cp's OEIS Frontend

A220484 Triangle read by rows: T(j,k) is the total number of appearances of the smallest parts in the j-th partition of n, with partitions as nonincreasing lists of parts in lexicographic order.

%I A220484 #17 Dec 13 2015 01:11:22
%S A220484 1,2,1,3,1,1,4,2,1,2,1,5,3,2,1,1,1,1,6,4,3,2,2,1,1,3,1,2,1,7,5,4,3,3,
%T A220484 2,2,1,1,1,1,2,1,1,1,8,6,5,4,4,3,3,2,2,2,2,1,1,1,1,4,2,1,1,1,2,1,9,7,
%U A220484 6,5,5,4,4,3,3,3,3,2,2,2,2,1,1,1,1,1,1,1,3,2,1,1,3,1,1,1
%N A220484 Triangle read by rows: T(j,k) is the total number of appearances of the smallest parts in the j-th partition of n, with partitions as nonincreasing lists of parts in lexicographic order.
%C A220484 The sum of row n equals spt(n) = A092269(n), the smallest part partition function.
%e A220484 For n = 5:
%e A220484 ------------------------------------------
%e A220484 .                         number of
%e A220484 Partitions of 5         smallest parts
%e A220484 ------------------------------------------
%e A220484 1 + 1 + 1 + 1 + 1              5
%e A220484 2 + 1 + 1 + 1                  3
%e A220484 3 + 1 + 1                      2
%e A220484 2 + 2 + 1                      1
%e A220484 4 + 1                          1
%e A220484 3 + 2                          1
%e A220484 5                              1
%e A220484 ------------------------------------------
%e A220484 So row 5 is [5, 3, 2, 1, 1, 1, 1]. The sum of row 5 is 5+3+2+1+1+1+1 = spt(5) = A092269(n) = 14.
%e A220484 .
%e A220484 Written as an irregular triangle begins:
%e A220484 1;
%e A220484 2,1;
%e A220484 3,1,1;
%e A220484 4,2,1,2,1;
%e A220484 5,3,2,1,1,1,1;
%e A220484 6,4,3,2,2,1,1,3,1,2,1;
%e A220484 7,5,4,3,3,2,2,1,1,1,1,2,1,1,1;
%e A220484 8,6,5,4,4,3,3,2,2,2,2,1,1,1,1,4,2,1,1,1,2,1;
%e A220484 9,7,6,5,5,4,4,3,3,3,3,2,2,2,2,1,1,1,1,1,1,1,3,2,1,1,3,1,1,1;
%Y A220484 Column 1 is A000027. Row n has length A000041(n). Row sums give A092269.
%Y A220484 Cf. A209514, A220504.
%K A220484 nonn,tabf
%O A220484 1,2
%A A220484 _Omar E. Pol_, Jan 20 2013