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Irregular triangle read by rows where T(n,d) is the number of set partitions of {1,...n} with all block-sums equal to d, where d is a divisor of 1 + ... + n.
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%I #8 Jan 09 2019 13:04:52

%S 1,1,1,1,1,1,1,1,1,1,1,4,1,1,3,7,1,1,9,1,1,1,1,43,35,1,1,102,62,1,1,1,

%T 1,68,595,1,1,17,187,871,1480,361,1,1,2650,657,1,1,9294,1,1,23728,1

%N Irregular triangle read by rows where T(n,d) is the number of set partitions of {1,...n} with all block-sums equal to d, where d is a divisor of 1 + ... + n.

%e Triangle begins:

%e 1

%e 1

%e 1 1

%e 1 1

%e 1 1

%e 1 1

%e 1 4 1

%e 1 3 7 1

%e 1 9 1

%e 1 1

%e 1 43 35 1

%e 1 102 62 1

%e 1 1

%e 1 68 595 1

%e 1 17 187 871 1480 361 1

%e 1 2650 657 1

%e Row 8 counts the following set partitions:

%e {{18}{27}{36}{45}} {{1236}{48}{57}} {{12348}{567}} {{12345678}}

%e {{138}{246}{57}} {{12357}{468}}

%e {{156}{237}{48}} {{12456}{378}}

%e {{1278}{3456}}

%e {{1368}{2457}}

%e {{1458}{2367}}

%e {{1467}{2358}}

%t spsu[_,{}]:={{}};spsu[foo_,set:{i_,___}]:=Join@@Function[s,Prepend[#,s]&/@spsu[Select[foo,Complement[#,Complement[set,s]]=={}&],Complement[set,s]]]/@Cases[foo,{i,___}];

%t Table[Length[spsu[Select[Subsets[Range[n]],Total[#]==d&],Range[n]]],{n,12},{d,Select[Divisors[n*(n+1)/2],#>=n&]}]

%Y Row lengths are A164978. Row sums are A035470.

%Y Cf. A000110, A000258, A008277, A112956, A164977, A275714, A279375, A300335, A320423, A320424, A321455, A321469.

%K nonn,tabf,more

%O 1,12

%A _Gus Wiseman_, Jan 08 2019