0,3

a(n) = Sum_p=1^P(n) Sum_i=1^D(p) Sum_j=1^D(p) 1 [subject to: i <> j and d(i,p) <= d(j,p) and d(i,p) <> d(i-1,p) (if i > 1) and d(j,i) <> d(j-1,i) (if j > 1 and if d(j-1,p) has given a contribution to the sum) ]; P(n) = number of partitions of n, D(p) = number of digits in partition p, d(i,p) and d(j,p) = digits number i and j in partition p of integer n.

See the corresponding formula for a(n) for the labeled case A094533.

T. Wieder, Home Page.

T. Wieder, (old) Home Page.

a(n) = Sum_i=1^P(n+1) S(i, n+1)^2 - S(i, n+1), where P(n+1) is the number of integer partitions of n+1 and S(i, n+1) is the number of digits in the set of digits of the i-th partition of n+1. (E.g. the partition [1111233] has the set of digits {1, 2, 3} and would contribute 3^2 - 3 = 6 to the sum.)

In the unlabeled case we have 10 one-element transitions among all partitions of n=4: [1,1,1,1] -> [1,1,2]; [1,1,2] -> [2,2]; [1,1,2] -> [1,3]; [2,2] -> [1,3]; [1,3] -> [4] and [1,1,2] -> [1,1,1,1]; [2,2] -> [1,1,2]; [1,3] -> [1,1,2]; [1,3] -> [2,2]; [4] -> [1,3].

n=5:

partition number p=1 is [1,1,1,1,1], digits d(1,1)=1, d(2,1)=1 contribute 1;

partition number p=2 is [1,1,1,2], digits d(1,1)=1, d(2,2)=1 contribute 1, digits d(1,2)=2, d(4,2)=2 contribute 1;

partition number p=3 is [1,2,2], digits d(1,3)=1, d(2,3)=2 contribute 1, digits d(2,3)=2, d(3,3)=2 contribute 1;

partition number p=4 is [1,1,3], digits d(1,4)=1, d(2,4)=1 contribute 1, digits d(1,4)=1, d(3,4)=3 contribute 1;

partition number p=5 is [2,3], digits d(1,5)=2, d(2,5)=3 contribute 1;

partition number p=6 is [1,4], digits d(1,6)=1, d(2,6)=4 contribute 1;

partition number p=7 is [5], digits d(1,7)=5 contributes 0;

==> a(5)=2*9=18 (factor 2 if we accept up and down transitions).

a(5) = 18 because the 11 partitions of n=5+1=6 have the following sets of digits:

{1} contributes 0, {1, 2} contributes 2, {1, 2} contributes 2,

{2} contributes 0, {1, 3} contributes 2, {1, 2, 3} contributes 6,

{3} contributes 0, {1, 4} contributes 2, {2, 4} contributes 2,

{1, 5} contributes 2, {6} contributes 0,

which gives 0 + 2 + 2 + 0 + 2 + 6 + 0 + 2 + 2 + 2 + 0 = 18.

A093695 := proc(n::integer) local a, ndxp, ListOfPartitions, APartition, PartOfAPartition, SetOfParts, iverbose; with(combinat): iverbose:=1; ListOfPartitions:=partition(n+1); a:=0; for ndxp from 1 to nops(ListOfPartitions) do APartition := ListOfPartitions[ndxp]; SetOfParts := convert(APartition, set); a := a + nops(SetOfParts)^2 - nops(SetOfParts); if iverbose = 1 then print ("ndxp, SetOfParts, nops(SetOfParts)^2 - nops(SetOfParts): ", ndxp, SetOfParts, nops(SetOfParts)^2 - nops(SetOfParts)); fi; # End of do-loop *** ndxp ***. end do; print("n, a(n):", n, a); end proc;

(* first do *) Needs["DiscreteMath`Combinatorica`"] (* then *) a[n_] := Block[{p = Partitions[n + 1], l = PartitionsP[n + 1]}, Sum[ Length[ Union[ p[[k]]]]^2 - Length[ Union[ p[[k]] ]], {k, l}]]; Table[ a[n], {n, 2, 40}] (from Robert G. Wilson v Jul 13 2004)

Cf. A093694, A089378.

Cf. A094533.

Sequence in context: A057491 A005541 A045955 this_sequence A104723 A079162 A043330

Adjacent sequences: A093692 A093693 A093694 this_sequence A093696 A093697 A093698

nonn

Thomas Wieder (wieder.thomas(AT)t-online.de), Apr 10 2004

More terms from Robert G. Wilson v (rgwv(AT)rgwv.com), Jul 13 2004