A049087 Ceiling of mean distance between successive distinct prime divisors of n.
0, 0, 0, 0, 0, 1, 0, 0, 0, 3, 0, 1, 0, 5, 2, 0, 0, 1, 0, 3, 4, 9, 0, 1, 0, 11, 0, 5, 0, 2, 0, 0, 8, 15, 2, 1, 0, 17, 10, 3, 0, 3, 0, 9, 2, 21, 0, 1, 0, 3, 14, 11, 0, 1, 6, 5, 16, 27, 0, 2, 0, 29, 4, 0, 8, 5, 0, 15, 20, 3, 0, 1, 0, 35, 2, 17, 4, 6, 0, 3, 0, 39, 0, 3, 12, 41, 26, 9, 0, 2, 6, 21, 28, 45
Offset: 1
Examples
a(105)=2: prime divisors of 105 are 3,5,7; distance between successive pairs: 5-3=2 and 7-5=2; mean of 2 and 2 is 2. If n=30 the mean is 1.5 and rounding up, a(30)=2.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..65537
Crossrefs
Cf. also A178921.
Programs
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Mathematica
Table[Ceiling[Mean[Differences[FactorInteger[n][[All,1]]]/.{}->{0,0}]],{n,100}] (* Harvey P. Dale, Jun 26 2021 *) -
PARI
A049087(n) = if(1>=omega(n), 0, my(ps = factor(n)[,1], s = 0); for(i=2, #ps, s += (ps[i]-ps[i-1])); ceil(s/(#ps-1))); \\ Antti Karttunen, Sep 07 2018