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Number of distinct products ij with 1 <= i <= n, 1 <= j <= n, (i,j)=1.
3

%I #28 Nov 18 2018 09:04:32

%S 1,2,4,6,10,11,17,21,27,29,39,42,54,57,62,70,86,89,107,113,120,125,

%T 147,152,172,178,196,204,232,236,266,282,294,302,320,329,365,374,388,

%U 400,440,446,488,501,518,529,575,586,628,638,657,672,724,733,758,778

%N Number of distinct products ij with 1 <= i <= n, 1 <= j <= n, (i,j)=1.

%C S. W. Golomb, personal communication, Svalbard, Norway, 7/97.

%H Andrew Howroyd, <a href="/A027435/b027435.txt">Table of n, a(n) for n = 1..10000</a>

%H Harri Hakula, Pauliina Ilmonen, Vesa Kaarnioja, <a href="https://arxiv.org/abs/1705.05163">Computation of extremal eigenvalues of high-dimensional lattice-theoretic tensors via tensor-train decompositions</a>, arXiv:1705.05163 [math.NA], 2017. See Table 2, d=4,5.

%F a(n) = Sum_{k=1..n} A014665(n). - _Sean A. Irvine_, Nov 15 2018

%F For n>1: # of positive integers u <= n(n-1) such that p^H_p(u)<=n for all p<=u, where H_p(u) = highest power of p dividing u.

%F a(n) = A236309(n) + 1. - _Andrew Howroyd_, Nov 16 2018

%p A027435 := proc(n)

%p local L, i, j ;

%p L := {};

%p for i from 1 to n do

%p for j from 1 to n do

%p if igcd(i,j) = 1 then

%p L := L union {i*j};

%p end if;

%p end do:

%p end do:

%p nops(L);

%p end proc: # _R. J. Mathar_, Jun 09 2016

%t Array[-Boole[# > 1] + Length@ Union@ Apply[Join, Table[If[CoprimeQ @@ #, i j, 0] &@ {i, j}, {i, #}, {j, #}]] &, 56] (* _Michael De Vlieger_, Nov 01 2017 *)

%o (PARI) a(n)={#Set(concat(vector(n, i, [i*j | j<-[1..n], gcd(i,j)==1])))} \\ _Andrew Howroyd_, Nov 15 2018

%o (PARI) seq(n)={my(v=vector(n),t=1);for(n=1, n, t+=sum(i=1, n-1, gcd(i,n) == 1 && 0==sumdiv(i*n, d, my(t=i*n/d); gcd(t,d)==1 && d<n && t<d)); v[n]=t); v} \\ _Andrew Howroyd_, Nov 16 2018

%Y Cf. A014665, A236309.

%K nonn

%O 1,2

%A _N. J. A. Sloane_

%E More terms from _Olivier GĂ©rard_, Nov 15 1997