|
| |
|
|
A211441
|
|
Number of ordered triples (w,x,y) with all terms in {-n,...,0,...,n} and w + x + y = 2.
|
|
2
|
|
|
|
0, 3, 15, 33, 57, 87, 123, 165, 213, 267, 327, 393, 465, 543, 627, 717, 813, 915, 1023, 1137, 1257, 1383, 1515, 1653, 1797, 1947, 2103, 2265, 2433, 2607, 2787, 2973, 3165, 3363, 3567, 3777, 3993, 4215, 4443, 4677, 4917, 5163, 5415, 5673, 5937
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
|
OFFSET
|
0,2
|
|
|
COMMENTS
|
For a guide to related sequences, see A211422.
Number of dots in a hexagon in which the sides are alternately n-1 and n+3 dots long for n >= 3. a(n) = 3*k^2 + 9*k + 3, where k = n - 1 denotes the number of dots at the shorter side of the hexagon.
Let p be a prime number > 5. Then p divides exactly 2 terms out of any in p consecutive terms. If a(i) and a(k) contain the prime factor p, then i + k == -3 (mod p). (End)
Proof of Colin Barker's formulas: The first differences of the sequence for n > 0 are equal to 6*n. We assume that the recurrence a(n) = a(n-1) + 6*n is valid for any n > 1. In it, a(n-1) denotes the number of all ordered triples with -n < w, x, y < n according to the definition of the sequence. Consequently, 6*n must denote the number of all ordered triples containing -n and/or n, which must be proved.
The statement is empirically confirmed for the first n. We show that this applies to any n. With even n we determine all triples {w,x,y} with w = -n, x + y = n + 2 and start the enumeration of the triples with the highest possible value for x and end with the triple where x = y is: {-n, n, 2}, {-n, n-1, 3}, ..., {-n, n-i, 2+i}, ..., {-n, n-(n-2)/2, 2+(n-2)/2}. These are (n-2)/2 + 1 triples.
Now we set w = n and determine all triples {w,x,y} with x + y = 2 - n. We start the enumeration with the smallest possible x and end again, when x = y: {n, -n, 2}, {n, 1-n, 1}, ..., {n, i-n, 2-i}, ..., {n, (n+2)/2-n, 2-(n+2)/2}. These are (n+2)/2 +1 triples. Altogether we get n + 2 triples. Since the first triples of the two enumerations are identical, n + 1 triples remain. To get the ordered triples, they have to be permuted. We take into account that the respectively last triples contain two identical components and that only n - 1 triples consist of three distinct components. Thus the number of ordered triples (w,x,y) totals (n-1)* 3! + 2* 3!/2! = 6*n, which had to be proven.
With odd n we proceed in the same way with the difference that we end the two enumerations when |x - y| = 1. With w = -n we start with the largest possible x: {-n, n, 2}, {-n, n-1, 3}, ..., {-n, n-i, 2+i}, ..., {-n, n-(n-3)/2, 2+(n-3)/2}. These are (n-3)/2 + 1 triples.
For w = n we start with the smallest possible x: {n, -n, 2}, {n, 1-n, 1}, ..., {n, i-n, 2-i}, ..., {n, (n+1)/2-n, 2-(n+1)/2}. These are (n+1)/2 + 1 triples. Altogether these are n + 1 triples. Because the respectively first triples are identical here, n triples remain, and by permutation this results in 6*n ordered triples. Thus the proof is complete.
(End)
|
|
|
LINKS
|
|
|
|
FORMULA
|
a(n) = 3*(n^2 + n - 1) for n > 0.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3.
G.f.: 3*x*(1 + 2*x - x^2)/(1 - x)^3. (End)
a(n) + a(n+1) + a(n+2) + a(n+3) = 3*(2*n+4)^2 = 12*(n+2)^2 for n > 0.
a(n) + a(n+1) + a(n+2) = 3*A003215(n+1) - 6 for n > 0. (End)
|
|
|
EXAMPLE
|
Illustration of initial terms for n >= 3:
.
. o o o o o o o o
. o o o o o o o o o o o o o o o o
. o o o o o o o o o o o o o o o o o o o o o o o o
. o o o o o o o o o o o o o o o o o o o o o o o o o o o
. o o o o o o o o o o o o o o o o o o o o o o o o
. o o o o o o o o o o o o o o o o o o o o o
. o o o o o o o o o o o o o o o o o o
. o o o o o o o o o o o o o o o
. o o o o o o o o o o o o
. o o o o o o o o
. o o o o
.
. a(3) = 33 a(4) = 57 a(5) = 87
(End)
|
|
|
MATHEMATICA
|
t[n_]:= t[n]= Flatten[Table[w+x+y-2, {w, -n, n}, {x, -n, n}, {y, -n, n}]]
c[n_]:= Count[t[n], 0]
t = Table[c[n], {n, 0, 60}] (* A211441 *)
Join[{0}, LinearRecurrence[{3, -3, 1}, {3, 15, 33}, 50]] (* Harvey P. Dale, May 10 2012 *)
|
|
|
PROG
|
(GAP) b:=[3, 15, 33];; for n in [4..50] do b[n]:=3*b[n-1]-3*b[n-2]+b[n-3]; od; a:=Concatenation([0], b);; Print(a); # Muniru A Asiru, Jan 23 2019
(PARI) vector(50, n, n--; if(n==0, 0, 3*(n^2+n-1))) \\ G. C. Greubel, Feb 10 2019
(Magma) [n le 0 select 0 else 3*(n^2+n-1): n in [0..50]]; // G. C. Greubel, Feb 10 2019
(Sage) [0] + [3*(n^2+n-1) for n in (1..50)] # G. C. Greubel, Feb 10 2019
|
|
|
CROSSREFS
|
|
|
|
KEYWORD
|
nonn,easy
|
|
|
AUTHOR
|
|
|
|
STATUS
|
approved
|
| |
|
|
