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A355759
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Sums of the first ceiling((n+1)/2) entries on the diagonals of a square spiral with a starting value of 1 in the center, where the diagonal and the antidiagonal are used alternately.
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1
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1, 4, 6, 11, 15, 24, 32, 45, 57, 76, 94, 119, 143, 176, 208, 249, 289, 340, 390, 451, 511, 584, 656, 741, 825, 924, 1022, 1135, 1247, 1376, 1504, 1649, 1793, 1956, 2118, 2299, 2479, 2680, 2880, 3101, 3321, 3564, 3806, 4071, 4335, 4624, 4912, 5225, 5537, 5876, 6214, 6579, 6943
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OFFSET
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1,2
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LINKS
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FORMULA
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G.f.: -(x^7 - 2*x^6 + x^4 - x^3 + 2*x^2 - 2*x-1)/((x - 1)^4 * (x + 1)^2 * (x^2 +1)).
a(n) = (24 + 20*n + 6*n^2 + n^3) / 24 for n even.
a(n) = (12 + 17*n + 6*n^2 + n^3) / 24 for n odd and n (mod 4) == 3.
a(n) = (17*n + 6*n^2 + n^3) / 24 for n odd and n (mod 4) == 1.
E.g.f.: ((30 + 45*x + 12*x^2 + x^3)*cosh(x) + (51 + 42*x + 12*x^2 + x^3)*sinh(x) - 6*cos(x))/24. - Stefano Spezia, Aug 19 2022
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EXAMPLE
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See the PDF in links.
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MATHEMATICA
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CoefficientList[Series[-(x^7 - 2*x^6 + x^4 - x^3 + 2*x^2 - 2*x - 1)/((x - 1)^4*(x + 1)^2*(x^2 + 1)), {x, 0, 50}], x] (* Amiram Eldar, Aug 19 2022 *)
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PROG
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(Python)
if n % 2 == 0: return((24 + 20*n + 6*n**2 + n**3)//24)
elif n % 4 == 3: return((12 + 17*n + 6*n**2 + n**3)//24)
elif n % 4 == 1: return(( 17*n + 6*n**2 + n**3)//24)
(PARI) a(n) = (n^2 + 6*n + if(n%2, 17, 20))*n \ 24 + (n%4!=1); \\ Kevin Ryde, Aug 19 2022
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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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