OFFSET
1,5
COMMENTS
The right diagonal labeled by the prime power of the form j:k = (prime(j))^k contains the j^th power primes in the factorization raised to the k^th power. For example, the right diagonal labeled by the number 2 = 1:1 = (prime(1))^1 contains the power-free parts of each positive integer, specifically A055231 and the right diagonal labeled by the number 4 = 1:2 = (prime(1))^2 contains the squares of the squarefree parts of positive integers.
In general, then the right diagonal labeled by m = (j_i : k_i)_i = Product_i prime(j_i)^(k_i) contains the product over i of the (j_i)th power primes in the factorization raised to the (k_i)th powers.
For example, the operator 5 = 3:1 extracts the 3rd power primes in the factorization of each n and raises them to the first power, thus sending 8 = 1:3 to 2 = 1:1, 27 = 2:3 to 3 = 2:1 and so on.
LINKS
J. Awbrey, Riffs and Rotes
FORMULA
If k = Product p_i^e_i, A(n,k) = p_i^A286561(n, A000040(e_i)), where A286561(x,y) gives the y-valuation of x. - Antti Karttunen, Nov 16 2019
EXAMPLE
` ` ` ` ` ` ` ` ` ` `n o m
` ` ` ` ` ` ` ` ` ` ` \ /
` ` ` ` ` ` ` ` ` ` `1 . 1
` ` ` ` ` ` ` ` ` ` \ / \ /
` ` ` ` ` ` ` ` ` `2 . 1 . 2
` ` ` ` ` ` ` ` ` \ / \ / \ /
` ` ` ` ` ` ` ` `3 . 1 . 1 . 3
` ` ` ` ` ` ` ` \ / \ / \ / \ /
` ` ` ` ` ` ` `4 . 1 . 2 . 1 . 4
` ` ` ` ` ` ` \ / \ / \ / \ / \ /
` ` ` ` ` ` `5 . 1 . 3 . 1 . 1 . 5
` ` ` ` ` ` \ / \ / \ / \ / \ / \ /
` ` ` ` ` `6 . 1 . 1 . 1 . 4 . 1 . 6
` ` ` ` ` \ / \ / \ / \ / \ / \ / \ /
` ` ` ` `7 . 1 . 5 . 2 . 9 . 1 . 1 . 7
` ` ` ` \ / \ / \ / \ / \ / \ / \ / \ /
` ` ` `8 . 1 . 6 . 1 . 1 . 1 . 2 . 1 . 8
` ` ` \ / \ / \ / \ / \ / \ / \ / \ / \ /
` ` `9 . 1 . 7 . 1 . 25. 1 . 3 . 1 . 1 . 9
` ` \ / \ / \ / \ / \ / \ / \ / \ / \ / \ /
` 10 . 1 . 1 . 1 . 36. 1 . 2 . 1 . 8 . 1 . 10
Primal codes of finite partial functions on positive integers:
1 = { }
2 = 1:1
3 = 2:1
4 = 1:2
5 = 3:1
6 = 1:1 2:1
7 = 4:1
8 = 1:3
9 = 2:2
10 = 1:1 3:1
11 = 5:1
12 = 1:2 2:1
13 = 6:1
14 = 1:1 4:1
15 = 2:1 3:1
16 = 1:4
17 = 7:1
18 = 1:1 2:2
19 = 8:1
20 = 1:2 3:1
From Antti Karttunen, Nov 16 2019: (Start)
When the sequence is viewed as a square array read by falling antidiagonals, the top left 15 X 15 corner looks like this:
k= | 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
----+--------------------------------------------------------------------
n= 1| 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
2| 1, 2, 3, 1, 5, 6, 7, 1, 1, 10, 11, 3, 13, 14, 15,
3| 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 1,
4| 1, 4, 9, 1, 25, 36, 49, 1, 1, 100, 121, 9, 169, 196, 225,
5| 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1,
6| 1, 2, 3, 2, 5, 6, 7, 1, 3, 10, 11, 6, 13, 14, 15,
7| 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
8| 1, 8, 27, 1, 125, 216, 343, 1, 1, 1000, 1331, 27, 2197, 2744, 3375,
9| 1, 1, 1, 4, 1, 1, 1, 1, 9, 1, 1, 4, 1, 1, 1,
10| 1, 2, 3, 1, 5, 6, 7, 2, 1, 10, 11, 3, 13, 14, 15,
11| 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
12| 1, 4, 9, 2, 25, 36, 49, 1, 3, 100, 121, 18, 169, 196, 225,
13| 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
14| 1, 2, 3, 1, 5, 6, 7, 1, 1, 10, 11, 3, 13, 14, 15,
15| 1, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 2, 1, 1, 1,
(End)
PROG
(PARI)
up_to = 105;
A106177sq(n, k) = { my(f = factor(k)); prod(i=1, #f~, f[i, 1]^valuation(n, prime(f[i, 2]))); };
A106177list(up_to) = { my(v = vector(up_to), i=0); for(a=1, oo, for(col=1, a, i++; if(i > up_to, return(v)); v[i] = A106177sq(col, (a-(col-1))))); (v); };
v106177 = A106177list(up_to);
A106177(n) = v106177[n]; \\ Antti Karttunen, Nov 16 2019
CROSSREFS
AUTHOR
Jon Awbrey, May 23 2005
STATUS
approved