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A195310
Triangle read by rows with T(n,k) = n - A001318(k), n >= 1, k >= 1, if (n - A001318(k)) >= 0.
24
0, 1, 0, 2, 1, 3, 2, 4, 3, 0, 5, 4, 1, 6, 5, 2, 0, 7, 6, 3, 1, 8, 7, 4, 2, 9, 8, 5, 3, 10, 9, 6, 4, 11, 10, 7, 5, 0, 12, 11, 8, 6, 1, 13, 12, 9, 7, 2, 14, 13, 10, 8, 3, 0, 15, 14, 11, 9, 4, 1, 16, 15, 12, 10, 5, 2, 17, 16, 13, 11, 6, 3, 18, 17, 14, 12, 7, 4
OFFSET
1,4
COMMENTS
Also triangle read by rows in which column k lists the nonnegative integers A001477 starting at the row A001318(k). This sequence is related to Euler's Pentagonal Number Theorem. A000041(a(n)) gives the absolute value of A175003(n). To get the number of partitions of n see the example.
FORMULA
A175003(n,k) = A057077(k-1)*A000041(T(n,k)), n >= 1, k >= 1.
EXAMPLE
Written as a triangle:
0;
1, 0;
2, 1;
3, 2;
4, 3, 0;
5, 4, 1;
6, 5, 2, 0;
7, 6, 3, 1;
8, 7, 4, 2;
9, 8, 5, 3;
10, 9, 6, 4;
11, 10, 7, 5, 0;
12, 11, 8, 6, 1;
13, 12, 9, 7, 2;
14, 13, 10, 8, 3, 0;
.
For n = 15, consider row 15 which lists the numbers 14, 13, 10, 8, 3, 0. From Euler's Pentagonal Number Theorem we have that the number of partitions of 15 is p(15) = p(14) + p(13) - p(10) - p(8) + p(3) + p(0) = 135 + 101 - 42 - 22 + 3 + 1 = 176.
MATHEMATICA
rows = 20;
a1318[n_] := If[EvenQ[n], n(3n/2+1)/4, (n+1)(3n+1)/8];
T[n_, k_] := n - a1318[k];
Table[DeleteCases[Table[T[n, k], {k, 1, n}], _?Negative], {n, 1, rows}] // Flatten (* Jean-François Alcover, Sep 22 2018 *)
CROSSREFS
Row sums give A195311.
Sequence in context: A185314 A285120 A282744 * A051282 A274121 A052306
KEYWORD
nonn,tabf
AUTHOR
Omar E. Pol, Sep 21 2011
EXTENSIONS
Name essentially suggested by Franklin T. Adams-Watters (see history), Sep 21 2011
STATUS
approved