# [seqfan] Avoiding Backwards Sums

Hagen von EItzen math at von-eitzen.de
Mon Jun 1 11:05:46 CEST 2009

```>
> Consider the sequence defined as:
>
> a(1) = 1.
> For n >=2,
> a(n) = the smallest integer > a(n-1) such that
> sum{k=1 to r} a(k)
> doesn't equal any value of
> sum{k=1 to q} a(n+1-k),
> for all positive integers n, any positive integer r, and any positive integers q <= n-1.
>
> (The second sum obviously does equal the first sum for q = n = r.)
>
> I get that the sequence {a(k)} starts:
> 1,2,4,5,6,8,9,10,11,13,14,15,17,...
> (Unless I made an error.)
>
> Hmmm. Looks like a Beatty sequence, perhaps?
>
> My question is:
> Is {a(k)} the same as sequence A141204? (Attn: Paul Hannah.)
>
> This seems like it should be absolutely trivial to prove, if it is, but I don't know where to begin.
>

Actually, you stopped just in time to avoid disproving your conjecture ;)

In PARI,

leftsums(v) = local(a=[], i, ac=0); for(i=1,#v,ac+=v[i];a=concat(a,ac));a
rightsums(v) = local(a=[0], i, ac=0);
forstep(i=#v,2,-1,ac+=v[i];a=concat(a,ac));a
A=[1];while(#A<100,L=Set(leftsums(A));R=rightsums(A);J=vector(#R,x,1);n=A[#A]+1;while(#setintersect(L,Set(R+n*J))>0,n+=1);A=concat(A,n));A

produces

[1, 2, 4, 5, 6, 8, 9, 10, 11, 13, 14, 15, 17, 19, 21, 22, 24, 25, 28,
29, 30, 31, 32, 34, 36, 37, 39, 40, 41,43, 44, 46, 47, 48, 49, 50, 51,
52, 54, 55, 57, 61, 62, 63, 64, 66, 67, 70, 72, 73, 74, 75, 78, 79, 80,
81, 84, 88, 90, 91, 94, 95, 97, 99, 100, 102, 104, 105, 106, 107, 109,
110, 111, 112, 116, 119, 120, 121, 122, 123, 124, 125, 126, 127, 128,
129, 130, 133, 138, 140, 142, 143, 144, 145, 146, 147, 148, 149, 150, 151]

Hence your sequence has ..,17, 19, 21, ... where A141204 has ...,17, 18,
20, ...

Hagen

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