

A236970


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


3



0, 0, 1, 2, 2, 3, 5, 6, 7, 13, 16, 19, 29, 38, 49, 72, 84, 108, 155, 195, 234, 331, 410, 501, 672, 824, 1006, 1341, 1621, 1981, 2583, 3111, 3740, 4846, 5819, 6957, 8787, 10582, 12606, 15840, 18762, 22386, 27851, 32934, 38824, 47961, 56633, 66577, 81168, 95612
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OFFSET

1,4


COMMENTS

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


LINKS



EXAMPLE

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.


MATHEMATICA

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 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



