[seqfan] Re: Symmetric and Antisymmetric Linear Recurrences?

[Ron Hardin]

Polynomial of degree 6 not 3.  I was mixing it up with another similar problem.

Herewith both problems
/tmp/ejo
T(n,k)=Number of length n+3 0..k arrays with no disjoint pairs in any consecutive four terms having the same sum

Table starts
.8....48.....168......440.......960.......1848........3248.........5328
.8....90.....456.....1592......4344......10098.......20816........39264
.8...172....1248.....5796.....19744......55372......133780.......290004
.8...334....3424....21152.....89836.....303924......860360......2143214
.8...656....9392....77236....408644....1668072.....5532212.....15837692
.8..1300...25822...282384...1859736....9157806....35577396....117045466
.8..2584...71060..1032952...8465936...50284864...228817500....865051288
.8..5148..195536..3779018..38539276..276119316..1471661464...6393427268
.8.10272..537880.13825712.175434372.1516191100..9465023576..47252411120
.8.20520.1480026.50587924.798617096.8325624724.60874728614.349232818280

formulas for row 2 depending on the index modulo 2

Empirical: a(n)=4*a(n-1)-4*a(n-2)-4*a(n-3)+10*a(n-4)-4*a(n-5)-4*a(n-6)+4*a(n-7)-a(n-8)

0 Empirical: a(n) = 1*n^5 + 1*n^4 + (9/2)*n^3 + (3/2)*n^2
1 Empirical: a(n) = 1*n^5 + 1*n^4 + (9/2)*n^3 + (3/2)*n^2 - (3/2)*n + (3/2)

formulas for row 3 depending on the index modulo 12

Empirical: a(n)=2*a(n-1)+a(n-2)-2*a(n-3)-2*a(n-4)-2*a(n-5)+5*a(n-6)+2*a(n-7)-2*a(n-9)-5*a(n-10)+2*a(n-11)+2*a(n-12)+2*a(n-13)-a(n-14)-2*a(n-15)+a(n-16)

0 Empirical: a(n) = 1*n^6 + (43/6)*n^4 - (203/54)*n^3 + (62/9)*n^2 - (5/3)*n
1 Empirical: a(n) = 1*n^6 + (43/6)*n^4 - (203/54)*n^3 + (35/9)*n^2 + (23/6)*n - (223/54)
2 Empirical: a(n) = 1*n^6 + (43/6)*n^4 - (203/54)*n^3 + (62/9)*n^2 - (7/9)*n - (70/27)
3 Empirical: a(n) = 1*n^6 + (43/6)*n^4 - (203/54)*n^3 + (35/9)*n^2 + (23/6)*n - (13/2)
4 Empirical: a(n) = 1*n^6 + (43/6)*n^4 - (203/54)*n^3 + (62/9)*n^2 - (5/3)*n + (64/27)
5 Empirical: a(n) = 1*n^6 + (43/6)*n^4 - (203/54)*n^3 + (35/9)*n^2 + (85/18)*n - (599/54)
6 Empirical: a(n) = 1*n^6 + (43/6)*n^4 - (203/54)*n^3 + (62/9)*n^2 - (5/3)*n + 2
7 Empirical: a(n) = 1*n^6 + (43/6)*n^4 - (203/54)*n^3 + (35/9)*n^2 + (23/6)*n - (223/54)
8 Empirical: a(n) = 1*n^6 + (43/6)*n^4 - (203/54)*n^3 + (62/9)*n^2 - (7/9)*n - (124/27)
9 Empirical: a(n) = 1*n^6 + (43/6)*n^4 - (203/54)*n^3 + (35/9)*n^2 + (23/6)*n - (13/2)
10 Empirical: a(n) = 1*n^6 + (43/6)*n^4 - (203/54)*n^3 + (62/9)*n^2 - (5/3)*n + (118/27)
11 Empirical: a(n) = 1*n^6 + (43/6)*n^4 - (203/54)*n^3 + (35/9)*n^2 + (85/18)*n - (599/54)


/tmp/ejp
T(n,k)=Number of length n+3 0..k arrays with some disjoint pairs in every consecutive four terms having the same sum

Table starts
.8..33...88...185....336....553....848....1233....1720....2321....3048....3913
.8..45..136...317....600...1033...1616....2409....3400....4661....6168....8005
.8..61..220...561...1124...2009...3220....4901....7016....9737...13000...17025
.8..81..364..1007...2164...3997...6584...10219...14852...20847...28108...37095
.8.105..604..1823...4228...8051..13668...21609...31924...45309...61740...82067
.8.153.1018..3455...8440..16683..29012...47061...70374..101211..139098..186709
.8.217.1732..6495..16932..34695..62108..103013..156308..227701..316236..428111
.8.297.2956.12105..34068..72269.133716..226309..349160..515043..723892..987667
.8.393.5050.22459..68688.150677.288996..498569..783568.1170169.1665908.2290065
.8.585.8638.43255.139040.318575.627654.1111891.1772920.2686215.3862654.5366083

formulas for row 2 depending on the index modulo 2

Empirical: a(n)=2*a(n-1)+a(n-2)-4*a(n-3)+a(n-4)+2*a(n-5)-a(n-6)

0 Empirical: a(n) = (9/2)*n^3 + (3/2)*n^2 + 1*n + 1
1 Empirical: a(n) = (9/2)*n^3 + (3/2)*n^2 - (1/2)*n + (5/2)

formulas for row 3 depending on the index modulo 12

Empirical: a(n)=a(n-2)+2*a(n-3)+a(n-4)-2*a(n-5)-2*a(n-6)-2*a(n-7)+a(n-8)+2*a(n-9)+a(n-10)-a(n-12)

0 Empirical: a(n) = (563/54)*n^3 - (127/18)*n^2 + 2*n + 1
1 Empirical: a(n) = (563/54)*n^3 - (127/18)*n^2 - (5/2)*n + (385/54)
2 Empirical: a(n) = (563/54)*n^3 - (127/18)*n^2 + (10/9)*n + (97/27)
3 Empirical: a(n) = (563/54)*n^3 - (127/18)*n^2 - (5/2)*n + (19/2)
4 Empirical: a(n) = (563/54)*n^3 - (127/18)*n^2 + 2*n - (37/27)
5 Empirical: a(n) = (563/54)*n^3 - (127/18)*n^2 - (61/18)*n + (761/54)
6 Empirical: a(n) = (563/54)*n^3 - (127/18)*n^2 + 2*n - 1
7 Empirical: a(n) = (563/54)*n^3 - (127/18)*n^2 - (5/2)*n + (385/54)
8 Empirical: a(n) = (563/54)*n^3 - (127/18)*n^2 + (10/9)*n + (151/27)
9 Empirical: a(n) = (563/54)*n^3 - (127/18)*n^2 - (5/2)*n + (19/2)
10 Empirical: a(n) = (563/54)*n^3 - (127/18)*n^2 + 2*n - (91/27)
11 Empirical: a(n) = (563/54)*n^3 - (127/18)*n^2 - (61/18)*n + (761/54)



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


>________________________________
> From: Ron Hardin <rhhardin at att.net>
>To: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu> 
>Sent: Monday, September 8, 2014 9:57 PM
>Subject: [seqfan] Re: Symmetric and Antisymmetric Linear Recurrences?
> 
>
>The (-1)^n is a way of saying that there are different polynomials for even and odd points.
>
>Generalizing a little, I find that if you take row 3, and categorize points by their index being 0..11 modulo 12, you get a polynomial of degree 3 in each category; but not for fewer than 12 categories.  6 categories finds 3 polynomials and 3 nonpolynomials.
>
>