

A283736


Semisimple numbers: positive integers having exactly one reduced phipartition.


1



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
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OFFSET

1,1


COMMENTS

A phipartition 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_1p)*...*(q_kp), see examples.


LINKS



EXAMPLE

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 A002210(i) as (p1)# := A034386(p1) with p := prime(i+1).
In the case k=1, we have multiples n = a q (p1)# 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*(q3) has only this "trivial" solution, a = 1, q = 5, n = 5*2 = 10 = a(9).
For i = 2, p = 5 > a*(q5) 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*(q7) 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.


PROG

(PARI) is_semisimple(n, Q, m)={if(bittest(n, 0), isprime(n)n==9, n\=2; forprime(p=3, , n<p&&return(n1); Q=factor(n)[, 1]; Q[#Q]>p && for(k=1, #Qm=#select(q>q<=p, Q), forvec(q=vector(k, j, [m+1, #Q]), prod(i=1, k, 1p/Q[q[i]], n)<p&&return([p, q]), 2)); n%p && return; n\=p))} \\ if n = a*q_1*...*q_k*(p1)# is semisimple, return a1 if k=0, or if k>0, p and the indices [ i_1 ... i_k ] such that q_m is the ( i_m )th prime factor of n/(p1)#.


CROSSREFS

See A283320 for the composite semisimple numbers.


KEYWORD

nonn


AUTHOR



STATUS

approved



