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.

A276427 Irregular triangle read by rows: T(n,k) = number of partitions of n having k distinct parts i of multiplicity i; 0 <= k <= A328806(n)-1 = largest index of a nonzero value; n >= 0.

Original entry on oeis.org

1, 0, 1, 2, 2, 1, 3, 2, 5, 1, 1, 8, 3, 9, 6, 16, 5, 1, 19, 10, 1, 29, 11, 2, 36, 18, 2, 53, 21, 3, 65, 32, 4, 92, 38, 4, 1, 115, 54, 7, 154, 67, 10, 195, 88, 14, 257, 112, 15, 1, 318, 152, 19, 1, 419, 178, 29, 1, 516, 243, 31, 2, 663, 293, 44, 2, 821, 376, 56, 2, 1039, 465, 67, 4, 1277, 589, 89, 3, 1606, 715, 108, 7
Offset: 0

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Author

Emeric Deutsch, Sep 19 2016

Keywords

Comments

The sum of entries in row n is A000041(n): the partition numbers. [This allows us to know the row length, i.e., when the largest value of k is reached for which T(n,k) is nonzero. The row lengths are now listed as A328806. - M. F. Hasler, Oct 28 2019]

Examples

			Triangle starts:
1;       (n=0: partition [] has k=0 parts i of multiplicity i: T(0,0) = 1.)
0, 1;    (n=1: partition [1] has k=1 part i of multiplicity i: T(1,1) = 1.)
2;       (n=2: partitions [1,1] and [2] have k=0 parts i occurring i times.)
2, 1;    (n=3: [1,1,1] and [3] have 0, [1,2] has 1 part i occurring i times)
3, 2;    (n=4: [4], [1,1,2] and [1,1,1,1] for k=0; [1,3] & [2,2] for k=1.)
5, 1, 1; (n=5: [1,4] has i=1, [1,2,2] has i=1 and i=2 occurring i times.)
(...)
The partition [1,2,3,3,3,4] has 2 parts i of multiplicity i: i=1 and i=3.
T(14,3) = 1, since [1,2,2,3,3,3] is the only partition of 14 having k=3 parts i with multiplicity i, namely i = 1, 2 and 3.
T(14,2) = 4, counting [1,2,2,3,6], [1,2,2,4,5], [1,2,2,9] (with i=1 and i=2), and [1,3,3,3,4] (with i=1 and i=3).
		

Crossrefs

Cf. A000041 (row sums), A276428, A276429, A328806 (row lengths).

Programs

  • Maple
    G := mul((t-1)*x^(i^2)+1/(1-x^i), i = 1 .. 100): Gser := simplify(series(G, x = 0, 35)): for n from 0 to 30 do P[n] := sort(coeff(Gser, x, n)) end do: for n from 0 to 30 do seq(coeff(P[n], t, k), k = 0 .. degree(P[n])) end do; # yields sequence in triangular form
    # second Maple program:
    b:= proc(n, i) option remember; expand(
          `if`(n=0, 1, `if`(i<1, 0, add(
          `if`(i=j, x, 1)*b(n-i*j, i-1), j=0..n/i))))
        end:
    T:= n->(p->seq(coeff(p, x, i), i=0..degree(p)))(b(n$2)):
    seq(T(n), n=0..30);  # Alois P. Heinz, Sep 19 2016
  • Mathematica
    b[n_, i_] := b[n, i] = Expand[If[n==0, 1, If[i<1, 0, Sum[If[i==j, x, 1] * b[n-i*j, i-1], {j, 0, n/i}]]]]; T[n_] := Function[p, Table[ Coefficient[ p, x, i], {i, 0, Exponent[p, x]}]][b[n, n]]; Table[T[n], {n, 0, 30}] // Flatten (* Jean-François Alcover, Oct 20 2016, after Alois P. Heinz *)
  • PARI
    apply( A276427_row(n, r=List(0))={forpart(p=n, my(s, c=1); for(i=1, #p, p[i]==if(i<#p, p[i+1]) && c++ && next; c==p[i] && s++; c=1); while(#r<=s, listput(r,0)); r[s+1]++);Vec(r)}, [0..20]) \\ M. F. Hasler, Oct 27 2019

Formula

G.f.: G(t,x) = Product_{i>=1} ((t-1)*x^{i^2} + 1/(1-x^i)).
T(n,0) = A276429(n).
Sum(k*T(n,k), k>=0) = A276428(n).

Extensions

Name edited by M. F. Hasler, Oct 27 2019