A290532 - OEIS

%I #43 Dec 28 2021 04:18:02

%S 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,

%T 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,

%U 2,4,1,2,2,4,1,2,1,2,2,3,4,4,6,8,1,2,3

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

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

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

%C 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:

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

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

%C 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;

%C ...

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

%C 1, 1;

%C 1, 2, 1;

%C 1, 3, 3, 1;

%C 1, 4, 6, 4, 1;

%C ...

%C 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;

%C 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;

%C 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;

%C ...

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

%C 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.

%H Robert Israel, <a href="/A290532/b290532.txt">Table of n, a(n) for n = 1..10006</a> (rows 1 to 1358, flattened)

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

%e 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.

%e Triangle begins:

%e   1;

%e   1, 2;

%e   1, 2;

%e   1, 2, 3;

%e   1, 2;

%e   1, 2, 2, 4;

%e   1, 2;

%e   1, 2, 3, 4;

%e   1, 2, 3;

%e   1, 2, 2, 4;

%e   ...

%p with(numtheory):nn:=100:

%p for n from 1 to nn do:

%p   d1:=divisors(n):n1:=nops(d1):

%p    for i from 1 to n1 do:

%p      n2:=tau(d1[i]):

%p      printf(`%d, `,n2):

%p    od:

%p od:

%t Table[DivisorSigma[0, Divisors@ n], {n, 25}] // Flatten (* _Michael De Vlieger_, Aug 07 2017 *)

%o (PARI) row(n) = apply(numdiv, divisors(n)); \\ _Michel Marcus_, Dec 27 2021

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

%K nonn,tabf

%O 1,3

%A _Michel Lagneau_, Aug 05 2017