A297003 a(n) = 2^(n-1), n=1,2,3; for n >= 4, a(n) is the number of the previous terms dividing n.

1, 2, 4, 3, 1, 4, 2, 6, 3, 4, 2, 11, 2, 6, 4, 10, 2, 11, 2, 13, 4, 10, 2, 18, 2, 11, 4, 16, 2, 17, 2, 19, 7, 13, 3, 24, 2, 14, 7, 21, 2, 23, 2, 24, 5, 16, 2, 31, 4, 19, 6, 25, 2, 24, 6, 27, 7, 17, 2, 35, 2, 20, 9, 28, 5, 29, 2, 29, 6, 29, 2, 41, 2, 22, 8, 31

Offset: 1

Author

Vladimir Shevelev, Dec 23 2017

Keywords

• nonn
• look

Examples

			1-3) a(1)=1, a(2)=2 a(3)=4 by the definition;
4) Let n=4. From the previous terms {1,2,4} everyone divides 4, so a(4)=3;
5) Let n=5. From the previous terms {1,2,4,3} only 1 divides 5. So a(5)=1;
6) Let n=6. From the previous terms {1,2,4,3,1} exactly four divide 6. So a(6)=4; etc.
		

Links

• Peter J. C. Moses, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A088167.

Programs

Mathematica

first[n_] := Fold[Append[#1, Count[#1, k_ /; Divisible[#2, k]]] &,
  2^Range[0, Min[n - 1, 2]], Range[4, n]] (* Michael De Vlieger, Dec 23 2017 *)

PARI

first(n) = my(res = vector(n), c = 0); for(x = 1, min(n, 3), res[x] = 1<<(x-1)); for(x=4, n, for(k=1, x-1, if(x%res[k]==0, c++)); res[x] = c; c = 0); res \\ Iain Fox, Dec 23 2017

Sage

def A297003_list(leng):
    L = [1, 2, 4]
    if leng < 4: return L[0:leng]
    for n in (4..leng) :
        count = 0
        for l in L: count += int(l.divides(n))
        L.append(count)
    return L
print(A297003_list(76)) # Peter Luschny, Dec 24 2017

Formula

a(p) = 2, where p is prime, other than 3 and 5.

Extensions

More terms from Peter J. C. Moses, Dec 23 2017