OFFSET
1,2
COMMENTS
Subsequence of A225110 where we find classes of numbers having the same first q divisors; for example, each of the numbers 6, 18, 42, 54, 66, ... has {1, 2, 3, 6} as its first four divisors, and 1/1 + 1/2 + 1/3 + 1/6 = 2; similarly, each of the numbers 28, 196, 812, 868, ... has {1, 2, 4, 7, 14, 28} as its first six divisors, and 1/1 + 1/2 + 1/4 + 1/7 + 1/14 + 1/28 = 2.
This sequence includes only the smallest number having any given set of first divisors {d(1), d(2), ..., d(q)}, i.e., the set of first divisors corresponding to each term occurs only once.
The sets of first divisors (such that Sum_{i = 1..q} 1/d(i) is an integer) corresponding to the first few terms are as follows:
a(1) = 1: [1];
a(2) = 6: [1, 2, 3, 6];
a(3) = 28: [1, 2, 4, 7, 14, 28];
a(4) = 120: [1, 2, 3, 4, 5, 6, 8, 10, 12, 15, 20, 24, 30, 40, 60, 120];
a(5) = 180: [1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45];
a(6) = 496: [1, 2, 4, 8, 16, 31, 62, 124, 248, 496];
a(7) = 672: [1, 2, 3, 4, 6, 7, 8, 12, 14, 16, 21, 24, 28, 32, 42, 48, 56, 84, 96, 112, 168, 224, 336, 672].
EXAMPLE
180 is in the sequence because the divisors are {1, 2, 3, 4, 5, 6, 9, 10, 12, 15, 18, 20, 30, 36, 45, 60, 90, 180} and the sum of the reciprocals of the first q = 15 divisors is 1/1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/9 + 1/10 + 1/12 + 1/15 + 1/18 + 1/20 + 1/30 + 1/36 + 1/45 = 3, which is an integer.
Although the first 4 divisors of 18 are {1, 2, 3, 6} and the sum of their reciprocals is 1/1 + 1/2 + 1/3 + 1/6 = 2, 18 is not in the sequence because 6 has those same first four divisors and is the smallest (i.e., primitive) number having that set of first 4 divisors. Thus, the primitive number 6 is in the sequence, so the non-primitive number 18 is not.
MAPLE
with(numtheory): printf ( "%d %d \n", 1, 6):lst:={6}:for n from 1 to 10000 do:x:=divisors(n):n1:=nops(x):s:=0:ii:=0:for q from 1 to n1 while(ii=0) do:s:=s+1/x[q]:if s=floor(s) and q>1 and {x[q]} intersect lst <>{x[q]} then lst:=lst union {x[q]}:ii:=1: printf(`%d, `, n):else fi:od:od:
PROG
(PARI) isok(k) = if (k==1, return([1])); my(d=divisors(k), s=1); for (i=2, #d, s += 1/d[i]; if (denominator(s)==1, return(Vec(d, i)); ));
already(list, v) = for (i=1, #list, if (list[i] == v, return(1)); );
lista(nn) = my(listv=List(), listi=List()); for (n=1, nn, my(v=isok(n)); if (v && !already(listv, v), listput(listi, n); listput(listv, v); ); ); Vec(listi); \\ Michel Marcus, Feb 22 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
Michel Lagneau, Jun 19 2013
EXTENSIONS
Edited by Jon E. Schoenfield, Oct 02 2017
More terms from Michel Marcus, Feb 22 2025
STATUS
approved