# [seqfan] Re: Formulas needed for new sequence.

Alexander P-sky apovolot at gmail.com
Wed Jun 29 20:52:14 CEST 2011

```Paolo's recurrence seems to be could be slightly simplified as

a(1)=1
a(n+1) = a(n)+floor(1/2 (sqrt(8 n-15)+1)) (1-floor(sin^2(1/2 sqrt(8 n-7) pi)))+1

On 6/29/11, Paolo Lava <paoloplava at gmail.com> wrote:
> Dear Ed,
>
> A191748={1,2,5,6,10,14,...},
>
> a(n+1)=a(n)+(1-b)*c+b, where b is one of the formula in A010054 and c one of
> the formula in A002024. Of course you need to correct the offset.
>
> For example, taking the formula of Carl R. White in A010054 and the formula
> of Neil Sloane in the comment section of A002024 you have:
>a(1)=1
> a(n+1)=a(n)+(1-floor((1-cos(Pi*sqrt(8*(n-1)+1)))/2))*floor((1+sqrt(1+8*(n-2)))/2+1)+floor((1-cos(Pi*sqrt(8*(n-1)+1)))/2)
>
> This is not a closed form but just a first attempt to find a formula:
> probably there is something less contrived than this...
>
> Paolo P. Lava
>
> 2011/6/29 Ed Jeffery <ed.jeffery at yahoo.com>
>
>> Fellow sequence fanatics,
>>
>> I recently submitted the following seemingly trivial sequence to OEIS:
>>
>> A191747={1,1,0,0,1,1,0,0,0,1,0,0,0,1,...} (offset=1).
>>
>> This sequence is just a concatenation of row entries of the N X N identity
>> matrices and hasn't been reviewed yet.
>>
>> However, of interest is the sequence of all m in {1,2,3,...} for which
>> A191747(m)=1. This definition gives the sequence
>>
>> A191748={1,2,5,6,10,14,...},
>>
>> which can be read by anti-diagonals from the table
>>
>> 1,  5,  14,  30,  55,  ..
>> 2,  10, 25,  49,  84,  ..
>> 6,  20, 43,  77,  124, ..
>> 15, 37, 70,  116, 177, ..
>> 31, 63, 108, 168, 245, ..
>> etc..
>>
>> I haven't submitted this sequence yet, since I am in the midst of some
>> adversity and need help with a formula for the general term: I haven't
>> been
>> able to concentrate long enough to get this done myself, so if some of you
>> would like to help, it would be greatly appreciated.
>>
>> Some analysis:
>>
>> The first row of the table is the core sequence A000330, omitting the
>> initial 0, and the first column is A056520. The j-th row can be found from
>> the generating functions
>>
>> (1)  (-j+(2*j+1)*x-(j-1)*x^2)/(1-x)^4, j in {0,1,2,...},
>>
>> by taking from the (j+1)-th term on, but this contradicts the definition
>> of
>> A000330 (corresponding to j=0) which has the initial term set to zero,
>> although the generating functions are in agreement: the reason for the
>> initial zero evidently is the alternative definition
>>
>> A056520(n)=A000300(n)+1.
>>
>> Here, j=1 in (1) gives essentially A058373, ignoring the two initial
>> zeros.
>> (A058373, by the way, was defined with offset=-2 which is bewildering to
>> me,
>> and perhaps it should be changed to offset=0 after dropping the two
>> zeros.)
>> Putting j=2 gives -A058372 which leads me to believe that the signs should
>> be reversed, that is, A058372 should be negated and rewritten in the
>> database. As far as I can determine, none of the other rows or columns of
>> the table appear in the OEIS database.
>>
>> Continuing with analysis of the above table, similarly, the k-th column
>> can
>> be found from the generating functions
>>
>> (2*k+1-(5*k+2)*x+4*(k+1)*x^2-(k+1)*x^3)/(1-x)^4, k in {0,1,2,...},
>>
>> by taking the k-th term on.
>>
>> Likewise, for j and k as above, for the j-th row R_j, we have the
>> closed-form expression
>>
>> R_j(n)=(n+1)*(2*n^2+n-6*j)/6, n=j+1,j+2,j+3,...;
>>
>> and for the k-th column C_k,
>>
>> C_k(n)=(n+2)*(2*n^2-n+6*k+3), n=j,j+1,j+2,....
>>
>> In conclusion, I see no way of defining a generating function for this
>> sequence, even though the rows and columns have generating functions with
>> a
>> common denominator, namely (1-x)^4. Maybe you all would like to try to
>> find
>> this function, if it exists. Furthermore, I suspect that a closed-form
>> expression for a general term of the sequence exists, but I haven't been
>> able to work it out. At present, my alleged mind can't handle
>> complications
>> with multiple indices (j and k) and varying offsets. So, I'll leave it to
>> those with more experience and hope to get a unified expression soon, at
>> which point I'll go ahead and submit the sequence.
>>
>> Thanks in advance to everyone,
>>
>> Ed Jeffery
>>
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>>
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>>
>
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