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A299768
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Triangle read by rows: T(n,k) = sum of all squares of the parts k in all partitions of n, with n >= 1, 1 <= k <= n.
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5
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1, 2, 4, 4, 4, 9, 7, 12, 9, 16, 12, 16, 18, 16, 25, 19, 32, 36, 32, 25, 36, 30, 44, 54, 48, 50, 36, 49, 45, 76, 81, 96, 75, 72, 49, 64, 67, 104, 135, 128, 125, 108, 98, 64, 81, 97, 164, 189, 208, 200, 180, 147, 128, 81, 100, 139, 224, 279, 288, 300, 252, 245, 192, 162, 100, 121
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OFFSET
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1,2
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LINKS
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FORMULA
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EXAMPLE
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Triangle begins:
1;
2, 4;
4, 4, 9;
7, 12, 9, 16;
12, 16, 18, 16, 25,
19, 32, 36, 32, 25, 36;
30, 44, 54, 48, 50, 36, 49;
...
For n = 4 the partitions of 4 are [4], [2, 2], [3, 1], [2, 1, 1], [1, 1, 1, 1], so the squares of the parts are respectively [16], [4, 4], [9, 1], [4, 1, 1], [1, 1, 1, 1]. The sum of the squares of the parts 1 is 1 + 1 + 1 + 1 + 1 + 1 + 1 = 7. The sum of the squares of the parts 2 is 4 + 4 + 4 = 12. The sum of the squares of the parts 3 is 9. The sum of the squares of the parts 4 is 16. So the fourth row of triangle is [7, 12, 9, 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)):
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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] + # + (Coefficient[#, x, 0]*i^2*x^i)&[b[n - i, Min[n - i, i]]]];
T[n_] := Table[Coefficient[#, x, i], {i, 1, n}]&[b[n, n]];
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PROG
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(PARI) row(n) = {v = vector(n); forpart(p=n, for(k=1, #p, v[p[k]] += p[k]^2; ); ); v; } \\ Michel Marcus, Mar 20 2018
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CROSSREFS
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Leading diagonal is A000290, n >= 1.
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KEYWORD
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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