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A175110
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a(n) = ((2*n+1)^4+1)/2.
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7
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1, 41, 313, 1201, 3281, 7321, 14281, 25313, 41761, 65161, 97241, 139921, 195313, 265721, 353641, 461761, 592961, 750313, 937081, 1156721, 1412881, 1709401, 2050313, 2439841, 2882401, 3382601, 3945241, 4575313, 5278001, 6058681
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OFFSET
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0,2
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COMMENTS
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Partial sums of A117216. Binomial transform of 1,40,232,384,192,0,0,.. (0 continued). Convolution of the finite sequence 1,36,118,36,1 with A000332, dropping zeros.
Hypotenuse of Pythagorean triangles with smallest side a square: A016754(n)^2 + (a(n)-1)^2 = a(n)^2. - Martin Renner, Nov 12 2011
a(n) is also the first integer in a sum of (2*n + 1)^4 consecutive integers that equal (2*n + 1)^8. See A016756 and A016760. - Patrick J. McNab, Dec 26 2016
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REFERENCES
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Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 106, table 54.
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LINKS
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FORMULA
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a(n) = 5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5).
G.f.: (1+36*x+118*x^2+36*x^3+x^4)/ (1-x)^5.
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MAPLE
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MATHEMATICA
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CoefficientList[Series[(1 + 36*x + 118*x^2 + 36*x^3 + x^4)/(1-x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 19 2012 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {1, 41, 313, 1201, 3281}, 40] (* Harvey P. Dale, Jan 01 2022 *)
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PROG
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(Magma) I:=[1, 41, 313, 1201, 3281]; [n le 5 select I[n] else 5*Self(n-1) - 10*Self(n-2) + 10*Self(n-3) - 5*Self(n-4) + Self(n-5): n in [1..40]]; // Vincenzo Librandi, Dec 19 2012
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CROSSREFS
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KEYWORD
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easy,nonn
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AUTHOR
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STATUS
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approved
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