A290532 Irregular triangle read by rows in which row n lists the number of divisors of each divisor of n.

1, 1, 2, 1, 2, 1, 2, 3, 1, 2, 1, 2, 2, 4, 1, 2, 1, 2, 3, 4, 1, 2, 3, 1, 2, 2, 4, 1, 2, 1, 2, 2, 3, 4, 6, 1, 2, 1, 2, 2, 4, 1, 2, 2, 4, 1, 2, 3, 4, 5, 1, 2, 1, 2, 2, 4, 3, 6, 1, 2, 1, 2, 3, 2, 4, 6, 1, 2, 2, 4, 1, 2, 2, 4, 1, 2, 1, 2, 2, 3, 4, 4, 6, 8, 1, 2, 3

Offset

1,3

Comments

Or, in the triangle A027750, replace each element with the number of its divisors.

The row of index n = p^m (p prime and m >= 0) is equal to (1, 2, ..., m + 1);

We observe an interesting property when the index n of the row n is the product of k distinct primes, k = 1,2,... For example:

The index n is prime => row n = (1, 2);

The index n equals the product of two distinct primes => row n = (1, 2, 2, 4);

The index n equals the product of three distinct primes => row n = (1, 2, 2, 2, 4, 4, 4, 8) or a permutation of the elements;

...

Let us now consider Pascal's triangle (A007318(n) for n > 0):

1, 1;

1, 2, 1;

1, 3, 3, 1;

1, 4, 6, 4, 1;

...

Row 1 of Pascal's triangle gives the number of "1" and the number of "2" respectively belonging to the row of index n = prime(m) of the sequence;

Row 2 of Pascal's triangle gives the number of "1", the number of "2" and the number of "4" respectively belonging to the row of index n = p*q of the sequence, where p and q are distinct primes;

Row 3 of Pascal's triangle gives the number of "1", the number of "2", the number of "4" and the number of "8" respectively belonging to the row of index n = p*q*r of the sequence, where p, q and r are distinct primes;

...

It is now easy to generalize this process by the following proposition.

Proposition: binomial(m,k) is the number of terms of the form 2^k belonging to the row of index n in the sequence when n is the product of m distinct primes.

Links

Robert Israel, Table of n, a(n) for n = 1..10006 (rows 1 to 1358, flattened)

Formula

T(n, k) = tau(A027750(n, k)).

Example

Row 6 is (1, 2, 2, 4) because the 6th row of A027750 is [1, 2, 3, 6] and tau(1) = 1, tau(2) = 2, tau(3) = 2 and tau(6) = 4.
Triangle begins:
  1;
  1, 2;
  1, 2;
  1, 2, 3;
  1, 2;
  1, 2, 2, 4;
  1, 2;
  1, 2, 3, 4;
  1, 2, 3;
  1, 2, 2, 4;
  ...

Maple

with(numtheory):nn:=100:
for n from 1 to nn do:
  d1:=divisors(n):n1:=nops(d1):
   for i from 1 to n1 do:
     n2:=tau(d1[i]):
     printf(`%d, `, n2):
   od:
od:

Mathematica

Table[DivisorSigma[0, Divisors@ n], {n, 25}] // Flatten (* Michael De Vlieger, Aug 07 2017 *)

Prog

(PARI) row(n) = apply(numdiv, divisors(n)); \\ Michel Marcus, Dec 27 2021

Crossrefs

Cf. A000005, A007318, A027750, A084997, A290478.

Sequence in context: A067815 A133780 A270808 * A080237 A136109 A105265

Adjacent sequences: A290529 A290530 A290531 * A290533 A290534 A290535

Keyword

nonn,tabf

Author

Michel Lagneau, Aug 05 2017

Status

approved