A347494 a(n) is the number of instances where tau(k)|tau(n); n>=1, kA000005.

0, 1, 2, 1, 3, 4, 4, 6, 2, 7, 5, 8, 6, 10, 11, 1, 7, 11, 8, 13, 14, 15, 9, 17, 3, 17, 18, 16, 10, 21, 11, 19, 21, 22, 23, 4, 12, 25, 26, 29, 13, 31, 14, 23, 24, 29, 15, 17, 4, 27, 31, 28, 16, 37, 33, 39, 34, 35, 17, 50, 18, 38, 32, 1, 39, 46, 19, 34, 41, 49, 20, 59

Offset: 1

Author

David James Sycamore, Aug 31 2021

Keywords

• nonn

Comments

a(n) >= 1, for n >= 2; equality only when n = 2^(p-1) for any prime p. More generally, if p is prime(m), q any prime, and n=p^(q-1) then tau(n) = q, and the only numbers k < m such that tau(k)|tau(n) are 1 and q. Every prime < p contributes 1 to the count of a(n), and so does 1 itself, therefore a(n) = m-1+1 = m; see formula. Since for a given m, this holds for all primes q, it follows that every m > 0 appears in the sequence infinitely many times.

Examples

			a(1) = 0 because there is no k < 1 such that tau(k)|tau(1).
a(2) = 1, since there is only one instance of tau(k)|tau(2), namely k=1.
a(3) = 2, since there are two instances of tau(k)|tau(3), namely k=1 and k=2.
a(4) = 3, since there is only one instance of tau(k)|tau(4), namely k=1, etc.
		

Links

• David A. Corneth, Table of n, a(n) for n = 1..10001

Crossrefs

Cf: A000005, A010553, A047983, A010553.

Programs

Mathematica

With[{s = DivisorSigma[0, Range[72]]}, Array[Count[Mod[#2, s[[Range[#1 - 1]]]], 0] & @@ {#, s[[#]]} &, Length[s] - 1, 2]] (* Michael De Vlieger, Sep 09 2021 *)

PARI

first(n) = {my(l = List(), res = vector(n)); for(i = 1, n, nd = numdiv(i); if(nd > #l, for(i = #l + 1, nd, listput(l, 0) ) ); d = divisors(nd); for(j = 1, #d, res[i] += l[d[j]] ); l[nd]++; ); res } \\ David A. Corneth, Sep 03 2021

Formula

a(prime(m)^(q-1)) = m for m >= 1 and any prime q.