A226853 Numbers n such that Sum_{i = 1..q} 1/d(i) is an integer where d(i) are the divisors of n for some q and n is primitive (the set {d(1), d(2), ..., d(q)} appears only once).
1, 6, 28, 120, 180, 496, 672, 1890, 8128, 8415, 20482, 20496, 25410, 30240, 32760, 33345, 34155, 38430, 40128, 47804, 72800, 90720, 103530, 407715, 523776, 806190, 979992, 1084160, 1273725, 1274100, 2178540, 3571050, 7441280, 10782216, 12499150, 23569920, 28360464, 33550336, 45532800
Offset: 1
Keywords
Examples
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.
Crossrefs
Subsequence of A225110.
Programs
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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:
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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
Extensions
Edited by Jon E. Schoenfield, Oct 02 2017
More terms from Michel Marcus, Feb 22 2025
Comments