

A287352


Irregular triangle T(n,k) = A112798(n,1) followed by first differences of A112798(n).


22



0, 1, 2, 1, 0, 3, 1, 1, 4, 1, 0, 0, 2, 0, 1, 2, 5, 1, 0, 1, 6, 1, 3, 2, 1, 1, 0, 0, 0, 7, 1, 1, 0, 8, 1, 0, 2, 2, 2, 1, 4, 9, 1, 0, 0, 1, 3, 0, 1, 5, 2, 0, 0, 1, 0, 3, 10, 1, 1, 1, 11, 1, 0, 0, 0, 0, 2, 3, 1, 6, 3, 1, 1, 0, 1, 0, 12, 1, 7, 2, 4, 1, 0, 0, 2, 13
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OFFSET

1,3


COMMENTS

Irregular triangle T(n,k) = first differences of indices of prime divisors p of n.
Row lengths = (big) Omega(n) = A001222(n).
We can concatenate the rows 1 <= n <= 28 as none of the values of k in this range exceed 9: {0, 1, 2, 10, 3, 11, 4, 100, 20, 12, 5, 101, 6, 13, 21, 1000, 7, 110, 8, 102, 22, 14, 9, 1001, 30, 15, 200, 103}; a(29) = {10}, which would require a digit greater than 9.
a(1) = 0 by convention.
a(0) is not defined (i.e., null set). a(n) is defined for positive nonzero n.
a(p^e) = A000720(p) followed by (e  1) zeros.
a(Product(p^e)) is the concatenation of the a(p^e) of the unitary prime power divisors p^e of n, sorted by the prime p (i.e. the function a(n) mapped across the terms of row n of A141809).
T(n,k) could be used to furnish A054841(n). We read data in row n of T(n,k). If T(n,1) = 0, then write 0. If T(n,1) > 0, then increment the kth place from the right. For k > 1, increment the kth place to the right of the lastincremented place.
T(n,k) can be used to render n in decimal. If T(n,1) = 0, then write 1. If T(n,1) > 0, then multiply 1 by A000720(T(n,1)). For k > 1, multiply the previous product by pi(x) = A000720(x) of the running total of T(n,k) for each k.
Ignoring zeros in row n > 1 and decoding the remaining values of T(n,k) as immediately above yields the squarefree kernel of n = A007947(n).
Leading zeros of a(n) are trimmed, but as in decimal notation numbers that include leading zeros symbolize the same n as without them. Zeros that precede nonzero values merely multiply implicit 1 by itself until we encounter nonzero values. Thus, {0,0,2} = 1*1*pi(2) = 3, as {2} = pi(2) = 3. Because of this no row n > 1 has 0 for k = 1 of T(n,k).
Interpreting n written in binary as a row of a(n) yields A057335(n).


LINKS



FORMULA



EXAMPLE

a(1) = {0} by convention.
a(2) = {pi(2)} = {1}.
a(4) = {pi(2), pi(2)  pi(2)}, = {1, 0} since 4 = 2 * 2.
a(6) = {pi(2), pi(3)  pi(2)} = {1, 1} since 6 = 2 * 3.
a(12) = {pi(2), pi(2)  pi(2), pi(3)  pi(2)  pi(2)} = {1, 0, 1}, since 12 = 2 * 2 * 3.
The triangle starts:
1: 0;
2: 1;
3: 2;
4: 1, 0;
5: 3;
6: 1, 1;
7: 4;
8: 1, 0, 0;
9: 2, 0;
10: 1, 2;
11: 5;
12: 1, 0, 1;
13: 6;
14: 1, 3;
15: 2, 1;
16: 1, 0, 0, 0;
17: 7;
18: 1, 1, 0;
19: 8;
20: 1, 0, 2;
...


MATHEMATICA

Table[Prepend[Differences@ #, First@ #] & Flatten[FactorInteger[n] /. {p_, e_} /; p > 0 :> ConstantArray[PrimePi@ p, e]], {n, 41}] // Flatten (* Michael De Vlieger, May 23 2017 *)


CROSSREFS



KEYWORD

nonn,tabf,easy


AUTHOR



STATUS

approved



