# [seqfan] Re: Symmetric and Antisymmetric Linear Recurrences?

Ron Hardin rhhardin at att.net
Thu Sep 11 15:36:08 CEST 2014

```Extending to the fourth row, using the linear recurrence to produce the values above n=360 that are needed :

Empirical for row 4 : a(n)=-a(n-1)-a(n-2)+a(n-4)+2*a(n-5)+3*a(n-6)+3*a(n-7)+2*a(n-8)-2*a(n-10)-4*a(n-11)-4*a(n-12)-4*a(n-13)-2*a(n-14)+2*a(n-16)+3*a(n-17)+3*a(n-18)+2*a(n-19)+a(n-20)-a(n-22)-a(n-23)-a(n-24)

which comes from polynomials of degree 3 with linear and constant terms depending on the index modulo 420
Here are the first 10 of 420

Empirical for n mod 420 = 0: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (4243/210)*n + 1
Empirical for n mod 420 = 1: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (2003/210)*n + (622/45)
Empirical for n mod 420 = 2: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (3263/210)*n + (3998/315)
Empirical for n mod 420 = 3: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (2003/210)*n + (148/5)
Empirical for n mod 420 = 4: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (4243/210)*n - (3821/315)
Empirical for n mod 420 = 5: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (341/70)*n + (3580/63)
Empirical for n mod 420 = 6: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (4243/210)*n - (404/35)
Empirical for n mod 420 = 7: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (2003/210)*n + (784/45)
Empirical for n mod 420 = 8: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (3263/210)*n + (1187/45)
Empirical for n mod 420 = 9: a(n) = (15541/630)*n^3 - (1401/35)*n^2 + (2003/210)*n + (886/35)

there are four linear terms (with counts)
140 (2003/210)*n
70 (3263/210)*n
70 (341/70)*n
140 (4243/210)*n

and 180 absolute values of constant terms, each of which occurs one or three times

This gets out of hand pretty fast for higher row numbers.

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

/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)
>
>

```