What next in 0,1,6,25,96,361,1350,5041?
Rainer Rosenthal
r.rosenthal at web.de
Sat Mar 13 23:45:41 CET 2004
Another way of putting the question above:
What have the following sequences in common:
S_4 = A000290 the squares
S_5 = A004146 Alternate Lucas Numbers - 2
S_7 = A054493 A Pellian-related sequence
S_8 = A001108 a(n)-th triangular number is a square
S_9 = A049684 F(2n)^2 where F() = Fibonacci numbers
S_20 = A049683 a(n)=(L(6n)-2)/16, L=Lucas Sequence
S_25 = A089927* Expansion of 1/((1-x^2)(1-5x+x^2))
S_36 = A001110 Both triangular and square
S_49 = A049682 a(n)=(L(8n)-2)/45, L=Lucas sequence
S_144 = A004191^2 a(n)=S(n,12) (Chebyshev's poly 2. kind)
where A089927*(n) = A089927(2n-2)
Answer: for any three successive members a, b, c we have
a * c = (b-1)^2, i.e. the obey the recurrence
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
a(n-1) * a(n+1) = ( a(n) - 1 )^2
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
All S_r sequences start with 0, 1, r.
I will enter sequence S_6 in the next few days, which is
the first S_r not in the OEIS.
But A002531 is closely related to S_6, as can be seen from
the following table:
S_6 A002531^2 2*A002531^2-2
----------------------------------------
0 0
1 1
6 6
25 25
96 96
361 361
1350 1350
5041 5041
18816 18816
70225 70225
262086 262086
I think it is a fine observation that so many sequences in
the OEIS obey the same recurrence rule a c = (b-1)^2.
And funny enough this rule is never mentioned for any of
these sequences! Shall I add this as a remark for them?
I recall that this recurrence has been stated some days ago
by "zaphod" alias Colin Dickson for the squaretriangles, i.e.
for sequence A001110.
Let me finish my lengthy posting with the answer to the
question in the subject line: 18816 (see table).
Best regards
Rainer Rosenthal
r.rosenthal at web.de
More information about the SeqFan
mailing list