

A290532


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


2



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
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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



