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

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, 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, 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

Offset

2,5

Comments

Records at indices {(1), 2, 6, 15, 30, 105, 210, 1155, 2310, 15015, 30030, ...}. - Michael De Vlieger, Sep 22 2017

Links

Michael De Vlieger, Table of n, a(n) for n = 2..30030

Formula

From Michael De Vlieger, Sep 22 2017: (Start)

Conjectures:

a(p^e) = 1, with e > 0.

a(A002110(n)) = 2^(n - 1) and a(A002110(n)/2) = 2^(n - 1) - 1 for n > 0.

a(n) = 2^(A001221(n) - 1) for n even, a(n) = 2^A001221(n) - 1 for n odd. (End)

Example

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.

Mathematica

s[k_] := k (k + 1)/2;
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}]   (* A206824 *)
(* Second program: *)
Table[Total@ Flatten@ Table[Boole@ Divisible[Subtract @@ Thread[PolygonalNumber[{n, k}]], n ], {k, n - 1}], {n, 2, 105}] (* Michael De Vlieger, Sep 22 2017 *)

Crossrefs

Cf. A001221, A002110.

Sequence in context: A316556 A187279 A076820 * A293810 A356553 A324369

Adjacent sequences: A206821 A206822 A206823 * A206825 A206826 A206827

Keyword

nonn

Author

Clark Kimberling, Feb 12 2012

Status

approved