A166104 -id:A166104 - cp's OEIS frontend

A166103 Squares of A166104.

1, 9, 25, 81, 121, 225, 289, 529, 625, 729, 841, 1089, 1681, 2025, 2209, 2601, 2809, 3025, 3481, 4761, 5041, 5625, 6561, 6889, 7225, 7569, 7921, 9801, 10201, 11449, 12769, 13225, 14641, 15129, 15625, 17161, 18225, 18769, 19881, 21025, 22201

Offset: 1

Antti Karttunen, Oct 13 2009

nonn

Conjecture: a(n)=A166102(n)/3. (Checked for terms a(1)-a(92).)

a(n) = A000290(A166104(n)).

A166100 Sum of those positive i <= 2n+1, for which J(i,2n+1)=+1. Here J(i,k) is the Jacobi symbol.

Original entry on oeis.org

1, 1, 5, 7, 27, 22, 39, 15, 68, 76, 63, 92, 250, 117, 203, 186, 165, 175, 333, 156, 410, 430, 270, 423, 1029, 357, 689, 440, 513, 767, 915, 504, 780, 1072, 759, 994, 1314, 725, 1155, 1343, 2187, 1577, 1360, 957, 1958, 1547, 1395, 1330, 2328, 1485, 2525

Offset: 0

Antti Karttunen, Oct 13 2009. Erroneous name corrected Oct 20 2009

nonn

Note that this sequence is not equal to the sum of the quadratic residues of 2n+1 in range [1,2n+1], and thus NOT a bisection of A165898.

			For n=5, we get odd number 11 (2*5+1), and J(i,11) = 1,-1,1,1,1,-1,-1,-1,1,-1,0 when i ranges from 1 to 11, J(i,11) getting value 1 when i=1, 3, 4, 5 and 9, thus a(5)=22.

A. Karttunen, Table of n, a(n) for n = 0..131071
Wikipedia, Jacobi symbol

A076409(n)=a(A102781(n)). Cf. A166101-A166104, A166405, A166406.

Scheme-code for jacobi-symbol is given at A165601.

Table[Total[Flatten[Position[JacobiSymbol[Range[2n+1],2n+1],1]]],{n,0,50}] (* Harvey P. Dale, Jun 19 2013 *)

from sympy import jacobi_symbol as J
def a(n): return sum([i for i in range(1, 2*n + 2) if J(i, 2*n + 1)==1]) # Indranil Ghosh, Jun 12 2017

cp's OEIS Frontend

A166103 Squares of A166104.

Views

Author

Keywords

Comments

Crossrefs

A166100 Sum of those positive i <= 2n+1, for which J(i,2n+1)=+1. Here J(i,k) is the Jacobi symbol.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Python