A285321 Square array A(1,k) = A019565(k), A(n,k) = A065642(A(n-1,k)), read by descending antidiagonals.

2, 3, 4, 6, 9, 8, 5, 12, 27, 16, 10, 25, 18, 81, 32, 15, 20, 125, 24, 243, 64, 30, 45, 40, 625, 36, 729, 128, 7, 60, 75, 50, 3125, 48, 2187, 256, 14, 49, 90, 135, 80, 15625, 54, 6561, 512, 21, 28, 343, 120, 225, 100, 78125, 72, 19683, 1024

Offset: 1

Author

Antti Karttunen, Apr 17 2017

Keywords

• nonn
• tabl

Comments

A permutation of the natural numbers > 1.

Otherwise like array A284311, but the columns come in different order.

Examples

			The top left 12x6 corner of the array:
   2,   3,  6,     5,  10,  15,  30,      7,  14,  21,  42,   35
   4,   9, 12,    25,  20,  45,  60,     49,  28,  63,  84,  175
   8,  27, 18,   125,  40,  75,  90,    343,  56, 147, 126,  245
  16,  81, 24,   625,  50, 135, 120,   2401,  98, 189, 168,  875
  32, 243, 36,  3125,  80, 225, 150,  16807, 112, 441, 252, 1225
  64, 729, 48, 15625, 100, 375, 180, 117649, 196, 567, 294, 1715
		

Links

• Antti Karttunen, Table of n, a(n) for n = 1..120; the first 15 antidiagonals of array

Crossrefs

Transpose: A285322.

Cf. A019565, A065642.

Cf. A008479 (index of the row where n is located), A087207 (of the column).

Cf. arrays A284311, A285325, also A285332.

Programs

Mathematica

a065642[n_] := Module[{k}, If[n == 1, Return[1], k = n + 1; While[ EulerPhi[k]/k != EulerPhi[n]/n, k++]]; k];
A[1, k_] := Times @@ Prime[Flatten[Position[#, 1]]]&[Reverse[ IntegerDigits[k, 2]]];
A[n_ /; n > 1, k_] := A[n, k] = a065642[A[n - 1, k]];
Table[A[n - k + 1, k], {n, 1, 10}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Nov 17 2019 *)

Python

from operator import mul
from sympy import prime, primefactors
def a019565(n): return reduce(mul, (prime(i+1) for i, v in enumerate(bin(n)[:1:-1]) if v == '1')) if n > 0 else 1 # This function from Chai Wah Wu
def a007947(n): return 1 if n<2 else reduce(mul, primefactors(n))
def a065642(n):
    if n==1: return 1
    r=a007947(n)
    n = n + r
    while a007947(n)!=r:
        n+=r
    return n
def A(n, k): return a019565(k) if n==1 else a065642(A(n - 1, k))
for n in range(1, 11): print([A(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, Apr 18 2017

Scheme

(define (A285321 n) (A285321bi (A002260 n) (A004736 n)))
(define (A285321bi row col) (if (= 1 row) (A019565 col) (A065642 (A285321bi (- row 1) col))))

Formula

A(1,k) = A019565(k), A(n,k) = A065642(A(n-1,k)).

For all n >= 2: A(A008479(n), A087207(n)) = n.