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A256666
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a(n) = ( 2*n*(2*n^2 + 11*n + 26) - (-1)^n + 1 )/16.
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2
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0, 5, 14, 29, 51, 82, 123, 176, 242, 323, 420, 535, 669, 824, 1001, 1202, 1428, 1681, 1962, 2273, 2615, 2990, 3399, 3844, 4326, 4847, 5408, 6011, 6657, 7348, 8085, 8870, 9704, 10589, 11526, 12517, 13563, 14666, 15827, 17048, 18330, 19675, 21084, 22559
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OFFSET
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0,2
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COMMENTS
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Consider a grid of small triangles of side 1 forming polygon with side n*(n+3): a(n) is the number of equilateral triangles of side length >=1 in this figure that are oriented with the sides of figure.
This sequence gives the number of triangles of all sizes in a ((n^2+3*n))-iamond with a 3*(2*n-1)-gon n>=1.
Equals (1/2)*Sum_{i=0..n-1} (n-i)*(n+1-i) + (-3 + (1/8)*Sum_{j=0..(2*n+3+(-1)^n)/4} (2*n+5-(-1)^n-4*j)*(2 n+5+(-1)^n-4*j) ) numbers of triangles in a direction and in the opposite direction.
It is also a way (3 stages) to surround triangular n^2-iamonds by 3*n triangles side 1: in first stage we obtain A045947, in second stage A248851, in third stage this sequence.
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LINKS
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FORMULA
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G.f.: x*(2*x^3-3*x^2-x+5) / ((x-1)^4*(x+1)). - Colin Barker, Apr 07 2015
a(n) = 3*a(n-1)-2*a(n-2)-2*a(n-3)+3*a(n-4)-a(n-5) for n>4. - Colin Barker, Apr 07 2015
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EXAMPLE
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From third comment: a(0)=0, a(1)=1+4, a(2)=4+10, a(3)=10+19, a(4)=20+31, a(5)=35+47, a(6)=56+67.
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MATHEMATICA
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Table[(4 n^3 + 22 n^2 + 52 n + 1 - (-1)^n)/16, {n, 0, 50}] (* Vincenzo Librandi, Apr 08 2015 *)
LinearRecurrence[{3, -2, -2, 3, -1}, {0, 5, 14, 29, 51}, 50] (* Harvey P. Dale, Aug 18 2020 *)
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PROG
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(PARI) concat(0, Vec(x*(2*x^3-3*x^2-x+5)/((x-1)^4*(x+1)) + O(x^100))) \\ Colin Barker, Apr 07 2015
(Magma) [(4*n^3+22*n^2+52*n+1-(-1)^n)/16: n in [0..50]]; // Vincenzo Librandi, Apr 08 2015
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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