Scaled Chebyshev U-polynomials evaluated at sqrt(3)/2

Fri Apr 22 17:23:56 CEST 2005

Gerald McGarvey wrote:

I get the following:
A105660: For n > 11, a(n) = - 970*a(n-6) - a(n-12)
A057083: For n > 5, a(n) = -27*a(n-6)

Sincerely,
Gerald

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Thank you for that information. Would you care to send it in as a
comment? 

[ Correction to my first message: The "Smith College Diploma Sequence"
is 
http://www.research.att.com/projects/OEIS?Anum=A057681 ]

I wrote:
"Can someone find more in common between these two sequences
 (or confirm that they belong to a more general class of sequences with
those properties)?" 

Last year, I sent the Seqfan message, enclosed in "****", at the bottom
of the page. The sequence property described is:

for m > n:  if  s | a(n) and  s | a(m)  then   s | a(2m - n)   

It may be that I am overlooking something obvious as I have only done a
bit of empircal checking, but I guess that A105660 has this property (if
Gerald's recurrence relation for A057083 is true, then it is clear that
that sequence also has the property after checking the first few terms).

Just to point out... my sequence began:
3,10,27,49,0,-485,-2643,-9602,-26163,-47525,0,470449,
2563707,9313930,25378083,46099201,0,-456335045,-2486793147,
-9034502498,-24616714347,-44716177445

a(3) = 49 = 7^2. The next term divisible by 7 is a(11) = 470449 =
(7^2)(6901). It follows that if the sequence is to have the property,
a(2*11-3) = a(19) = 9034502498 = 2*(7^2)*(92188801) should be divisible
by 7, etc.  

IMHO, I propose that the above simple property is given a name (which I
would be more than happy to let Gerald choose...) if it does not already
have one. 

By the way, I just submitted an unrelated sequence, below, which also
appears to have the property. [ Sorry, I forgot to factor the
denominator
 (-x^12-1+2*x^6) = (x-1)^2*(x+1)^2*(x^2+x+1)^2*(x^2-x+1)^2 ]

-1 1 2 2 9 9 4 -4 5 5 18 18 7 -7 8 8 27 27 10 -10 11 11 36 36 13 -13 14
14 45 45 16 -16 17 17 54 54 19 -19 20 20 63 63 22 -22 23 23 72 72 25 -25
26 26 81 81 28 -28 29 29 90 90 31 -31 32 32 99 99 34 -34 35 35 108 108
37 -37 38 38 117 117 40 -40 41 41 126 126 43 -43 44 44 135 135 46 -46 47
47 144 144 49 -49 50 50 153 153 52 -52 53 53 162 162 55 -55 56 56 171
171
%N A000001 Expansion of
(1-x-2*x^2-2*x^3-9*x^4-9*x^5-6*x^6+6*x^7-x^8-x^9-2*x^13+2*x^12)/(-x^12-1+2*x^6)
%H A000001 C. Dement, <a
href="http://www.crowdog.de/13829/home.html">The
Floretions</a>.
%H A000001 C. Dement, <a
href="http://www.crowdog.de/SeqContext/Plush.html">Sequences in
Context</a>.
%o A000001 Floretion Algebra Multiplication Program, FAMP Code:
4kbasesigcycrokseq[+ .25'j - .25'k + .25j' - .25k' + .5'ii' + .25'ij' +
.25'ik' + .25'ji' + .25'ki' + .5e]. See "Sequences in Context" for
details on the "roktype" used.

Sincerely, 
Creighton   

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[snip]
a(1) = 2
a(2) = 7
a(3) = 11
a(4) = 25 = 5*5
a(5) = 47 
a(6) = 97
a(7) = 191
a(8) = 385 = 5*7*11
a(9) = 767 = 13*59
a(10) = 1537 = 29*53
a(11) = 3071 = 37*83
a(12) = 6145 = 5*1229
a(13) = 12287 = 11*1117
a(14) = 24577 = 7*3511
a(15) = 49151 = 23*2137
a(16) = 98305 = 5*19661
a(17) = 196607 = 421*467
a(18) = 393217 = 11*35747
a(19) = 786431 
a(20) = 1572865 = 5*7*44939
a(21) = 3145727 = 13*241979
a(22) = 6291457 = 347*18131

One could assume (law of small numbers) that the following holds: for m
> n
( s | a(n) ) and ( s | a(m) ) -> ( s | a(2m - n) )  

For ex. ( 7 divides a(2) = 7 ) and ( 7 divides a(8) = 385 = 5*7*11 )  
 and ( 7 divides a(2*8 - 2) = a(14) = 24577 = 7*3511 ) 
and ( 7 divides a(2*14 - 8) = a(20) = 1572865 = 5*7*44939 )

( If this is true, then 13 should divide a(21*2 - 9) =  a(33) ) 

Do these types of sequences have a name (if so, what would be the best
way to prove it...)?
In any case, the above sequence (unlisted in OEIS) is connected with the

Jacobsthal-Lucas numbers 
http://www.research.att.com/projects/OEIS?Anum=A014551 via
 c(n) +  b(n) + A014551(n+1) = 4*a(n) 

(b(n)) =  (5, 19, 29, 67, 125, 259, 509, 1027, 2045, 4099, 8189, ) ;
unlisted
(c(n)) = (2, 4, 8, 16, 32, 64, )

Apparently, the above "law of small numbers" holds for A014551 as well, 
ex.  A014551(5) = 31 and A014551(15) = 7*31*151 and 
A014551(25) = 31*601*1801
It probably also holds for (b(n)) [and trivially for (c(n)) ]

Now, for (Fib(n)), one has ( s | Fib(n) ) -> ( s | Fib(kn) ) which is 
readily seen to be a stronger condition than the above statement. 

Proof:  
( s | Fib(n) ) and ( s | Fib(m) ) ->   ( s | Fib(kn) ) and ( s |
Fib(k'm) ) 

A simple calculation shows:

F(2m - n) = | 2F(2m - n + 2) - F(2m - n + 3) | 
                = | -3F(2m - n + 3) + 2F(2m - n + 4) |
                = | 5F(2m - n + 4) - 3F(2m - n + 5) |
                = | F(n) F(2m-1) - F(n-1)F(2m) |

Thus, setting k' = 2 and k = 1, it follows that s |  F(2m - n),   q.e.d.

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