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A206827
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Number of solutions (n,k) of s(k) == s(n) (mod n), where 1 <= k < n and s(k) = k*(k+1)*(2*k+1)/6.
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4
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1, 1, 1, 2, 2, 2, 1, 1, 3, 2, 2, 2, 3, 3, 1, 2, 1, 2, 3, 3, 3, 2, 1, 2, 3, 1, 3, 2, 3, 2, 1, 3, 3, 8, 1, 2, 3, 3, 3, 2, 4, 2, 3, 3, 3, 2, 2, 2, 3, 3, 3, 2, 2, 8, 3, 3, 3, 2, 2, 2, 3, 3, 1, 8, 3, 2, 3, 3, 9, 2, 1, 2, 3, 3, 3, 8, 2, 2, 3, 1, 3, 2, 4, 8, 3, 3, 3, 2, 5, 8, 3, 3, 3, 8, 2, 2, 3, 3, 3
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OFFSET
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2,4
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COMMENTS
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For a guide to related sequences, see A206588.
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LINKS
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EXAMPLE
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5 divides exactly two of the numbers s(n)-s(k) for k=1,2,3,4, so a(5)=2.
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MAPLE
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f:= n -> numboccur(S[n] mod n, S[1..n-1] mod n):
S:= [seq(k*(k+1)*(2*k+1)/6, k=1..100)]:
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MATHEMATICA
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s[k_] := k (k + 1) (2 k + 1)/6;
f[n_, k_] := If[Mod[s[n] - s[k], n] == 0, 1, 0];
t[n_] := Flatten[Table[f[n, k], {k, 1, n - 1}]]
a[n_] := Count[Flatten[t[n]], 1]
Table[a[n], {n, 2, 120}] (* A206827 *)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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