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A206824 Number of solutions (n,k) of s(k) = s(n) (mod n), where 1 <= k < n and s(k) = k(k+1)/2.

%I A206824 #12 Sep 23 2017 03:22:55
%S A206824 1,1,1,1,2,1,1,1,2,1,2,1,2,3,1,1,2,1,2,3,2,1,2,1,2,1,2,1,4,1,1,3,2,3,
%T A206824 2,1,2,3,2,1,4,1,2,3,2,1,2,1,2,3,2,1,2,3,2,3,2,1,4,1,2,3,1,3,4,1,2,3,
%U A206824 4,1,2,1,2,3,2,3,4,1,2,1,2,1,4,3,2,3,2,1,4,3,2,3,2,3,2,1,2,3,2
%N A206824 Number of solutions (n,k) of s(k) = s(n) (mod n), where 1 <= k < n and s(k) = k(k+1)/2.
%C A206824 Records at indices {(1), 2, 6, 15, 30, 105, 210, 1155, 2310, 15015, 30030, ...}. - _Michael De Vlieger_, Sep 22 2017
%H A206824 Michael De Vlieger, <a href="/A206824/b206824.txt">Table of n, a(n) for n = 2..30030</a>
%F A206824 From _Michael De Vlieger_, Sep 22 2017: (Start)
%F A206824 Conjectures:
%F A206824 a(p^e) = 1, with e > 0.
%F A206824 a(A002110(n)) = 2^(n - 1) and a(A002110(n)/2) = 2^(n - 1) - 1 for n > 0.
%F A206824 a(n) = 2^(A001221(n) - 1) for n even, a(n) = 2^A001221(n) - 1 for n odd. (End)
%e A206824 s(6) = 21, and 6 divides exactly two of the numbers 21 - s(k) = {20, 18, 15, 11, 6} for k = 1, 2, ..., 5, so that a(6) = 2.
%t A206824 s[k_] := k (k + 1)/2;
%t A206824 f[n_, k_] := If[Mod[s[n] - s[k], n] == 0, 1, 0];
%t A206824 t[n_] := Flatten[Table[f[n, k], {k, 1, n - 1}]]
%t A206824 a[n_] := Count[Flatten[t[n]], 1]
%t A206824 Table[a[n], {n, 2, 120}]   (* A206824 *)
%t A206824 (* Second program: *)
%t A206824 Table[Total@ Flatten@ Table[Boole@ Divisible[Subtract @@ Thread[PolygonalNumber[{n, k}]], n ], {k, n - 1}], {n, 2, 105}] (* _Michael De Vlieger_, Sep 22 2017 *)
%Y A206824 Cf. A001221, A002110.
%K A206824 nonn
%O A206824 2,5
%A A206824 _Clark Kimberling_, Feb 12 2012