# [seqfan] Re: Linear recurrences with large gaps

Ron Hardin rhhardin at att.net
Tue Jan 11 14:15:35 CET 2011

```That looks like a likely guess!  (owing to an array limit in bc(1) I can't do
recurrences over a(n-99))

An additional not yet complete set having a recurrence with a huge gap
T(n,k)=Number of strings of numbers x(i=1..n) in 0..k with sum i^4*x(i) equal to
n^4*k

Table starts
.1.1.1..1...1....1.....1.....1.....1......1......1......1.......1.......1
.1.1.1..1...1....1.....1.....1.....1......1......1......1.......1.......1
.1.1.1..1...2....2.....2.....2.....2......3......3......3.......3.......3
.1.1.1..1...1....1.....1.....2.....3......4......6......9......11......14
.1.2.2..3...3....4.....7.....8....12.....18.....29.....40......60......80
.1.1.1..1...2....7....16....47....90....137....203....295.....412.....584
.1.1.2..5...9...17....45...124...307....654...1211...2023....3151....4750
.1.1.1..2..26..200...628..1371..2578...4737...8602..15219...25591...41530
.1.2.4.20..79..353..1612..5439.14366..31892..63524.118564..212692..367954
.1.2.2.45.860.4751.14387.34449.81289.193997.434662.897102.1742199.3234015

Row 3:
Number of strings of numbers x(i=1..3) in 0..n with sum i^4*x(i) equal to 81*n
1 1 1 1 2 2 2 2 2 3 3 3 3 3 4 4 5 6 7 9 9 9 9 10 11 11 11 11 12 13 13 13 14 16
18 19 20 21 23 25 25 25 26 27 28

Empirical: a(n)=2*a(n-1)-a(n-2)+a(n-81)-2*a(n-82)+a(n-83)

rhhardin at mindspring.com
rhhardin at att.net (either)

----- Original Message ----
> From: Joerg Arndt <arndt at jjj.de>
> To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
> Sent: Tue, January 11, 2011 7:47:45 AM
> Subject: [seqfan] Re: Linear recurrences with large gaps
>
> * Ron Hardin <rhhardin at att.net> [Jan 11. 2011  13:28]:
> > Rows of  http://oeis.org/A184257
> > T(n,k)=Number of strings of numbers  x(i=1..n) in 0..k with sum i^3*x(i)
>equal to
>
> > n^3*k
>
> Soemwhat  easier (?) to read:
> "Number of length-n strings of numbers ..."
>
> > seem to have linear recurrences with large  gaps
> >
> > http://oeis.org/A184258
> > Number of strings of numbers x(i=1..3) in  0..n with sum i^3*x(i) equal to
>27*n
> > Empirical:  a(n)=2*a(n-1)-a(n-2)+a(n-27)-2*a(n-28)+a(n-29)
> >
> > http://oeis.org/A184259
> > Number of strings of numbers x(i=1..4) in  0..n with sum i^3*x(i) equal to
>64*n
> > Empirical:
> >
>a(n)=2*a(n-1)-a(n-2)+a(n-8)-2*a(n-9)+a(n-10)+a(n-64)-2*a(n-65)+a(n-66)-a(n-72)+2*a(n-73)-a(n-74)
>
> >
> >
> > probably for some simple reason that hasn't occured to me  yet.
> >
>
> Note the factorization:
>
> ?  p=-1+2*a^1-a^2+a^8-2*a^9+a^10+a^64-2*a^65+a^66-a^72+2*a^73-a^74
> -a^74 +  2*a^73 - a^72 + a^66 - 2*a^65 + a^64 + a^10 - 2*a^9 + a^8 - a^2 + 2*a
>- 1
> ?  factor(p)
> [a - 1 4]
> [a + 1 2]
> [a^2 + 1 2]
> [a^4 + 1 2]
> [a^8 + 1  1]
> [a^16 + 1 1]
> [a^32 + 1 1]
>
> /* == */
> ?  -(1-a)^2*(1-a^8)*(1-a^64)
> -a^74 + 2*a^73 - a^72 + a^66 - 2*a^65 + a^64 + a^10  - 2*a^9 + a^8 - a^2 + 2*a
>- 1
>
> Is the recurrence complete?
> Else more  terms could pop up, so that the recurrence really would be
>  -(1-a) *  prod(n=0, infinity, (1-a^(8^n)) )
>
> This is a a lacunary product:   e.g., for infinity=3:
> a^586 - 2*a^585 + a^584 - a^578 + 2*a^577 - a^576 -  a^522 + 2*a^521 - a^520 +
>a^514 - 2*a^513 + a^512 - a^74 + 2*a^73 - a^72 + a^66  - 2*a^65 + a^64 + a^10 -
>2*a^9 + a^8 - a^2 + 2*a - 1
>
> >
> >
> >  rhhardin at mindspring.com
> > rhhardin at att.net (either)
> >
> >
> > _______________________________________________
> >
> > Seqfan  Mailing list - http://list.seqfan.eu/
>
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>
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>

```