A234923 - OEIS

%I #7 Mar 14 2015 01:26:18

%S 1,2,3,5,4,7,5,9,8,6,11,10,14,7,13,12,11,17,8,15,14,13,20,19,9,17,16,

%T 15,23,14,22,20,10,19,18,17,26,16,25,24,23,30,11,21,20,19,29,18,28,17,

%U 27,26,25,34,12,23,22,21,32,20,31,19,30,29,29,28,38,26

%N Array w(n,h), in which row n shows the weights, as defined in Comments, of the distinct-parts partitions of n, arranged in Mathematica order.

%C The weight of a partition P = x (1)+ x(2)+...+x(k) of n is given at A234094 as k*x(1) + (k-1)*x(2) + ... + x(k), which is the number of steps needed to make P from the sum 1+1+...+1 = n by moving dividers into the sum; see the Example section.

%F w(n,h) = dot product of (partition # h of n) and (k, k-1, ..., 1), where k = length of (partition # h of n).

%e Represent 1+1+1+1+1 as _1_1_1_1_1_. The partition 3+2+1 matches the placement of dividers d indicated by _1 _1_1d1_1_d_1_d. To place the 1st d takes 3 steps (starting at the 1st '_'); to place the 2nd d takes 3+2 steps (starting at the 1st '_'); to place the 3rd d takes 3+2+1 steps. The total number of steps is 3+5+6 = 14, the 4th number in row 4 because 3+2+1 is the 4th distinct-parts partition of 6 in Mathematica ordering. The first 9 rows:

%e 1

%e 2

%e 3    5

%e 4    7

%e 5    9    8

%e 6   11   10   14

%e 7   13   12   11   17

%e 8   15   14   13   20   19

%e 9   17   16   15   23   14   22   20

%t p[n_] := p[n] = Select[IntegerPartitions[n], Max[Length /@ Split@#] == 1 &]; q[n_] := q[n] = Length[p[n]]; v[n_] := v[n] = Table[n + 1 - i, {i, 1, n}]; w[n_, h_] := w[n, h] = Dot[p[n][[h]], v[Length[p[n][[h]]]]]; Flatten[Table[w[n, h], {n, 1, 9}, {h, 1, q[n]}]]

%t TableForm[Table[w[n, h], {n, 1, 9}, {h, 1, q[n]}]]

%Y Cf. A234094, A234924.

%K nonn,easy,tabf

%O 1,2

%A _Clark Kimberling_, Jan 01 2014