is the number of distinct products which can be formed using commutative, nonassociative multiplication and a nonempty subset of n given variables).
Constant terms of polynomials in A098503. First row of array A099020.
Subsequence of A248652.
Cf. A082161 (relaxed compacted binary trees of unbounded right height).
Cf. A053871 (binomial transform).

Programs

  • GAP
    A001147 := function(n) local i, s, t; t := 1; i := 0; Print(t, ", "); for i in [1 .. n] do t := t*(2*i-1); Print(t, ", "); od; end; A001147(100); # Stefano Spezia, Nov 13 2018
    
  • Haskell
    a001147 n = product [1, 3 .. 2 * n - 1]
    a001147_list = 1 : zipWith (*) [1, 3 ..] a001147_list
    -- Reinhard Zumkeller, Feb 15 2015, Dec 03 2011
    
  • Magma
    A001147:=func< n | n eq 0 select 1 else &*[ k: k in [1..2*n-1 by 2] ] >; [ A001147(n): n in [0..20] ]; // Klaus Brockhaus, Jun 22 2011
    
  • Magma
    I:=[1,3]; [1] cat [n le 2 select I[n]  else (3*n-2)*Self(n-1)-(n-1)*(2*n-3)*Self(n-2): n in [1..25] ]; // Vincenzo Librandi, Feb 19 2015
    
  • Maple
    f := n->(2*n)!/(n!*2^n);
    A001147 := proc(n) doublefactorial(2*n-1); end: # R. J. Mathar, Jul 04 2009
    A001147 := n -> 2^n*pochhammer(1/2, n); # Peter Luschny, Aug 09 2009
    G(x):=(1-2*x)^(-1/2): f[0]:=G(x): for n from 1 to 29 do f[n]:=diff(f[n-1],x) od: x:=0: seq(f[n],n=0..19); # Zerinvary Lajos, Apr 03 2009; aligned with offset by Johannes W. Meijer, Aug 11 2009
    series(hypergeom([1,1/2],[],2*x),x=0,20); # Mark van Hoeij, Apr 07 2013
  • Mathematica
    Table[(2 n - 1)!!, {n, 0, 19}] (* Robert G. Wilson v, Oct 12 2005 *)
    a[ n_] := 2^n Gamma[n + 1/2] / Gamma[1/2]; (* Michael Somos, Sep 18 2014 *)
    Join[{1}, Range[1, 41, 2]!!] (* Harvey P. Dale, Jan 28 2017 *)
    a[ n_] := If[ n < 0, (-1)^n / a[-n], SeriesCoefficient[ Product[1 - (1 - x)^(2 k - 1), {k, n}], {x, 0, n}]]; (* Michael Somos, Jun 27 2017 *)
    (2 Range[0, 20] - 1)!! (* Eric W. Weisstein, Jul 22 2017 *)
  • Maxima
    a(n):=if n=0 then 1 else sum(sum(binomial(n-1,i)*binomial(n-i-1,j)*a(i)*a(j)*a(n-i-j-1),j,0,n-i-1),i,0,n-1); /* Vladimir Kruchinin, May 06 2020 */
  • PARI
    {a(n) = if( n<0, (-1)^n / a(-n), (2*n)! / n! / 2^n)}; /* Michael Somos, Sep 18 2014 */
    
  • PARI
    x='x+O('x^33); Vec(serlaplace((1-2*x)^(-1/2))) \\ Joerg Arndt, Apr 24 2011
    
  • Python
    from sympy import factorial2
    def a(n): return factorial2(2 * n - 1)
    print([a(n) for n in range(101)])  # Indranil Ghosh, Jul 22 2017
    
  • Sage
    [rising_factorial(n+1,n)/2^n for n in (0..15)] # Peter Luschny, Jun 26 2012
    

Formula

E.g.f.: 1 / sqrt(1 - 2*x).
D-finite with recurrence: a(n) = a(n-1)*(2*n-1) = (2*n)!/(n!*2^n) = A010050(n)/A000165(n).
a(n) ~ sqrt(2) * 2^n * (n/e)^n.
Rational part of numerator of Gamma(n+1/2): a(n) * sqrt(Pi) / 2^n = Gamma(n+1/2). - Yuriy Brun, Ewa Dominowska (brun(AT)mit.edu), May 12 2001
With interpolated zeros, the sequence has e.g.f. exp(x^2/2). - Paul Barry, Jun 27 2003
The Ramanujan polynomial psi(n+1, n) has value a(n). - Ralf Stephan, Apr 16 2004
a(n) = Sum_{k=0..n} (-2)^(n-k)*A048994(n, k). - Philippe Deléham, Oct 29 2005
Log(1 + x + 3*x^2 + 15*x^3 + 105*x^4 + 945*x^5 + 10395*x^6 + ...) = x + 5/2*x^2 + 37/3*x^3 + 353/4*x^4 + 4081/5*x^5 + 55205/6*x^6 + ..., where [1, 5, 37, 353, 4081, 55205, ...] = A004208. - Philippe Deléham, Jun 20 2006
1/3 + 2/15 + 3/105 + ... = 1/2. [Jolley eq. 216]
Sum_{j=1..n} j/a(j+1) = (1 - 1/a(n+1))/2. [Jolley eq. 216]
1/1 + 1/3 + 2/15 + 6/105 + 24/945 + ... = Pi/2. - Gary W. Adamson, Dec 21 2006
a(n) = (1/sqrt(2*Pi))*Integral_{x>=0} x^n*exp(-x/2)/sqrt(x). - Paul Barry, Jan 28 2008
a(n) = A006882(2n-1). - R. J. Mathar, Jul 04 2009
G.f.: 1/(1-x-2x^2/(1-5x-12x^2/(1-9x-30x^2/(1-13x-56x^2/(1- ... (continued fraction). - Paul Barry, Sep 18 2009
a(n) = (-1)^n*subs({log(e)=1,x=0},coeff(simplify(series(e^(x*t-t^2/2),t,2*n+1)),t^(2*n))*(2*n)!). - Leonid Bedratyuk, Oct 31 2009
a(n) = 2^n*gamma(n+1/2)/gamma(1/2). - Jaume Oliver Lafont, Nov 09 2009
G.f.: 1/(1-x/(1-2x/(1-3x/(1-4x/(1-5x/(1- ...(continued fraction). - Aoife Hennessy (aoife.hennessy(AT)gmail.com), Dec 02 2009
The g.f. of a(n+1) is 1/(1-3x/(1-2x/(1-5x/(1-4x/(1-7x/(1-6x/(1-.... (continued fraction). - Paul Barry, Dec 04 2009
a(n) = Sum_{i=1..n} binomial(n,i)*a(i-1)*a(n-i). - Vladimir Shevelev, Sep 30 2010
E.g.f.: A(x) = 1 - sqrt(1-2*x) satisfies the differential equation A'(x) - A'(x)*A(x) - 1 = 0. - Vladimir Kruchinin, Jan 17 2011
a(n) = A123023(2*n). - Michael Somos, Jul 24 2011
a(n) = (1/2)*Sum_{i=1..n} binomial(n+1,i)*a(i-1)*a(n-i). See link above. - Dennis P. Walsh, Dec 02 2011
a(n) = Sum_{k=0..n} (-1)^k*binomial(2*n,n+k)*Stirling_1(n+k,k) [Kauers and Ko].
a(n) = A035342(n, 1), n >= 1 (first column of triangle).
a(n) = A001497(n, 0) = A001498(n, n), first column, resp. main diagonal, of Bessel triangle.
From Gary W. Adamson, Jul 19 2011: (Start)
a(n) = upper left term of M^n and sum of top row terms of M^(n-1), where M = a variant of the (1,2) Pascal triangle (Cf. A029635) as the following production matrix:
1, 2, 0, 0, 0, ...
1, 3, 2, 0, 0, ...
1, 4, 5, 2, 0, ...
1, 5, 9, 7, 2, ...
...
For example, a(3) = 15 is the left term in top row of M^3: (15, 46, 36, 8) and a(4) = 105 = (15 + 46 + 36 + 8).
(End)
G.f.: A(x) = 1 + x/(W(0) - x); W(k) = 1 + x + x*2*k - x*(2*k + 3)/W(k+1); (continued fraction). - Sergei N. Gladkovskii, Nov 17 2011
a(n) = Sum_{i=1..n} binomial(n,i-1)*a(i-1)*a(n-i). - Dennis P. Walsh, Dec 02 2011
a(n) = A009445(n) / A014481(n). - Reinhard Zumkeller, Dec 03 2011
a(n) = (-1)^n*Sum_{k=0..n} 2^(n-k)*s(n+1,k+1), where s(n,k) are the Stirling numbers of the first kind, A048994. - Mircea Merca, May 03 2012
a(n) = (2*n)4! = Gauss_factorial(2*n,4) = Product{j=1..2*n, gcd(j,4)=1} j. - Peter Luschny, Oct 01 2012
G.f.: (1 - 1/Q(0))/x where Q(k) = 1 - x*(2*k - 1)/(1 - x*(2*k + 2)/Q(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Mar 19 2013
G.f.: 1 + x/Q(0), where Q(k) = 1 + (2*k - 1)*x - 2*x*(k + 1)/Q(k+1); (continued fraction). - Sergei N. Gladkovskii, May 01 2013
G.f.: 2/G(0), where G(k) = 1 + 1/(1 - 2*x*(2*k + 1)/(2*x*(2*k + 1) - 1 + 2*x*(2*k + 2)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, May 31 2013
G.f.: G(0)/2, where G(k) = 1 + 1/(1 - x/(x + 1/(2*k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 01 2013
G.f.: G(0), where G(k) = 1 + 2*x*(4*k + 1)/(4*k + 2 - 2*x*(2*k + 1)*(4*k + 3)/(x*(4*k + 3) + 2*(k + 1)/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 22 2013
a(n) = (2*n - 3)*a(n-2) + (2*n - 2)*a(n-1), n > 1. - Ivan N. Ianakiev, Jul 08 2013
G.f.: G(0), where G(k) = 1 - x*(k+1)/(x*(k+1) - 1/G(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Aug 04 2013
a(n) = 2*a(n-1) + (2n-3)^2*a(n-2), a(0) = a(1) = 1. - Philippe Deléham, Oct 27 2013
G.f. of reciprocals: Sum_{n>=0} x^n/a(n) = 1F1(1; 1/2; x/2), confluent hypergeometric Function. - R. J. Mathar, Jul 25 2014
0 = a(n)*(+2*a(n+1) - a(n+2)) + a(n+1)*(+a(n+1)) for all n in Z. - Michael Somos, Sep 18 2014
a(n) = (-1)^n / a(-n) = 2*a(n-1) + a(n-1)^2 / a(n-2) for all n in Z. - Michael Somos, Sep 18 2014
From Peter Bala, Feb 18 2015: (Start)
Recurrence equation: a(n) = (3*n - 2)*a(n-1) - (n - 1)*(2*n - 3)*a(n-2) with a(1) = 1 and a(2) = 3.
The sequence b(n) = A087547(n), beginning [1, 4, 52, 608, 12624, ... ], satisfies the same second-order recurrence equation. This leads to the generalized continued fraction expansion lim_{n -> infinity} b(n)/a(n) = Pi/2 = 1 + 1/(3 - 6/(7 - 15/(10 - ... - n*(2*n - 1)/((3*n + 1) - ... )))). (End)
E.g.f of the sequence whose n-th element (n = 1,2,...) equals a(n-1) is 1-sqrt(1-2*x). - Stanislav Sykora, Jan 06 2017
Sum_{n >= 1} a(n)/(2*n-1)! = exp(1/2). - Daniel Suteu, Feb 06 2017
a(n) = A028338(n, 0), n >= 0. - Wolfdieter Lang, May 27 2017
a(n) = (Product_{k=0..n-2} binomial(2*(n-k),2))/n!. - Stefano Spezia, Nov 13 2018
a(n) = Sum_{i=0..n-1} Sum_{j=0..n-i-1} C(n-1,i)*C(n-i-1,j)*a(i)*a(j)*a(n-i-j-1), a(0)=1, - Vladimir Kruchinin, May 06 2020
From Amiram Eldar, Jun 29 2020: (Start)
Sum_{n>=1} 1/a(n) = sqrt(e*Pi/2)*erf(1/sqrt(2)), where erf is the error function.
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(Pi/(2*e))*erfi(1/sqrt(2)), where erfi is the imaginary error function. (End)
G.f. of reciprocals: R(x) = Sum_{n>=0} x^n/a(n) satisfies (1 + x)*R(x) = 1 + 2*x*R'(x). - Werner Schulte, Nov 04 2024

Extensions

Removed erroneous comments: neither the number of n X n binary matrices A such that A^2 = 0 nor the number of simple directed graphs on n vertices with no directed path of length two are counted by this sequence (for n = 3, both are 13). - Dan Drake, Jun 02 2009

A008292 Triangle of Eulerian numbers T(n,k) (n >= 1, 1 <= k <= n) read by rows.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 1, 11, 11, 1, 1, 26, 66, 26, 1, 1, 57, 302, 302, 57, 1, 1, 120, 1191, 2416, 1191, 120, 1, 1, 247, 4293, 15619, 15619, 4293, 247, 1, 1, 502, 14608, 88234, 156190, 88234, 14608, 502, 1, 1, 1013, 47840, 455192, 1310354, 1310354, 455192, 47840, 1013, 1
Offset: 1

Author

N. J. A. Sloane, Mar 15 1996

Comments

The indexing used here follows that given in the classic books by Riordan and Comtet. For two other versions see A173018 and A123125. - N. J. A. Sloane, Nov 21 2010
Coefficients of Eulerian polynomials. Number of permutations of n objects with k-1 rises. Number of increasing rooted trees with n+1 nodes and k leaves.
T(n,k) = number of permutations of [n] with k runs. T(n,k) = number of permutations of [n] requiring k readings (see the Knuth reference). T(n,k) = number of permutations of [n] having k distinct entries in its inversion table. - Emeric Deutsch, Jun 09 2004
T(n,k) = number of ways to write the Coxeter element s_{e1}s_{e1-e2}s_{e2-e3}s_{e3-e4}...s_{e_{n-1}-e_n} of the reflection group of type B_n, using s_{e_k} and as few reflections of the form s_{e_i+e_j}, where i = 1, 2, ..., n and j is not equal to i, as possible. - Pramook Khungurn (pramook(AT)mit.edu), Jul 07 2004
Subtriangle for k>=1 and n>=1 of triangle A123125. - Philippe Deléham, Oct 22 2006
T(n,k)/n! also represents the n-dimensional volume of the portion of the n-dimensional hypercube cut by the (n-1)-dimensional hyperplanes x_1 + x_2 + ... x_n = k, x_1 + x_2 + ... x_n = k-1; or, equivalently, it represents the probability that the sum of n independent random variables with uniform distribution between 0 and 1 is between k-1 and k. - Stefano Zunino, Oct 25 2006
[E(.,t)/(1-t)]^n = n!*Lag[n,-P(.,t)/(1-t)] and [-P(.,t)/(1-t)]^n = n!*Lag[n, E(.,t)/(1-t)] umbrally comprise a combinatorial Laguerre transform pair, where E(n,t) are the Eulerian polynomials and P(n,t) are the polynomials in A131758. - Tom Copeland, Sep 30 2007
From Tom Copeland, Oct 07 2008: (Start)
G(x,t) = 1/(1 + (1-exp(x*t))/t) = 1 + 1*x + (2+t)*x^2/2! + (6+6*t+t^2)*x^3/3! + ... gives row polynomials for A090582, the reverse f-polynomials for the permutohedra (see A019538).
G(x,t-1) = 1 + 1*x + (1+t)*x^2/2! + (1+4*t+t^2)*x^3/3! + ... gives row polynomials for A008292, the h-polynomials for permutohedra (Postnikov et al.).
G((t+1)*x, -1/(t+1)) = 1 + (1+t)*x + (1+3*t+2*t^2)*x^2/2! + ... gives row polynomials for A028246.
(End)
A subexceedant function f on [n] is a map f:[n] -> [n] such that 1 <= f(i) <= i for all i, 1 <= i <= n. T(n,k) equals the number of subexceedant functions f of [n] such that the image of f has cardinality k [Mantaci & Rakotondrajao]. Example T(3,2) = 4: if we identify a subexceedant function f with the word f(1)f(2)...f(n) then the subexceedant functions on [3] are 111, 112, 113, 121, 122 and 123 and four of these functions have an image set of cardinality 2. - Peter Bala, Oct 21 2008
Further to the comments of Tom Copeland above, the n-th row of this triangle is the h-vector of the simplicial complex dual to a permutohedron of type A_(n-1). The corresponding f-vectors are the rows of A019538. For example, 1 + 4*x + x^2 = y^2 + 6*y + 6 and 1 + 11*x + 11*x^2 + x^3 = y^3 + 14*y^2 + 36*y + 24, where x = y + 1, give [1,6,6] and [1,14,36,24] as the third and fourth rows of A019538. The Hilbert transform of this triangle (see A145905 for the definition) is A047969. See A060187 for the triangle of Eulerian numbers of type B (the h-vectors of the simplicial complexes dual to permutohedra of type B). See A066094 for the array of h-vectors of type D. For tables of restricted Eulerian numbers see A144696 - A144699. - Peter Bala, Oct 26 2008
For a natural refinement of A008292 with connections to compositional inversion and iterated derivatives, see A145271. - Tom Copeland, Nov 06 2008
The polynomials E(z,n) = numerator(Sum_{k>=1} (-1)^(n+1)*k^n*z^(k-1)) for n >=1 lead directly to the triangle of Eulerian numbers. - Johannes W. Meijer, May 24 2009
From Walther Janous (walther.janous(AT)tirol.com), Nov 01 2009: (Start)
The (Eulerian) polynomials e(n,x) = Sum_{k=0..n-1} T(n,k+1)*x^k turn out to be also the numerators of the closed-form expressions of the infinite sums:
S(p,x) = Sum_{j>=0} (j+1)^p*x^j, that is
S(p,x) = e(p,x)/(1-x)^(p+1), whenever |x| < 1 and p is a positive integer.
(Note the inconsistent use of T(n,k) in the section listing the formula section. I adhere tacitly to the first one.) (End)
If n is an odd prime, then all numbers of the (n-2)-th and (n-1)-th rows are in the progression k*n+1. - Vladimir Shevelev, Jul 01 2011
The Eulerian triangle is an element of the formula for the r-th successive summation of Sum_{k=1..n} k^j which appears to be Sum_{k=1..n} T(j,k-1) * binomial(j-k+n+r, j+r). - Gary Detlefs, Nov 11 2011
Li and Wong show that T(n,k) counts the combinatorially inequivalent star polygons with n+1 vertices and sum of angles (2*k-n-1)*Pi. An equivalent formulation is: define the total sign change S(p) of a permutation p in the symmetric group S_n to be equal to Sum_{i=1..n} sign(p(i)-p(i+1)), where we take p(n+1) = p(1). T(n,k) gives the number of permutations q in S_(n+1) with q(1) = 1 and S(q) = 2*k-n-1. For example, T(3,2) = 4 since in S_4 the permutations (1243), (1324), (1342) and (1423) have total sign change 0. - Peter Bala, Dec 27 2011
Xiong, Hall and Tsao refer to Riordan and mention that a traditional Eulerian number A(n,k) is the number of permutations of (1,2...n) with k weak exceedances. - Susanne Wienand, Aug 25 2014
Connections to algebraic geometry/topology and characteristic classes are discussed in the Buchstaber and Bunkova, the Copeland, the Hirzebruch, the Lenart and Zainoulline, the Losev and Manin, and the Sheppeard links; to the Grassmannian, in the Copeland, the Farber and Postnikov, the Sheppeard, and the Williams links; and to compositional inversion and differential operators, in the Copeland and the Parker links. - Tom Copeland, Oct 20 2015
The bivariate e.g.f. noted in the formulas is related to multiplying edges in certain graphs discussed in the Aluffi-Marcolli link. See p. 42. - Tom Copeland, Dec 18 2016
Distribution of left children in treeshelves is given by a shift of the Eulerian numbers. Treeshelves are ordered binary (0-1-2) increasing trees where every child is connected to its parent by a left or a right link. See A278677, A278678 or A278679 for more definitions and examples. - Sergey Kirgizov, Dec 24 2016
The row polynomial P(n, x) = Sum_{k=1..n} T(n, k)*x^k appears in the numerator of 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} P(n, x)/(1 - x)^(n+2) for n >= 0 (with 0^0=1). See also triangle A131689 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
For relations to Ehrhart polynomials, volumes of polytopes, polylogarithms, the Todd operator, and other special functions, polynomials, and sequences, see A131758 and the references therein. - Tom Copeland, Jun 20 2017
For relations to values of the Riemann zeta function at integral arguments, see A131758 and the Dupont reference. - Tom Copeland, Mar 19 2018
Normalized volumes of the hypersimplices, attributed to Laplace. (Cf. the De Loera et al. reference, p. 327.) - Tom Copeland, Jun 25 2018

Examples

			The triangle T(n, k) begins:
n\k 1    2     3      4       5       6      7     8    9 10 ...
1:  1
2:  1    1
3:  1    4     1
4:  1   11    11      1
5:  1   26    66     26       1
6:  1   57   302    302      57       1
7:  1  120  1191   2416    1191     120      1
8:  1  247  4293  15619   15619    4293    247     1
9:  1  502 14608  88234  156190   88234  14608   502    1
10: 1 1013 47840 455192 1310354 1310354 455192 47840 1013  1
... Reformatted. - _Wolfdieter Lang_, Feb 14 2015
-----------------------------------------------------------------
E.g.f. = (y) * x^1 / 1! + (y + y^2) * x^2 / 2! + (y + 4*y^2 + y^3) * x^3 / 3! + ... - _Michael Somos_, Mar 17 2011
Let n=7. Then the following 2*7+1=15 consecutive terms are 1(mod 7): a(15+i), i=0..14. - _Vladimir Shevelev_, Jul 01 2011
Row 3: The plane increasing 0-1-2 trees on 3 vertices (with the number of colored vertices shown to the right of a vertex) are
.
.   1o (1+t)         1o t         1o t
.   |                / \          / \
.   |               /   \        /   \
.   2o (1+t)      2o     3o    3o    2o
.   |
.   |
.   3o
.
The total number of trees is (1+t)^2 + t + t = 1 + 4*t + t^2.
		

References

  • Mohammad K. Azarian, Geometric Series, Problem 329, Mathematics and Computer Education, Vol. 30, No. 1, Winter 1996, p. 101. Solution published in Vol. 31, No. 2, Spring 1997, pp. 196-197.
  • Miklos Bona, editor, Handbook of Enumerative Combinatorics, CRC Press, 2015, page 106.
  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 243.
  • F. N. David, M. G. Kendall and D. E. Barton, Symmetric Function and Allied Tables, Cambridge, 1966, p. 260.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 1990, p. 254; 2nd. ed., p. 268.[Worpitzky's identity (6.37)]
  • D. E. Knuth, The Art of Computer Programming. Addison-Wesley, Reading, MA, 1998, Vol. 3, p. 47 (exercise 5.1.4 Nr. 20) and p. 605 (solution).
  • Meng Li and Ron Goldman. "Limits of sums for binomial and Eulerian numbers and their associated distributions." Discrete Mathematics 343.7 (2020): 111870.
  • Anthony Mendes and Jeffrey Remmel, Generating functions from symmetric functions, Preliminary version of book, available from Jeffrey Remmel's home page http://math.ucsd.edu/~remmel/
  • K. Mittelstaedt, A stochastic approach to Eulerian numbers, Amer. Math. Mnthly, 127:7 (2020), 618-628.
  • T. K. Petersen, Eulerian Numbers, Birkhauser, 2015.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p. 215.
  • R. Sedgewick and P. Flajolet, An Introduction to the Analysis of Algorithms, Addison-Wesley, Reading, MA, 1996.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Figure M3416, Academic Press, 1995.
  • H. S. Wall, Analytic Theory of Continued Fractions, Chelsea, 1973, see p. 208.
  • D. B. West, Combinatorial Mathematics, Cambridge, 2021, p. 101.

Programs

  • GAP
    Flat(List([1..10],n->List([1..n],k->Sum([0..k],j->(-1)^j*(k-j)^n*Binomial(n+1,j))))); # Muniru A Asiru, Jun 29 2018
    
  • Haskell
    import Data.List (genericLength)
    a008292 n k = a008292_tabl !! (n-1) !! (k-1)
    a008292_row n = a008292_tabl !! (n-1)
    a008292_tabl = iterate f [1] where
       f xs = zipWith (+)
         (zipWith (*) ([0] ++ xs) (reverse ks)) (zipWith (*) (xs ++ [0]) ks)
         where ks = [1 .. 1 + genericLength xs]
    -- Reinhard Zumkeller, May 07 2013
    
  • Magma
    Eulerian:= func< n,k | (&+[(-1)^j*Binomial(n+1,j)*(k-j+1)^n: j in [0..k+1]]) >; [[Eulerian(n,k): k in [0..n-1]]: n in [1..10]]; // G. C. Greubel, Apr 15 2019
  • Maple
    A008292 := proc(n,k) option remember; if k < 1 or k > n then 0; elif k = 1 or k = n then 1; else k*procname(n-1,k)+(n-k+1)*procname(n-1,k-1) ; end if; end proc:
  • Mathematica
    t[n_, k_] = Sum[(-1)^j*(k-j)^n*Binomial[n+1, j], {j, 0, k}];
    Flatten[Table[t[n, k], {n, 1, 10}, {k, 1, n}]] (* Jean-François Alcover, May 31 2011, after Michael Somos *)
    Flatten[Table[CoefficientList[(1-x)^(k+1)*PolyLog[-k, x]/x, x], {k, 1, 10}]] (* Vaclav Kotesovec, Aug 27 2015 *)
    Table[Tally[
       Count[#, x_ /; x > 0] & /@ (Differences /@
          Permutations[Range[n]])][[;; , 2]], {n, 10}] (* Li Han, Oct 11 2020 *)
  • PARI
    {T(n, k) = if( k<1 || k>n, 0, if( n==1, 1, k * T(n-1, k) + (n-k+1) * T(n-1, k-1)))}; /* Michael Somos, Jul 19 1999 */
    
  • PARI
    {T(n, k) = sum( j=0, k, (-1)^j * (k-j)^n * binomial( n+1, j))}; /* Michael Somos, Jul 19 1999 */
    
  • PARI
    {A(n,c)=c^(n+c-1)+sum(i=1,c-1,(-1)^i/i!*(c-i)^(n+c-1)*prod(j=1,i,n+c+1-j))}
    
  • Python
    from sympy import binomial
    def T(n, k): return sum([(-1)**j*(k - j)**n*binomial(n + 1, j) for j in range(k + 1)])
    for n in range(1, 11): print([T(n, k) for k in range(1, n + 1)]) # Indranil Ghosh, Nov 08 2017
    
  • R
    T <- function(n, k) {
      S <- numeric()
      for (j in 0:k) S <- c(S, (-1)^j*(k-j)^n*choose(n+1, j))
      return(sum(S))
    }
    for (n in 1:10){
      for (k in 1:n) print(T(n,k))
    } # Indranil Ghosh, Nov 08 2017
    
  • Sage
    [[sum((-1)^j*binomial(n+1, j)*(k-j)^n for j in (0..k)) for k in (1..n)] for n in (1..12)] # G. C. Greubel, Feb 23 2019
    

Formula

T(n, k) = k * T(n-1, k) + (n-k+1) * T(n-1, k-1), T(1, 1) = 1.
T(n, k) = Sum_{j=0..k} (-1)^j * (k-j)^n * binomial(n+1, j).
Row sums = n! = A000142(n) unless n=0. - Michael Somos, Mar 17 2011
E.g.f. A(x, q) = Sum_{n>0} (Sum_{k=1..n} T(n, k) * q^k) * x^n / n! = q * ( e^(q*x) - e^x ) / ( q*e^x - e^(q*x) ) satisfies dA / dx = (A + 1) * (A + q). - Michael Somos, Mar 17 2011
For a column listing, n-th term: T(c, n) = c^(n+c-1) + Sum_{i=1..c-1} (-1)^i/i!*(c-i)^(n+c-1)*Product_{j=1..i} (n+c+1-j). - Randall L Rathbun, Jan 23 2002
From John Robertson (jpr2718(AT)aol.com), Sep 02 2002: (Start)
Four characterizations of Eulerian numbers T(i, n):
1. T(0, n)=1 for n>=1, T(i, 1)=0 for i>=1, T(i, n) = (n-i)T(i-1, n-1) + (i+1)T(i, n-1).
2. T(i, n) = Sum_{j=0..i} (-1)^j*binomial(n+1,j)*(i-j+1)^n for n>=1, i>=0.
3. Let C_n be the unit cube in R^n with vertices (e_1, e_2, ..., e_n) where each e_i is 0 or 1 and all 2^n combinations are used. Then T(i, n)/n! is the volume of C_n between the hyperplanes x_1 + x_2 + ... + x_n = i and x_1 + x_2 + ... + x_n = i+1. Hence T(i, n)/n! is the probability that i <= X_1 + X_2 + ... + X_n < i+1 where the X_j are independent uniform [0, 1] distributions. - See Ehrenborg & Readdy reference.
4. Let f(i, n) = T(i, n)/n!. The f(i, n) are the unique coefficients so that (1/(r-1)^(n+1)) Sum_{i=0..n-1} f(i, n) r^{i+1} = Sum_{j>=0} (j^n)/(r^j) whenever n>=1 and abs(r)>1. (End)
O.g.f. for n-th row: (1-x)^(n+1)*polylog(-n, x)/x. - Vladeta Jovovic, Sep 02 2002
Triangle T(n, k), n>0 and k>0, read by rows; given by [0, 1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] DELTA [1, 0, 2, 0, 3, 0, 4, 0, 5, 0, 6, ...] (positive integers interspersed with 0's) where DELTA is Deléham's operator defined in A084938.
Sum_{k=1..n} T(n, k)*2^k = A000629(n). - Philippe Deléham, Jun 05 2004
From Tom Copeland, Oct 10 2007: (Start)
Bell_n(x) = Sum_{j=0..n} S2(n,j) * x^j = Sum_{j=0..n} E(n,j) * Lag(n,-x, j-n) = Sum_{j=0..n} (E(n,j)/n!) * (n!*Lag(n,-x, j-n)) = Sum_{j=0..n} E(n,j) * binomial(Bell.(x)+j, n) umbrally where Bell_n(x) are the Bell / Touchard / exponential polynomials; S2(n,j), the Stirling numbers of the second kind; E(n,j), the Eulerian numbers; and Lag(n,x,m), the associated Laguerre polynomials of order m.
For x = 0, the equation gives Sum_{j=0..n} E(n,j) * binomial(j,n) = 1 for n=0 and 0 for all other n. By substituting the umbral compositional inverse of the Bell polynomials, the lower factorial n!*binomial(y,n), for x in the equation, the Worpitzky identity is obtained; y^n = Sum_{j=0..n} E(n,j) * binomial(y+j,n).
Note that E(n,j)/n! = E(n,j)/(Sum_{k=0..n} E(n,k)). Also (n!*Lag(n, -1, j-n)) is A086885 with a simple combinatorial interpretation in terms of seating arrangements, giving a combinatorial interpretation to the equation for x=1; n!*Bell_n(1) = n!*Sum_{j=0..n} S2(n,j) = Sum_{j=0..n} E(n,j) * (n!*Lag(n, -1, j-n)).
(Appended Sep 16 2020) For connections to the Bernoulli numbers, extensions, proofs, and a clear presentation of the number arrays involved in the identities above, see my post Reciprocity and Umbral Witchcraft. (End)
From the relations between the h- and f-polynomials of permutohedra and reciprocals of e.g.f.s described in A049019: (t-1)((t-1)d/dx)^n 1/(t-exp(x)) evaluated at x=0 gives the n-th Eulerian row polynomial in t and the n-th row polynomial in (t-1) of A019538 and A090582. From the Comtet and Copeland references in A139605: ((t+exp(x)-1)d/dx)^(n+1) x gives pairs of the Eulerian polynomials in t as the coefficients of x^0 and x^1 in its Taylor series expansion in x. - Tom Copeland, Oct 05 2008
G.f: 1/(1-x/(1-x*y/1-2*x/(1-2*x*y/(1-3*x/(1-3*x*y/(1-... (continued fraction). - Paul Barry, Mar 24 2010
If n is odd prime, then the following consecutive 2*n+1 terms are 1 modulo n: a((n-1)*(n-2)/2+i), i=0..2*n. This chain of terms is maximal in the sense that neither the previous term nor the following one are 1 modulo n. - _Vladimir Shevelev, Jul 01 2011
From Peter Bala, Sep 29 2011: (Start)
For k = 0,1,2,... put G(k,x,t) := x -(1+2^k*t)*x^2/2 +(1+2^k*t+3^k*t^2)*x^3/3-(1+2^k*t+3^k*t^2+4^k*t^3)*x^4/4+.... Then the series reversion of G(k,x,t) with respect to x gives an e.g.f. for the present table when k = 0 and for A008517 when k = 1.
The e.g.f. B(x,t) := compositional inverse with respect to x of G(0,x,t) = (exp(x)-exp(x*t))/(exp(x*t)-t*exp(x)) = x + (1+t)*x^2/2! + (1+4*t+t^2)*x^3/3! + ... satisfies the autonomous differential equation dB/dx = (1+B)*(1+t*B) = 1 + (1+t)*B + t*B^2.
Applying [Bergeron et al., Theorem 1] gives a combinatorial interpretation for the Eulerian polynomials: A(n,t) counts plane increasing trees on n vertices where each vertex has outdegree <= 2, the vertices of outdegree 1 come in 1+t colors and the vertices of outdegree 2 come in t colors. An example is given below. Cf. A008517. Applying [Dominici, Theorem 4.1] gives the following method for calculating the Eulerian polynomials: Let f(x,t) = (1+x)*(1+t*x) and let D be the operator f(x,t)*d/dx. Then A(n+1,t) = D^n(f(x,t)) evaluated at x = 0.
(End)
With e.g.f. A(x,t) = G[x,(t-1)]-1 in Copeland's 2008 comment, the compositional inverse is Ainv(x,t) = log(t-(t-1)/(1+x))/(t-1). - Tom Copeland, Oct 11 2011
T(2*n+1,n+1) = (2*n+2)*T(2*n,n). (E.g., 66 = 6*11, 2416 = 8*302, ...) - Gary Detlefs, Nov 11 2011
E.g.f.: (1-y) / (1 - y*exp( (1-y)*x )). - Geoffrey Critzer, Nov 10 2012
From Peter Bala, Mar 12 2013: (Start)
Let {A(n,x)} n>=1 denote the sequence of Eulerian polynomials beginning [1, 1 + x, 1 + 4*x + x^2, ...]. Given two complex numbers a and b, the polynomial sequence defined by R(n,x) := (x+b)^n*A(n+1,(x+a)/(x+b)), n >= 0, satisfies the recurrence equation R(n+1,x) = d/dx((x+a)*(x+b)*R(n,x)). These polynomials give the row generating polynomials for several triangles in the database including A019538 (a = 0, b = 1), A156992 (a = 1, b = 1), A185421 (a = (1+i)/2, b = (1-i)/2), A185423 (a = exp(i*Pi/3), b = exp(-i*Pi/3)) and A185896 (a = i, b = -i).
(End)
E.g.f.: 1 + x/(T(0) - x*y), where T(k) = 1 + x*(y-1)/(1 + (k+1)/T(k+1) ); (continued fraction). - Sergei N. Gladkovskii, Nov 07 2013
From Tom Copeland, Sep 18 2014: (Start)
A) Bivariate e.g.f. A(x,a,b)= (e^(ax)-e^(bx))/(a*e^(bx)-b*e^(ax)) = x + (a+b)*x^2/2! + (a^2+4ab+b^2)*x^3/3! + (a^3+11a^2b+11ab^2+b^3)x^4/4! + ...
B) B(x,a,b)= log((1+ax)/(1+bx))/(a-b) = x - (a+b)x^2/2 + (a^2+ab+b^2)x^3/3 - (a^3+a^2b+ab^2+b^3)x^4/4 + ... = log(1+u.*x), with (u.)^n = u_n = h_(n-1)(a,b) a complete homogeneous polynomial, is the compositional inverse of A(x,a,b) in x (see Drake, p. 56).
C) A(x) satisfies dA/dx = (1+a*A)(1+b*A) and can be written in terms of a Weierstrass elliptic function (see Buchstaber & Bunkova).
D) The bivariate Eulerian row polynomials are generated by the iterated derivative ((1+ax)(1+bx)d/dx)^n x evaluated at x=0 (see A145271).
E) A(x,a,b)= -(e^(-ax)-e^(-bx))/(a*e^(-ax)-b*e^(-bx)), A(x,-1,-1) = x/(1+x), and B(x,-1,-1) = x/(1-x).
F) FGL(x,y) = A(B(x,a,b) + B(y,a,b),a,b) = (x+y+(a+b)xy)/(1-ab*xy) is called the hyperbolic formal group law and related to a generalized cohomology theory by Lenart and Zainoulline. (End)
For x > 1, the n-th Eulerian polynomial A(n,x) = (x - 1)^n * log(x) * Integral_{u>=0} (ceiling(u))^n * x^(-u) du. - Peter Bala, Feb 06 2015
Sum_{j>=0} j^n/e^j, for n>=0, equals Sum_{k=1..n} T(n,k)e^k/(e-1)^(n+1), a rational function in the variable "e" which evaluates, approximately, to n! when e = A001113 = 2.71828... - Richard R. Forberg, Feb 15 2015
For a fixed k, T(n,k) ~ k^n, proved by induction. - Ran Pan, Oct 12 2015
From A145271, multiply the n-th diagonal (with n=0 the main diagonal) of the lower triangular Pascal matrix by g_n = (d/dx)^n (1+a*x)*(1+b*x) evaluated at x= 0, i.e., g_0 = 1, g_1 = (a+b), g_2 = 2ab, and g_n = 0 otherwise, to obtain the tridiagonal matrix VP with VP(n,k) = binomial(n,k) g_(n-k). Then the m-th bivariate row polynomial of this entry is P(m,a,b) = (1, 0, 0, 0, ...) [VP * S]^(m-1) (1, a+b, 2ab, 0, ...)^T, where S is the shift matrix A129185, representing differentiation in the divided powers basis x^n/n!. Also, P(m,a,b) = (1, 0, 0, 0, ...) [VP * S]^m (0, 1, 0, ...)^T. - Tom Copeland, Aug 02 2016
Cumulatively summing a row generates the n starting terms of the n-th differences of the n-th powers. Applying the finite difference method to x^n, these terms correspond to those before constant n! in the lowest difference row. E.g., T(4,k) is summed as 0+1=1, 1+11=12, 12+11=23, 23+1=4!. See A101101, A101104, A101100, A179457. - Andy Nicol, May 25 2024

Extensions

Thanks to Michael Somos for additional comments.
Further comments from Christian G. Bower, May 12 2000
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