A256581 Number of conditions on m under which m^n + (m+1)^n + ... + (m+k)^n is composite for every k>=1 (see comment).

2, 3, 2, 7, 5, 7, 7, 11, 5, 7, 7, 31, 23, 11, 9, 15, 17, 31, 31, 47, 23, 15, 29, 47, 23, 15, 7, 15, 11, 31, 47, 95, 47, 15, 11, 127, 95, 47, 39, 63, 71, 63, 63, 95, 47, 31, 71, 95, 71, 47, 31, 31, 47, 63, 39, 47, 23, 15, 23, 255, 191, 127, 111, 95, 71, 127

Offset: 1

Author

Vladimir Shevelev, Apr 02 2015

Keywords

• nonn

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

Cf. A000005, A064538, A089306 (a(1)=2), A256385 (a(2)=3), A256546 (a(4)=7).

Formula

For n>=2, a(n) = A000005(A064538(n))-1.

Extensions

More terms from Peter J. C. Moses, Apr 02 2015