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A085047
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a(n) is the least number not already used such that the arithmetic mean of the first n terms is a square.
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3
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1, 7, 4, 24, 9, 51, 16, 88, 25, 135, 36, 192, 49, 259, 64, 336, 81, 423, 100, 520, 121, 627, 144, 744, 169, 871, 196, 1008, 225, 1155, 256, 1312, 289, 1479, 324, 1656, 361, 1843, 400, 2040, 441, 2247, 484, 2464, 529, 2691, 576, 2928, 625, 3175, 676, 3432, 729
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OFFSET
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1,2
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COMMENTS
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This is very nearly a linear recurrence, but the distinctness requirement occasionally foils it. - Charles R Greathouse IV, Nov 07 2014
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LINKS
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FORMULA
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a(2*n-1) = n^2; a(2*n) = n*(2+5*n). Also, (1/(2*n))*(Sum_{i=1..n} i^2 + i*(2+5*i)) = (n+1)^2 and (1/(2*n-1))*(Sum_{i=1..n} i^2 + (i-1)*(5*i-3)) = k^2. Thus the arithmetic mean of the first 2*n terms is (n+1)^2 and the arithmetic mean of the first 2*n-1 terms is n^2. - Derek Orr, Jun 26 2015
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EXAMPLE
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(a(1) + a(2) + a(3) + a(4) + a(5))/5 = (1+7+4+24+9)/5 = 9 = 3^2.
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MAPLE
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b:= proc(n) is(n>1) end:
s:= proc(n) option remember;
`if`(n=1, 1, s(n-1)+a(n))
end:
a:= proc(n) option remember; local k;
if n=1 then 1
else for k from n-irem(s(n-1), n) by n
do if b(k) and issqr((s(n-1)+k)/n)
then b(k):=false; return k
fi
od
fi
end:
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MATHEMATICA
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Clear[a, b, s]; b[n_] := n>1; s[n_] := s[n] = If[n == 1, 1, s[n-1] + a[n]]; a[n_] := a[n] = Module[{k}, If [n == 1, 1, For[k = n - Mod[s[n-1], n], True, k = k+n, If[b[k] && IntegerQ[Sqrt[(s[n-1]+k)/n]], b[k] = False; Return[k]]]]]; Table[a[n], {n, 1, 150}] (* Jean-François Alcover, Jun 10 2015, after Alois P. Heinz *)
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PROG
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(PARI) v=[1]; n=1; while(#v<50, s=(n+vecsum(v))/(#v+1); if(type(s)=="t_INT", if(issquare(s)&&!vecsearch(vecsort(v), n), v=concat(v, n); n=0)); n++); v \\ Derek Orr, Nov 05 2014, edited Jun 26 2015
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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