A297217 Most common value of the number of divisors function among all composites up to composite(n) inclusive, or 0 if there is a tie.

3, 0, 4, 0, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4

Offset: 1

Author

Felix Fröhlich, Mar 02 2018

Keywords

• nonn
• easy

Comments

Is a(n) = 4 for all n > 4?

Examples

			          n  |  1 |  2 |  3 |  4 |  5 |  6 |  7 |  8 |  9 | 10 | 11 | 12
-------------------------------------------------------------------------
  A002808(n) |  4 |  6 |  8 |  9 | 10 | 12 | 14 | 15 | 16 | 18 | 20 | 21
-------------------------------------------------------------------------
A035004(n+1) |  3 |  4 |  4 |  3 |  4 |  6 |  4 |  4 |  5 |  6 |  6 |  4
-------------------------------------------------------------------------
        a(n) |  3 |  0 |  4 |  0 |  4 |  4 |  4 |  4 |  4 |  4 |  4 |  4
		

Crossrefs

Cf. A002808, A035004.

Programs

PARI

composite(n) = my(i=0); forcomposite(c=1, , i++; if(i==n, return(c)))
mcv(v) = my(w=vecsort(v, , 8), count=vector(#w), ind=0, i=0); for(x=1, #w, for(y=1, #v, if(w[x]==v[y], count[x]++))); for(k=1, #count, if(count[k]==vecmax(count), ind=k; i++)); if(i > 1, return(0), return(w[ind]))
a(n) = my(v=[]); for(k=1, n, v=concat(v, numdiv(composite(k)))); mcv(v)