mobius of binomials

[Paul D. Hanna]

     Do you have any additional conditions?
For I get 149 terms under 1000 that satisfy:
 
sum(k=0,n,moebius(binomial(n,k))) = 0
 
Here is my crude PARI code:
{for(n=1,1000,
if(sum(k=0,n,moebius(binomial(n,k)))==0,print1(n,",")))}
  
3,12,24,29,34,40,54,60,67,68,75,86,93,97,102,119,125,131,133,
142,152,157,160,163,164,168,170,172,189,193,197,208,210,220,
221,228,229,246,251,255,257,261,270,275,280,293,296,307,308,
313,315,332,337,338,340,341,345,350,353,370,374,381,385,388,
394,402,406,429,438,442,459,460,468,471,473,487,491,493,503,
505,516,530,536,541,547,564,573,577,580,595,596,606,609,615,
616,622,628,645,651,652,660,662,672,677,680,691,700,705,707,
712,730,732,745,751,757,760,763,771,777,781,793,805,808,813,
828,832,834,836,853,866,870,879,881,883,897,899,907,908,916,
933,938,945,946,957,962,964,970,977,991,
 
Paul
 
On Sun, 6 Feb 2005 16:33:37 -0500 (EST) Emeric Deutsch
<deutsch at duke.poly.edu> writes:
> Dear seqfans,
> The values of n between 0 and 1000 for which 
>         sum(mobius(binom(n,k),k=0..n))=0 
> are 
>           3,12,24,29,34,40  (not yet in OEIS).
> For example mobius(1)+mobius(3)+mobius(3)+mobius(1)=1-1-1+1=0.
> Is this a finite sequence?
> Thanks.
> Emeric