A322034 Let p1 <= p2 <= ... <= pk be the prime factors of n, with repetition; let s = 1/p1 + 1/(p1*p2) + 1/(p1*p2*p3) + ... + 1/(p1*p2*...*pk); a(n) = numerator of s. a(1)=0 by convention.
0, 1, 1, 3, 1, 2, 1, 7, 4, 3, 1, 5, 1, 4, 2, 15, 1, 13, 1, 4, 8, 6, 1, 11, 6, 7, 13, 11, 1, 7, 1, 31, 4, 9, 8, 31, 1, 10, 14, 9, 1, 29, 1, 17, 7, 12, 1, 23, 8, 31, 6, 10, 1, 20, 12, 25, 20, 15, 1, 17, 1, 16, 29, 63, 14, 15, 1, 13, 8, 43, 1, 67
Offset: 1
Examples
If n=12 we get the prime factors 2,2,3, and s = 1/2 + 1/4 + 1/12 = 5/6. So a(12) = 5. The fractions s for n >= 2 are 1/2, 1/3, 3/4, 1/5, 2/3, 1/7, 7/8, 4/9, 3/5, 1/11, 5/6, 1/13, 4/7, 2/5, 15/16, 1/17, 13/18, 1/19, 4/5, 8/21, ...
Links
- Antti Karttunen, Table of n, a(n) for n = 1..16384
- Antti Karttunen, Data supplement: n, a(n) computed for n = 1..65537
Crossrefs
Programs
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Maple
# This generates the terms starting at n=2: P:=proc(n) local FM: FM:=ifactors(n)[2]: seq(seq(FM[j][1], k=1..FM[j][2]), j=1..nops(FM)) end: # A027746 f0:=[]; f1:=[]; f2:=[]; for n from 2 to 120 do a:=0; b:=1; t1:=[P(n)]; for i from 1 to nops(t1) do b:=b/t1[i]; a:=a+b; od; f0:=[op(f0),a]; f1:=[op(f1), numer(a)]; f2:=[op(f2),denom(a)]; od: f0; # s f1; # A322034 f2; # A322035 f2-f1; # A322036
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Mathematica
f[x_] := Flatten[ConstantArray[#1, #2] & @@@ FactorInteger[x]]; {0}~Join~Table[Numerator@ Total@ Table[1/Times @@ #[[;; i]], {i, Length[#]}] &@ f[n], {n, 2, 72}] (* Michael De Vlieger, Jun 20 2025 *)
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PARI
A322034(n) = if(1==n,0,my(f=factor(n),pm=1,s=0); for(i=1,#f~,while(f[i,2],pm *= f[i,1]; f[i,2]--; s += 1/pm)); numerator(s)); \\ Antti Karttunen, Feb 28 2019
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