# Number of subsets of {1,2,3,...,n} that sum to 0 mod 17

Antti Karttunen Antti.Karttunen at iki.fi
Mon Jul 25 00:23:48 CEST 2005

```Max wrote:

> Creighton Dement wrote:
>
>>With the exception of the 17th and 33rd terms of
>>http://www.research.att.com/projects/OEIS?Anum=A068038
>>I get a floretion sequence that matches numbers from this sequence:
>>

>> To be on the safe side, I would suggest that the 17th term of A068038 is
>> recalculated.
>
>
> My calculations in PARI/GP:
>
> {A068038(n)=local(v,v1);v=vector(17);v[1]=1;for(i=1,n,v1=vector(17);for(j=0,16,v1[j+1]=v[j+1]+v[(j-i)%17+1]);v=v1);v[1]}
>
>
> for(n=0,50,print1(" ",A068038(n)))
>
>  1 1 1 1 1 1 3 8 15 30 60 120 241 482 964 1928 3856 7712 15422 30842
> 61682 123362 246722 493446 986896 1973790 3947580 7895160 15790320
> 31580642 63161284 126322568 252645136 505290272 1010580544 2021161084
> 4042322164 8084644324 16169288644 32338577284 64677154572 129354309152
> 258708618300 517417236600 1034834473200 2069668946400 4139337892804
> 8278675785608 16557351571216 33114703142432 66229406284864
>
> So OEIS content is correct.
>
> Max
>
Moreover, 17 is prime, thus

Sophie LeBlanc's formula given in:

http://www.research.att.com/projects/OEIS?Anum=A068009

T(p,k+p) = 2*T(p,k) + (2^k * ((2^p) - 2)/p) for all k* *>= 0

should work. So maybe you can check the problematic
values A068038(17) and A068038(33) also with that formula.

It should be possible also to work out similar formulae
as given for the case p=3 in:
http://www.research.att.com/projects/OEIS?Anum=A068010
by Galvin et al.

See the original discussion at: