A283736 Semisimple numbers: positive integers having exactly one reduced phi-partition.

3, 4, 5, 7, 9, 10, 11, 12, 13, 17, 18, 19, 23, 24, 29, 31, 37, 41, 42, 43, 47, 53, 59, 60, 61, 67, 71, 73, 79, 83, 84, 89, 90, 97, 101, 103, 107, 109, 113, 120, 127, 131, 137, 139, 149, 150, 151, 157, 163, 167, 173, 179, 180, 181, 191, 193, 197

Offset: 1

Author

M. F. Hasler, Mar 15 2017

Keywords

• nonn

Comments

A phi-partition of n is a sum x_1 + ... + x_r = n, 1 <= x_1 <= ... <= x_r with r > 1, such that phi(x_1) + ... + phi(x_r) = phi(n), where phi = A000010 is Euler's totient function.

A partition is reduced iff each summand is a primorial number A002110(k) = product of the first k primes, k >= 0.

It is known that semisimple numbers are the union of odd primes, {9} and numbers of the form n = a*q_1*...q_k*A002110(i) with k >= 0, i >= 1, q_1 > ... > q_k > p := prime(i+1) > a*(q_1-p)*...*(q_k-p), see examples.

Examples

			As said in comments, this sequence contains the odd primes A065091, 9, and elements of A060735: multiples of primorials A002110 not larger than the next primorial, except for the primorials themselves. These could be called trivial solutions and include all numbers up to 13 except for 1, 2, 6 (primorials), 8 (not semisimple) and 10 (semisimple, see below).
Let us call nontrivial the terms that can only be written in the form a*q_1*...*q_k*A002110(i) with k >= 1. It will be convenient to write A002110(i) as (p-1)# := A034386(p-1) with p := prime(i+1).
In the case k=1, we have multiples n = a q (p-1)# such that a*(q - p) < p.
Here, a = 1 and q = prime(i+2) always yields a solution (since prime(i+2) < 2 prime(i+1) for all i), so these could also be considered as "trivial" solutions.
For i = 1, p = 3 > a*(q-3) has only this "trivial" solution, a = 1, q = 5, n = 5*2 = 10 = a(9).
For i = 2, p = 5 > a*(q-5) for q = 7, a = 1, n = 7*3*2 = 42 ("trivial") and a = 2, n = 2*7*6 = 84, no other solution with q > 7, i.e., q >= 11.
For i = 3, p = 7 > a*(q-7) has solutions q = 11, a = 1, n = 11*5*3*2 ("trivial"), and q = 13, a = 1 : n = 13*5# = 390.
For i = 4, p = 11 > a*(q - 11) has solutions:
   q = 13, a = 1,2,3,4,5 : n = a*13*7# = a*2730, and
   q = 17 and 19, a = 1 : n = 17*7# = 3570  and n = 19*7# = 3990.
Concerning the solutions with k=2, one can easily check that (prime(i+2)-prime(i+1))*(prime(i+3)-prime(i+1)) < prime(i+1) for i >= 6 but not i = 7, 8, 10, 14, 22, 23, 29, 45. Thus, prime(i+2)*prime(i+3)*A002110(i) = A002110(i+3)/prime(i+1) is a solution for all these values of i, the smallest term of this form being prime(8)*prime(9)*prime(6)# = prime(9)# / prime(7) = 13123110.
		

Links

• Jun Wang and Xin Wang, On the set of reduced φ-partitions of a positive integer, Fibonacci Quarterly 44, no. 2 (May 2006), p. 98

Crossrefs

Cf. A002110, A034386, A060735, A283528, A283530.

See A283320 for the composite semisimple numbers.

Programs

PARI

is_semisimple(n,Q,m)={if(bittest(n,0),isprime(n)||n==9,n\=2;forprime(p=3,,np && for(k=1,#Q-m=#select(q->q<=p,Q),forvec(q=vector(k,j,[m+1,#Q]),prod(i=1,k,1-p/Q[q[i]],n)0, p and the indices [ i_1 ... i_k ] such that q_m is the ( i_m )-th prime factor of n/(p-1)#.