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A175393 a(n) = number of pairs (j,k), j <= k, of divisors of 2n such that j+k divides 2n. 2

%I #16 Apr 09 2023 10:03:37

%S 1,2,3,3,2,7,2,4,5,5,2,11,2,4,8,5,2,12,2,8,7,4,2,15,3,4,7,7,2,20,2,6,

%T 6,4,5,20,2,4,6,11,2,17,2,6,15,4,2,19,3,8,6,6,2,17,5,10,6,4,2,33,2,4,

%U 13,7,4,15,2,6,6,12,2,28,2,4,13,6,4,15,2,14,9,4,2,29,4,4,6,8,2,38,5,6,6,4,4

%N a(n) = number of pairs (j,k), j <= k, of divisors of 2n such that j+k divides 2n.

%C The sum of any two divisors of an odd number n never divides n.

%C From _Robert Israel_, Oct 09 2017: (Start)

%C (1,1) and (n,n) are always such pairs, so a(n) >= 2 for n >= 2.

%C a(n)=2 if and only if n is a prime other than 3.

%C a(3^k) = 2*k+1.

%C a(p^k) = k+1 for primes other than 3. (End)

%H Robert Israel, <a href="/A175393/b175393.txt">Table of n, a(n) for n = 1..10000</a>

%e The divisors of 2*6 = 12 are 1,2,3,4,6,12. The pairs of these divisors that each sum to a divisor of 12 are 1+1 = 2, 1+2 = 3, 1+3 = 4, 2+2 = 4, 2+4 = 6, 3+3 = 6, and 6+6 = 12. There are seven of these pairs, so a(6) = 7.

%p f:= proc(n) local D,k,t;

%p D:= numtheory:-divisors(2*n);

%p t:= 0;

%p for k from 1 to nops(D) do

%p t:= t + nops(select(j -> 2*n mod (D[j]+D[k])=0, [$1..k]))

%p od;

%p t

%p end proc:

%p map(f, [$1..100]); # _Robert Israel_, Oct 09 2017

%t Table[Length[Select[Total/@Union[Sort/@Tuples[Divisors[2 n],2]],Mod[2 n,#]==0&]],{n,100}] (* _Harvey P. Dale_, Apr 09 2023 *)

%K nonn

%O 1,2

%A _Leroy Quet_, Apr 29 2010

%E More terms from _D. S. McNeil_, May 09 2010

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Last modified March 28 10:55 EDT 2024. Contains 371241 sequences. (Running on oeis4.)