

A256581


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


8



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
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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


FORMULA

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


CROSSREFS

Cf. A000005, A064538, A089306 (a(1)=2), A256385 (a(2)=3), A256546 (a(4)=7).
Sequence in context: A195401 A158747 A260724 * A122697 A129022 A210564
Adjacent sequences: A256578 A256579 A256580 * A256582 A256583 A256584


KEYWORD

nonn


AUTHOR

Vladimir Shevelev, Apr 02 2015


EXTENSIONS

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


STATUS

approved



