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A103115
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a(n) = 6*n*(n-1)-1.
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3
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-1, -1, 11, 35, 71, 119, 179, 251, 335, 431, 539, 659, 791, 935, 1091, 1259, 1439, 1631, 1835, 2051, 2279, 2519, 2771, 3035, 3311, 3599, 3899, 4211, 4535, 4871, 5219, 5579, 5951, 6335, 6731, 7139, 7559, 7991, 8435, 8891, 9359, 9839, 10331, 10835, 11351, 11879, 12419
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OFFSET
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0,3
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COMMENTS
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What A163433 does for a triangle, this sequence is doing for a square but giving one-half the results. Take a square with vertices n, n+1, n+2, and n+3 and find the sum of the four products of each four vertices times the sum of the other three; at n you have n((n+1)+(n+2)+(n+3)) and so on for the other three vertices. The result of all four is 12*n^2+36*n+22; half this is 6*n^2+18*n+11 and gives the numbers in this sequence starting with n=0. - J. M. Bergot, May 23 2012
Multiplying a(n) by 16 gives the sum of the convolution with itself of each of the 24 permutations of four consecutive numbers. - J. M. Bergot, May 15 2017
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LINKS
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FORMULA
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a(n) = 3*a(n-1)-3*a(n-2)+a(n-3), with a(0)=-1, a(1)=-1, a(2)=11. - Harvey P. Dale, Nov 14 2011
a(n) = (n-2)*(n-1 + n + n+1) + (n-1)*(n + n+1) + n*(n+1), which is applying A000914 to four consecutive numbers. - J. M. Bergot, May 15 2017
Sum_{n>=1} 1/a(n) = tan(sqrt(5/3)*Pi/2)*Pi/(2*sqrt(15)). Amiram Eldar, Aug 20 2022
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MATHEMATICA
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Table[6n(n-1)-1, {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {-1, -1, 11}, 50] (* Harvey P. Dale, Nov 14 2011 *)
CoefficientList[Series[(1 - 2 x - 11 x^2) / (x - 1)^3, {x, 0, 50}], x] (* Vincenzo Librandi, May 16 2017 *)
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PROG
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CROSSREFS
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KEYWORD
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easy,sign
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AUTHOR
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Jacob Landon (jacoblandon(AT)aol.com), May 09 2009
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EXTENSIONS
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STATUS
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approved
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