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%I #21 May 30 2021 07:27:44
%S 1,1,3,4,9,12,22,30,50,68,105,142,210,281,400,531,736,967,1311,1707,
%T 2274,2935,3851,4930,6389,8116,10402,13121,16658,20872,26275,32719,
%U 40880,50613,62807,77343,95389,116874,143331,174789,213251,258903,314367,380079,459462
%N Sum of binomial(Y(2,p), 2) over the partitions p of n, where Y(2,p) is the number of part sizes with multiplicity 2 or greater in p.
%F a(n) = (A301313(n) - A024788(n))/4.
%F G.f.: q^6 /((1-q^2)*(1-q^4))*Product_{j>=1} 1/(1-q^j).
%e For a(8), we sum over the partitions of eight. For each partition p, we take binomial(Y(2,p),2): that is, the number of parts with multiplicity at least two choose 2.
%e 8................B(0,2) = 0
%e 7,1..............B(0,2) = 0
%e 6,2..............B(0,2) = 0
%e 6,1,1............B(1,2) = 0
%e 5,3..............B(0,2) = 0
%e 5,2,1............B(0,2) = 0
%e 5,1,1,1..........B(1,2) = 0
%e 4,4..............B(1,2) = 0
%e 4,3,1............B(0,2) = 0
%e 4,2,2............B(1,2) = 0
%e 4,2,1,1..........B(1,2) = 0
%e 4,1,1,1,1........B(1,2) = 0
%e 3,3,2............B(1,2) = 0
%e 3,3,1,1..........B(2,2) = 1
%e 3,2,2,1..........B(1,2) = 0
%e 3,2,1,1,1........B(1,2) = 0
%e 3,1,1,1,1,1......B(1,2) = 0
%e 2,2,2,2..........B(1,2) = 0
%e 2,2,2,1,1........B(2,2) = 1
%e 2,2,1,1,1,1......B(2,2) = 1
%e 2,1,1,1,1,1,1....B(1,2) = 0
%e 1,1,1,1,1,1,1,1..B(1,2) = 0
%e ---------------------------
%e Total.....................3
%p b:= proc(n, i, p) option remember; `if`(n=0 or i=1,
%p binomial(`if`(n>1, 1, 0)+p, 2), add(
%p b(n-i*j, i-1, `if`(j>1, 1, 0)+p), j=0..n/i))
%p end:
%p a:= n-> b(n$2, 0):
%p seq(a(n), n=6..60); # _Alois P. Heinz_, May 19 2018
%t Array[Total[Binomial[Count[Split@#, _?(Length@# >= 2 &)], 2] & /@IntegerPartitions[#]] &, 50]
%t (* Second program: *)
%t b[n_, i_, p_] := b[n, i, p] = If[n == 0 || i == 1,
%t Binomial[If[n > 1, 1, 0] + p, 2], Sum[
%t b[n-i*j, i-1, If[j>1, 1, 0]+p], {j, 0, n/i}]];
%t a[n_] := b[n, n, 0];
%t a /@ Range[6, 60] (* _Jean-François Alcover_, May 30 2021, after _Alois P. Heinz_ *)
%Y Cf. A024786, A302347.
%K nonn
%O 6,3
%A _Emily Anible_, May 19 2018
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