A010553 a(n) = tau(tau(n)).

1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 2, 4, 2, 3, 3, 2, 2, 4, 2, 4, 3, 3, 2, 4, 2, 3, 3, 4, 2, 4, 2, 4, 3, 3, 3, 3, 2, 3, 3, 4, 2, 4, 2, 4, 4, 3, 2, 4, 2, 4, 3, 4, 2, 4, 3, 4, 3, 3, 2, 6, 2, 3, 4, 2, 3, 4, 2, 4, 3, 4, 2, 6, 2, 3, 4, 4, 3, 4, 2, 4, 2

Offset: 1

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

Ramanujan (1915) posed the problem of finding the extreme large values of a(n). Buttkewitz et al. determined the maximal order of log a(n).

Every number eventually appears. Sequence A193987 gives the least term where each number appears. - T. D. Noe, Aug 10 2011

References

• S. Ramanujan, Highly composite numbers. Proc. London Math. Soc., series 2, 14 (1915), 347-409. Republished in Collected papers of Srinivasa Ramanujan, AMS Chelsea Publ., Providence, RI, 2000, pp. 78-128.

Links

• Indranil Ghosh, Table of n, a(n) for n = 1..10000 (first 2000 terms from Enrique Pérez Herrero)
• M. Abramowitz and I. A. Stegun, eds., Handbook of Mathematical Functions, National Bureau of Standards, Applied Math. Series 55, Tenth Printing, 1972.
• R. Bellman and H. N. Shapiro, On a problem in additive number theory, Annals Math., 49 (1948), 333-340. See Eq. 1.5.
• Yvonne Buttkewitz, Christian Elsholtz, Kevin Ford, Jan-Christoph Schlage-Puchta, A problem of Ramanujan, Erdos and Katai on the iterated divisor function, arXiv:1108.1815 [math.NT], Aug 08 2011.

Crossrefs

Cf. A000005, A036450, A193987 (least number k such that tau(tau(k)) = n), A335831.

Programs

Maple

with(numtheory): f := n->tau(tau(n));

Mathematica

Table[Nest[DivisorSigma[0, #] &, n, 2], {n, 81}] (* Michael De Vlieger, Dec 24 2015 *)

PARI

A010553(n)=numdiv(numdiv(n)); \\ Enrique Pérez Herrero, Jul 13 2010

Formula

a(n) = A000005(A000005(n)). a(1) = 1, a(p) = 2 for p = primes (A000040), a(pq) = 3 for pq = product of two distinct primes (A006881), a(pq...z) = k + 1 for pq...z = product of k (k > 2) distinct primes p,q,...,z (A120944), a(p^k) = A000005(k+1) for p^k = prime powers (A000961(n) for n > 1), k = natural numbers (A000027). - Jaroslav Krizek, Jul 17 2009

a(A007947(n)) = 1 + A001221(n); (n>1). - Enrique Pérez Herrero, May 30 2010

Asymptotically, Max_{i<=n} log(tau(tau(i))) = sqrt(log(n))/log_2(n) * (c + O(log_3(n)/log_2(n)) where c = 8*Sum_{j>=1} log^2 (1 + 1/j)) ~ 2.7959802335... [Buttkewitz et al.].