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A118061
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9800*n^2-5740*n-4059
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5
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1, 23661, 66921, 129781, 212241, 314301, 435961, 577221, 738081, 918541, 1118601, 1338261, 1577521, 1836381, 2114841, 2412901, 2730561, 3067821, 3424681, 3801141, 4197201, 4612861, 5048121, 5502981, 5977441, 6471501
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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(5)=A001652(5)=4059 and z(5)=A001653(5)=5741; cf. A000290,A118057-A118060.
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LINKS
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FORMULA
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a(n)+(a(n)+1)+...+(a(n)+9800n+6929)=(a(n)+9800n+6930)+...+(a(n)+19600n+4059); a(n)+19600n+4059=a(n+1)-1; a(n+1)-1=a(n)+19600n+4059.
a(n)+(a(n)+1)+...+(a(n)+576n+203)=35(140n-41)(140n+29)(140n+99); e.g., 66921+66922+...+103250=3091156215=35*379*449*519.
a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). G.f.: x*(1+23658*x-4059*x^2)/(1-x)^3. - Colin Barker, Jul 01 2012
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EXAMPLE
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a(3)= 9800*3^2-5740*3-4059=66921, a(4)=9800*4^2-5740*4-4059=129781 and 66921+66922+...+103250=103251+...+129780
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MATHEMATICA
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CoefficientList[Series[(1+23658*x-4059*x^2)/(1-x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Jul 09 2012 *)
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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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