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A231071
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Number of solutions to n = +- 1^2 +- 2^2 +- 3^2 +- 4^2 +- ... +- k^2 for minimal k giving at least one solution.
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4
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2, 1, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 1, 6, 1, 1, 1, 1, 1, 2, 2, 2, 2, 1, 1, 1, 1, 2, 3, 2, 2, 2, 1, 1, 2, 1, 2, 1, 2, 1, 9, 1, 3, 1, 1, 1, 2, 1, 1, 6, 1, 1, 1, 1, 1, 2, 1, 5, 1, 1, 1, 1, 4, 3, 1, 2, 1, 2, 2, 1, 2, 1, 14, 2, 1, 3, 2, 1, 2, 1, 1, 7, 1, 3, 2, 5, 1, 2, 1
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OFFSET
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0,1
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COMMENTS
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This type of sequence was first studied by Andrica and Vacaretu. - Jonathan Sondow, Nov 06 2013
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LINKS
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FORMULA
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a(n(n+1)(2n+1)/6) = 1 for n > 0: n(n+1)(2n+1)/6 = 1+4+9+...+n^2. See A000330.
a(n(n+1)(2n+1)/6 - 2) = 1 for n > 1: n(n+1)(2n+1)/6 - 2 = -1+4+9+...+n^2. (End)
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EXAMPLE
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a(8) = 3: 8 = -1-4-9-16+25-36+49 = -1-4+9+16-25-36+49 = -1+4+9-16+25+36-49.
a(9) = 2: 9 = -1-4+9+16+25-36 = 1+4+9-16-25+36.
a(10) = 1: 10 = -1+4-9+16.
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MAPLE
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b:= proc(n, i) option remember; (m->`if`(n>m, 0, `if`(n=m, 1,
b(n+i^2, i-1) +b(abs(n-i^2), i-1))))((1+(3+2*i)*i)*i/6)
end:
a:= proc(n) local k; for k while b(n, k)=0 do od; b(n, k) end:
seq(a(n), n=0..100);
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MATHEMATICA
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b[n_, i_] := b[n, i] = Function[m, If[n > m, 0, If[n == m, 1,
b[n+i^2, i-1] + b[Abs[n-i^2], i-1]]]][(1+(3+2*i)*i)*i/6];
a[n_] := Module[{k}, For[k = 1, b[n, k] == 0, k++]; b[n, k]];
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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