[seqfan] Re: Symmetric and Antisymmetric Linear Recurrences?

[Ron Hardin]

Row 5, extending it from n=210 using its recurrence
Empirical: a(n)=-3*a(n-1)-6*a(n-2)-10*a(n-3)-15*a(n-4)-19*a(n-5)-21*a(n-6)-20*a(n-7)-15*a(n-8)-5*a(n-9)+9*a(n-10)+26*a(n-11)+44*a(n-12)+60*a(n-13)+71*a(n-14)+75*a(n-15)+70*a(n-16)+55*a(n-17)+32*a(n-18)+3*a(n-19)-29*a(n-20)-60*a(n-21)-85*a(n-22)-102*a(n-23)-108*a(n-24)-102*a(n-25)-85*a(n-26)-60*a(n-27)-29*a(n-28)+3*a(n-29)+32*a(n-30)+55*a(n-31)+70*a(n-32)+75*a(n-33)+71*a(n-34)+60*a(n-35)+44*a(n-36)+26*a(n-37)+9*a(n-38)-5*a(n-39)-15*a(n-40)-20*a(n-41)-21*a(n-42)-19*a(n-43)-15*a(n-44)-10*a(n-45)-6*a(n-46)-3*a(n-47)-a(n-48)


has 27720 different cubics.  Taking the sequence of number of cubics for rows 2,3,4,5 we get
2, 12, 420, 27720 different cubics, which is
http://oeis.org/A060942

The first 10 cubics of row 5 are
Empirical for n mod 27720 = 0: a(n) = (1027297/17325)*n^3 - (3502537/23100)*n^2 + (878029/6930)*n + 1
Empirical for n mod 27720 = 1: a(n) = (1027297/17325)*n^3 - (3502537/23100)*n^2 + (3587189/34650)*n - (44311/13860)
Empirical for n mod 27720 = 2: a(n) = (1027297/17325)*n^3 - (3502537/23100)*n^2 + (420821/3850)*n + (45853/2475)
Empirical for n mod 27720 = 3: a(n) = (1027297/17325)*n^3 - (3502537/23100)*n^2 + (716329/6930)*n + (17263/300)
Empirical for n mod 27720 = 4: a(n) = (1027297/17325)*n^3 - (3502537/23100)*n^2 + (4379057/34650)*n - (891203/17325)
Empirical for n mod 27720 = 5: a(n) = (1027297/17325)*n^3 - (3502537/23100)*n^2 + (198223/2310)*n + (44759/252)
Empirical for n mod 27720 = 6: a(n) = (1027297/17325)*n^3 - (3502537/23100)*n^2 + (4395689/34650)*n - (68743/1155)
Empirical for n mod 27720 = 7: a(n) = (1027297/17325)*n^3 - (3502537/23100)*n^2 + (3587189/34650)*n + (341887/9900)
Empirical for n mod 27720 = 8: a(n) = (1027297/17325)*n^3 - (3502537/23100)*n^2 + (84041/770)*n + (1394257/17325)
Empirical for n mod 27720 = 9: a(n) = (1027297/17325)*n^3 - (3502537/23100)*n^2 + (3570557/34650)*n + (56851/1100)



 
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: Thursday, September 11, 2014 9:36 AM
>Subject: [seqfan] Re: Symmetric and Antisymmetric Linear Recurrences?
> 
>
>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)
>>
>>
>
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