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A299769
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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.
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1
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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
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OFFSET
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1,3
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COMMENTS
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The partial sums of the k-th column of this triangle give the k-th column of triangle A299768.
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.
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LINKS
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FORMULA
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EXAMPLE
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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].
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MAPLE
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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)):
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MATHEMATICA
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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]];
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CROSSREFS
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Leading diagonal gives A000290, n >= 1.
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KEYWORD
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AUTHOR
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STATUS
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approved
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