A178638 -id:A178638 - cp's OEIS frontend

A286561 Square array A(n,k): A(n,1) = 1, and for k > 1, A(n,k) = the highest exponent e such that k^e divides n, read by descending antidiagonals as A(1,1), A(1,2), A(2,1), etc.

1, 0, 1, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 2, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 1, 0, 0, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 1, 0, 1, 2, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 1, 1

Offset: 1

Antti Karttunen, May 20 2017

			The top left 18 X 18 corner of the array:
  n \k 1  2  3  4  5  6  7  8  9 10 11 12 13 14 15 16 17 18
     .-----------------------------------------------------
   1 | 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
   2 | 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
   3 | 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
   4 | 1, 2, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
   5 | 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
   6 | 1, 1, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
   7 | 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
   8 | 1, 3, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
   9 | 1, 0, 2, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0
  10 | 1, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0
  11 | 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0
  12 | 1, 2, 1, 1, 0, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0
  13 | 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0
  14 | 1, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0
  15 | 1, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0
  16 | 1, 4, 0, 2, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0
  17 | 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0
  18 | 1, 1, 2, 0, 0, 1, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1
  ---------------------------------------------------------
A(18,2) = 1, because 2^1 divides 18, but 2^2 does not. A(18,3) = 2, because 3^2 divides 18 (but 3^3 does not). A(18,4) = 0, because 4^0 (= 1) divides 18, but 4^1 does not. A(18,18) = 1, because 18^1 divides 18, but 18^2 does not.
A(2,18) = 0, because 18^0 divides 2, but 18^1 does not.

Antti Karttunen, Table of n, a(n) for n = 1..10585; the first 145 antidiagonals of array

Cf. A286562 (transpose), A286563 (lower triangular region), A286564 (lower triangular region reversed).
Cf. A169594 (row sums), also A168512, A178638, A186643.
Cf. also array A286156.

Table[Function[m, If[k == 1, 1, IntegerExponent[m, k]]][n - k + 1], {n, 15}, {k, n}] // TableForm (* Michael De Vlieger, May 20 2017 *)

A286561(n,k) = if(1==k, 1, valuation(n, k)); \\ Antti Karttunen, May 27 2017

def a(n, k):
    i=1
    if k==1: return 1
    while n%(k**i)==0:
        i+=1
    return i-1
for n in range(1, 21): print([a(k, n - k + 1) for k in range(1, n + 1)]) # Indranil Ghosh, May 20 2017

(define (A286561 n) (A286561bi (A002260 n) (A004736 n)))
(define (A286561bi row col) (if (= 1 col) 1 (let loop ((i 1)) (if (not (zero? (modulo row (expt col i)))) (- i 1) (loop (+ 1 i))))))

A186643 The number of divisors d of n which are either d=1 or for which the highest power d^k dividing n has odd exponent k.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 4, 2, 4, 2, 5, 2, 4, 4, 3, 2, 5, 2, 5, 4, 4, 2, 8, 2, 4, 4, 5, 2, 8, 2, 5, 4, 4, 4, 6, 2, 4, 4, 8, 2, 8, 2, 5, 5, 4, 2, 8, 2, 5, 4, 5, 2, 8, 4, 8, 4, 4, 2, 11, 2, 4, 5, 5, 4, 8, 2, 5, 4, 8, 2, 10, 2, 4, 5, 5, 4, 8, 2, 8

Offset: 1

Vladimir Shevelev, Feb 25 2011

A divisor d of n is called an "oex divisor" if d=1 or if the highest power d^k dividing n has odd exponent k. a(n) is the number of oex divisors of n.
If q is in A050376, then it is an infinitary divisor of n iff it is an oex divisor of n.
Moreover, every infinitary divisor of n is an oex divisor of n. The converse statement is not generally true.
If d_1 and d_2 are oex divisors of n, then lcm(d_1,d_2) is an oex divisor of n as well.
Not multiplicative: a(2)*a(9) <> a(18), for example. - R. J. Mathar, Mar 25 2012

			For n=16, the oex divisors are 1, 8 with 8^1|16, and 16 with 16^1|16. Therefore, a(16)=3.

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000005, A037445, A050376, A178638.

highpp := proc(n,d) if n mod d <> 0 then 0; else nshf := n ; a := 0 ; while nshf mod d = 0 do nshf := nshf /d ; a := a+1 ; end do: a; end if; end proc:
isoex := proc(d,n) d= 1 or (n mod d = 0 and type(highpp(n,d),'odd') ) ; end proc:
A186643 := proc(n) a := 0 ; for d in numtheory[divisors](n) do if isoex(d,n) then a := a+1 ; end if; end do: a ; end proc: # R. J. Mathar, Mar 18 2011

Table[DivisorSum[n, 1 &, Or[# == 1, OddQ@ IntegerExponent[n, #]] &], {n, 80}] (* Michael De Vlieger, May 28 2017 *)

a(n) = sumdiv(n, d, (d==1) || (valuation(n, d) % 2)); \\ Michel Marcus, Feb 06 2016

a(n) >= A037445(n).

cp's OEIS Frontend

A286561 Square array A(n,k): A(n,1) = 1, and for k > 1, A(n,k) = the highest exponent e such that k^e divides n, read by descending antidiagonals as A(1,1), A(1,2), A(2,1), etc.

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