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A264405
Triangle read by rows: T(n,k) is the number of integer partitions of n having k repeated parts (each occurrence is counted).
2
1, 1, 0, 1, 0, 1, 2, 0, 0, 1, 2, 0, 2, 0, 1, 3, 0, 2, 1, 0, 1, 4, 0, 2, 2, 2, 0, 1, 5, 0, 4, 2, 1, 2, 0, 1, 6, 0, 6, 2, 3, 2, 2, 0, 1, 8, 0, 7, 4, 4, 2, 2, 2, 0, 1, 10, 0, 8, 6, 6, 4, 3, 2, 2, 0, 1, 12, 0, 13, 6, 6, 8, 3, 3, 2, 2, 0, 1, 15, 0, 15, 9, 11, 6, 9, 4, 3, 2, 2, 0, 1, 18, 0, 21, 10, 13, 12, 7, 8, 4, 3, 2, 2, 0, 1
OFFSET
0,7
COMMENTS
Compare with A264052 where only one occurrence of a repeated part is counted.
Sum of entries in row n = number of partitions of n = A000041(n).
Sum_{k>=0} k*T(n,k) = A194452(n).
LINKS
FORMULA
G.f.: G(t,x) = Product_{j>=1}(1 + x^j + t^2*x^{2j}/(1 - tx^j)).
EXAMPLE
T(4,2) = 2 because each of the partitions [2,2] and [2,1,1] have 2 repeated parts, while [4], [3,1], [1,1,1,1] have 0 or 4 repeated parts.
Triangle starts:
1;
1, 0;
1, 0, 1;
2, 0, 0, 1;
2, 0, 2, 0, 1;
3, 0, 2, 1, 0, 1;
MAPLE
g := product(1+x^j+t^2*x^(2*j)/(1-t*x^j), j = 1 .. 100): gser := simplify(series(g, x = 0, 30)): for n from 0 to 20 do P[n] := sort(coeff(gser, x, n)) end do: for n from 0 to 20 do seq(coeff(P[n], t, k), k = 0 .. n) end do; # yields sequence in triangular form
# second Maple program:
b:= proc(n, i) option remember; `if`(n=0, 1, `if`(i<1, 0,
add(expand(b(n-i*j, i-1)*`if`(j>1, x^j, 1)), j=0..n/i)))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=0..n))(b(n$2)):
seq(T(n), n=0..14); # Alois P. Heinz, Dec 07 2015
MATHEMATICA
b[n_, i_] := b[n, i] = If[n == 0, 1, If[i < 1, 0, Sum[Expand[b[n - i*j, i - 1]*If[j > 1, x^j, 1]], {j, 0, n/i}]]]; T[n_] := Function[p, Table[ Coefficient[p, x, i], {i, 0, n}]][b[n, n]]; Table[T[n], {n, 0, 14}] // Flatten (* Jean-François Alcover, Jan 23 2016, after Alois P. Heinz *)
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
Emeric Deutsch, Dec 07 2015
STATUS
approved