OFFSET
1,12
COMMENTS
To base the triangle on (0, 0) a column (1, 0, 0, ...) has to be appended to the left hand side of the triangle. To compute this triangle with Michael De Vlieger's Mathematica program only the ranges of the indices have to be adapted. The SageMath program computes the extended triangle by default. - Peter Luschny, Aug 24 2019
LINKS
Michael De Vlieger, Table of n, a(n) for n = 1..5050 (rows 1 <= n <= 100)
FORMULA
T(n, k) = Sum_{d|n} Moebius(d) * A008284(n/d, k) for n >= 1, T(0, 0) = 1. - Peter Luschny, Aug 24 2019
EXAMPLE
Triangle begins:
n/k: 1, 2, 3, 4, 5, 6, 7, 8, ...
1: 1;
2: 0, 1;
3: 0, 1, 1;
4: 0, 1, 1, 1;
5: 0, 2, 2, 1, 1;
6: 0, 1, 2, 2, 1, 1;
7: 0, 3, 4, 3, 2, 1, 1;
8: 0, 2, 4, 4, 3, 2, 1, 1;
9: 0, 3, 6, 6, 5, 3, 2, 1, 1;
10: 0, 2, 6, 8, 6, 5, 3, 2, 1, 1;
11: 0, 5, 10, 11, 10, 7, 5, 3, 2, 1, 1;
12: 0, 2, 8, 12, 12, 10, 7, 5, 3, 2, 1, 1;
...
The partitions with their gcd value for n=8, k=2..5:
(1, 7)=1, (2, 6)=2, (3, 5)=1, (4, 4)=4, so T(8,2)=2.
(1, 1, 6)=1, (1, 2, 5)=1, (1, 3, 4)=1, (2, 2, 4)=2, (2, 3, 3)=1, so T(8,2)=4.
(1, 1, 1, 5)=1, (1, 1, 2, 4)=1, (1, 1, 3, 3)=1, (1, 2, 2, 3)=1, (2, 2, 2, 2)=2, so T(8,3)=4.
(1, 1, 1, 1, 4)=1, (1, 1, 1, 2, 3)=1, (1, 1, 2, 2, 2)=1, so T(8,4)=3.
(1, 1, 1, 1, 1, 3)=1, (1, 1, 1, 1, 2, 2)=1, so T(8,5)=2.
MATHEMATICA
Table[Length@ Select[IntegerPartitions[n, {k}], GCD @@ # == 1 &], {n, 13}, {k, n}] // Flatten (* Michael De Vlieger, Mar 08 2017 *)
PROG
DivisorTriangle(moebius, A008284, 13) # Peter Luschny, Aug 24 2019
CROSSREFS
KEYWORD
nonn,tabl
AUTHOR
N. J. A. Sloane, Mar 05 2017
EXTENSIONS
Corrected a(30)-a(32) and more terms from Lars Blomberg, Mar 08 2017
STATUS
approved