%I #21 Jul 19 2018 03:07:00

%S 0,0,1,2,2,3,5,6,7,13,16,19,29,38,49,72,84,108,155,195,234,331,410,

%T 501,672,824,1006,1341,1621,1981,2583,3111,3740,4846,5819,6957,8787,

%U 10582,12606,15840,18762,22386,27851,32934,38824,47961,56633,66577,81168,95612

%N The number of partitions of n into at least 3 parts from which we can form every partition of n into 3 parts by summing elements

%C The corresponding partitions with 2 in the definition instead of 3 are the complete partitions, which are counted by A126796.

%C The qualifier 'into at least 3 parts' is only relevant for n = 1 or 2. It is included because otherwise the condition would be vacuously true for all partitions of 1 and 2. It seems neater to consider that there are no partitions of 1 and 2 of this form.

%H Jack W Grahl, <a href="/A236970/a236970.hs.txt">Haskell code for generating this sequence</a>

%e The valid partitions of 5 are (2,1,1,1) and (1,1,1,1,1). Given any partition of 5 into 3 parts, it contains one part of at least 2. Therefore we can make any partition of 5 into 3 parts by joining (2,1,1,1) into three sums. (3, 1, 1) is not a valid partition, since (2,2,1) is a partition of 5 into 3 parts which cannot be made by summing elements from (3,1,1). Therefore a(5) = 2.

%t ric[p_, {x_,y_}] := If[x==0, If[y > Total[p], False, y==0 || AnyTrue[ Reverse@ Union[p], y>=# && ric[ DeleteCases[p, #, 1, 1], {0, y-#}] &]], If[x >= Total[p], False, AnyTrue[ Reverse@ Union@ p, x>=# && ric[ DeleteCases[p, #, 1, 1], {x-#, y}] &]]]; chk[p_] := AllTrue[ Rest /@ IntegerPartitions[Plus @@ p, {3}], ric[p,#] &]; a[n_] := Length@ Select[ IntegerPartitions[n, {3, Infinity}], chk]; Array[a, 24] (* _Giovanni Resta_, Jul 18 2018 *)

%Y Cf. A000041. A126796 is the case for 2 instead of 3, A236971 and A236972 are the cases for 4 and 5.

%K nonn

%O 1,4

%A _Jack W Grahl_, Feb 02 2014