A285730 Square array: If A001222(n) < k, then A(n,k) = n, otherwise A(n,k) = product of k largest prime factors of n (taken with multiplicity), read by descending antidiagonals.

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

Offset: 1

Author

Antti Karttunen, Apr 28 2017

Keywords

• nonn
• tabl

Comments

Square array A(n,k) [where n is row and k is column] is read by descending antidiagonals: A(1,1), A(1,2), A(2,1), A(1,3), A(2,2), A(3,1), etc

Examples

			The top left 5x18 corner of the array:
   1,  1,  1,  1,  1
   2,  2,  2,  2,  2
   3,  3,  3,  3,  3
   2,  4,  4,  4,  4
   5,  5,  5,  5,  5
   3,  6,  6,  6,  6
   7,  7,  7,  7,  7
   2,  4,  8,  8,  8
   3,  9,  9,  9,  9
   5, 10, 10, 10, 10
  11, 11, 11, 11, 11
   3,  6, 12, 12, 12
  13, 13, 13, 13, 13
   7, 14, 14, 14, 14
   5, 15, 15, 15, 15
   2,  4,  8, 16, 16
  17, 17, 17, 17, 17
   3,  9, 18, 18, 18
For A(18,1) we take just the largest prime factor of 18 = 2*3*3, thus A(18,1) = 3.
For A(18,2) we take the product of two largest prime factors of 18 (duplicates not discarded), thus A(18,2) = 3*3 = 9.
For A(18,3) we take the product of three largest prime factors of 18, thus A(18,2) = 3*3*2 = 18.
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..10440; the first 144 antidiagonals of array

Crossrefs

Transpose: A285731.

Cf. A006530 (the leftmost column).

Cf. A001222, A052126.

Programs

Mathematica

With[{nn = 14}, Function[s, Table[s[[#, k]] &[n - k + 1], {n, nn}, {k, n, 1, -1}]]@ MapIndexed[PadRight[#1, nn, First@ #2] &, Table[FoldList[Times, Reverse@ Flatten[FactorInteger[n] /. {p_, e_} /; e > 0 :> ConstantArray[p, e]]], {n, nn}]]] // Flatten (* Michael De Vlieger, Apr 28 2017 *)

Python

from sympy import primefactors
def a006530(n): return 1 if n==1 else max(primefactors(n))
def A(n, k): return a006530(n) if k==1 else a006530(n)*A(n//a006530(n), k - 1)
for n in range(1, 21): print([A(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Apr 28 2017

Scheme

(define (A285730 n) (A285730bi (A002260 n) (A004736 n)))
(define (A285730bi row col) (let loop ((n row) (k col) (m 1)) (if (zero? k) m (loop (/ n (A006530 n)) (- k 1) (* m (A006530 n))))))
;; Alternatively, implemented with the given recurrence formula:
(define (A285730bi row col) (if (= 1 col) (A006530 row) (* (A006530 row) (A285730bi (A052126 row) (- col 1)))))

Formula

A(n,1) = A006530(n), for k > 1, A(n,k) = A006530(n) * A(n/A006530(n),k-1).