A242378 Square array read by antidiagonals: to obtain A(i,j), replace each prime factor prime(k) in prime factorization of j with prime(k+i).

0, 1, 0, 2, 1, 0, 3, 3, 1, 0, 4, 5, 5, 1, 0, 5, 9, 7, 7, 1, 0, 6, 7, 25, 11, 11, 1, 0, 7, 15, 11, 49, 13, 13, 1, 0, 8, 11, 35, 13, 121, 17, 17, 1, 0, 9, 27, 13, 77, 17, 169, 19, 19, 1, 0, 10, 25, 125, 17, 143, 19, 289, 23, 23, 1, 0, 11, 21, 49, 343, 19, 221, 23, 361, 29, 29, 1, 0

Offset: 0

Author

Antti Karttunen, May 12 2014

Keywords

• nonn
• tabl
• changed

Comments

Each row i is a multiplicative function, being in essence "the i-th power" of A003961, i.e., A(i,j) = A003961^i (j). Zeroth power gives an identity function, A001477, which occurs as the row zero.

The terms in the same column have the same prime signature.

The array is read by antidiagonals: A(0,0), A(0,1), A(1,0), A(0,2), A(1,1), A(2,0), ... .

Examples

			The top-left corner of the array:
  0,   1,   2,   3,   4,   5,   6,   7,   8, ...
  0,   1,   3,   5,   9,   7,  15,  11,  27, ...
  0,   1,   5,   7,  25,  11,  35,  13, 125, ...
  0,   1,   7,  11,  49,  13,  77,  17, 343, ...
  0,   1,  11,  13, 121,  17, 143,  19,1331, ...
  0,   1,  13,  17, 169,  19, 221,  23,2197, ...
...
A(2,6) = A003961(A003961(6)) = p_{1+2} * p_{2+2} = p_3 * p_4 = 5 * 7 = 35, because 6 = 2*3 = p_1 * p_2.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 0..10439; Antidiagonals n = 0..143, flattened

Crossrefs

Taking every second column from column 2 onward gives array A246278 which is a permutation of natural numbers larger than 1.

Transpose: A242379.

Row 0: A001477, Row 1: A003961 (from 1 onward), Row 2: A357852 (from 1 onward), Row 3: A045968 (from 7 onward), Row 4: A045970 (from 11 onward).

Column 2: A000040 (primes), Column 3: A065091 (odd primes), Column 4: A001248 (squares of primes), Column 6: A006094 (products of two successive primes), Column 8: A030078 (cubes of primes).

Excluding column 0, a subtable of A297845.

Permutations whose formulas refer to this array: A122111, A241909, A242415, A242419, A246676, A246678, A246684.

Formula

A(0,j) = j, A(i,0) = 0, A(i > 0, j > 0) = A003961(A(i-1,j)).

For j > 0, A(i,j) = A297845(A000040(i+1),j) = A297845(j,A000040(i+1)). - Peter Munn, Sep 02 2025