A206588 -id:A206588 - cp's OEIS frontend

A206827 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.

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

Offset: 2

Clark Kimberling, Feb 15 2012

nonn

For a guide to related sequences, see A206588.

If n is a prime > 3, a(n) = 2. - Robert Israel, Jun 04 2023

			5 divides exactly two of the numbers s(n)-s(k) for k=1,2,3,4, so a(5)=2.

Robert Israel, Table of n, a(n) for n = 2..10000

Cf. A206588.

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)]:
map(f,[$2..100]); # Robert Israel, Jun 04 2023

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 *)

A206828 Number of solutions k of C(2k,k)=C(2n,n) (mod n), where 1<=k

Original entry on oeis.org

1, 1, 2, 1, 2, 1, 2, 2, 1, 1, 1, 1, 2, 2, 3, 2, 2, 1, 8, 2, 1, 1, 5, 2, 1, 2, 11, 1, 0, 1, 4, 1, 3, 4, 1, 2, 1, 1, 8, 1, 15, 1, 2, 12, 1, 1, 5, 2, 3, 0, 1, 3, 3, 3, 1, 0, 1, 1, 2, 1, 1, 0, 5, 2, 23, 1, 4, 0, 4, 1, 7, 3, 1, 12, 2, 24, 2, 1, 8, 3, 3, 1, 6, 0, 3, 1, 37, 1, 3, 26, 1, 1, 1, 0, 4

Offset: 2

Clark Kimberling, Feb 15 2012

nonn

For a guide to related sequences, see A206588.

			2 divides exactly two of the numbers 20-1, 20-2, 20-6, so that a(3)-2.

Cf. A206588.

s[k_] := Binomial[2 k, k];
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}]  (* A206828 *)

A206589 Number of solutions (n,k) of p(k+1)=p(n+1) (mod n), where 1<=k

Original entry on oeis.org

1, 0, 2, 1, 2, 1, 1, 1, 1, 0, 3, 1, 2, 1, 2, 0, 2, 0, 2, 1, 2, 1, 1, 0, 0, 1, 1, 0, 4, 1, 2, 2, 2, 1, 1, 1, 1, 1, 3, 1, 3, 1, 1, 2, 1, 0, 3, 1, 1, 1, 1, 0, 2, 1, 2, 2, 1, 1, 4, 0, 1, 1, 0, 0, 2, 0, 2, 2, 3, 0, 4, 1, 2, 2, 1, 1, 3, 1, 2, 1, 2, 1, 3, 1, 3, 2, 3, 1, 3, 0, 1, 0, 2, 1, 2, 0, 2, 0, 2

Offset: 2

Clark Kimberling, Feb 09 2012

nonn

Related to A206588, which includes differences p-2.

			For k=1 to 5, the numbers p(7)-p(k+1) are 14,12,10,6,4, so that a(6)=2.

Cf. A206588.

f[n_,k_]:=If[Mod[Prime[n+1]-Prime[k+1],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}]  (* A206589 *)

cp's OEIS Frontend

A206827 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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A206828 Number of solutions k of C(2k,k)=C(2n,n) (mod n), where 1<=k

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A206589 Number of solutions (n,k) of p(k+1)=p(n+1) (mod n), where 1<=k

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cp's OEIS Frontend

A206827 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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Author

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Maple

Mathematica

A206828 Number of solutions k of C(2k,k)=C(2n,n) (mod n), where 1<=k

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A206589 Number of solutions (n,k) of p(k+1)=p(n+1) (mod n), where 1<=k

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

A206827 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.