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 1-3 of 3 results.

A210590 Triangle of numbers generated by the Nekrasov-Okounkov formula.

Original entry on oeis.org

1, 1, 1, 4, 5, 1, 18, 29, 12, 1, 120, 218, 119, 22, 1, 840, 1814, 1285, 345, 35, 1, 7920, 18144, 14674, 5205, 805, 51, 1, 75600, 196356, 185080, 79219, 16450, 1624, 70, 1, 887040, 2427312, 2515036, 1258628, 324569, 43568, 2954, 92, 1, 10886400, 32304240, 37012572, 21034376, 6431733, 1088409, 101178, 4974, 117, 1
Offset: 0

Views

Author

Wouter Meeussen, Mar 24 2012

Keywords

Comments

Row sums are A000712, alternating sign row sums are zero (except for first row); application of the Nekrasov-Okounkov formula; see A138782.

Examples

			Table starts as:
     1;
     1,     1;
     4,     5,     1;
    18,    29,    12,    1;
   120,   218,   119,   22,   1;
   840,  1814,  1285,  345,  35,  1;
  7920, 18144, 14674, 5205, 805, 51,  1;
  ...
		

Crossrefs

T(2n,n) gives A338755.

Programs

  • Mathematica
    w=9; MapIndexed[ CoefficientList[#1,t] Tr[#2-1]! &, CoefficientList[Series[Product[(1-x^i)^(-1-t), {i,w}], {x,0,w}], x]];
    or alternatively:
    CoefficientList[#, t] & /@ Table[1/n! Tr[(NumberOfTableaux[#1]^2 Apply[Times, t + Flatten[hooklength[#1]]^2] &) /@ Partitions[n]], {n,0,9}]
    or alternatively:
    Table[1/n!Tr[NumberOfTableaux[#]^2 f[ Flatten[hooklength[#]]^2,e,k,n ]&/@ Partitions[n] ],{n,0,9},{k,0,n}]
    with e and f defined as:
    e[n_,v_]:= Tr[Times @@@ Select[Subsets[Table[Subscript[x,j],{j,v}]],Length[#]==n&]];
    f[li_List,fun_,par_,k_]:=fun[par,k]/.Thread[Array[Subscript[x,#1]&,Length[li]]->li];

Formula

E.g.f.: Product_{i=1..n} (1 - x^i)^(-1 - t).

A234937 Triangle read by rows of coefficients of polynomials generated by the Han/Nekrasov-Okounkov formula.

Original entry on oeis.org

1, 1, -1, 4, -5, 1, 18, -29, 12, -1, 120, -218, 119, -22, 1, 840, -1814, 1285, -345, 35, -1, 7920, -18144, 14674, -5205, 805, -51, 1, 75600, -196356, 185080, -79219, 16450, -1624, 70, -1, 887040, -2427312, 2515036, -1258628, 324569, -43568, 2954, -92, 1
Offset: 0

Views

Author

William J. Keith, Jan 01 2014

Keywords

Comments

Coefficients of the polynomials p_n(b) defined by Product_{k>0} (1-q^k)^(b-1) = Sum n! p_n(b) q^n.
Each row is length 1+n, starting from n=0, and consists of the coefficients of one of the p_n(b).
A210590 is an unsigned version using the form preferred by Nekrasov and Okounkov. This is the form for which Guo-Niu Han's reference below gives the hooklength formula:
p_n(b) = Sum_{lambda partitioning n} Product_{h_{ij} in lambda} (1-b/(h_{ij}^2)).
Coefficients reduced mod 5 are those of 2 times Pascal's triangle and an alternating sign. Other primes have slightly more complex reduction behavior. See second link.
Lehmer's conjecture on the tau function states that the evaluation at b=25 (A000594) is never 0.
The general diagonal and column are probably of combinatorial interest.

Examples

			The coefficient of q^3 in the indeterminate power is (1/6) (18-29b+12b^2-b^3).
		

Crossrefs

Row entries sum to 0.
A210590 is the unsigned version.
Starting from row 0: final entry of row n, (-1)^n (A033999).
From row 1: next-to-last entry of row n, (-1)^(n-1) * n(3n-1)/2 (signed version of A000326).
First entry of row n, n! * p(n) (A053529).
Second entry of row n, -1 * n! * (sum of reciprocals of all parts in partitions of n) (negatives of A057623).
(Sum of absolute values of row entries)/n!: A000712.
Evaluations at various powers of b, divided by n!, enumerate multipartitions or powers of the eta function. Some special cases that appear in the OEIS:
b=0: A000041, the partition numbers,
b=2: A010815, from Euler's Pentagonal Number Theorem,
b=-1: A000712, partitions into 2 colors,
b=-11: A005758, reciprocal of the square root of the tau function,
b=-23: A006922, reciprocal of the tau function,
b=13: A000735, square root of the tau function,
b=25: A000594, Ramanujan's tau function.

Programs

  • Mathematica
    nn=10;
    Clear[b]; PolyTable = Table[0, {n, 1, nn}];
    PolyTable[[1]]=1-b;
    For[n = 2, n <= nn, n++,
    PolyTable[[n]] = Simplify[(((n - 1)!)*(b - 1))*(Sum[
           PolyTable[[n - m]]*(-1*DivisorSigma[1, m]/((n - m)!)), {m, 1,
            n - 1}] + (-1*DivisorSigma[1, n]))]];
    LongTable = Table[Table[
       Which[k == 0, PartitionsP[n]*n!, k > 0,
        Coefficient[Expand[PolyTable[[n]]], b^k]], {k, 0, n}], {n, 1, nn}];
    Flatten[PrependTo[LongTable,1]]

Formula

E.g.f.: Product_{k>0} (1-q^k)^(b-1).
Recurrence: With p_0(b) = 1, p_n(b) = (n-1)!*(b-1)*Sum_{m=1..n} -sigma(m)*p_{n-m}(b) / (n-m)!, sigma being the divisor function.

A103738 a(n) = n! * (sum of reciprocals of parts in all partitions of n into distinct parts).

Original entry on oeis.org

1, 1, 11, 38, 274, 2844, 21888, 231888, 2580912, 37879200, 459884160, 7372650240, 112624905600, 2002334100480, 37047155846400, 721997863372800, 14458523340441600, 320885263596441600, 7222523219238297600, 172441642330718208000, 4367517061604788224000
Offset: 1

Views

Author

Vladeta Jovovic, Mar 27 2005

Keywords

Crossrefs

Cf. A057623.

Programs

  • Maple
    gf:=sum(x^k/k/(1+x^k), k=1..50)*product((1+x^k), k=1..50): s:=series(gf, x, 50): for n from 1 to 30 do printf(`%d,`,coeff(s, x, n)*n!) od: # James Sellers, Apr 10 2005
    # second Maple program:
    b:= proc(n, i) option remember; `if`(n=0, [1, 0], `if`(i<1, 0,
           b(n, i-1)+`if`(i>n, 0, (p-> p+[0, p[1]/i])(b(n-i, i-1)))))
        end:
    a:= n-> n!*b(n$2)[2]:
    seq(a(n), n=1..30);  # Alois P. Heinz, Sep 11 2014
  • Mathematica
    b[n_, i_] := b[n, i] = If[n == 0, {1, 0}, If[i < 1, {0, 0}, b[n, i - 1] + If[i > n, 0, Function[p, p + {0, p[[1]]/i}][b[n - i, i - 1]]]]]; a[n_] := n!*b[n, n][[2]]; Table[a[n], {n, 1, 30}] (* Jean-François Alcover, Jan 10 2016, after Alois P. Heinz *)

Formula

E.g.f.: A(x)*B(x), where A(x) = Sum_{k>0} x^k/(k*(1+x^k)) and B(x) = Product_{k>0} (1 + x^k).

Extensions

More terms from James Sellers, Apr 10 2005
Showing 1-3 of 3 results.