0,1

a(n) is least integer k such that at least one signed sum of the first k n-th powers equals zero.

Contribution from Pietro Majer (majer(AT)dm.unipi.it), Mar 14 2009: (Start)

a(n)<2^(n+1).

The partition of the set {k: 0=<k<2^(n+1)} into two sets A,B according

to the parity of the number of 1s in the binary expansion of k, has the

property that sum_{k in A}p(k) = sum_{k in B}p(k) for any polynomial p

of degree <= n. Equivalently, if e(k) is the Thue-Morse

sequence A106400, then sum_{0=<k<2^m} e(k)p(k)=0 for any

polynomial p with deg(p)<m. (End)

Posting to sci.math Nov 11 1996 by fredh(AT)ix.netcom.com (Fred W. Helenius).

a(n) == 0 or 3 (mod 4) for n >= 1 - David W. Wilson, Oct 20 2005.

For n=1 and 2 we have: 1+2-3 = 0 (so a(1)=3), 1+4-9+16-25-36+49 = 0 (so a(2)=7).

The sum of the ninth powers of 3 5 9 10 14 19 20 21 25 26 28 31 35 36 37 38 40 41 42 is half the sum of the ninth powers of 1..44, so a(9)=44 - Don Reble, Oct 21 2005.

Contribution from Pietro Majer (majer(AT)dm.unipi.it), Mar 14 2009: (Start)

Example: the signs (+--+-++--++-+--+) in (+0)-1-8+27-64+125+216-343=0 are

those of the expansion of Q(x):=(1-x)(1-x^2)(1-x^4)(1-x^8)=

=+1-x-x^2+x^3-..+x^15. Since (1-x)^4 divides Q(x), if S is the shift

operator on sequences, the operator Q(S) has the fourth discrete

difference (I-S)^4 as factor, hence annihilates the sequence of cubes. (End)

Sequence in context: A046480 A137767 A080140 this_sequence A128458 A066733 A049623

Adjacent sequences: A019565 A019566 A019567 this_sequence A019569 A019570 A019571

nonn

Robert G. Wilson v (rgwv(AT)rgwv.com)

More from Don Reble, Oct 21 2005

Definition simplified by Pietro Majer (majer(AT)dm.unipi.it), Mar 15 2009