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%I A264402 #17 Jan 15 2016 15:42:54
%S A264402 1,2,2,1,3,2,2,4,1,4,5,2,2,8,4,1,4,9,7,2,3,12,10,4,1,4,14,15,7,2,2,17,
%T A264402 20,12,4,1,6,18,27,17,7,2,2,23,33,26,12,4,1,4,24,44,35,19,7,2,4,27,51,
%U A264402 49,28,12,4,1,5,30,64,63,41,19,7,2,2
%N A264402 Triangle read by rows: T(n,k) is the number of partitions of n that have k parts larger than the smallest part (n>=1, k>=0).
%C A264402 T(n,k) = number of partitions of n in which the 2nd largest part is k (0 if all parts are equal). Example: T(7,2) = 4 because we have [3,2,1,1], [3,2,2], [4,2,1], and [5,2].
%C A264402 The fact that the above two statistics (in Name and in 1st Comment) have the same distribution follows at once by conjugation. - _Emeric Deutsch_, Dec 11 2015
%C A264402 Row sums yield the partition numbers (A000041).
%C A264402 T(n,0) = A000005(n) = number of divisors of n.
%C A264402 Sum_{k>=0} k*T(n,k) = A182984(n).
%H A264402 Alois P. Heinz, <a href="/A264402/b264402.txt">Rows n = 1..350, flattened</a>
%F A264402 G.f.: G(t,x) = Sum_{i>=1} (x^i/((1 - x^i)*Product_{j>=i+1}(1-t*x^j))).
%e A264402 T(7,2) = 4 because we have [2,2,1,1,1], [3,2,1,1], [3,3,1], and [4,2,1].
%e A264402 Triangle starts:
%e A264402 1;
%e A264402 2;
%e A264402 2,1;
%e A264402 3,2;
%e A264402 2,4,1;
%e A264402 4,5,2;
%e A264402 2,8,4,1;
%p A264402 g := sum(x^i/((1-x^i)*(product(1-t*x^j, j = i+1 .. 100))), i = 1 .. 100): gser := simplify(series(g, x = 0, 30)): for n to 27 do P[n] := sort(coeff(gser, x, n)) end do: for n to 27 do seq(coeff(P[n], t, j), j = 0 .. degree(P[n])) end do; # yields sequence in triangular form
%p A264402 # second Maple program:
%p A264402 b:= proc(n, i) option remember; `if`(n=0, [1, 0],
%p A264402       `if`(i<1, 0, b(n, i-1) +add((p->[0, p[1]+
%p A264402        expand(p[2]*x^j)])(b(n-i*j, i-1)) , j=1..n/i)))
%p A264402     end:
%p A264402 T:= n->(p->seq(coeff(p, x, i), i=0..degree(p)))(b(n$2)[2]):
%p A264402 seq(T(n), n=1..20);  # _Alois P. Heinz_, Nov 29 2015
%t A264402 b[n_, i_] := b[n, i] = If[n==0, {1, 0}, If[i<1, {0, 0}, b[n, i-1] + Sum[ Function[p, {0, p[[1]] + Expand[p[[2]]*x^j]}][b[n-i*j, i-1]], {j, 1, n/i} ]]]; T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 0, Exponent[p, x]}]][b[n, n][[2]]]; Table[T[n], {n, 1, 20}] // Flatten (* _Jean-François Alcover_, Jan 15 2016, after _Alois P. Heinz_ *)
%Y A264402 Cf. A000005, A000041, A116685, A182984, A002541.
%K A264402 nonn,tabf
%O A264402 1,2
%A A264402 _Emeric Deutsch_, Nov 21 2015

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