A206440
Volume of the last section of the set of partitions of n from the shell model of partitions version "Boxes".
Original entry on oeis.org
1, 5, 11, 27, 43, 93, 131, 247, 352, 584, 808, 1306, 1735, 2643, 3568, 5160, 6835, 9721, 12672, 17564, 22832, 30818, 39743, 53027, 67594, 88740, 112752, 145944, 183979, 236059, 295370, 375208, 467363, 588007, 728437, 910339, 1121009, 1391083, 1706003, 2103013
Offset: 1
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b:= proc(n, i) option remember; `if`(n=0 or i=1, [1, n],
b(n, i-1)+(p-> p+[0, p[1]*i^2])(b(n-i, min(n-i, i))))
end:
a:= n-> (b(n$2)-b(n-1$2))[2]:
seq(a(n), n=1..40); # Alois P. Heinz, Feb 23 2022
-
b[n_, i_] := b[n, i] = If[n == 0 || i == 1, {1, n},
b[n, i-1] + Function[p, p + {0, p[[1]]*i^2}][b[n-i, Min[n-i, i]]]];
a[n_] := (b[n, n] - b[n-1, n-1])[[2]];
Table[a[n], {n, 1, 40}] (* Jean-François Alcover, Apr 25 2022, after Alois P. Heinz *)
A299769
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.
Original entry on oeis.org
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, 15, 32, 27, 48, 25, 36, 0, 64, 22, 28, 54, 32, 50, 36, 49, 0, 81, 30, 60, 54, 80, 75, 72, 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
Offset: 1
Triangle begins:
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;
15, 32, 27, 48, 25, 36, 0, 64;
22, 28, 54, 32, 50, 36, 49, 0, 81;
30, 60, 54, 80, 75, 72, 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;
...
Illustration for the 4th row of triangle:
.
. Last section of the set
. Partitions of 4. of the partitions of 4.
. _ _ _ _ _
. |_| | | | [1,1,1,1] | | [1]
. |_ _| | | [2,1,1] | | [1]
. |_ _ _| | [3,1] _ _ _| | [1]
. |_ _| | [2,2] |_ _| | [2,2]
. |_ _ _ _| [4] |_ _ _ _| [4]
.
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].
Leading diagonal gives
A000290, n >= 1.
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b:= proc(n, i) option remember; `if`(n=0 or i=1, 1+n*x, b(n, i-1)+
(p-> p+(coeff(p, x, 0)*i^2)*x^i)(b(n-i, min(n-i, i))))
end:
T:= n-> (p-> seq(coeff(p, x, i), i=1..n))(b(n$2)-b(n-1$2)):
seq(T(n), n=1..14); # Alois P. Heinz, Jul 23 2018
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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[n_] := Function[p, Table[Coefficient[p, x, i], {i, 1, n}]][b[n, n] - b[n-1, n-1]];
T /@ Range[14] // Flatten (* Jean-François Alcover, Dec 10 2019, after Alois P.heinz *)
Showing 1-2 of 2 results.
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