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).
Row sums = A061395(n).
Row maxima = A286469(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) = A000720(p) for p prime.
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).
a(A002110(n)) = an array of n 1s.
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 k-th place from the right. For k > 1, increment the k-th place to the right of the last-incremented 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
Michael De Vlieger, Table of n, a(n) for n = 1..15568 (rows 1 <= n <= 5000).
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
Michael De Vlieger, May 23 2017
STATUS
approved