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Irregular triangle read by rows listing divisors d of n in order of appearance in a matrix of products that arranges the powers of prime divisors p of n along independent axes.
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%I #12 Jul 28 2016 21:47:50

%S 1,1,2,1,3,1,2,4,1,5,1,2,3,6,1,7,1,2,4,8,1,3,9,1,2,5,10,1,11,1,2,4,3,

%T 6,12,1,13,1,2,7,14,1,3,5,15,1,2,4,8,16,1,17,1,2,3,6,9,18,1,19,1,2,4,

%U 5,10,20,1,3,7,21,1,2,11,22,1,23,1,2,4,8,3,6,12,24,1,5,25,1,2,13,26,1,3

%N Irregular triangle read by rows listing divisors d of n in order of appearance in a matrix of products that arranges the powers of prime divisors p of n along independent axes.

%C a(p^e) = A027750(p^e) for e >= 1.

%C The matrix of products that are divisors of n is arranged such that the powers of the prime divisors range across an axis, one axis per prime divisor. Thus a squarefree semiprime has a 2-dimensional matrix, a sphenic number has 3 dimensions, etc.

%C Generally, the number of dimensions for the matrix of divisors = omega(n) = A001221(n). Because of this, tau(n)*(mod omega(n)) = 0 for n > 1.

%C This follows from the formula for tau(n).

%C Prime divisors p of n are considered in numerical order.

%C Product matrix of tensors T = 1,p,p^2,...,p^e that include the powers 1 <= e of the prime divisor p that divide n.

%H Michael De Vlieger, <a href="/A275055/b275055.txt">Table of n, a(n) for n = 1..11214</a> (Rows 1 <= n <= 1500)

%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/Divisor.html">Divisor</a>

%e Triangle begins:

%e 1;

%e 1, 2;

%e 1, 3;

%e 1, 2, 4;

%e 1, 5;

%e 1, 2, 3, 6;

%e 1, 7;

%e 1, 2, 4, 8;

%e 1, 3, 9;

%e 1, 2, 5, 10;

%e 1, 11;

%e 1, 2, 4, 3, 6, 12;

%e 1, 13;

%e 1, 2, 7, 14;

%e 1, 3, 5, 15;

%e 1 2, 4, 8, 16;

%e 1, 17;

%e 1, 2, 3, 6, 9, 18;

%e ...

%e 2 prime divisors: n = 72

%e 1 2 4 8

%e 3 6 12 24

%e 9 18 36 72

%e thus a(72) = {1, 2, 4, 8, 3, 6, 12, 24, 9, 18, 36, 72}

%e 3 prime divisors: n = 60

%e (the 3 dimensional levels correspond with powers of 5)

%e level 5^0: level 5^1:

%e 1 2 4 | 5 10 20

%e 3 6 12 | 15 30 60

%e thus a(60) = {1, 2, 4, 3, 6, 12, 5, 10, 20, 15, 30, 60}

%e 4 prime divisors: n = 210

%e (the 3 dimensional levels correspond with powers of 5,

%e the 4 dimensional levels correspond with powers of 7)

%e level 5^0*7^0: level 5^1*7^0:

%e 1 2 | 5 10

%e 3 6 | 15 30

%e level 5^0*7^1: level 5^1*7^1:

%e 7 14 | 35 70

%e 21 42 | 105 210

%e thus a(210) = {1,2,3,6,5,10,15,30,7,14,21,42,35,70,105,210}

%t {{1}}~Join~Table[TensorProduct @@ Reverse@ Apply[PowerRange[1, #1^#2, #1] &, # &@ FactorInteger@ n, 1], {n, 2, 30}] // Flatten

%Y Cf. A027750, A000005 (row length), A000203 (row sums), A056538.

%K nonn,easy,tabf

%O 1,3

%A _Michael De Vlieger_, Jul 14 2016