A103314 Total number of subsets of the n-th roots of 1 that add to zero.

1, 1, 2, 2, 4, 2, 10, 2, 16, 8, 34, 2, 100, 2, 130, 38, 256, 2, 1000, 2, 1156, 134, 2050, 2, 10000, 32, 8194, 512, 16900, 2, 146854, 2, 65536, 2054, 131074, 158, 1000000, 2, 524290, 8198, 1336336, 2, 11680390, 2, 4202500, 54872, 8388610, 2, 100000000, 128

Offset: 0

Author

Wouter Meeussen, Mar 11 2005

Keywords

• nonn
• nice
• hard

Comments

The term a(0) = 1 counts the single zero-sum subset of the (by convention) empty set of zeroth roots of 1.

I am inclined to believe that if S is a zero-sum subset of the n-th roots of 1, that n can be built up from (zero-sum) cyclically balanced subsets via the following operations: 1. A U B, where A and B are disjoint. 2. A - B, where B is a subset of A. - David W. Wilson, May 19 2005

Lam and Leung's paper, though interesting, does not apply directly to this sequence because it allows repetitions of the roots in the sums.

Observe that 2^n=a(n) (mod n). Sequence A107847 is the quotient (2^n-a(n))/n. - T. D. Noe, May 25 2005

From Max Alekseyev, Jan 31 2008: (Start)

Every subset of the set U(n) = { 1=r^0, r^1, ..., r^(n-1) } of the n-th power roots of 1 (where r is a fixed primitive root) defines a binary word w of the length n where the j-th bit is 1 iff the root r^j is included in the subset.

If d is the period of w with respect to cyclic rotations (thus d|n) then the periodic part of w uniquely defines some binary Lyndon word of the length d (see A001037). In turn, each binary Lyndon word of the length d, where d

The binary Lyndon words of the length n are different in this respect: only some of them correspond to n distinct zero-sum subsets of U(n) while the others do not correspond to such subsets at all. A110981(n) gives the number of binary Lyndon words of the length n that correspond to zero-sum subsets of U(n). (End)

Links

• Max Alekseyev and M. F. Hasler, Table of n, a(n) for n = 0..164
• T. Y. Lam and K. H. Leung, On vanishing sums for roots of unity, arXiv:math/9511209 [math.NT], 1995.
• T. D. Noe, Sums of Roots of Unity Plots
• T. D. Noe, Proof a(n)=a(s(n))^(n/ s(n))
• Sasha Rybak, Zero sums of roots of unity (in Russian), forum dxdy.ru.
• Gary Sivek, On vanishing sums of distinct roots of unity, #A31, Integers 10 (2010), 365-368.

Crossrefs

Equals A070894 + 1. A107847(n) = (2^n - A103314(n))/n, A110981 = A001037 - A107847.

Row sums of A103306. See also A006533, A006561, A006600, A007569, A007678.

Cf. A070925, A107753 (number of primitive subsets of the n-th roots of unity summing to zero), A107754 (number of subsets of the n-th roots of unity summing to one), A107861 (number of distinct values in the sums of all subsets of the n-th roots of unity).

Cf. A322366.

Programs

Mathematica

Needs["DiscreteMath`Combinatorica`"]; Table[Plus@@Table[Count[ (KSubsets[ Range[n], k]), q_List/;Chop[ Abs[Plus@@(E^(2.*Pi*I*q/n))]]==0], {k, 0, n}], {n, 15}] (* T. D. Noe *)

PARI

/* This program implements all known results; it works for all n except for 165, 195, 210, 231, 255, 273, 285, 330, 345, ... */
A103314(n) = { local(f=factor(n)); n<2 & return(1); n==f[1,1] & return(2);
vecmax(f[,2])>1 & return(A103314(f=prod(i=1,#f~,f[i,1]))^(n/f));
if( 2==#f=f[,1], return(2^f[1]+2^f[2]-2));
#f==3 & f[1]==2 & return(sum(j=0,f[2],binomial(f[2],j)*(2^j+2^(f[2]-j))^f[3])
+(2^f[2]+2)^f[3]+(2^f[3]+2)^f[2]-2*((2^f[2]+1)^f[3]+(2^f[3]+1)^f[2])+2^(f[2]*f[3]));
n==105 & return(166093023482); error("A103314(n) is unknown for n=",n) }
/* Max Alekseyev and M. F. Hasler, Jan 31 2008 */

Formula

a(n) = A070894(n)+1.

a(2^n) = 2^(2^(n-1)). - Dan Asimov and Gareth McCaughan, Mar 11 2005

a(2n) = a(n)^2 if n is even. If p, q are primes, a(pq) = 2^p+2^q-2. In particular, if p is prime, a(2p) = 2^p + 2. - Gareth McCaughan, Mar 12 2005

a(n) == 2^n (mod n), a(p) = 2 (p prime). - David W. Wilson, May 08 2005

It appears that a(n) = a(s(n))^(n/s(n)) where s(n) = A007947(n) is the squarefree kernel of n. This is true if all zero-sum subsets of the n-th roots of 1 are formed by set operations on cyclic subsets. If true, A103314 is determined by its values on squarefree numbers (A005117). Some consequences would be a(p^n) = 2^p^(n-1), a(p^m q^n) = (2^p+2^q+2)^(p^(m-1) q^(n-1)) and a(p^2 n) = a(pn)^p for primes p and q. - David W. Wilson, May 08 2005

a(pn) = a(n)^p when p is prime and p|n; a(2p) = 2^p+2 when p is an odd prime. More generally a(pq) = 2^p+2^q-2 when p, q are distinct primes. - Gareth McCaughan, Mar 12 2005

For distinct odd primes p and q, a(2pq) = (2^p+2)^q + (2^q+2)^p - 2(2^p+1)^q - 2(2^q+1)^p + 2^(pq) + SUM[j=0..p] binomial(p,j)(2^j+2^(p-j))^q. - Sasha Rybak, Sep 21 2007.

a(n) = n*A110981(n) + 2^n - n*A001037(n). - Max Alekseyev, Jan 14 2008

Extensions

More terms from David W. Wilson, Mar 12 2005

Scott Huddleston (scotth(AT)ichips.intel.com) finds that a(30) >= 146854 and conjectures that is the true value of a(30). - Mar 24 2005. Confirmed by Meeussen and Wilson.

More terms from T. D. Noe, May 25 2005

Further terms from Max Alekseyev and M. F. Hasler, Jan 07 2008

Edited by M. F. Hasler, Feb 06 2008

Duplicate Mathematica program deleted by Harvey P. Dale, Jun 28 2021

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