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A093916
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a(2*k-1) = (2*k-1)^2 + 2 - k, a(2*k) = 6*k^2 + 2 - k: First column of the triangle A093915.
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4
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2, 7, 9, 24, 24, 53, 47, 94, 78, 147, 117, 212, 164, 289, 219, 378, 282, 479, 353, 592, 432, 717, 519, 854, 614, 1003, 717, 1164, 828, 1337, 947, 1522, 1074, 1719, 1209, 1928, 1352, 2149, 1503, 2382, 1662, 2627, 1829, 2884, 2004, 3153, 2187, 3434, 2378, 3727, 2577, 4032, 2784, 4349, 2999, 4678, 3222, 5019, 3453, 5372
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OFFSET
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1,1
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COMMENTS
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The sequence was initially defined as the first column of the triangle A093915, constructed by trial and error. It is however easy to prove that the sum of the r-th row of A093915, A093917(r), equals twice A006003(r) when r is odd, and three times A006003(r) when r is even. Given the expression of the row sum A093917(r) in terms of the first element a(r), one obtains the explicit formula for a(r). - M. F. Hasler, Apr 04 2009
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LINKS
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FORMULA
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a(n) = ((n^2+1)*b(n) - n + 1)/2 where b(n) = 3 - (n mod 2) = 2 if n odd, = 3 if n even. - M. F. Hasler, Apr 04 2009
a(n) = (n*(5*n-2) + (n^2+1)*(-1)^n + 7)/4.
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6).
G.f.: x*(2+7*x+3*x^2+3*x^3+3*x^4+2*x^5)/((1-x)^3*(1+x)^3). (End)
E.g.f.: (1/4)*( (7 +3*x +5*x^2)*exp(x) - 8 + (1 -x +x^2)*exp(-x) ). - G. C. Greubel, Dec 30 2021
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MATHEMATICA
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LinearRecurrence[{0, 3, 0, -3, 0, 1}, {2, 7, 9, 24, 24, 53}, 80] (* Harvey P. Dale, Nov 24 2017 *)
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PROG
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(PARI)
/* or the "experimental" version, trying out all allowed values */
A093916(n)={ local( s=(n^3+n)/2, d=(n^2-n)/2, k=ceil((2*s-d)/n)); while( (n*k+d)%s, k++ ); k } \\ M. F. Hasler, Apr 04 2009
(Magma) [(n*(5*n-2) + (-1)^n*(n^2+1) + 7)/4: n in [1..70]]; // G. C. Greubel, Dec 30 2021
(SageMath) [(5*n^2 -2*n +7 +(-1)^n*(n^2 +1))/4 for n in (1..70)] # G. C. Greubel, Dec 30 2021
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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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EXTENSIONS
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STATUS
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approved
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