A187778 Numbers k dividing psi(k), where psi is the Dedekind psi function (A001615).

1, 6, 12, 18, 24, 36, 48, 54, 72, 96, 108, 144, 162, 192, 216, 288, 324, 384, 432, 486, 576, 648, 768, 864, 972, 1152, 1296, 1458, 1536, 1728, 1944, 2304, 2592, 2916, 3072, 3456, 3888, 4374, 4608, 5184, 5832, 6144, 6912, 7776, 8748, 9216, 10368, 11664, 12288, 13122, 13824, 15552, 17496, 18432, 20736, 23328

Offset: 1

Author

Enrique Pérez Herrero, Jan 05 2013

Keywords

• nonn

Comments

This sequence is closed under multiplication.

Also 1 and the numbers where psi(n)/n = 2, or n/phi(n)=3, or psi(n)/phi(n)=6.

Also 1 and the numbers of the form 2^i*3^j with i, j >= 1 (A033845).

If M(n) is the n X n matrix whose elements m(i,j) = 2^i*3^j, with i, j >= 1, then det(M(n))=0.

Numbers n such that Product_{i=1..q} (1 + 1/d(i)) is an integer where q is the number of the distinct prime divisors d(i) of n. - Michel Lagneau, Jun 17 2016

Examples

			psi(48) = 96 and 96/48 = 2 so 48 is in this sequence.
		

References

• S. Ramanujan, Collected Papers, Ed. G. H. Hardy et al., Cambridge 1927; Chelsea, NY, 1962, p. xxiv.

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..191 from Vincenzo Librandi)
• R. Blecksmith, M. McCallum and J. L. Selfridge, 3-smooth representations of integers, Amer. Math. Monthly, 105 (1998), 529-543.
• E. Deutsch, Tree statistics from Matula numbers, arXiv:1111.4288 [math.CO], 2011.
• Eric Weisstein's World of Mathematics, Smooth Number
• Wikipedia, Closure

Crossrefs

Cf. A003586, A001615, A007694, A033950, A074946, A075592.

Programs

Magma

[6*n: n in [1..3000] | PrimeDivisors(n) subset [2, 3]]; // Vincenzo Librandi, Jan 11 2019

Mathematica

Select[Range[10^4],#/EulerPhi[#]==3 || #==1&]
Join[{1}, 6 Select[Range@4000, Last@Map[First, FactorInteger@#]<=3 &]] (* Vincenzo Librandi, Jan 11 2019 *)

PARI

dedekindpsi(n) = if( n<1, 0, direuler( p=2, n, (1 + X) / (1 - p*X)) [n]);
k=0; n=0; while(k<10000,n++; if( dedekindpsi(n) % n== 0, k++; print1(n, ", ")));

Formula

For n > 1, a(n) = 6 * A003586(n).

Sum_{n>0} 1/a(n)^k = 1 + Sum_{i>0} Sum_{j>0} 1/(2^i * 3^j)^k = 1 + 1/((2^k-1)*(3^k-1)).