A085047 a(n) is the least number not already used such that the arithmetic mean of the first n terms is a square.
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
Offset: 1
Keywords
Examples
(a(1) + a(2) + a(3) + a(4) + a(5))/5 = (1+7+4+24+9)/5 = 9 = 3^2.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..10000
Crossrefs
Cf. A168668.
Programs
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Maple
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: seq(a(n), n=1..150); # Alois P. Heinz, Nov 07 2014
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Mathematica
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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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
Formula
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
Extensions
More terms from David Wasserman, Jan 11 2005
Incorrect formulas and programs removed by Charles R Greathouse IV, Nov 07 2014
Comments