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A340583
Triangle read by rows: T(n,k) = A002865(n-k)*A000203(k), 1 <= k <= n.
4
1, 0, 3, 1, 0, 4, 1, 3, 0, 7, 2, 3, 4, 0, 6, 2, 6, 4, 7, 0, 12, 4, 6, 8, 7, 6, 0, 8, 4, 12, 8, 14, 6, 12, 0, 15, 7, 12, 16, 14, 12, 12, 8, 0, 13, 8, 21, 16, 28, 12, 24, 8, 15, 0, 18, 12, 24, 28, 28, 24, 24, 16, 15, 13, 0, 12, 14, 36, 32, 49, 24, 48, 16, 30, 13, 18, 0, 28
OFFSET
1,3
COMMENTS
T(n,k) is the total number of cubic cells added at n-th stage to the right prisms whose bases are the parts of the symmetric representation of sigma(k) in the polycube described in A221529.
Partial sums of column k gives the column k of A221529.
EXAMPLE
Triangle begins:
1;
0, 3;
1, 0, 4;
1, 3, 0, 7;
2, 3, 4, 0, 6;
2, 6, 4, 7, 0, 12;
4, 6, 8, 7, 6, 0, 8;
4, 12, 8, 14, 6, 12, 0, 15;
7, 12, 16, 14, 12, 12, 8, 0, 13;
8, 21, 16, 28, 12, 24, 8, 15, 0, 18;
12, 24, 28, 28, 24, 24, 16, 15, 13, 0, 12;
14, 36, 32, 49, 24, 48, 16, 30, 13, 18, 0, 28;
...
For n = 6 the calculation of every term of row 6 is as follows:
--------------------------
k A000203 T(6,k)
--------------------------
1 1 * 2 = 2
2 3 * 2 = 6
3 4 * 1 = 4
4 7 * 1 = 7
5 6 * 0 = 0
6 12 * 1 = 12
--------------------------
The sum of row 6 is 2 + 6 + 4 + 7 + 0 + 12 = 31, equaling A138879(6).
MATHEMATICA
A340583[n_, k_] := (PartitionsP[n - k] - PartitionsP[(n - k) - 1])*
DivisorSigma[1, k];
Table[A340583[n, k], {n, 1, 12}, {k, 1, n}] // Flatten (* Robert P. P. McKone, Jan 25 2021 *)
CROSSREFS
Row sums give A138879.
Column 1 gives A002865.
Diagonals 1, 3 and 4 give A000203.
Diagonal 2 gives A000004.
Diagonals 5 and 6 give A074400.
Diagonals 7 and 8 give A239050.
Diagonal 9 gives A319527.
Diagonal 10 gives A319528.
Cf. A221529 (partial column sums).
Cf. A340426 (mirror).
Sequence in context: A127569 A117372 A127570 * A292506 A212186 A274662
KEYWORD
nonn,tabl
AUTHOR
Omar E. Pol, Jan 15 2021
STATUS
approved