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A257093
a(n) = n*(n+1)*(13*n+2)/6.
1
0, 5, 28, 82, 180, 335, 560, 868, 1272, 1785, 2420, 3190, 4108, 5187, 6440, 7880, 9520, 11373, 13452, 15770, 18340, 21175, 24288, 27692, 31400, 35425, 39780, 44478, 49532, 54955, 60760, 66960, 73568, 80597, 88060, 95970, 104340, 113183, 122512, 132340
OFFSET
0,2
COMMENTS
This sequence gives the number of triangles of all sizes in (5*n^2)-polyiamonds in a tetragonal or hexagonal or heptagonal configuration.
It is the sum of (1/2)*Sum{j=0..n-1}(n-j)*(5*n+1-j) triangles oriented in one direction and (1/2)*Sum{j-0..n-1}(n-j)*(5*n-1-3*j) oriented in the opposite direction.
Shäfli's notation: 3.3.3.3.3 for a(1).
The difference between this sequence and A050409(n) equals A000292(n-1).
Also, (1/3)*(A002717(2*n) + A255211(n) - 2*A000330(n) gives A033994(n): a (5*n^2)-polyiamond in pentagonal configuration that does not belong to this sequence because a(1)=6.
a(n) is odd only when n mod 4 = 1.
FORMULA
a(n) = Sum_{j=0..n-1}(n-j)*(5*n-2*j).
G.f.: x*(5+8*x)/(1-x)^4. - Vincenzo Librandi, Apr 16 2015
a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4). - Vincenzo Librandi, Apr 16 2015
EXAMPLE
Second comment a(0) = 0; a(1) = 3 + 2; a(2) = 16 + 12; a(3) = 46 + 36; a(4) = 100 + 80; a(5) = 185 + 150; a(6) = 308 + 252.
MATHEMATICA
Table[n (n + 1) (13 n + 2)/6, {n, 0, 40}] (* Vincenzo Librandi, Apr 16 2015 *)
CoefficientList[Series[x (5+8x)/(1-x)^4, {x, 0, 50}], x] (* or *) LinearRecurrence[{4, -6, 4, -1}, {0, 5, 28, 82}, 60] (* Harvey P. Dale, Feb 12 2023 *)
PROG
(Magma) [n*(n+1)*(13*n+2)/6: n in [0..40]]; // Vincenzo Librandi, Apr 16 2015
KEYWORD
nonn,easy
AUTHOR
Luce ETIENNE, Apr 16 2015
EXTENSIONS
Corrected by Harvey P. Dale, Feb 12 2023
STATUS
approved