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%I #42 Mar 05 2021 21:43:56
%S 0,1,2,5,7,16,20,39,52,86,113,184,232,353,462,661,851,1202,1526,2098,
%T 2670,3565,4514,5967,7473,9715,12162,15583,19373,24625,30410,38274,
%U 47112,58725,71951,89129,108599,133612,162259,198346,239825,291718,351269,425102
%N Sum of all parts minus the total number of parts of the last section of the set of partitions of n.
%H Alois P. Heinz, <a href="/A207035/b207035.txt">Table of n, a(n) for n = 1..1000</a>
%F a(n) = A138879(n) - A138137(n) = A138880(n) - A138135(n). - _Omar E. Pol_, Apr 21 2012
%F G.f.: Sum_{k>=1} x^(2*k)/(1 - x^k)^2 / Product_{j>=2} (1 - x^j). - _Ilya Gutkovskiy_, Mar 05 2021
%e For n = 7 the last section of the set of partitions of 7 looks like this:
%e .
%e . (. . . . . . 7)
%e . (. . . 4 . . 3)
%e . (. . . . 5 . 2)
%e . (. . 3 . 2 . 2)
%e . (1)
%e . (1)
%e . (1)
%e . (1)
%e . (1)
%e . (1)
%e . (1)
%e . (1)
%e . (1)
%e . (1)
%e . (1)
%e .
%e The sum of all parts = 7+4+3+5+2+3+2+2+1*11 = 39, on the other hand the total number of parts is 1+2+2+3+1*11 = 19, so a(7) = 39 - 19 = 20. Note that the number of dots in the picture is also equal to a(7) = 6+5+5+4 = 20.
%p b:= proc(n, i) option remember; local f, g;
%p if n=0 then [1, 0]
%p elif i<2 then [0, 0]
%p elif i>n then b(n, i-1)
%p else f:= b(n, i-1); g:= b(n-i, i);
%p [f[1]+g[1], f[2]+g[2] +g[1]*(i-1)]
%p fi
%p end:
%p a:= n-> b(n, n)[2]:
%p seq (a(n), n=1..50); # _Alois P. Heinz_, Feb 20 2012
%t b[n_, i_] := b[n, i] = Module[{f, g}, Which[n==0, {1, 0}, i<2, {0, 0}, i>n , b[n, i-1], True, f = b[n, i-1]; g = b[n-i, i]; {f[[1]] + g[[1]], f[[2]] + g[[2]] + g[[1]]*(i-1)}]]; a[n_] := b[n, n][[2]]; Table[a[n], {n, 1, 50}] (* _Jean-François Alcover_, Sep 13 2015, after _Alois P. Heinz_ *)
%Y Row sums of triangle A207034. Partial sums give A196087.
%Y Cf. A006128, A066186, A135010, A138121, A138135, A138137, A138879, A138880, A187219, A194548, A207038.
%K nonn
%O 1,3
%A _Omar E. Pol_, Feb 20 2012
%E More terms from _Alois P. Heinz_, Feb 20 2012
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