0,3

Vaclav Kotesovec, Table of n, a(n) for n = 0..10000 (terms 0..1000 from Alois P. Heinz)

Eric Weisstein's World of Mathematics, q-Polygamma Function, q-Pochhammer Symbol

a(n) = Sum_p=1^P(n) Sum_i=1^D(p) Sum_j=i^D(p) 1 [subject to: d(i, p) <= d(j, p) ]; P(n) = number of partitions of n, D(p) = number of parts in partition p, d(i, p) and d(j, p) = parts number i and j in partition p of integer n.

a(n) = Sum_i=1^P(n) p(i, n)^2 - p(i, n), where P(n) is the number of integer partitions of n and p(i, n) is the number of parts of the i-th partition of n.

G.f.: (log(1-x)^2 - log(1-x)*log(x) + psi_x(1)*(2*log(1-x) - log(x) + psi_x(1)) + psi^1_x(1))/((x; x)_inf * log(x)^2), where psi_q(z) is the q-digamma function, psi^1_q(z) is the q-trigamma function, and (a; q)_inf is the q-Pochhammer symbol (the Euler function). To get this g.f., take the derivative (d/da)^2 (x^2/(a; x)_inf) and let a = x. - Vladimir Reshetnikov, Nov 21 2016

In the labeled case we have 22 one-element transitions among all partitions of n=4:

[1,1,1,1] -> [1,1,2] arises 6 times (the first 1 added to the second 1 gives 2, the first 1 added to the third 1 gives 2, the first 1 added to the fourth 1 gives 2, the second 1 added to the third 1 gives 2, the second 1 added to the fourth 1 gives 2, the third 1 added to the fourth 1 gives 2),

[1,1,2] -> [2,2] arises 1 times,

[1,1,2] -> [1,3] arises 2 times,

[2,2] -> [1,3] arises 1 times,

[1,3] -> [4] arises 1 time,

which gives 11 upwards transitions and 22 transitions in total if we include downwards transitions.

n=4: partition number p=1 is [1,1,1,1],

parts d(1,1)=1, d(2,1)=1 contribute 1,

parts d(1,1)=1, d(3,1)=1 contribute 1,

etc...

parts d(3,1)=1, d(4,1)=1 contribute 1,

(in total 6 contributions by [1,1,1,1]);

partition number p=2 is [1,1,2],

parts d(1,2)=1, d(2,2)=1 contribute 1,

parts d(1,2)=1, d(3,2)=2 contribute 1,

parts d(2,2)=1, d(3,2)=2 contribute 1;

partition number p=3 is [2,2],

parts d(1,3)=2, d(2,3)=2 contribute 1;

partition number p=4 is [1,3],

parts d(1,4)=1, d(2,4)=3 contribute 1;

partition number p=5 is [4],

part d(1,5)=4 contributes 0;

main := proc(n::integer) local a, ndxp, ListOfPartitions, APartition, PartOfAPartition; with(combinat): ListOfPartitions:=partition(n); a:=0; for ndxp from 1 to nops(ListOfPartitions) do APartition := ListOfPartitions[ndxp]; a := a + nops(APartition)^2 - nops(APartition); end do; print("n, a(n):", n, a); end proc;

# Alternative:

b:= proc(n, i) option remember; local f, g;

if n=0 then [1, [1]] elif i<1 then [0, [0]]

else f:= b(n, i-1); g:= `if`(i>n, [0, [0]], b(n-i, i));

[f[1]+g[1], zip((x, y)-> x+y, f[2], [0, g[2][]], 0)]

fi

end:

a:= n-> (l-> add(l[t+1]*t*(t-1), t=1..nops(l)-1))(b(n$2)[2]):

seq(a(n), n=0..50); # Alois P. Heinz, Apr 05 2012

a[n_] := Block[{p = IntegerPartitions[n], l = PartitionsP[n]}, Sum[ Length[p[[k]]]^2 - Length[p[[k]]], {k, l}]]; Table[ a[n], {n, 0, 37}] (* Robert G. Wilson v, Jul 13 2004, updated by Jean-François Alcover, Jan 29 2014 *)

Simplify@Table[SeriesCoefficient[(Log[1 - x]^2 - Log[1 - x] Log[x] + QPolyGamma[1, x] (2 Log[1 - x] - Log[x] + QPolyGamma[1, x]) + QPolyGamma[1, 1, x])/(QPochhammer[x] Log[x]^2), {x, 0, n}], {n, 0, 40}] (* Vladimir Reshetnikov, Nov 21 2016 *)

Simplify@Table[SeriesCoefficient[2 q^2/QPochhammer[q + a, q], {a, 0, 2}, {q, 0, n}], {n, 0, 40}] (* Vladimir Reshetnikov, Nov 22 2016 *)

Cf. A093695.

Sequence in context: A212970 A212683 A346586 * A006696 A094939 A006732

Adjacent sequences: A094530 A094531 A094532 * A094534 A094535 A094536

Thomas Wieder, Jun 05 2004

More terms from Robert G. Wilson v, Jul 13 2004

approved