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For a partition P of a positive integer, let f(P) be the product of k+1, over all parts k in P. Let a(n,r) be the sum of f(P) over all partitions P of n with smallest part r. Sequence gives table of a(n,r) for 1 <= r <= n, in the order a(1,1); a(2,1), a(2,2); a(3,1), a(3,2), a(3,3); ...
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%I #13 May 22 2015 14:35:10

%S 2,4,3,14,0,4,36,9,0,5,100,12,0,0,6,236,42,16,0,0,7,602,54,20,0,0,0,8,

%T 1368,195,24,25,0,0,0,9,3242,246,92,30,0,0,0,0,10,7240,759,112,35,36,

%U 0,0,0,0,11,16386,1134,232,40,42,0,0,0,0,0,12,35692,2859,528,170,48,49,0,0,0,0,0,13

%N For a partition P of a positive integer, let f(P) be the product of k+1, over all parts k in P. Let a(n,r) be the sum of f(P) over all partitions P of n with smallest part r. Sequence gives table of a(n,r) for 1 <= r <= n, in the order a(1,1); a(2,1), a(2,2); a(3,1), a(3,2), a(3,3); ...

%H Alois P. Heinz, <a href="/A079308/b079308.txt">Rows n = 1..141, flattened</a>

%e The partitions with minimal part 3 begin 3, 3+3, 4+3, 5+3, 6+3, 3+3+3, ... which yield the following values of f: 4, 16, 20, 24, 28, 64, ... therefore the 3rd column of our table begins 4,0,0,16,20,24,(28+64)=92,...

%e Triangle a(n,r) begins:

%e : 2;

%e : 4, 3;

%e : 14, 0, 4;

%e : 36, 9, 0, 5;

%e : 100, 12, 0, 0, 6;

%e : 236, 42, 16, 0, 0, 7;

%e : 602, 54, 20, 0, 0, 0, 8;

%e : 1368, 195, 24, 25, 0, 0, 0, 9;

%e : 3242, 246, 92, 30, 0, 0, 0, 0, 10;

%e : 7240, 759, 112, 35, 36, 0, 0, 0, 0, 11;

%p b:= proc(n, k) option remember; `if`(n=0, 1,

%p `if`(k>n, 0, b(n, k+1) +(k+1)*b(n-k, k)))

%p end:

%p a:= (n, k)-> b(n,k)-b(n, k+1):

%p seq(seq(a(n, k), k=1..n), n=1..12); # _Alois P. Heinz_, May 22 2015

%t a[n_, r_] := Which[r>n, 0, r==n, n+1, True, a[n, r]=(r+1)Sum[a[n-r, s], {s, r, n-r}]]; Flatten[Table[a[n, r], {n, 1, 12}, {r, 1, n}]]

%Y Cf. A074139, A074141 (row sums).

%K easy,nonn,tabl

%O 1,1

%A _Alford Arnold_, Feb 09 2003

%E Edited by _Dean Hickerson_, Feb 11 2003

%E Offset changed to 1 by _Alois P. Heinz_, May 22 2015