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

1,2

Links

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

Formula

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

Example

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.

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 *)

Prog

(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

Crossrefs

Cf. A088167.

Sequence in context: A275117 A243141 A243207 * A071284 A104753 A201759

Adjacent sequences: A297000 A297001 A297002 * A297004 A297005 A297006

Keyword

nonn,look

Author

Vladimir Shevelev, Dec 23 2017

Extensions

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

Status

approved