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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A319226 Irregular triangle where T(n,k) is the number of acyclic spanning subgraphs of a cycle graph, where the sizes of the connected components are given by the integer partition with Heinz number A215366(n,k).

Original entry on oeis.org

1, 2, 1, 3, 3, 1, 4, 2, 4, 4, 1, 5, 5, 5, 5, 5, 5, 1, 6, 6, 6, 3, 2, 6, 12, 9, 6, 6, 1, 7, 7, 7, 7, 14, 7, 7, 7, 7, 7, 21, 14, 7, 7, 1, 8, 8, 8, 4, 8, 8, 8, 16, 16, 8, 2, 24, 8, 24, 12, 16, 8, 32, 20, 8, 8, 1, 9, 9, 9, 9, 9, 9, 18, 9, 9, 9, 18, 18, 3, 27, 27
Offset: 1

Views

Author

Gus Wiseman, Sep 13 2018

Keywords

Comments

A refinement of A135278, up the sign these are the coefficients appearing in the expansion of power-sum symmetric functions in terms of elementary or homogeneous symmetric functions.

Examples

			Triangle begins:
  1
  2  1
  3  3  1
  4  2  4  4  1
  5  5  5  5  5  5  1
  6  6  6  3  2  6 12  9  6  6  1
The fourth row corresponds to the symmetric function identities:
  p(4) = -4 e(4) + 2 e(22) + 4 e(31) - 4 e(211) + e(1111)
  p(4) =  4 h(4) - 2 h(22) - 4 h(31) + 4 h(211) - h(1111).
		

Crossrefs

Signed versions with different row-orderings are A115131, A210258, A263916.

Programs

  • Mathematica
    primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]];
    csm[s_]:=With[{c=Select[Tuples[Range[Length[s]],2],And[OrderedQ[#],UnsameQ@@#,Length[Intersection@@s[[#]]]>0]&]},If[c=={},s,csm[Union[Append[Delete[s,List/@c[[1]]],Union@@s[[c[[1]]]]]]]]];
    Table[Length[Select[Subsets[Partition[Range[n],2,1,1],{n-PrimeOmega[m]}],Sort[Length/@csm[Union[#,List/@Range[n]]]]==primeMS[m]&]],{n,6},{m,Sort[Times@@Prime/@#&/@IntegerPartitions[n]]}]

A015540 a(n) = 5*a(n-1) + 6*a(n-2), a(0) = 0, a(1) = 1.

Original entry on oeis.org

0, 1, 5, 31, 185, 1111, 6665, 39991, 239945, 1439671, 8638025, 51828151, 310968905, 1865813431, 11194880585, 67169283511, 403015701065, 2418094206391, 14508565238345, 87051391430071, 522308348580425, 3133850091482551, 18803100548895305, 112818603293371831
Offset: 0

Views

Author

Keywords

Comments

Number of walks of length n between any two distinct vertices of the complete graph K_7. Example: a(2)=5 because the walks of length 2 between the vertices A and B of the complete graph ABCDEFG are ACB, ADB, AEB, AFB and AGB. - Emeric Deutsch, Apr 01 2004
Pisano period lengths: 1, 1, 2, 2, 2, 2, 14, 2, 2, 2, 10, 2, 12, 14, 2, 2, 16, 2, 18, 2, ... - R. J. Mathar, Aug 10 2012
Sum_{i=0..m} (-1)^(m+i)*6^i, for m >= 0, gives all terms after 0. - Bruno Berselli, Aug 28 2013
The ratio a(n+1)/a(n) converges to 6 as n approaches infinity. Also A053524, A080424, A051958. - Felix P. Muga II, Mar 09 2014

Examples

			G.f. = x + 5*x^2 + 31*x^3 + 185*x^4 + 1111*x^5 + 6665*x^6 + 39991*x^7 + ...
		

Crossrefs

Partial sums are in A033116. Cf. A014987.

Programs

  • Magma
    [Floor(6^n/7-(-1)^n/7): n in [0..30]]; // Vincenzo Librandi, Jun 24 2011
    
  • Maple
    seq(round(6^n/7),n=0..25); # Mircea Merca, Dec 28 2010
  • Mathematica
    k=0; lst={k}; Do[k = 6^n-k; AppendTo[lst, k], {n, 0, 5!}];lst (* Vladimir Joseph Stephan Orlovsky, Dec 11 2008 *)
    CoefficientList[Series[x / ((1 - 6 x) (1 + x)), {x, 0, 50}], x] (* Vincenzo Librandi, Mar 26 2014 *)
    LinearRecurrence[{5,6},{0,1},30] (* Harvey P. Dale, May 12 2015 *)
  • PARI
    my(x='x+O('x^30)); concat([0], Vec(x/((1-6*x)*(1+x)))) \\ G. C. Greubel, Jan 24 2018
    
  • PARI
    a(n) = round(6^n/7); \\ Altug Alkan, Sep 05 2018
  • Sage
    [lucas_number1(n,5,-6) for n in range(21)] # Zerinvary Lajos, Apr 24 2009
    

Formula

a(n) = 5*a(n-1) + 6*a(n-2).
From Paul Barry, Apr 20 2003: (Start)
a(n) = (6^n - (-1)^n)/7.
G.f.: x/((1-6*x)*(1+x)).
E.g.f.: (exp(6*x) - exp(-x))/7. (End)
a(n) = 6^(n-1) - a(n-1). - Emeric Deutsch, Apr 01 2004
a(n+1) = Sum_{k=0..n} binomial(n-k, k)*5^(n-2*k)*6^k. - Paul Barry, Jul 29 2004
a(n) = round(6^n/7). - Mircea Merca, Dec 28 2010
a(n) = (-1)^(n-1)*Sum_{k=0..n-1} A135278(n-1,k)*(-7)^k = (6^n - (-1)^n)/7 = (-1)^(n-1)*Sum_{k=0..n-1} (-6)^k. Equals (-1)^(n-1)*Phi(n,-6), where Phi is the cyclotomic polynomial when n is an odd prime. (For n > 0.) - Tom Copeland, Apr 14 2014

A110813 A triangle of pyramidal numbers.

Original entry on oeis.org

1, 3, 1, 5, 4, 1, 7, 9, 5, 1, 9, 16, 14, 6, 1, 11, 25, 30, 20, 7, 1, 13, 36, 55, 50, 27, 8, 1, 15, 49, 91, 105, 77, 35, 9, 1, 17, 64, 140, 196, 182, 112, 44, 10, 1, 19, 81, 204, 336, 378, 294, 156, 54, 11, 1, 21, 100, 285, 540, 714, 672, 450, 210, 65, 12, 1, 23, 121, 385, 825
Offset: 0

Views

Author

Paul Barry, Aug 05 2005

Keywords

Comments

Triangle A029653 less first column. In general, the product (1/(1-x),x/(1-x))*(1+m*x,x) yields the Riordan array ((1+(m-1)x)/(1-x)^2,x/(1-x)) with general term T(n,k)=(m*n-(m-1)*k+1)*C(n+1,k+1)/(n+1). This is the reversal of the (1,m)-Pascal triangle, less its first column. - Paul Barry, Mar 01 2006
The column sequences give, for k=0..10: A005408 (odd numbers), A000290 (squares), A000330, A002415, A005585, A040977, A050486, A053347, A054333, A054334, A057788.
Linked to Chebyshev polynomials by the fact that this triangle with interpolated zeros in the rows and columns is a scaled version of A053120.
Row sums are A033484. Diagonal sums are A001911(n+1) or F(n+4)-2. Factors as (1/(1-x),x/(1-x))*(1+2x,x). Inverse is A110814 or (-1)^(n-k)*A104709.
This triangle is a subtriangle of the [2,1] Pascal triangle A029653 (omit there the first column).
Subtriangle of triangles in A029653, A131084, A208510. - Philippe Deléham, Mar 02 2012
This is the iterated partial sums triangle of A005408 (odd numbers). Such iterated partial sums of arithmetic progression sequences have been considered by Narayana Pandit (see the Mar 20 2015 comment on A000580 where the MacTutor History of Mathematics archive link and the Gottwald et al. reference, p. 338, are given). - Wolfdieter Lang, Mar 23 2015

Examples

			The number triangle T(n, k) begins
n\k  0   1   2   3    4    5    6   7   8  9 10 11
0:   1
1:   3   1
2:   5   4   1
3:   7   9   5   1
4:   9  16  14   6    1
5:  11  25  30  20    7    1
6:  13  36  55  50   27    8    1
7:  15  49  91 105   77   35    9   1
8:  17  64 140 196  182  112   44  10   1
9:  19  81 204 336  378  294  156  54  11  1
10: 21 100 285 540  714  672  450 210  65 12  1
11: 23 121 385 825 1254 1386 1122 660 275 77 13  1
... reformatted by _Wolfdieter Lang_, Mar 23 2015
As a number square S(n, k) = T(n+k, k), rows begin
  1,   1,   1,   1,   1,   1, ...
  3,   4,   5,   6,   7,   8, ...
  5,   9,  14,  20,  27,  35, ...
  7,  16,  30,  50,  77, 112, ...
  9,  25,  55, 105, 182, 294, ...
		

Crossrefs

Programs

  • Mathematica
    Table[2*Binomial[n + 1, k + 1] - Binomial[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* G. C. Greubel, Oct 19 2017 *)
  • PARI
    for(n=0,10, for(k=0,n, print1(2*binomial(n+1, k+1) - binomial(n,k), ", "))) \\ G. C. Greubel, Oct 19 2017

Formula

Number triangle T(n, k) = C(n, k)*(2n-k+1)/(k+1) = 2*C(n+1, k+1) - C(n, k); Riordan array ((1+x)/(1-x)^2, x/(1-x)); As a number square read by antidiagonals, T(n, k)=C(n+k, k)(2n+k+1)/(k+1).
Equals A007318 * an infinite bidiagonal matrix with 1's in the main diagonal and 2's in the subdiagonal. - Gary W. Adamson, Dec 01 2007
Binomial transform of an infinite lower triangular matrix with all 1's in the main diagonal, all 2's in the subdiagonal and the rest zeros. - Gary W. Adamson, Dec 12 2007
T(n,k) = 2*T(n-1,k) + T(n-1,k-1) - T(n-2,k) - T(n-2,k-1), T(0,0)=T(1,1)=1, T(1,0)=3, T(n,k)=0 if k<0 or if k>n. - Philippe Deléham, Nov 30 2013
exp(x) * e.g.f. for row n = e.g.f. for diagonal n. For example, for n = 3 we have exp(x)*(7 + 9*x + 5*x^2/2! + x^3/3!) = 7 + 16*x + 30*x^2/2! + 50*x^3/3! + 77*x^4/4! + .... The same property holds more generally for Riordan arrays of the form ( f(x), x/(1 - x) ). - Peter Bala, Dec 21 2014
T(n, k) = ps(1, 2; k, n-k) with ps(a, d; k, n) = sum(ps(a, d; k-1, j), j=0..n) and input ps(a, d; 0, j) = a + d*j. See the iterated partial sums comment from Mar 23 2015 above. - Wolfdieter Lang, Mar 23 2015
From Franck Maminirina Ramaharo, May 21 2018: (Start)
T(n,k) = coefficients in the expansion of ((x + 2)*(x + 1)^n - 2)/x.
T(n,k) = A135278(n,k) + A135278(n-1,k).
T(n,k) = A097207(n,n-k).
G.f.: (y + 1)/((y - 1)*(x*y + y - 1)).
E.g.f.: ((x + 2)*exp(x*y + y) - 2*exp(y))/x.
(End)

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

A101391 Triangle read by rows: T(n,k) is the number of compositions of n into k parts x_1, x_2, ..., x_k such that gcd(x_1,x_2,...,x_k) = 1 (1<=k<=n).

Original entry on oeis.org

1, 0, 1, 0, 2, 1, 0, 2, 3, 1, 0, 4, 6, 4, 1, 0, 2, 9, 10, 5, 1, 0, 6, 15, 20, 15, 6, 1, 0, 4, 18, 34, 35, 21, 7, 1, 0, 6, 27, 56, 70, 56, 28, 8, 1, 0, 4, 30, 80, 125, 126, 84, 36, 9, 1, 0, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1, 0, 4, 42, 154, 325, 461, 462, 330, 165, 55, 11, 1, 0, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1
Offset: 1

Views

Author

Emeric Deutsch, Jan 26 2005

Keywords

Comments

If instead we require that the individual parts (x_i,x_j) be relatively prime, we get A282748. This is the question studied by Shonhiwa (2006). - N. J. A. Sloane, Mar 05 2017.

Examples

			T(6,3)=9 because we have 411,141,114 and the six permutations of 123 (222 does not qualify).
T(8,3)=18 because binomial(0,2)*mobius(8/1)+binomial(1,2)*mobius(8/2)+binomial(3,2)*mobius(8/4)+binomial(7,2)*mobius(8/8)=0+0+(-3)+21=18.
Triangle begins:
   1;
   0,  1;
   0,  2,  1;
   0,  2,  3,   1;
   0,  4,  6,   4,   1;
   0,  2,  9,  10,   5,   1;
   0,  6, 15,  20,  15,   6,   1;
   0,  4, 18,  34,  35,  21,   7,   1;
   0,  6, 27,  56,  70,  56,  28,   8,   1;
   0,  4, 30,  80, 125, 126,  84,  36,   9,   1;
   0, 10, 45, 120, 210, 252, 210, 120,  45,  10,  1;
   0,  4, 42, 154, 325, 461, 462, 330, 165,  55, 11,  1;
   0, 12, 66, 220, 495, 792, 924, 792, 495, 220, 66, 12, 1;
  ...
From _Gus Wiseman_, Oct 19 2020: (Start)
Row n = 6 counts the following compositions:
  (15)  (114)  (1113)  (11112)  (111111)
  (51)  (123)  (1122)  (11121)
        (132)  (1131)  (11211)
        (141)  (1212)  (12111)
        (213)  (1221)  (21111)
        (231)  (1311)
        (312)  (2112)
        (321)  (2121)
        (411)  (2211)
               (3111)
Missing are: (42), (24), (33), (222).
(End)
		

Crossrefs

Mirror image of A039911.
Row sums are A000740.
A000837 counts relatively prime partitions.
A135278 counts compositions by length.
A282748 is the pairwise coprime instead of relatively prime version.
A282750 is the unordered version.
A291166 ranks these compositions (evidently).
T(2n+1,n+1) gives A000984.

Programs

  • Maple
    with(numtheory): T:=proc(n,k) local d, j, b: d:=divisors(n): for j from 1 to tau(n) do b[j]:=binomial(d[j]-1,k-1)*mobius(n/d[j]) od: sum(b[i],i=1..tau(n)) end: for n from 1 to 14 do seq(T(n,k),k=1..n) od; # yields the sequence in triangular form
    # second Maple program:
    b:= proc(n, g) option remember; `if`(n=0, `if`(g=1, 1, 0),
          expand(add(b(n-j, igcd(g, j))*x, j=1..n)))
        end:
    T:= (n, k)-> coeff(b(n,0),x,k):
    seq(seq(T(n,k), k=1..n), n=1..14);  # Alois P. Heinz, May 05 2025
  • Mathematica
    t[n_, k_] := Sum[Binomial[d-1, k-1]*MoebiusMu[n/d], {d, Divisors[n]}]; Table[t[n, k], {n, 2, 14}, {k, 2, n}] // Flatten (* Jean-François Alcover, Jan 20 2014 *)
    Table[Length[Select[Join@@Permutations/@IntegerPartitions[n,{k}],GCD@@#==1&]],{n,10},{k,2,n}] (* change {k,2,n} to {k,1,n} for the version with zeros. - Gus Wiseman, Oct 19 2020 *)
  • PARI
    T(n, k) = sumdiv(n, d, binomial(d-1, k-1)*moebius(n/d)); \\ Michel Marcus, Mar 09 2016

Formula

T(n,k) = Sum_{d|n} binomial(d-1,k-1)*mobius(n/d).
Sum_{k=1..n} k * T(n,k) = A085411(n). - Alois P. Heinz, May 05 2025

Extensions

Definition clarified by N. J. A. Sloane, Mar 05 2017
Edited by Alois P. Heinz, May 05 2025

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)

A344678 Coefficients for normal ordering of (x + D)^n and for the unsigned, probabilist's (or Chebyshev) Hermite polynomials H_n(x+y).

Original entry on oeis.org

1, 1, 1, 1, 2, 1, 1, 1, 3, 3, 3, 3, 1, 1, 4, 6, 6, 12, 4, 3, 6, 1, 1, 5, 10, 10, 30, 10, 15, 30, 5, 15, 10, 1, 1, 6, 15, 15, 60, 20, 45, 90, 15, 90, 60, 6, 15, 45, 15, 1, 1, 7, 21, 21, 105, 35, 105, 210, 35, 315, 210, 21, 105, 315, 105, 7, 105, 105, 21, 1
Offset: 0

Views

Author

Tom Copeland, May 26 2021

Keywords

Comments

This irregular triangular array contains the integer coefficients of the normal ordering of the expansions of the differential operator R = (x + D)^n with D = d/dx. This operator is the raising/creation operator for the unsigned, modified Hermite polynomials H_n(x) of A099174; i.e., R H_n(x) = H_{n+1}(x). The lowering/annihilation operator L is D; i.e, L H_n(x) = D H_n(x) = n H_{n-1}(x).
Generalizing to the ladder operators for S_n(x) a general Sheffer polynomial sequence, (L + R)^n 1 = H_n(S.(x)), where the umbral composition is defined by H_n(S.(x)) = (h. + S.(x))^n = Sum_{k = 0..n} binomial(n,k) h_{n-k} S_k(x), where h_n are the Taylor series coefficients of e^{t^2/2}; i.e., e^{h.t} = e^{t^2/2}.
Another interpretation is that the n-th row contains the coefficients of the terms in the normal ordering of the 2^n permutations of two binary symbols given by (L + R)^n under the Leibniz condition LR = RL + 1 equivalent to the commutator relation [L,R] = LR - RL = 1. For example, (x + D)^2 = xx + xD + Dx + DD = x^2 + 2 xD + 1 + D^2.
As in the examples, the coefficients are ordered as those for the terms x^k D^m, ordered first from higher k to lower k and subordinately from lower m to higher m.
The number sequences associated to these polynomials are intimately related to the complete graphs K_n, which have n vertices and (n-1) edges. H_n(x) are the independence polynomials for K_n. The moments, h_n, of the H_n(x) Appell polynomials are the aerated double factorials A001147, the number of perfect matchings in the complete graph K_{n}, zero for odd n. The row lengths, A002620, give the number of maximal strokes on the complete graph K_n. The triangular numbers--the sum of two consecutive row lengths--give the number of edges of K_n. The row sums, A005425, are the number of matchings of the corona K'_n of the complete graph K_n and the complete graph K_1. K_n can be viewed as the projection onto a plane of the edges of the regular (n-1)-dimensional simplex, whose face polynomials are (x+1)^n - 1 (cf. A135278 and A074909).
The coefficients represent the combinatorics of a switchboard problem in which among n subscribers, a subscriber may talk to one of the others, someone on an outside line, or not at all--no conference calls allowed. E.g., the coefficient 12 for H_4(x+y) is the number of ways among 4 subscribers that a pair of subscribers can talk to each other while another is on an outside line and the remaining subscriber is disconnected. The coefficient 3 for the polynomial corresponds to the number of ways two pairs of subscribers can talk among themselves. This can be regarded as a dinner table problem also where a diner may exchange positions with another diner; remain seated; or get up, change plans, and sit back down in the same chair--no more than one exchange per diner.
From Peter Luschny, May 28 2021: (Start)
If we write the monomials of the row polynomials in degree-lexicographic order, we observe: The coefficient triangle appears as a series of concatenated subtriangles. The first one is Pascal's triangle A007318. Appending the rows of triangle A094305 begins in row 2. In row 4, the next triangle starts, which is A344565. This scheme seems to go on indefinitely. [This is now set out in A344911.] (End)
From Tom Copeland, May 29 2021: (Start)
Simplex model: H_n(x+y) = (h. + x + y)^n with h_n the aerated odd double factorials (h_0 = 1, h_1 = 0, h_2 = 1, h_3 = 0, h_4 = 3, h_5 = 0, h_6 = 15, ...), the number of perfect matchings for the n vertices of the (n-1)-Dimensional simplices, i.e., the hypertriangles, or hypertetrahedrons. This multinomial enumerates permutations of three families of objects--vertices labeled with either x or y, or perfect (pair) matchings of the unlabeled vertices for the n vertices of the (n-1)-D simplex. For example, (x+D)^2 = xx + xD + Dx + D^2 = x^2 + 2xD + D^2 + 1 corresponds to H_2(x+y) = (h.+ x + y)^2 = h_0 (x+y)^2 + 2 h_1 (x+y) + h_2 = x^2 + 2xy + y^2 + 1, which, in turn, corresponds to a line segment, the 1-D simplex, with both vertices labeled with x's; or one with an x, the other a y; or both vertices labeled with y's; or one unlabeled matched pair. The exponent k of x^k y^m represents the number of these vertices to which an x is assigned, and m, those with a y assigned. The remaining unlabeled vertices (a sub-simplex) form an independent edge set of (single color) colored edges, i.e., no colored edge is touching another colored edge (a perfect matching for the sub-simplex) nor the vertices assigned an x or y. Accordingly, the coefficients for k=m=0 represent the number of ways to assign a perfect matching to the simplex--zero for simplices with an odd number of vertices and the odd double factorials (1, 3, 15, 105, ...) for the simplices with an even number. For the simplices with an odd number of vertices, the coefficients for k+m=1 are the odd double factorials. (Note the polynomials are invariant upon interchange of the variables x and y.) (End)

Examples

			(x + D)^0 = 1,
(x + D)^1 = x + D,
(x + D)^2 = x^2 + 2 x D + 1 + D^2,
(x + D)^3 = x^3 + 3 x^2 D + 3 x + 3 x D^2 + 3 D + D^3,
(x + D)^4 = x^4 + 4 x^3 D + 6 x^2 + 6 x^2 D^2 + 12 x D + 4 x D^3 + 3 + 6 D^2 + D^4.
(x + D)^5 = x^5 + 5 x^4 D + 10 x^3 + 10 x^3 D^2 + 30 x^2 D + 10 x^2 D^3 + 15 x + 30 x D^2 + 5 x D^4 + 15 D + 10 D^3 + D^5
H_6(x + y) = x^6 + 6 x^5 y + 15 x^4 + 15 x^4 y^2 + 60 x^3 y + 20 x^3 y^3 + 45 x^2 + 90 x^2 y^2 + 15 x^2 y^4 + 90 x y + 60 x y^3 + 6 x y^5 + 15 + 45 y^2 + 15 y^4 + y^6
H_7(x + y) = x^7 + 7 x^6 y + 21 x^5 + 21 x^5 y^2 + 105 x^4 y + 35 x^4 y^3 + 105 x^3 + 210 x^3 y^2 + 35 x^3 y^4 + 315 x^2 y + 210 x^2 y^3 + 21 x^2 y^5 + 105 x + 315 x y^2 + 105 x y^4 + 7 x y^6 + 105 y + 105 y^3 + 21 y^5 + y^7
		

Crossrefs

Programs

  • Mathematica
    Last /@ CoefficientRules[#, {x, y}] & /@ Table[Simplify[(-y)^n (-2)^(-n/2) HermiteH[n, (x + 1/y)/Sqrt[-2]]], {n, 0, 7}] // Flatten (* Andrey Zabolotskiy, Mar 08 2024 *)

Formula

The bivariate e.g.f. is e^{t^2/2} e^{t(x + y)} = Sum_{n >= 0} H_n(x+y) t^n/n! = e^{t H.(x+y)} = e^{t (x + H.(y))}, as described below.
The coefficient of x^k D^m is n! h_{n-k-m} / [(n-k-m)! k! m!] with 0 <= k,m <= n and (k+m) <= n with h_n, as defined in the comments, aerated A001147.
Row lengths, r(n): 1, 2, 4, 6, 9, 12, 16, 20, 25, ... A002620(n).
Row sums: A005425 = 1, 2, 5, 14, 43, 142, ... .
The recursion H_{n+1}(x+y) = (x+y) H_n(x+y) + n H_{n-1}(x+y) follows from the differential raising and lowering operations of the Hermite polynomials.
The Baker-Campbell-Hausdorff-Dynkin expansion leads to the disentangling relation e^{t (x + D)} = e^{t^2/2} e^{tx} e^{tD} from which the formula above for the coefficients may be derived via differentiation with respect to t.
The row bivariate polynomials P_n(x,y) with y a commutative analog of D, or L, have the e.g.f. e^{-xy} e^{t(x + D)} e^{xy} = e^{-xy} e^{t^2/2} e^{tx} e{tD} e^{xy} = e^{t^2/2} e^{t(x + y)} = e^{t(h. + x + y)} = e^{t (x + H.(y))} = e^{t H.(x +y)}, so P_n(x,y) = H_n(x + y) = (x + H.(y))^n, the Hermite polynomials mentioned in the comments along with the umbral composition. The row sums are H_n(2), listed in A005425. For example, P_3(x,y) = (x + H.(y))^3 = x^3 H_0(y) + 3 x^2 H_1(y) + 3 x H_2(y) + H_3(y) = H_3(x+y) = (x+y)^3 + 3(x+y) = x^3 + 3 x^2 y + 3 x + 3 x y^2 + 3 y + y^2.
Alternatively, P_n(x,y) = H_n(x+y) = (z + d/dz)^n 1 with z replaced by (x+y) after the repeated differentiations since (x + D)^n 1 = H_n(x).
With initial index 1, the lengths r(n) of the rows of nonzero coefficients are the same as those for the polynomials given by 1 + (x+y)^2 + (x+y)^4 + ... + (x+y)^n for n even and for those for (x+y)^1 + (x+y)^3 + ... + (x+y)^n for n odd since the Hermite polynomials are even or odd polynomials. Consequently, r(n)= O(n) = 1 + 2 + 4 + ... n for n odd and r(n) = E(n) = 2 + 4 + ... + n for n even, so O(n) = ((n+1)/2)^2 and E(n) = (n/2)((n/2)+1) = n(n+2)/4 = 2 T(n/2) where T(k) are the triangular numbers defined by T(k) = 0 + 1 + 2 + 3 + ... + k = A000217(k). E(n) corresponds to A002378. Additionally, r(n) + r(n+1) = 1 + 2 + 3 + ... + n+1 = T(n+1).
From Tom Copeland, May 31 2021: (Start)
e^{D^2/2} = e^{h.D}, so e^{D^2/2} x^n = e^{h. D} x^n = (h. + x)^n = H_n(x) and e^{D^2/2} (x+y)^n = e^{h. D} (x+y)^n = (h. + x + y)^n = H_n(x+y).
From the Appell Sheffer polynomial calculus, the umbral compositional inverse of the sequence H_n(x+y), i.e., the sequence HI_n(x+y) such that H_n(HI.(x+y)) = (h. + HI.(x+y))^n = (h. + hi. + x + y)^n = (x+y)^n, is determined by e^{-t^2/2} = e^{hi. t}, so hi_n = -h_n and HI_n(x+y) = (-h. + x + y)^n = (-1)^n (h. - x - y)^n = (-1)^n H_n(-(x+y)). Then H_n(-H.(-(x+y))) = (x+y)^n.
In addition, HI_n(x) = (x - D) HI_{n-1}(x) = (x - D)^n 1 = e^{-D^2/2} x^n = e^{hi. D} x^n = e^{-h. D} x^n with the e.g.f. e^{-t^2/2} e^{xt}.
The umbral compositional inversion property follows from x^n = e^{-D^2/2} e^{D^2/2} x^n = e^{-D^2/2} H_n(x) = H_n(HI.(x)) = e^{D^2/2} e^{-D^2/2} x^n = HI_n(H.(x)). (End)
The umbral relations above reveal that H_n(x+y) = (h. + x + y)^n = Sum_{k = 0..n} binomial(n,k) h_k (x+y)^{n-k}, which gives, e.g., for n = 3, H_3(x+y) = h_0 * (x+y)^3 + 3 h_1 * (x+y)^2 + 3 h_2 * (x+y) + h_3 = (x+y)^3 + 3 (x+y), the n-th through 0th rows of the Pascal matrix embedded within the n-th row of the Pascal matrix modulated by h_k. - Tom Copeland, Jun 01 2021
Varvak gives the coefficients of x^(n-m-k) D^{m-k} as n! / ( 2^k k! (n-k-m)! (m-k)! ), referring to them as the Weyl binomial coefficients, and derives them from rook numbers on Ferrers boards. (No mention of Hermite polynomials nor matchings on simplices are made.) Another combinatorial model and equivalent formula are presented in Blasiak and Flajolet (p. 16). References to much earlier work are given in both papers. - Tom Copeland, Jun 03 2021

A249548 Coefficients of reduced partition polynomials of A134264 for computing Lagrange compositional inversion.

Original entry on oeis.org

1, 0, 1, 1, 1, 2, 1, 5, 1, 6, 3, 5, 1, 7, 7, 21, 1, 8, 8, 4, 28, 28, 14, 1, 9, 9, 9, 36, 72, 12, 84, 1, 10, 10, 10, 5, 45, 90, 45, 45, 120, 180, 42, 1, 11, 11, 11, 11, 55, 110, 110, 55, 55, 165, 495, 165, 330, 1, 12, 12, 12, 12, 6, 66, 132, 132, 66, 66, 132, 22, 220, 660, 330, 660, 55, 495, 990, 132
Offset: 0

Views

Author

Tom Copeland, Oct 31 2014

Keywords

Comments

Coefficients of reduced partition polynomials of A134264 for computing the complete partition polynomials for the Lagrange compositional inversion of A134264 (see Oct 2014 comment by Copeland there). Umbrally,
e^(x*t) * exp[Prt(.;1,0,h_2,..) * t] = exp[Prt(.;1,x,h_2,..) * t], where Prt(n;1,0,h_2,..,h_n) are the reduced (h_0 = 1 and h_1 = 0) partition polynomials of the complete polynomials Prt(n;h_0,h_1,h_2,..,h_n) of A134264.
Partitions are given in the order of those on page 831 of Abramowitz and Stegun. Formulas for the coefficients of the partitions are given in A134264.
Row sums are the Motzkin sums or Riordan numbers A005043. - Tom Copeland, Nov 09 2014
From Tom Copeland, Jul 03 2018: (Start)
The matrix and operator formalism for Sheffer Appell sequences leads to the following relations with D = d/dh_1.
Exp[Prt(.;1,0,h_2,..) * D] (h_1)^n = [h_1 + Prt(.;1,0,h_2,...)]^n = Prt(n;1,h_1,h_2,...), the partition polynomials of A134264 for g(t)/t with h_0 = 1.
For the umbral compositional inverses described below,
Exp[UPrt(.;1,0,h_2,..) * D] (h_1)^n = [h_1 + UPrt(.;1,0,h_2,...)]^n = UPrt(n;1,h_1,h_2,...).
The respective e.g.f.s are multiplicative inverses; that is, exp[Prt(.;1,0,h_2,..) * t] = 1/exp[UPrt(.;1,0,h_2,..) * t], so the formalism of A133314 applies.
The raising operator R such that R Prt(n;1,h_1,h_2,...) = Prt(n+1;1,h_1,h_2,...) is R = exp[Prt(.;1,0,h_2,...)*D] h_1 exp[UPrt(.;1,0,h_2,..)*D] since R Prt(n+1;1,h_1,h_2,...) = exp[Prt(.;1,0,h_2,...)*D] h_1 (h_1)^n = Prt(n+1;h_1,h_2,...) from the definition of the umbral compositional inverse. This may also be expressed as R = h_1 + d/dD log[exp[Prt(.;1,0,h_2,...) * D]], so, using A127671, R = h_1 + h_2 D + h_3 D^2/2! + (h_4 - h_2^2) D^3/3! + (h_5 - 5 h_2 h_3) D^4/4! + (h_6 - 9 h_2 h_4 + 5 h_2^3 - 7 h_3^2) D^5/5! + (h_7 - 28 h_3 h_4 - 14 h_2 h_5 + 56 h_2^2 h_3) D^6/6! + ... . (End)

Examples

			Prt(0) = 1
Prt(1;1,0) = 0
Prt(2;1,0,h_2) = 1 h_2
Prt(3;1,0,h_2,h_3) = 1 h_3
Prt(4;1,0,h_2,..,h_4) = 1 h_4 + 2 (h_2)^2
Prt(5;1,0,h_2,..,h_5) = 1 h_5 + 5 h_2 h_3
Prt(6;1,0,h_2,..,h_6) = 1 h_6 + 6 h_2 h_4 + 3 (h_3)^2 + 5 (h_2)^3
Prt(7;1,0,h_2,..,h_7) = 1 h_7 + 7 h_3 h_4 + 7 h_2 h_5 + 21 h_3 (h_2)^2
...
------------
With h_n denoted by (n'), the first seven partition polynomials of A134264 with h_0=1 are given by the first seven coefficients of the truncated Taylor series expansion of the Euler binomial transform
e^[(1') * t] * {1 + 1 (2') *  t^2/2! + 1 (3') *  t^3/3! + [1 (4') + 2 (2')^2] *  t^4/4! + [1 (5') + 5 (2')(3')] *  t^5/5! + [1 (6') + 6 (2')(4') + 3 (3')^2 + 5 (2')^3] *  t^6/6!}, giving the truncated expansion
1 + 1 (1') * t + [1 (2') + 1 (1')^2] * t^2/2! + ... + [1 (6') + 6 (1')(5') + 6 (2')(4') + 3 (3')^2 + 15 (1')^2(4') + 30 (1')(2')(3') + 5 (2')^3 + 20 (1')^3(3') + 30 (1')^2(2')^2 + 15 (1')^4(2') + 1 (1')^6] * t^6/6!.
Extending the number of reduced partition polynomials of the transform allows for further complete polynomials of A134264 to be computed.
		

Crossrefs

Cf. A127671.
Rows lengths are given by A002865 (except for row 1).

Programs

  • Mathematica
    rows[n_] := {{1}, {0}}~Join~Module[
        {g = 1 / D[t / (1 + Sum[h[k] t^k, {k, 2, n}] + O[t]^(n+1)), t], p = t, r},
        r = Reap[Do[p = g D[p, t]/k; Sow[Expand[Normal@p /. {t -> 0}]], {k, n+1}]][[2, 1, 2 ;;]];
        Table[Coefficient[r[[k]], Product[h[t], {t, p}]], {k, 2, n}, {p, Sort[Sort /@ IntegerPartitions[k, k, Range[2, k]]]}]];
    rows[12] // Flatten (* Andrey Zabolotskiy, Feb 18 2024 *)

Formula

From Tom Copeland, Nov 10 2014: (Start)
Terms may be computed symbolically up to order n by using an iterated derivative evaluated at t=0:
with g(t) = 1/{d/dt [t/(1 + 0 t + h_2 t^2 + h_3 t^3 + ... + h_n t^n)]},
evaluate 1/n! * [g(t) d/dt]^n t at t=0, i.e., ask a symbolic math app for the first term in a series expansion of this iterated derivative, to obtain Prt(n-1).
Alternatively, the explicit formula in A134264 for the numerical coefficients of each partition can be used. (End)
From Tom Copeland, Nov 12 2014: (Start)
The first few partitions polynomials formed by taking the reciprocal of the e.g.f. of this entry's e.g.f. (cf. A133314) are
UPrt(0) = 1
UPrt(1;1,0) = 0
UPrt(2;1,0,h_2) = -1 h_2
UPrt(3;1,0,h_2,h_3) = -1 h_3
UPrt(4;1,0,h_2,..,h_4) = -1 h_4 + 4 (h_2)^2
UPrt(5;1,0,h_2,..,h_5) = -1 h_5 + 15 h_2 h_3
UPrt(6;1,0,h_2,..,h_6) = -1 h_6 + 24 h_2 h_4 + 17 (h_3)^2 + -35 (h_2)^3
...
Therefore, umbrally, [Prt(.;1,0,h_2,...) + UPrt(.;1,0,h_2,...)]^n = 0 for n>0 and unity for n=0.
Example of the umbral operation:
(a. + b.)^2 = a.^2 + 2 a.* b. + b.^2 = a_2 + 2 a_1 * b_1 + b_2.
This implies that the umbral compositional inverses (see below) of the partition polynomials of the Lagrange inversion formula (LIF) of A134264 with h_0=1 are given by UPrt(n;1,h_1,h_2,...,h_n) = [UPrt(.;1,0,h_2,...,h_n) + h_1]^n, so that the sequence of polynomials UPrt(n;1,h_1,h_2,...,h_n) is an Appell sequence in the indeterminate h_1. So, if one calculates UPrt(n;1,h_1,...,h_n), the lower order UPrt(n-1;1,h_1,...,h_(n-1)) can be found by taking the derivative w.r.t. h_1 and dividing by n. Same applies for Prt(n;1,h_1,h_2,...,h_n).
This connects the combinatorics of the permutohedra through A133314 and A049019, or their duals, to the noncrossing partitions, Dyck lattice paths, etc. that are isomorphic with the LIF of A134264.
An Appell sequence P(.,x) with the e.g.f. e^(x*t)/w(t) possesses an umbral inverse sequence UP(.,x) with the e.g.f. w(t)e^(x*t), i.e., polynomials such that P(n,UP(.,x))= x^n = UP(n,P(.,x)) through umbral substitution, as in the binomial example. The Bernoulli polynomials with w(t) = t/(e^t - 1) are a good example with the umbral compositional inverse sequence UP(n,x) = [(x+1)^(n+1)-x^(n+1)] / (n+1) (cf. A074909 and A135278). (End)

Extensions

Formula for Prt(7,..) and a(12)-a(15) added by Tom Copeland, Jul 22 2016
Rows 8-12 added by Andrey Zabolotskiy, Feb 18 2024

A321752 Irregular triangle read by rows where T(H(u),H(v)) is the coefficient of e(v) in p(u), where H is Heinz number, e is elementary symmetric functions, and p is power sum symmetric functions.

Original entry on oeis.org

1, 1, -2, 1, 0, 1, 3, -3, 1, 0, -2, 1, -4, 2, 4, -4, 1, 0, 0, 1, 0, 4, 0, -4, 1, 0, 0, 3, -3, 1, 5, -5, -5, 5, 5, -5, 1, 0, 0, 0, -2, 1, -6, 6, 6, 3, -2, -6, -12, 9, 6, -6, 1, 0, -4, 0, 2, 4, -4, 1, 0, 0, -6, 6, 3, -5, 1, 0, 0, 0, 0, 1, 7, -7, -7, -7, 14, 7, 7
Offset: 1

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Author

Gus Wiseman, Nov 20 2018

Keywords

Comments

Row n has length A000041(A056239(n)).
The Heinz number of an integer partition (y_1, ..., y_k) is prime(y_1) * ... * prime(y_k).

Examples

			Triangle begins:
   1
   1
  -2   1
   0   1
   3  -3   1
   0  -2   1
  -4   2   4  -4   1
   0   0   1
   0   4   0  -4   1
   0   0   3  -3   1
   5  -5  -5   5   5  -5   1
   0   0   0  -2   1
  -6   6   6   3  -2  -6 -12   9   6  -6   1
   0  -4   0   2   4  -4   1
   0   0  -6   6   3  -5   1
   0   0   0   0   1
   7  -7  -7  -7  14   7   7   7  -7  -7 -21  14   7  -7   1
   0   0   0   4   0  -4   1
For example, row 15 gives: p(32) = -6e(32) + 6e(221) + 3e(311) - 5e(2111) + e(11111).
		

Crossrefs

A321918 Tetrangle where T(n,H(u),H(v)) is the coefficient of e(v) in p(u), where u and v are integer partitions of n, H is Heinz number, e is elementary symmetric functions, and p is power sum symmetric functions.

Original entry on oeis.org

1, -2, 1, 0, 1, 3, -3, 1, 0, -2, 1, 0, 0, 1, -4, 2, 4, -4, 1, 0, 4, 0, -4, 1, 0, 0, 3, -3, 1, 0, 0, 0, -2, 1, 0, 0, 0, 0, 1, 5, -5, -5, 5, 5, -5, 1, 0, -4, 0, 2, 4, -4, 1, 0, 0, -6, 6, 3, -5, 1, 0, 0, 0, 4, 0, -4, 1, 0, 0, 0, 0, 3, -3, 1, 0, 0, 0, 0, 0, -2, 1
Offset: 1

Views

Author

Gus Wiseman, Nov 22 2018

Keywords

Comments

The Heinz number of an integer partition (y_1, ..., y_k) is prime(y_1) * ... * prime(y_k).

Examples

			Tetrangle begins (zeroes not shown):
  (1):  1
.
  (2):  -2  1
  (11):     1
.
  (3):    3 -3  1
  (21):     -2  1
  (111):        1
.
  (4):    -4  2  4 -4  1
  (22):       4    -4  1
  (31):          3 -3  1
  (211):           -2  1
  (1111):              1
.
  (5):      5 -5 -5  5  5 -5  1
  (41):       -4     2  4 -4  1
  (32):          -6  6  3 -5  1
  (221):             4    -4  1
  (311):                3 -3  1
  (2111):                 -2  1
  (11111):                    1
For example, row 14 gives: p(32) = -6e(32) + 6e(221) + 3e(311) - 5e(2111) + e(11111).
		

Crossrefs

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