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A118060
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a(n) = 1681*n^2 - 984*n - 696.
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3
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1, 4060, 11481, 22264, 36409, 53916, 74785, 99016, 126609, 157564, 191881, 229560, 270601, 315004, 362769, 413896, 468385, 526236, 587449, 652024, 719961, 791260, 865921, 943944, 1025329, 1110076, 1198185, 1289656, 1384489, 1482684, 1584241
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OFFSET
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1,2
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COMMENTS
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In general, all sequences of equations which contain every positive integer in order exactly once (a pairwise equal summed, ordered partition of the positive integers) may be defined as follows: For all k, let x(k)=A001652(k) and z(k)=A001653(k). Then if we define a(n) to be (x(k)+z(k))n^2-(z(k)-1)n-x(k), the following equation is true: a(n)+(a(n)+1)+...+(a(n)+(x(k)+z(k))n+(2x(k)+z(k)-1)/2)=(a(n)+ (x(k)+z(k))n+(2x(k)+z(k)+1)/2)+...+(a(n)+2(x(k)+z(k))n+x(k)); a(n)+2(x(k)+z(k))n+x(k))=a(n+1)-1; e.g., in this sequence, x(4)=A001652(4)=696 and z(4)=A001653(4)=985; cf. A000290, A118057-A118059, A118061.
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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). G.f.: x*(1+4057*x-696*x^2)/(1-x)^3. - Colin Barker, Jul 01 2012
a(n)+(a(n)+1)+...+(a(n)+1681n+1188) = (a(n)+1681n+1189)+ ... +a(n+1)-1; a(n+1)-1 = a(n)+3362n+696.
a(n)+(a(n)+1)+...+(a(n)+1681n+1188)=41(41n-12)(41n+29)(82n+17)/2; e.g., 11481+11482+...+17712=90965388=41*111*152*263/2.
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EXAMPLE
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a(3)=1681*3^2-984*3-696=11481, a(4)=1681*4^2-984*4-696=22264 and 11481+11482+...+17712=17713+...+22263
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MATHEMATICA
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CoefficientList[Series[(1+4057*x-696*x^2)/(1-x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 09 2012 *)
LinearRecurrence[{3, -3, 1}, {1, 4060, 11481}, 40] (* Harvey P. Dale, Oct 28 2016 *)
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PROG
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CROSSREFS
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KEYWORD
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nonn,easy,less
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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