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%I #36 Dec 10 2019 12:10:57
%S 1,1,4,2,0,9,3,8,0,16,5,4,9,0,25,7,16,18,16,0,36,11,12,18,16,25,0,49,
%T 15,32,27,48,25,36,0,64,22,28,54,32,50,36,49,0,81,30,60,54,80,75,72,
%U 49,64,0,100,42,60,90,80,100,72,98,64,81,0,121,56,108,126,160,125,180,98,128,81,100,0,144
%N Triangle read by rows: T(n,k) is the sum of all squares of the parts k in the last section of the set of partitions of n, with n >= 1, 1 <= k <= n.
%C The partial sums of the k-th column of this triangle give the k-th column of triangle A299768.
%C Note that the last section of the set of partitions of n is also the n-th section of the set of partitions of any positive integer >= n.
%H Alois P. Heinz, <a href="/A299769/b299769.txt">Rows n = 1..200, flattened</a>
%F T(n,k) = A299768(n,k) - A299768(n-1,k). - _Alois P. Heinz_, Jul 23 2018
%e Triangle begins:
%e 1;
%e 1, 4;
%e 2, 0, 9;
%e 3, 8, 0, 16;
%e 5, 4, 9, 0, 25;
%e 7, 16, 18, 16, 0, 36;
%e 11, 12, 18, 16, 25, 0, 49;
%e 15, 32, 27, 48, 25, 36, 0, 64;
%e 22, 28, 54, 32, 50, 36, 49, 0, 81;
%e 30, 60, 54, 80, 75, 72, 49, 64, 0, 100;
%e 42, 60, 90, 80, 100, 72, 98, 64, 81, 0, 121;
%e 56, 108, 126, 160, 125, 180, 98, 128, 81, 100, 0, 144;
%e ...
%e Illustration for the 4th row of triangle:
%e .
%e . Last section of the set
%e . Partitions of 4. of the partitions of 4.
%e . _ _ _ _ _
%e . |_| | | | [1,1,1,1] | | [1]
%e . |_ _| | | [2,1,1] | | [1]
%e . |_ _ _| | [3,1] _ _ _| | [1]
%e . |_ _| | [2,2] |_ _| | [2,2]
%e . |_ _ _ _| [4] |_ _ _ _| [4]
%e .
%e For n = 4 the last section of the set of partitions of 4 is [4], [2, 2], [1], [1], [1], so the squares of the parts are respectively [16], [4, 4], [1], [1], [1]. The sum of the squares of the parts 1 is 1 + 1 + 1 = 3. The sum of the squares of the parts 2 is 4 + 4 = 8. The sum of the squares of the parts 3 is 0 because there are no parts 3. The sum of the squares of the parts 4 is 16. So the fourth row of triangle is [3, 8, 0, 16].
%p b:= proc(n, i) option remember; `if`(n=0 or i=1, 1+n*x, b(n, i-1)+
%p (p-> p+(coeff(p, x, 0)*i^2)*x^i)(b(n-i, min(n-i, i))))
%p end:
%p T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n$2)-b(n-1$2)):
%p seq(T(n), n=1..14); # _Alois P. Heinz_, Jul 23 2018
%t b[n_, i_] := b[n, i] = If[n==0 || i==1, 1 + n*x, b[n, i-1] + Function[p, p + (Coefficient[p, x, 0]*i^2)*x^i][b[n-i, Min[n-i, i]]]];
%t T[n_] := Function[p, Table[Coefficient[p, x, i], {i, 1, n}]][b[n, n] - b[n-1, n-1]];
%t T /@ Range[14] // Flatten (* _Jean-François Alcover_, Dec 10 2019, after _Alois P.heinz_ *)
%Y Column 1 is A000041.
%Y Leading diagonal gives A000290, n >= 1.
%Y Second diagonal gives A000007.
%Y Row sums give A206440.
%Y Cf. A066183, A135010, A206438, A299768.
%K nonn,tabl
%O 1,3
%A _Omar E. Pol_, Mar 20 2018
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