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A339304
Irregular triangle read by rows T(n,k) in which row n has length the partition number A000041(n-1) and columns k give the number of divisors function A000005, 1 <= k <= n.
13
1, 2, 2, 1, 3, 2, 1, 2, 2, 2, 1, 1, 4, 3, 2, 2, 2, 1, 1, 2, 2, 3, 2, 2, 2, 2, 1, 1, 1, 1, 4, 4, 2, 3, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 3, 2, 4, 2, 2, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 4, 4, 2, 4, 4, 2, 2, 3, 3, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1
OFFSET
1,2
COMMENTS
T(n,k) is also the number of divisors of A336811(n,k).
Conjecture: the sum of row n equals A138137(n), the total number of parts in the last section of the set of partitions of n.
LINKS
Paolo Xausa, Table of n, a(n) for n = 1..11732 (rows 1..27 of the triangle, flattened)
FORMULA
a(m) = A000005(A336811(m)).
T(n,k) = A000005(A336811(n,k).
EXAMPLE
Triangle begins:
1;
2;
2, 1;
3, 2, 1;
2, 2, 2, 1, 1;
4, 3, 2, 2, 2, 1, 1;
2, 2, 3, 2, 2, 2, 2, 1, 1, 1, 1;
4, 4, 2, 3, 3, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1;
3, 2, 4, 2, 2, 3, 3, 2, 2, 2, 2, 2, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1;
...
MATHEMATICA
A339304row[n_]:=Flatten[Table[ConstantArray[DivisorSigma[0, n-m], PartitionsP[m]-PartitionsP[m-1]], {m, 0, n-1}]]; Array[A339304row, 10] (* Paolo Xausa, Sep 01 2023 *)
CROSSREFS
Number of divisors of A336811.
Row n has length A000041(n-1).
Every column gives A000005.
Row sums give A138137 (conjectured).
Sequence in context: A281013 A190683 A181810 * A237578 A026146 A325519
KEYWORD
nonn,tabf
AUTHOR
Omar E. Pol, Nov 29 2020
STATUS
approved