A355228 - OEIS

A355228

a(n) is the smallest integer m such that there exist n of its distinct divisors (d_1, d_2, ..., d_n) with the property that m = d_1 + d_2 + ... + d_n = lcm(d_1, d_2, ..., d_n), or 0 if no such number m exists.

%I #37 Jun 27 2022 10:03:21

%S 1,0,6,18,28,24,48,60,84,120,120,120,180,180,240,360,360,360,360,672,

%T 720,720,720,840,840,1080,1260,1260,1260,1680,1680,1680,2160,2520,

%U 2520,2520,2520,2520,2520,3360,4320,5040,5040,5040,5040,5040,5040,5040,5040

%N a(n) is the smallest integer m such that there exist n of its distinct divisors (d_1, d_2, ..., d_n) with the property that m = d_1 + d_2 + ... + d_n = lcm(d_1, d_2, ..., d_n), or 0 if no such number m exists.

%C This sequence is the generalization of the problem A1737 proposed on French mathematical site Diophante (see link).

%C a(2) = 0 but all other terms are nonzero.

%C a(n) >= A081512(n) because in A081512, it is not required that m = lcm(d_1, d_2, ..., d_n). Currently, the strict inequality happens for n = 4 and n = 5; are there other such cases?

%H Diophante, <a href="http://www.diophante.fr/problemes-par-themes/arithmetique-et-algebre/a1-pot-pourri/4960-a1737-fideles-au-rendez-vous">A1737 - Fidèles au rendez-vous</a> (in French).

%e In the following triangle, the n-th row gives an example of a set of n divisors d_1, ..., d_n of a(n) such that a(n) = d_1 + ... + d_n = lcm(d_1, ..., d_n):

%e .

%e n m d_1 d_2 d_3 d_4 d_5 d_6 d_7 d_8 d_9 d10 d11 d12

%e -----------------------------------------------------------

%e 1 1 1

%e 2 0

%e 3 6 1 2 3

%e 4 18 1 2 6 9

%e 5 28 1 2 4 7 14

%e 6 24 1 2 3 4 6 8

%e 7 48 1 2 3 4 8 16 24

%e 8 60 1 2 3 4 5 10 15 20

%e 9 84 1 2 3 4 6 7 12 21 28

%e 10 120 1 2 3 4 5 6 15 20 24 40

%e 11 120 1 2 3 4 5 6 8 12 15 24 40

%e 12 120 1 2 3 4 5 6 8 10 12 15 24 30

%e However, for a given value of a(n) = m, there may be more than one way to choose d_1, ..., d_n. For example, for n=10, a(10)=120 and all seventeen solutions provided by _Jinyuan Wang_ in the Comments section of A081512 are also solutions here.

%o (PARI) isok(m, n) = {my(d=divisors(m)); if (#d<n, return(0)); forsubset([#d, n], s, my(vd = vector(n, k, d[s[k]])); if (lcm(vd) == vecsum(vd), return(1)););}

%o a(n) = {if (n==1, return(1)); if (n==2, return(0)); my(m=1); while (!isok(m, n), m++); m;} \\ _Michel Marcus_, Jun 25 2022

%Y Cf. A000396, A081512.

%K nonn

%O 1,3

%A _Bernard Schott_, Jun 25 2022

%E More terms from _Jinyuan Wang_, Jun 25 2022