cp's OEIS Frontend

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

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A028421 Triangle read by rows: T(n, k) = (k+1)*A132393(n+1, k+1), for 0 <= k <= n.

Original entry on oeis.org

1, 1, 2, 2, 6, 3, 6, 22, 18, 4, 24, 100, 105, 40, 5, 120, 548, 675, 340, 75, 6, 720, 3528, 4872, 2940, 875, 126, 7, 5040, 26136, 39396, 27076, 9800, 1932, 196, 8, 40320, 219168, 354372, 269136, 112245, 27216, 3822, 288, 9
Offset: 0

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Author

Peter Wiggen (wiggen(AT)math.psu.edu)

Keywords

Comments

Previous name was: Number triangle f(n, k) from n-th differences of the sequence {1/m^2}{m >= 1}, for n >= 0; the n-th difference sequence is {(-1)^n*n!*P(n, m)/D(n, m)^2}{m >= 1} where P(n, x) is the row polynomial P(n, x) = Sum_{k=0..n} f(n,k)*x^k and D(n, x) = x*(x+1)*...*(x+n).
From Johannes W. Meijer, Oct 07 2009: (Start)
The higher-order exponential integrals E(x,m,n) are defined in A163931 and the general formula of the asymptotic expansion of E(x,m,n) can be found in A163932.
We used the general formula and the asymptotic expansion of E(x,m=1,n), see A130534, to determine that E(x,m=2,n) ~ (exp(-x)/x^2)*(1 - (1+2*n)/x + (2 + 6*n + 3*n^2)/x^2 - (6 + 22*n + 18*n^2 + 4*n^3)/x^3 + ...) which can be verified with the EA(x,2,n) formula, see A163932. The coefficients in the denominators of this expansion lead to the sequence given above.
The asymptotic expansion of E(x,m=2,n) leads for n from one to ten to known sequences, see the cross-references. With these sequences one can form the triangles A165674 (left hand columns) and A093905 (right hand columns).
(End)
For connections to an operator relation between log(x) and x^n(d/dx)^n, see A238363. - Tom Copeland, Feb 28 2014
From Wolfdieter Lang, Nov 25 2018: (Start)
The signed triangle t(n, k) := (-1)^{n-k}*f(n, k) gives (n+1)*N(-1;n,x) = Sum_{k=0..n} t(n, k)*x^k, where N(-1;n,x) are the Narumi polynomials with parameter a = -1 (see the Weisstein link).
The members of the n-th difference sequence of the sequence {1/m^2}_{m>=1} mentioned above satisfies the recurrence delta(n, m) = delta(n-1, m+1) - delta(n-1, m), for n >= 1, m >= 1, with input delta(0, m) = 1/m^2. The solution is delta(n, m) = (n+1)!*N(-1;n,-m)/risefac(m, n+1)^2, with Narumi polynomials N(-1;n,x) and the rising factorials risefac(x, n+1) = D(n, x) = x*(x+1)*...*(x+n).
The above mentioned row polynomials P satisfy P(n, x) = (-1)^n*(n + 1)*N(-1;n,-x), for n >= 0. The recurrence is P(n, x) = (-x^2*P(n-1, x+1) + (n+x)^2*P(n-1, x))/n, for n >= 1, and P(0, x) = 1. (End)
The triangle is the exponential Riordan square (cf. A321620) of -log(1-x) with an additional main diagonal of zeros. - Peter Luschny, Jan 03 2019

Examples

			The triangle T(n, k) begins:
n\k       0        1        2        3        4       5       6      7     8   9 10
------------------------------------------------------------------------------------
0:        1
1:        1        2
2:        2        6        3
3:        6       22       18        4
4:       24      100      105       40        5
5:      120      548      675      340       75       6
6:      720     3528     4872     2940      875     126       7
7:     5040    26136    39396    27076     9800    1932     196      8
8:    40320   219168   354372   269136   112245   27216    3822    288     9
9:   362880  2053152  3518100  2894720  1346625  379638   66150   6960   405  10
10: 3628800 21257280 38260728 33638000 17084650 5412330 1104411 145200 11880 550 11
... - _Wolfdieter Lang_, Nov 23 2018
		

Crossrefs

Row sums give A000254(n+1), n >= 0.
Cf. A132393 (unsigned Stirling1), A061356, A139526, A321620.
From Johannes W. Meijer, Oct 07 2009: (Start)
A000142, A052517, 3*A000399, 5*A000482 are the first four left hand columns; A000027, A002411 are the first two right hand columns.
The asymptotic expansion of E(x,m=2,n) leads to A000254 (n=1), A001705 (n=2), A001711 (n=3), A001716 (n=4), A001721 (n=5), A051524 (n=6), A051545 (n=7), A051560 (n=8), A051562 (n=9), A051564 (n=10), A093905 (triangle) and A165674 (triangle).
Cf. A163931 (E(x,m,n)), A130534 (m=1), A163932 (m=3), A163934 (m=4), A074246 (E(x,m=2,n+1)). (End)

Programs

  • Maple
    A028421 := proc(n,k) (-1)^(n+k)*(k+1)*Stirling1(n+1,k+1) end:
    seq(seq(A028421(n,k), k=0..n), n=0..8);
    # Johannes W. Meijer, Oct 07 2009, Revised Sep 09 2012
    egf := (1 - t)^(-x - 1)*(1 - x*log(1 - t)):
    ser := series(egf, t, 16): coefft := n -> expand(coeff(ser,t,n)):
    seq(seq(n!*coeff(coefft(n), x, k), k = 0..n), n = 0..8); # Peter Luschny, Jun 12 2022
  • Mathematica
    f[n_, k_] = (k + 1) StirlingS1[n + 1, k + 1] // Abs; Flatten[Table[f[n, k], {n, 0, 9}, {k, 0, n}]][[1 ;; 47]] (* Jean-François Alcover, Jun 01 2011, after formula *)
  • Sage
    # uses[riordan_square from A321620]
    riordan_square(-ln(1 - x), 10, True) # Peter Luschny, Jan 03 2019

Formula

E.g.f.: d/dt(-log(1-t)/(1-t)^x). - Vladeta Jovovic, Oct 12 2003
The e.g.f. with offset 1: y = x + (1 + 2*t)*x^2/2! + (2 + 6*t + 3*t^2)*x^3/3! + ... has series reversion with respect to x equal to y - (1 + 2*t)*y^2/2! + (1 + 3*t)^2*y^3/3! - (1 + 4*t)^3*y^4/4! + .... This is an e.g.f. for a signed version of A139526. - Peter Bala, Jul 18 2013
Recurrence: T(n, k) = 0 if n < k; if k = 0 then T(0, 0) = 1 and T(n, 0) = n * T(n-1, 0) for n >= 1, otherwise T(n, k) = n*T(n-1, k) + ((k+1)/k)*T(n-1, k-1). From the unsigned Stirling1 recurrence. - Wolfdieter Lang, Nov 25 2018

Extensions

Edited by Wolfdieter Lang, Nov 23 2018

A033487 a(n) = n*(n+1)*(n+2)*(n+3)/4.

Original entry on oeis.org

0, 6, 30, 90, 210, 420, 756, 1260, 1980, 2970, 4290, 6006, 8190, 10920, 14280, 18360, 23256, 29070, 35910, 43890, 53130, 63756, 75900, 89700, 105300, 122850, 142506, 164430, 188790, 215760, 245520, 278256, 314160, 353430, 396270, 442890, 493506, 548340, 607620
Offset: 0

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Author

Keywords

Comments

Non-vanishing diagonal of (A132440)^4/4. Third subdiagonal of unsigned A238363 without the zero. Cf. A130534 for relations to colored forests, disposition of flags on flagpoles, and colorings of the vertices of the complete graph K_4. - Tom Copeland, Apr 05 2014
Total number of pips on a set of trominoes (3-armed dominoes) with up to n pips on each arm. - Alan Shore and N. J. A. Sloane, Jan 06 2016
Also the number of minimum connected dominating sets in the (n+2)-crown graph. - Eric W. Weisstein, Jun 29 2017
Crossing number of the (n+3)-cocktail party graph (conjectured). - Eric W. Weisstein, Apr 29 2019
Sum of all numbers in ordered triples (x,y,z) where 0 <= x <= y <= z <= n. - Edward Krogius, Jul 31 2022

Examples

			G.f. = 6*x + 30*x^2 + 90*x^3 + 210*x^4 + 420*x^5 + 756*x^6 + 1260*x^7 + ...
		

References

  • J. Riordan, Combinatorial Identities, Wiley, 1968, p. 77.

Crossrefs

Partial sums of A007531.
A row of the array in A129533.
A column of the triangle in A331430.
Sequences of the form binomial(n+k,k)*binomial(n+k+2,k): A000012 (k=0), A005563 (k=1), this sequence (k=2), A027790 (k=3), A107395 (k=4), A107396 (k=5), A107397 (k=6), A107398 (k=7), A107399 (k=8).

Programs

Formula

From Antonio G. Astudillo (afg_astudillo(AT)hotmail.com), Jun 10 2001: (Start)
G.f.: 6*x/(1-x)^5.
a(n) = 6*binomial(n+3, 4) = 6*A000332(n+3).
a(n) = a(n-1) + A007531(n+1).
a(n) = Sum_{i=0..n} i*(i+1)*(i+2). (End)
Constant term in Bessel polynomial {y_n(x)}''.
a(n) = binomial(n+1,2)*binomial(n+3,2) = A000217(n)*A000217(n+2). - Zerinvary Lajos, May 25 2005
a(n) = binomial(n+2,2)^2 - binomial(n+2,2). - Zerinvary Lajos, May 17 2006
From Zerinvary Lajos, May 11 2007: (Start)
a(n-1) = Sum_{j=1..n} Sum_{i=2..n} i*j.
a(n) = Sum_{j=1..n} j*(n+2)*(n-1)/2. (End)
Sum_{n>0} 1/a(n) = 2/9. - Enrique Pérez Herrero, Nov 10 2013
a(-3-n) = a(n) = 2 * binomial(binomial(n+2, 2), 2). - Michael Somos, Apr 06 2014
a(n) = A002378(binomial(n+2,2)-1). - Salvador Cerdá, Nov 04 2016
a(n) = Sum_{k=0..n} A007531(k+2). See Proof Without Words link. - Michel Marcus, Oct 29 2021
Sum_{n>=1} (-1)^(n+1)/a(n) = 16*log(2)/3 - 32/9. - Amiram Eldar, Nov 02 2021
E.g.f.: exp(x)*x*(24 + 36*x + 12*x^2 + x^3)/4. - Stefano Spezia, Jul 03 2025

A104712 Pascal's triangle, with the first two columns removed.

Original entry on oeis.org

1, 3, 1, 6, 4, 1, 10, 10, 5, 1, 15, 20, 15, 6, 1, 21, 35, 35, 21, 7, 1, 28, 56, 70, 56, 28, 8, 1, 36, 84, 126, 126, 84, 36, 9, 1, 45, 120, 210, 252, 210, 120, 45, 10, 1, 55, 165, 330, 462, 462, 330, 165, 55, 11, 1, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1, 78, 286, 715
Offset: 2

Views

Author

Gary W. Adamson, Mar 19 2005

Keywords

Comments

A000295 (Eulerian numbers) gives the row sums.
Write A004736 and Pascal's triangle as infinite lower triangular matrices A and B; then A*B is this triangle.
From Peter Luschny, Apr 10 2011: (Start)
A slight variation has a combinatorial interpretation: remove the last column and the second one from Pascal's triangle. Let P(m, k) denote the set partitions of {1,2,..,n} with the following properties:
(a) Each partition has at least one singleton block;
(c) k is the size of the largest block of the partition;
(b) m = n - k + 1 is the number of parts of the partition.
Then A000295(n) = Sum_{k=1..n} card(P(n-k+1,k)).
For instance, A000295(4) = P(4,1) + P(3,2) + P(2,3) + P(1,4) = card({1|2|3|4}) + card({1|2|34, 1|3|24,1|4|23, 2|3|14, 2|4|13, 3|4|12}) + card({1|234, 2|134, 3|124, 4|123}) = 1 + 6 + 4 = 11.
This interpretation can be superimposed on the sequence by changing the offset to 1 and adding the value 1 in front. The triangle then starts
1;
1, 3;
1, 6, 4;
1, 10, 10, 5;
1, 15, 20, 15, 6;
...
(End)
Diagonal sums are A001924(n+1). - Philippe Deléham, Jan 11 2014
Relation to K-theory: T acting on the column vector (d,-d^2,d^3,...) generates the Euler classes for a hypersurface of degree d in CP^n. Cf. Dugger p. 168, A111492, A238363, and A135278. - Tom Copeland, Apr 11 2014

Examples

			The triangle a(n, k) begins:
  n\k  2   3   4    5    6    7    8   9  10 11 12 13
  2:   1
  3:   3   1
  4:   6   4   1
  5:  10  10   5    1
  6:  15  20  15    6    1
  7:  21  35  35   21    7    1
  8:  28  56  70   56   28    8    1
  9:  36  84 126  126   84   36    9   1
  10: 45 120 210  252  210  120   45  10   1
  11: 55 165 330  462  462  330  165  55  11  1
  12: 66 220 495  792  924  792  495 220  66 12  1
  13: 78 286 715 1287 1716 1716 1287 715 286 78 13  1
... reformatted. - _Wolfdieter Lang_, Mar 20 2015
		

Crossrefs

Cf. A000295, A007318, A008292, A104713, A027641/A027642 (first Bernoulli numbers B-), A164555/A027642 (second Bernoulli numbers B+), A176327/A176289.

Programs

  • Magma
    /* As triangle */ [[Binomial(n, k): k in [2..n]]: n in [2..10]]; // G. C. Greubel, May 15 2018
  • Mathematica
    t[n_, k_] := Binomial[n, k]; Table[ t[n, k], {n, 2, 13}, {k, 2, n}] // Flatten (* Robert G. Wilson v, Apr 16 2011 *)
  • PARI
    for(n=2, 10, for(k=2,n, print1(binomial(n,k), ", "))) \\ G. C. Greubel, May 15 2018
    

Formula

T(n,k) = binomial(n,k), for 2 <= k <= n.
From Peter Bala, Jul 16 2013: (Start)
The following remarks assume an offset of 0.
Riordan array (1/(1 - x)^3, x/(1 - x)).
O.g.f.: 1/(1 - t)^2*1/(1 - (1 + x)*t) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2 + ....
E.g.f.: (1/x*d/dt)^2 (exp(t)*(exp(x*t) - 1 - x*t)) = 1 + (3 + x)*t + (6 + 4*x + x^2)*t^2/2! + ....
The infinitesimal generator for this triangle has the sequence [3,4,5,...] on the main subdiagonal and 0's elsewhere. (End)
As triangle T(n,k), 0<=k<=n: T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 3*T(n-2,k) - 2*T(n-2,k-1) + T(n-3,k) + T(n-3,k-1), T(0,0)=1, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Jan 11 2014
From Tom Copeland, Apr 11 2014: (Start)
A) The infinitesimal generator for this matrix is given in A132681 with m=2. See that entry for numerous relations to differential operators and the Laguerre polynomials of order m=2, i.e., Lag(n,t,2) = Sum_{j=0..n} binomial(n+2,n-j)*(-t)^j/j!.
B) O.g.f.: 1 / { [ 1 - t * x/(1-x) ] * (1-x)^3 }
C) O.g.f. of row e.g.f.s: exp[t*x/(1-x)]/(1-x)^3 = [Sum_{n>=0} x^n * Lag(n,-t,2)] = 1 + (3 + t)*x + (6 + 4t + t^2/2!)*x^2 + (10 + 10t + 5t^2/2! + t^3/3!)*x^3 + ....
D) E.g.f. of row o.g.f.s: [(1+t)*exp((1+t)*x) - (1+t+t*x)exp(x)]/t^2. (End)
O.g.f. for m-th row (m=n-2): [(1+x)^(m+2)-(1+(m+2)*x)]/x^2. - Tom Copeland, Apr 16 2014
Reverse T = [St2]*dP*[St1]- dP = [St2]*(exp(x*M)-I)*[St1]-(exp(x*M)-I) with top two rows of zeros removed, [St1]=padded A008275 just as [St2]=A048993=padded A008277, dP= A132440, M=A238385-I, and I=identity matrix. Cf. A238363. - Tom Copeland, Apr 26 2014
O.g.f. of column k (with k leading zeros): (x^k)/(1-x)^(k+1), k >= 2. - Wolfdieter Lang, Mar 20 2015

Extensions

Edited and extended by David Wasserman, Jul 03 2007

A208535 Square array read by descending antidiagonals: T(n,k) is the number of n-bead necklaces of k colors not allowing reversal, with no adjacent beads having the same color (n, k >= 1).

Original entry on oeis.org

1, 2, 0, 3, 1, 0, 4, 3, 0, 0, 5, 6, 2, 1, 0, 6, 10, 8, 6, 0, 0, 7, 15, 20, 24, 6, 1, 0, 8, 21, 40, 70, 48, 14, 0, 0, 9, 28, 70, 165, 204, 130, 18, 1, 0, 10, 36, 112, 336, 624, 700, 312, 36, 0, 0, 11, 45, 168, 616, 1554, 2635, 2340, 834, 58, 1, 0, 12, 55, 240, 1044, 3360, 7826, 11160
Offset: 1

Views

Author

R. H. Hardin, Feb 27 2012

Keywords

Comments

For prime rows, these appear to be evaluations of Moreau's necklace polynomials at the integers with several combinatorial interpretations (see Wikipedia link). - Tom Copeland, Oct 20 2014
From Petros Hadjicostas, Nov 05 2017: (Start)
The g.f. for column k follows easily from I. Gessel's formulas for this sequence. Since S(1,k) = k-1, we have T(1,k) = k + S(1,k) - (k - 1). Thus, Sum_{n >= 1} T(n,k)*x^n = k*x + Sum_{n >= 1} (1/n)*Sum_{d|n} (k - 1)^d*phi(n/d)*x^n - Sum_{s=0} (k-1)*x^{2*s+1}. Letting m = n/d, we get that (column k g.f.) = k*x - (k - 1)*x/(1 -x^2) + Sum_{m >= 1} (phi(m)/m)*Sum_{d >= 1}((k - 1)*x^m)^d/d. But Sum_{d>=1} z^d/d = -log(1 - z), and so (column k g.f.) = k*x - (k - 1)*x/(1 - x^2) - Sum_{m >= 1} (phi(m)/m)*log(1 - (k - 1)*x^m).
The other formula for the g.f. of column k of this sequence follows from the formula Sum_{n >= 1} (phi(n)/n)*log(1 + t^n) = t/(1 - t^2), which in turn follows from the well-known series Sum_{n >= 1} phi(n)*t^n/(1 + t^n) = t*(1 + t^2)/(1 - t^2)^2.
The extra term k*x in the g.f. for column k is due to the fact that we conventionally assume that a necklace with only one bead, colored with one of the k colors available, is such that there are "no adjacent beads having the same color" (even though, strictly speaking, a single bead is adjacent to itself when we go around the circle of the necklace).
One can use the g.f. for column k to derive the so-called "Empirical for row n" formulae that are denoted by a(k) and given in the formula section below (from n = 1 to n = 7). For example, for n = 3, a(k) = a(k, x=0), where a(k, x) = (1/3!)*d^3/dx^3 (column k g.f.). Here, d^3/dx^3 stands for "third derivative w.r.t. x". If we let f(x) = x/(1 - x^2) and g(x, m) = -log(1 - (k - 1)*x^m), then f^{(3)}(0) = 6, while g^{(3)}(0,m) = 2*(k - 1)^3 for m = 1, 0 for m=2, 6*(k - 1) for m = 3, and 0 for m >= 4. Then, a(k) = (1/6)*(-6*(k - 1) + 2*(k - 1)^3 + (2/3)*6*(k - 1)) = (1/3)*k^3 - k^2 + (2/3)*k. Using this method, one can derive an "Empirical for row n" formula for a(k) for any positive integer n. (End)

Examples

			Table T(n,k) (with rows n >= 1 and columns k >= 1) starts:
  1 2  3   4    5     6      7      8       9      10       11       12       13 ...
  0 1  3   6   10    15     21     28      36      45       55       66       78 ...
  0 0  2   8   20    40     70    112     168     240      330      440      572 ...
  0 1  6  24   70   165    336    616    1044    1665     2530     3696     5226 ...
  0 0  6  48  204   624   1554   3360    6552   11808    19998    32208    49764 ...
  0 1 14 130  700  2635   7826  19684   43800   88725   166870   295526   498004 ...
  0 0 18 312 2340 11160  39990 117648  299592  683280  1428570  2783880  5118828 ...
  0 1 36 834 8230 48915 210126 720916 2097684 5381685 12501280 26796726 53750346 ...
  ...
All solutions for n = 4 and k = 3:
  1    2    1    1    1    1
  3    3    2    2    3    2
  2    2    3    1    1    1
  3    3    2    2    3    3
		

Crossrefs

Columns 3..6: A106365, A106366, A106367, A106368.
Rows 2..7: A000217(n-1), A007290, A006528(n-1), A208536, A006565(n-1), A208537.

Programs

  • Mathematica
    T[n_, k_] := If[n == 1, k, Sum[ EulerPhi[n/d]*(k-1)^d, {d, Divisors[n]}]/n - If[OddQ[n], k-1, 0]]; Table[T[n-k+1, k], {n, 1, 12}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Oct 31 2017, after Andrew Howroyd *)
  • PARI
    T(n,k) = if(n==1, k, sumdiv(n,d,eulerphi(n/d)*(k-1)^d)/n - if(n%2, k-1));
    for(n=1, 10, for(k=1, 10, print1(T(n,k), ", ")); print) \\ Andrew Howroyd, Oct 14 2017

Formula

Let S(n,k) = (1/n) Sum_{d|n} (k-1)^d phi(n/d), where phi is Euler's function.
Then for n even, T(n,k) = S(n,k) and for n > 1 and odd, T(n,k) = S(n,k) - (k-1), and T(1,k) = k. - Ira M. Gessel, Oct 21 2014, Sep 25 2017
Empirical for row n:
n=1: a(k) = k
n=2: a(k) = (1/2)*k^2 - (1/2)*k
n=3: a(k) = (1/3)*k^3 - k^2 + (2/3)*k
n=4: a(k) = (1/4)*k^4 - k^3 + (7/4)*k^2 - k
n=5: a(k) = (1/5)*k^5 - k^4 + 2*k^3 - 2*k^2 + (4/5)*k
n=6: a(k) = (1/6)*k^6 - k^5 + (5/2)*k^4 - (19/6)*k^3 + (7/3)*k^2 - (5/6)*k
n=7: a(k) = (1/7)*k^7 - k^6 + 3*k^5 - 5*k^4 + 5*k^3 - 3*k^2 + (6/7)*k
-----------
From Tom Copeland, Oct 20 2014: (Start)
The first three numbers in each row of the triangular array are given by T(n,k) = (1/k)*(n-k+1)! / (n-2*k+1)!.
For the table here, the first three rows, aside from initial zeros, are given by a(n,k) = (1/n)*(k + 1 - n)! / (k + 1 - 2*n)! or, with no leading zeros, by a(n,k) = (1/n)*(n+k-1)! / (k-1)!. The first three elements of each column correspond to the last three elements of a row in A238363 and the first three of A111492.
Prime rows (> 1) appear to be a(m,n) = (n^m - n)/m. See Wikipedia link. (End)
G.f. for column k: Sum_{n >= 1} T(n,k)*x^n = k*x - Sum_{n >= 1} (phi(n)/n)*((k - 1)*log(1 + x^n) + log(1 - (k - 1)*x^n)) = k*x - (k - 1)*x/(1 - x^2) - Sum_{n >= 1} (phi(n)/n)*log(1 - (k - 1)*x^n). - Petros Hadjicostas, Nov 05 2017

Extensions

Name edited by Petros Hadjicostas, Jun 24 2020

A055137 Regard triangle of rencontres numbers (see A008290) as infinite matrix, compute inverse, read by rows.

Original entry on oeis.org

1, 0, 1, -1, 0, 1, -2, -3, 0, 1, -3, -8, -6, 0, 1, -4, -15, -20, -10, 0, 1, -5, -24, -45, -40, -15, 0, 1, -6, -35, -84, -105, -70, -21, 0, 1, -7, -48, -140, -224, -210, -112, -28, 0, 1, -8, -63, -216, -420, -504, -378, -168, -36, 0, 1, -9, -80, -315, -720
Offset: 0

Views

Author

Christian G. Bower, Apr 25 2000

Keywords

Comments

The n-th row consists of coefficients of the characteristic polynomial of the adjacency matrix of the complete n-graph.
Triangle of coefficients of det(M(n)) where M(n) is the n X n matrix m(i,j)=x if i=j, m(i,j)=i/j otherwise. - Benoit Cloitre, Feb 01 2003
T is an example of the group of matrices outlined in the table in A132382--the associated matrix for rB(0,1). The e.g.f. for the row polynomials is exp(x*t) * exp(x) *(1-x). T(n,k) = Binomial(n,k)* s(n-k) where s = (1,0,-1,-2,-3,...) with an e.g.f. of exp(x)*(1-x) which is the reciprocal of the e.g.f. of A000166. - Tom Copeland, Sep 10 2008
Row sums are: {1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, ...}. - Roger L. Bagula, Feb 20 2009
T is related to an operational calculus connecting an infinitesimal generator for fractional integro-derivatives with the values of the Riemann zeta function at positive integers (see MathOverflow links). - Tom Copeland, Nov 02 2012
The submatrix below the null subdiagonal is signed and row reversed A127717. The submatrix below the diagonal is A074909(n,k)s(n-k) where s(n)= -n, i.e., multiply the n-th diagonal by -n. A074909 and its reverse A135278 have several combinatorial interpretations. - Tom Copeland, Nov 04 2012
T(n,k) is the difference between the number of even (A145224) and odd (A145225) permutations (of an n-set) with exactly k fixed points. - Julian Hatfield Iacoponi, Aug 08 2024

Examples

			1; 0,1; -1,0,1; -2,-3,0,1; -3,-8,-6,0,1; ...
(Bagula's matrix has a different sign convention from the list.)
From _Roger L. Bagula_, Feb 20 2009: (Start)
  { 1},
  { 0,   1},
  {-1,   0,    1},
  { 2,  -3,    0,    1},
  {-3,   8,   -6,    0,     1},
  { 4, -15,   20,  -10,     0,    1},
  {-5,  24,  -45,   40,   -15,    0,    1},
  { 6, -35,   84, -105,    70,  -21,    0,   1},
  {-7,  48, -140,  224,  -210,  112,  -28,   0,   1},
  { 8, -63,  216, -420,   504, -378,  168, -36,   0, 1},
  {-9,  80, -315,  720, -1050, 1008, -630, 240, -45, 0, 1}
(End)
R(3,x) = (-1)^3*Sum_{permutations p in S_3} sign(p)*(-x)^(fix(p)).
    p   | fix(p) | sign(p) | (-1)^3*sign(p)*(-x)^fix(p)
========+========+=========+===========================
  (123) |    3   |    +1   |      x^3
  (132) |    1   |    -1   |       -x
  (213) |    1   |    -1   |       -x
  (231) |    0   |    +1   |       -1
  (312) |    0   |    +1   |       -1
  (321) |    1   |    -1   |       -x
========+========+=========+===========================
                           | R(3,x) = x^3 - 3*x - 2
- _Peter Bala_, Aug 08 2011
		

References

  • Norman Biggs, Algebraic Graph Theory, 2nd ed. Cambridge University Press, 1993. p. 17.
  • J. Riordan, An Introduction to Combinatorial Analysis, Wiley, 1958, p.184 problem 3.

Crossrefs

Cf. A005563, A005564 (absolute values of columns 1, 2).
Cf. A000312.

Programs

  • Mathematica
    M[n_] := Table[If[i == j, x, 1], {i, 1, n}, {j, 1, n}]; a = Join[{{1}}, Flatten[Table[CoefficientList[Det[M[n]], x], {n, 1, 10}]]] (* Roger L. Bagula, Feb 20 2009 *)
    t[n_, k_] := (k-n+1)*Binomial[n, k]; Table[t[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Nov 29 2013, after Pari *)
  • PARI
    T(n,k)=(1-n+k)*if(k<0 || k>n,0,n!/k!/(n-k)!)

Formula

G.f.: (x-n+1)*(x+1)^(n-1) = Sum_(k=0..n) T(n,k) x^k.
T(n, k) = (1-n+k)*binomial(n, k).
k-th column has o.g.f. x^k(1-(k+2)x)/(1-x)^(k+2). k-th row gives coefficients of (x-k)(x+1)^k. - Paul Barry, Jan 25 2004
T(n,k) = Coefficientslist[Det[Table[If[i == j, x, 1], {i, 1, n}, {k, 1, n}],x]. - Roger L. Bagula, Feb 20 2009
From Peter Bala, Aug 08 2011: (Start)
Given a permutation p belonging to the symmetric group S_n, let fix(p) be the number of fixed points of p and sign(p) its parity. The row polynomials R(n,x) have a combinatorial interpretation as R(n,x) = (-1)^n*Sum_{permutations p in S_n} sign(p)*(-x)^(fix(p)). An example is given below.
Note: The polynomials P(n,x) = Sum_{permutations p in S_n} x^(fix(p)) are the row polynomials of the rencontres numbers A008290. The integral results Integral_{x = 0..n} R(n,x) dx = n/(n+1) = Integral_{x = 0..-1} R(n,x) dx lead to the identities Sum_{p in S_n} sign(p)*(-n)^(1 + fix(p))/(1 + fix(p)) = (-1)^(n+1)*n/(n+1) = Sum_{p in S_n} sign(p)/(1 + fix(p)). The latter equality was Problem B6 in the 66th William Lowell Putnam Mathematical Competition 2005. (End)
From Tom Copeland, Jul 26 2017: (Start)
The e.g.f. in Copeland's 2008 comment implies this entry is an Appell sequence of polynomials P(n,x) with lowering and raising operators L = d/dx and R = x + d/dL log[exp(L)(1-L)] = x+1 - 1/(1-L) = x - L - L^2 - ... such that L P(n,x) = n P(n-1,x) and R P(n,x) = P(n+1,x).
P(n,x) = (1-L) exp(L) x^n = (1-L) (x+1)^n = (x+1)^n - n (x+1)^(n-1) = (x+1-n)(x+1)^(n-1) = (x+s.)^n umbrally, where (s.)^n = s_n = P(n,0).
The formalism of A133314 applies to the pair of entries A008290 and A055137.
The polynomials of this pair P_n(x) and Q_n(x) are umbral compositional inverses; i.e., P_n(Q.(x)) = x^n = Q_n(P.(x)), where, e.g., (Q.(x))^n = Q_n(x).
The exponential infinitesimal generator (infinigen) of this entry is the negated infinigen of A008290, the matrix (M) noted by Bala, related to A238363. Then e^M = [the lower triangular A008290], and e^(-M) = [the lower triangular A055137]. For more on the infinigens, see A238385. (End)
From the row g.f.s corresponding to Bagula's matrix example below, the n-th row polynomial has a zero of multiplicity n-1 at x = 1 and a zero at x = -n+1. Since this is an Appell sequence dP_n(x)/dx = n P_{n-1}(x), the critical points of P_n(x) have the same abscissas as the zeros of P_{n-1}(x); therefore, x = 1 is an inflection point for the polynomials of degree > 2 with P_n(1) = 0, and the one local extremum of P_n has the abscissa x = -n + 2 with the value (-n+1)^{n-1}, signed values of A000312. - Tom Copeland, Nov 15 2019
From Julian Hatfield Iacoponi, Aug 08 2024: (Start)
T(n,k) = A145224(n,k) - A145225(n,k).
T(n,k) = binomial(n,k)*(A003221(n-k)-A000387(n-k)). (End)

Extensions

Additional comments from Michael Somos, Jul 04 2002

A111492 Triangle read by rows: a(n,k) = (k-1)! * C(n,k).

Original entry on oeis.org

1, 2, 1, 3, 3, 2, 4, 6, 8, 6, 5, 10, 20, 30, 24, 6, 15, 40, 90, 144, 120, 7, 21, 70, 210, 504, 840, 720, 8, 28, 112, 420, 1344, 3360, 5760, 5040, 9, 36, 168, 756, 3024, 10080, 25920, 45360, 40320, 10, 45, 240, 1260, 6048, 25200, 86400, 226800, 403200, 362880
Offset: 1

Views

Author

Ross La Haye, Nov 15 2005

Keywords

Comments

For k > 1, a(n,k) = the number of permutations of the symmetric group S_n that are pure k-cycles.
Reverse signed array is A238363. For a relation to (Cauchy-Euler) derivatives of the Vandermonde determinant, see Chervov link. - Tom Copeland, Apr 10 2014
Dividing the k-th column of T by (k-1)! for each column generates A135278 (the f-vectors, or face-vectors for the n-simplices). Then ignoring the first column gives A104712, so T acting on the column vector (-0,d,-d^2/2!,d^3/3!,...) gives the Euler classes for hypersurfaces of degree d in CP^n. Cf. A104712 and Dugger link therein. - Tom Copeland, Apr 11 2014
With initial i,j,n=1, given the n X n Vandermonde matrix V_n(x_1,...,x_n) with elements a(i=row,j=column)=(x_j)^(i-1), its determinant |V_n|, and the column vector of n ones C=(1,1,...,1), the n-th row of the lower triangular matrix T is given by the column vector determined by (1/|V_n|) * V_n(:x_1*d/dx_1:,...,:x_n*d/dx_n:)|V_n| * C, where :x_j*d/dx_j:^n = (x_j)^n*(d/dx_j)^n. - Tom Copeland, May 20 2014
For some other combinatorial interpretations of the first three columns of T, see A208535 and the link to necklace polynomials therein. Because of the simple relation of the array to the Pascal triangle, it can easily be related to many other arrays, e.g., T(p,k)/(p*(k-1)!) with p prime gives the prime rows of A185158 and A051168 when the non-integers are rounded to 0. - Tom Copeland, Oct 23 2014

Examples

			a(3,3) = 2 because (3-1)!C(3,3) = 2.
1;
2 1;
3 3 2;
4 6 8 6;
5 10 20 30 24;
6 15 40 90 144 120;
7 21 70 210 504 840 720;
8 28 112 420 1344 3360 5760 5040;
9 36 168 756 3024 10080 25920 45360 40320;
		

Programs

  • Magma
    /* As triangle: */ [[Factorial(k-1)*Binomial(n,k): k in [1..n]]: n in [1.. 15]]; // Vincenzo Librandi, Oct 21 2014
  • Mathematica
    Flatten[Table[(k - 1)!Binomial[n, k], {n, 10}, {k, n}]]

Formula

a(n, k) = (k-1)!C(n, k) = P(n, k)/k.
E.g.f. (by columns) = exp(x)((x^k)/k).
a(n, 1) = A000027(n);
a(n, 2) = A000217(n-1);
a(n, 3) = A007290(n);
a(n, 4) = A033487(n-3).
a(n, n) = A000142(n-1);
a(n, n-1) = A001048(n-1) for n > 1.
Sum[a(n, k), {k, 1, n}] = A002104(n);
Sum[a(n, k), {k, 2, n}] = A006231(n).
a(n,k) = sum(j=k..n-1, j!/(j-k)!) (cf. Chervov link). - Tom Copeland, Apr 10 2014
From Tom Copeland, Apr 28 2014: (Start)
E.g.f. by row: [(1+t)^n-1]/t.
E.g.f. of row e.g.f.s: {exp[(1+t)*x]-exp(x)}/t.
O.g.f. of row e.g.f.s: {1/[1-(1+t)*x] - 1/(1-x)}/t.
E.g.f. of row o.g.f.s: -exp(x) * log(1-t*x). (End)

A092271 Triangle read by rows. First in a series of triangular arrays counting permutations of partitions.

Original entry on oeis.org

1, 1, 1, 2, 3, 1, 6, 8, 6, 1, 24, 30, 20, 10, 1, 120, 144, 90, 40, 15, 1, 720, 840, 504, 210, 70, 21, 1, 5040, 5760, 3360, 1344, 420, 112, 28, 1, 40320, 45360, 25920, 10080, 3024, 756, 168, 36, 1, 362880, 403200, 226800, 86400, 25200, 6048, 1260, 240, 45, 1, 3628800, 3991680, 2217600, 831600, 237600, 55440, 11088, 1980, 330, 55, 1
Offset: 1

Views

Author

Alford Arnold, Feb 14 2004

Keywords

Comments

Generate signatures in accordance with A086141. Map to partitions in accordance with A025487. Calculate the number of permutations in accordance with Abramowitz and Stegun, p. 831 (reference M2). Display the results as illustrated by A090774. The second array is:
3
20 15
90 120 45
504 630 420 105
...
Apart from the main diagonal, appears to be the same as A211603 (see also A238363) and is related to the infinitesimal generator of A008290; if so, the off-diagonal triangle entries are given by binomial(n,k)*(n-k-1)! for n >= 2 and 0 <= k <= n-2. - Peter Bala, Feb 13 2017
Let aut(p) denote the size of the centralizer of the partition p (see A339016 for the definition). Then row(n) = [n!/aut(p) for p in P], where P are the partitions of n with largest part k and length n + 1 - k. Row sums are A121726. - Peter Luschny, Nov 19 2020

Examples

			The triangle begins:
1:    1
2:    1   1
3:    2   3  1
4:    6   8  6  1
5:   24  30 20 10  1
6:  120 144 90 40 15 1
  ...
From _Peter Luschny_, Nov 19 2020: (Start):
The combinatorial interpretation is illustrated by this computation of row 6:
6! / aut([6])                = 720 / A339033(6, 1) = 720/6   = 120 = T(6, 1)
6! / aut([5, 1])             = 720 / A339033(6, 2) = 720/5   = 144 = T(6, 2)
6! / aut([4, 1, 1])          = 720 / A339033(6, 3) = 720/8   =  90 = T(6, 3)
6! / aut([3, 1, 1, 1])       = 720 / A339033(6, 4) = 720/18  =  40 = T(6, 4)
6! / aut([2, 1, 1, 1, 1])    = 720 / A339033(6, 5) = 720/48  =  15 = T(6, 5)
6! / aut([1, 1, 1, 1, 1, 1]) = 720 / A339033(6, 6) = 720/720 =   1 = T(6, 6)
-------------------------------------------------------------------------------
                                                         Sum:  410 = A121726(6)
(End)
		

References

  • Abramowitz and Stegun, p. 831.

Crossrefs

Programs

  • Mathematica
    f[list_] :=Total[list]!/Apply[Times, list]/Apply[Times, Map[Length, Split[list]]!];Table[Append[Map[f, Select[Partitions[n], Count[#, Except[1]] == 1 &]], 1], {n,1, 10}] // Grid (* Geoffrey Critzer, Nov 07 2015 *)
  • SageMath
    def A092271(n, k):
        if n == k: return 1
        return factorial(n) // ((n + 1 - k)*factorial(k - 1))
    for n in (1..9): print(n, [A092271(n, k) for k in (1..n)])
    def A092271Row(n):
        if n == 0: return [1]
        f = factorial(n); S = []
        for k in range(n,0,-1):
            for p in Partitions(n, max_part=k, inner=[k], length=n+1-k):
                S.append(f // p.aut())
        return S
    for n in (1..9): print(A092271Row(n)) # Peter Luschny, Nov 20 2020

Extensions

More terms from Geoffrey Critzer, Nov 10 2015

A235706 (I + A132440)^3: Coefficients for normalized generalized Laguerre polynomials n!*Lag(n, 3-n, -x).

Original entry on oeis.org

1, 3, 1, 6, 6, 1, 6, 18, 9, 1, 0, 24, 36, 12, 1, 0, 0, 60, 60, 15, 1, 0, 0, 0, 120, 90, 18, 1, 0, 0, 0, 0, 210, 126, 21, 1, 0, 0, 0, 0, 0, 336, 168, 24, 1, 0, 0, 0, 0, 0, 0, 504, 216, 27, 1, 0, 0, 0, 0, 0, 0, 0, 720, 270, 30, 1
Offset: 0

Views

Author

Tom Copeland, Apr 20 2014

Keywords

Comments

The associated Laguerre polynomials n!*Lag(n,3-n,-x) are related to the rook polynomials of a rectangular 3 X n chessboard by R(3,n,x) = n!*x^n*Lag(n,3-n,-1/x), which are also the matching polynomials, or generating function of the number of k-edge matchings, of the complete bipartite graph K(m,n), or biclique (cf. Wikipedia for details).
The formulas here and below can be naturally extended with 3 replaced by any positive integer m. For m = 1 and 2, see unsigned A132013 and A132014. The formulas there can be extrapolated to apply to this matrix.

Examples

			Triangle begins:
  1;
  3,  1;
  6,  6,  1;
  6, 18,  9,  1;
  0, 24, 36, 12,  1;
  0,  0, 60, 60, 15, 1;
  ...
		

Crossrefs

Cf. A007318, A008306 for a generalization, A132013, A132014, A132440, A238363, A238385.
....................................
With 0th row: 1
n-th row: n!*Lag(n,3-n,-x)
....................................
1st: 1!*Lag(1,2,-x) = A062139(1,k,-x)
2nd: 2!*Lag(2,1,-x) = A105278(2,k,x)
3rd: 3!*Lag(3,0,-x) = A021009(3,k,-x)
4th: 4!*Lag(4,-1,-x) = A111596(4,k,-x)
5th: 5!*Lag(5,-2,-x) = cf. x^2*A062139(3,k,x)
6th: 6!*Lag(6,-3,-x) = cf. x^3*A062137(3,k,-x)
....................................
n-th row: x^(n-3)*3!*Lag(3,n-3,-x)
....................................
1st: x^(-2)*3!Lag(3,-2,-x) = cf. x^(-2)*[x^2*A062139(1,k,x)]
2nd: x^(-1)*3!Lag(3,-1,-x) = x^(-1)*A111596(3,k,-x)
3rd: x^0*3!Lag(3,0,-x) = x^0*A021009(3,k,-x)
4th: x^1*3!Lag(3,1,-x) = x^1*A105278(3,k,x)
5th: x^2*3!Lag(3,2,-x) = x^2*A062139(3,k,-x)
6th: x^3*3!Lag(3,3,-x) = x^3*A062137(3,k,-x)

Programs

  • Magma
    /* As triangle */ [[Binomial(3, n-k)*Factorial(n)/Factorial(k): k in [0..n]]: n in [0.. 15]]; // Vincenzo Librandi, Jul 28 2017
  • Mathematica
    Table[Binomial[3, n - k] n! / k!, {n, 0, 9}, {k, 0, n}]//Flatten (* Vincenzo Librandi, Jul 28 2017 *)
  • PARI
    T(n,k) = binomial(3,n-k)*n!/k!
    tabl(nn) = for (n=0, nn, for (k=0, n, print1(T(n, k), ", ")); print); \\ Michel Marcus, Jul 28 2017
    
  • PARI
    row(n) = Vecrev(n!*pollaguerre(n, 3-n, -x)); \\ Michel Marcus, Feb 06 2021
    

Formula

T(n,k) = binomial(3,n-k)*n!/k! = binomial(n,k)*3!/(3-n+k)!.
E.g.f.: exp(y*x)(1+y)^3, so this is an Appell sequence of polynomials with lowering operator L= D= d/dx and raising operator R = x + 3/(1+D).
E.g.f. of inverse matrix is exp(x*y)/(1+y)^3.
Multiply the n-th diagonal of the Pascal matrix A007318 by d(0)=1, d(1)=3, d(2)=6, d(3)=6, and d(n)=0 for n>3 to obtain T.
Row polynomials: n!*Lag(n,3-n,-x) = x^(n-3)*3!*Lag(3,n-3,-x) =
(3!/(3-n)!)*K(-n,3-n+1,-x) where K is Kummer's confluent hypergeometric function (as a limit of n+c as c tends to zero).
T = (I + A132440)^3 = exp[3*(A238385-I)]. I = identity matrix.
Operationally, n!Lag(n,3-n,-:xD:) = x^(n-3)*:Dx:^n*x^(3-n) = x^(-3)*:xD:^n*x^3 = n!*binomial(xD+3,n) = n!*binomial(3,n)*K(-n,3-n+1,-:xD:) where :AB:^n = A^n*B^n for any two operators.
n-th row polynomial: n!*Sum_{k = 0..n} (-1)^(n-k)*binomial(n,k)*Lag(k,3,-x). - Peter Bala, Jul 25 2021

A211603 Triangular array read by rows: T(n,k) is the number of n-permutations that are pure cycles having exactly k fixed points; n>=2, 0<=k<=n-2.

Original entry on oeis.org

1, 2, 3, 6, 8, 6, 24, 30, 20, 10, 120, 144, 90, 40, 15, 720, 840, 504, 210, 70, 21, 5040, 5760, 3360, 1344, 420, 112, 28, 40320, 45360, 25920, 10080, 3024, 756, 168, 36, 362880, 403200, 226800, 86400, 25200, 6048, 1260, 240, 45, 3628800, 3991680, 2217600, 831600, 237600, 55440, 11088, 1980, 330, 55
Offset: 2

Views

Author

Geoffrey Critzer, Feb 10 2013

Keywords

Comments

Equivalently, T(n,k) is the number of n-permutations that are pure cycles of length n-k.
Row sums = A006231.
With a different row and column indexing, this triangle equals the infinitesimal generator of A008290. Equals the unsigned version of A238363, omitting its main diagonal. See also A092271. - Peter Bala, Feb 13 2017

Examples

			T(3,1) = 3 because we have (1)(2,3), (2)(1,3), (3)(1,2).
1;
2, 3;
6, 8, 6;
24, 30, 20, 10;
120, 144, 90, 40, 15;
720, 840, 504, 210, 70, 21;
5040, 5760, 3360, 1344, 420, 112, 28;
40320, 45360, 25920, 10080, 3024, 756, 168, 36;
362880, 403200, 226800, 86400, 25200, 6048, 1260, 240, 45;
		

Crossrefs

Cf. A006231 (row sums), A008290, A092271, A111492, A238363.

Programs

  • Maple
    T:= (n, k)-> binomial(n, k)*(n-k-1)!:
    seq(seq(T(n,k), k=0..n-2), n=2..12);  # Alois P. Heinz, Feb 10 2013
  • Mathematica
    nn=10;f[list_]:=Select[list,#>0&];Map[f,Range[0,nn]!CoefficientList[ Series[Exp[y x](Log[1/(1-x)]-x),{x,0,nn}],{x,y}]]//Grid

Formula

E.g.f.: exp(y*x)*(log(1/(1-x))-x).
T(n,k) = C(n,k)*(n-k-1)!. - Alois P. Heinz, Feb 10 2013
T(n,k) = A111492(n,n-k). - R. J. Mathar, Mar 07 2013

A238474 a(n) = (-1)^n*(n+3)!/(2*(n+1)) for n >= 0.

Original entry on oeis.org

3, -6, 20, -90, 504, -3360, 25920, -226800, 2217600, -23950080, 283046400, -3632428800, 50295168000, -747242496000, 11856247603200, -200074178304000, 3577797070848000, -67580611338240000, 1344498478202880000, -28100018194440192000, 615524208068689920000
Offset: 0

Views

Author

Wolfdieter Lang, Mar 01 2014

Keywords

Comments

Third column of Sheffer type triangle A238363 (without leading zeros).

Crossrefs

Cf. A238363.

Programs

  • Mathematica
    Fold[Append[#1, (-1)^#2 (#2 + 3)!/(2 (#2 + 1))] &, {3}, Range@ 20] (* Michael De Vlieger, Dec 18 2017 *)

Formula

a(n) = (-1)^n*(n+3)!/(2*(n+1)) for n >= 0.
E.g.f.: (d^3/dx^3)(log(1+x)*x^2/2!) = (x^2+3*x+3)/(1+x)^3.
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