OFFSET
1,1
COMMENTS
The squares>1 and the prime numbers are in the sequence.
EXAMPLE
The prime numbers q are in the sequence because P(q,z) = qz + 1.
6 is not in the sequence because P(6,z)=(2z+1)*(3z^2+1).
The following table gives the irreducible polynomials.
+-----------------------------------------------------------+
| m | P(m,z) |
+-----------------------------------------------------------+
| 4 | 1 + 2z + 4z^2 |
+-----------------------------------------------------------+
| 9 | 1 + 3z + 9z^2 |
+-----------------------------------------------------------+
| 12 | 1 + 2z + 3z^2 + 4z^3 + 6z^4 + 12z^5 |
+-----------------------------------------------------------+
| 16 | 1 + 2z + 4z^2 + 8z^3 + 16z^4 |
+----------------------------+------------------------------+
| 24 | 1 + 2z + 3z^2 + 4z^3 + 6z^4 + 8z^5 + 12z^6 + 24z^7 |
+-----------------------------------------------------------+
| 25 | 1 + 5z + 25z^2 |
+-----------------------------------------------------------+
| 30 | 1 + 2z + 3z^2 + 5z^3 + 6z^4 + 10z^5 + 15z^6 + 30z^7 |
+-----------------------------------------------------------+
MAPLE
with(numtheory):
for n from 1 to 135 do :
it:=0:d:=divisors(n):P:=add(op(i, d)*x^(i-1), i=1..nops(d)):
y:=fsolve(P, x, complex):z:=evalf({%}):k:=nops(z):
if irreduc(P) then printf(`%d, `, n):else fi:
od:
PROG
(PARI) isok(n) = my(d=divisors(n)); polisirreducible(sum(i=1, #d, d[i]*z^(i-1))); \\ Michel Marcus, Feb 14 2025
CROSSREFS
KEYWORD
nonn,new
AUTHOR
Michel Lagneau, Feb 14 2025
STATUS
approved