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A147973
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a(n) = -2*n^2 + 12*n - 14.
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24
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-4, 2, 4, 2, -4, -14, -28, -46, -68, -94, -124, -158, -196, -238, -284, -334, -388, -446, -508, -574, -644, -718, -796, -878, -964, -1054, -1148, -1246, -1348, -1454, -1564, -1678, -1796, -1918, -2044, -2174, -2308, -2446, -2588, -2734, -2884
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OFFSET
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1,1
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COMMENTS
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-a(n+3) = 2*n^2 - 4, n >= 0, [-4,-2, 4, 14, ...] appears as the first member of the quartet for the square of [n, n+1, n+2, n+3], for n >= 0, in the Clifford algebra Cl_2. The other members are given in A046092(n), A054000(n+1) and A139570(n). The basis of Cl_2 is <1, s1, s2, s12> with s1.s1 = s2.s2 = 1, s12.s12 = -1, s1.s2 = -s2.s1 = s12. See e.g., pp. 5-6, eqs. (2.4)-(2.13) of the S. Gull et al. reference. - Wolfdieter Lang, Oct 15 2014
Related to the previous comment: if one uses the exterior (Grassmann) product with s1.s1 = s2.s2 = = s12.s12 = 0 and s1.s2 = -s2.s1 = s12, then the four components of the square of [n, n+1, n+2, n+3] are [A000290(n), A046092(n), A054000(n+1), A139570(n)], n >= 0. - Wolfdieter Lang, Nov 13 2014
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LINKS
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FORMULA
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G.f.: -2*x*(2 - 7*x + 7*x^2) / (1 - x)^3. - Colin Barker, Feb 12 2019
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MAPLE
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MATHEMATICA
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lst={}; Do[k=n^2-((n-1)^2+(n-2)^2+(n-3)^2); AppendTo[lst, k], {n, 5!}]; lst
LinearRecurrence[{3, -3, 1}, {-4, 2, 4}, 50] (* Harvey P. Dale, Mar 02 2020 *)
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PROG
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(PARI) Vec(-2*x*(2 - 7*x + 7*x^2) / (1 - x)^3 + O(x^40)) \\ Colin Barker, Feb 12 2019
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CROSSREFS
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KEYWORD
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sign,easy
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AUTHOR
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STATUS
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approved
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