OFFSET
1,1
COMMENTS
We consider a(n), n>=2, conditions of the form: all numbers P_i(m) are composite, i = 1, ..., a(n), where P_i(m) is a polynomial of power n+1. It could be proved that S_k(m)= m^n + (m+1)^n + ... + (m+k)^n, as a polynomial in m of degree n+1, is divisible by k+1. Let S*_k(m) = S_k(m)/(k+1). So we have
S_k(m)=S*_k(m)*(k+1)=(T_k(m)/b(n))*(k+1), (1)
where b(n)=A064538(n) and, by the definition of A064538, T_k(m) = b(n)*S*_k(m) is a polynomial with integer coefficients.
It is clear that (1) could be prime only if k+1>=2 is a divisor of b(n). In this case we should require that (1) be a composite number. We have exactly A000005(b(n))-1 such requirements. In case of n=1, a(n)=2 (see A089306, A077654).
Remark. Sometimes some considered conditions satisfy trivially. For example, both a(3)=2 conditions for every m>=2 evidently hold, such that every number of the form m^3 + (m+1)^3 + ... +(m+k)^3 is composite.
Note that essentially this method is useful only in case of even n. Indeed, according to our comment in A001017, in case of odd n>=3 the number m^n + (m+1)^n + ... + (m+k)^n is composite for every k>=1. - Vladimir Shevelev, Apr 06 2015
LINKS
Peter J. C. Moses, Table of n, a(n) for n = 1..1000
CROSSREFS
KEYWORD
nonn
AUTHOR
Vladimir Shevelev, Apr 02 2015
EXTENSIONS
More terms from Peter J. C. Moses, Apr 02 2015
STATUS
approved