A011583 -id:A011583 - cp's OEIS frontend

A101455 a(n) = 0 for even n, a(n) = (-1)^((n-1)/2) for odd n. Periodic sequence 1,0,-1,0,...

0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0, -1, 0, 1, 0

Offset: 0

Gerald McGarvey, Jan 20 2005

Called X(n) (i.e., Chi(n)) in Hardy and Wright (p. 241), who show that X(n*m) = X(n)*X(m) for all n and m (i.e., X(n) is completely multiplicative) since (n*m - 1)/2 - (n - 1)/2 - (m - 1)/2 = (n - 1)*(m - 1)/2 == 0 (mod 2) when n and m are odd.
Same as A056594 but with offset 1.
From R. J. Mathar, Jul 15 2010: (Start)
The sequence is the non-principal Dirichlet character mod 4. (The principal character is A000035.)
Associated Dirichlet L-functions are for example L(1,chi) = Sum_{n>=1} a(n)/n = A003881, or L(2,chi) = Sum_{n>=1} a(n)/n^2 = A006752, or L(3,chi) = Sum_{n>=1} a(n)/n^3 = A153071. (End)
a(n) is a strong elliptic divisibility sequence t_n as given in [Kimberling, p. 16] where x = 0, y = -1, z is arbitrary. - Michael Somos, Nov 27 2019

			G.f. = x - x^3 + x^5 - x^7 + x^9 - x^11 + x^13 - x^15 + x^17 - x^19 + x^21 + ...

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1986, page 139, k=4, Chi_2(n).
G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 5th ed., Oxford Univ. Press, 1979, p. 241.

Muniru A Asiru, Table of n, a(n) for n = 0..1000 (a(0) prepended by Jianing Song)
Étienne Fouvry, Claude Levesque, Michel Waldschmidt, Representation of integers by cyclotomic binary forms, arXiv:1712.09019 [math.NT], 2017.
Clark Kimberling, Strong divisibility sequences and some conjectures, Fib. Quart., 17 (1979), 13-17.
Grant Sanderson, Pi hiding in prime regularities, 3Blue1Brown video (2017).
Index to divisibility sequences
Index entries for linear recurrences with constant coefficients, signature (0,-1).

Cf. A002654, A009116, A009454, A056594, A108520, A111335, A126120, A146559.

Kronecker symbols {(d/n)} where d is a fundamental discriminant with |d| <= 24: A109017 (d=-24), A011586 (d=-23), A289741 (d=-20), A011585 (d=-19), A316569 (d=-15), A011582 (d=-11), A188510 (d=-8), A175629 (d=-7), this sequence (d=-4), A102283 (d=-3), A080891 (d=5), A091337 (d=8), A110161 (d=12), A011583 (d=13), A011584 (d=17), A322829 (d=21), A322796 (d=24).

a := [1, 0];; for n in [3..10^2] do a[n] := a[n-2]; od; a; # Muniru A Asiru, Feb 02 2018

m:=75; R:=PowerSeriesRing(Integers(), m); Coefficients(R!(x/(1+x^2))); // G. C. Greubel, Aug 23 2018

a := n -> `if`(n mod 2=0, 0, (-1)^((n-1)/2)):
seq(a(n), n=1..10^3); # Muniru A Asiru, Feb 02 2018

a[ n_] := {1, 0, -1, 0}[[ Mod[ n, 4, 1]]]; (* Michael Somos, Jan 13 2014 *)
LinearRecurrence[{0, -1}, {1, 0}, 75] (* G. C. Greubel, Aug 23 2018 *)

{a(n) = if( n%2, (-1)^(n\2))}; /* Michael Somos, Sep 02 2005 */

{a(n) = kronecker( -4, n)}; /* Michael Somos, Mar 30 2012 */

def A101455(n): return (0,1,0,-1)[n&3] # Chai Wah Wu, Jun 21 2024

Multiplicative with a(2^e) = 0, a(p^e) = (-1)^((p^e-1)/2) otherwise. - Mitch Harris May 17 2005
Euler transform of length 4 sequence [0, -1, 0, 1]. - Michael Somos, Sep 02 2005
G.f.: (x - x^3)/(1 - x^4) = x/(1 + x^2). - Michael Somos, Sep 02 2005
G.f. A(x) satisfies: 0 = f(A(x), A(x^2)) where f(u, v) = v - u^2 * (1 + 2*v). - Michael Somos, Aug 04 2011
a(n + 4) = a(n), a(n + 2) = a(-n) = -a(n), a(2*n) = 0, a(2*n + 1) = (-1)^n for all n in Z. - Michael Somos, Aug 04 2011
a(n + 1) = A056594(n). - Michael Somos, Jan 13 2014
REVERT transform is A126120. STIRLING transform of A009454. BINOMIAL transform is A146559. BINOMIAL transform of A009116. BIN1 transform is A108520. MOBIUS transform of A002654. EULER transform is A111335. - Michael Somos, Mar 30 2012
Completely multiplicative with a(p) = 2 - (p mod 4). - Werner Schulte, Feb 01 2018
a(n) = (-(n mod 2))^binomial(n, 2). - Peter Luschny, Sep 08 2018
a(n) = sin(n*Pi/2) = Im(i^n) where i is the imaginary unit. - Jianing Song, Sep 09 2018
From Jianing Song, Nov 14 2018: (Start)
a(n) = ((-4)/n) (or more generally, ((-4^i)/n) for i > 0), where (k/n) is the Kronecker symbol.
E.g.f.: sin(x).
Dirichlet g.f. is the Dirichlet beta function.
a(n) = A091337(n)*A188510(n). (End)

a(0) prepended by Jianing Song, Nov 14 2024

A091337 a(n) = (2/n), where (k/n) is the Kronecker symbol.

Original entry on oeis.org

0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1, 0, -1, 0, -1, 0, 1, 0, 1

Offset: 0

Eric W. Weisstein, Dec 30 2003

Sinh(1) in 'reflected factorial' base is 1.01010101010101010101010101010101010101010101... see A073097 for cosh(1). - Robert G. Wilson v, May 04 2005
A non-principal character for the Dirichlet L-series modulo 8, see arXiv:1008.2547 and L-values Sum_{n >= 1} a(n)/n^s in eq (318) by Jolley. - R. J. Mathar, Oct 06 2011 [The other two non-principal characters are A101455 = {(-4/n)} and A188510 = {(-2/n)}. - Jianing Song, Nov 14 2024]
Period 8: repeat [0, 1, 0, -1, 0, -1, 0, 1]. - Wesley Ivan Hurt, Sep 07 2015 [Adapted by Jianing Song, Nov 14 2024 to include a(0) = 0.]
a(n) = (2^(2i+1)/n), where (k/n) is the Kronecker symbol and i >= 0. - A.H.M. Smeets, Jan 23 2018

			G.f. = x - x^3 - x^5 + x^7 + x^9 - x^11 - x^13 + x^15 + x^17 - x^19 - x^21 + ...

L. B. W. Jolley, Summation of series, Dover (1961).

Jianing Song, Table of n, a(n) for n = 0..1000
John M. Campbell, An Integral Representation of Kekulé Numbers, and Double Integrals Related to Smarandache Sequences, arXiv:1105.3399 [math.GM], 2011.
R. J. Mathar, Table of Dirichlet L-series..., arXiv:1008.2547 [math.NT], 2010, 2015, L(m=8,r=2,s).
Michael Somos, Rational Function Multiplicative Coefficients
Eric Weisstein's World of Mathematics, Kronecker Symbol
Index entries for linear recurrences with constant coefficients, signature (0,0,0,-1).

Cf. A000129, A035185, A073097, A087960, A102587.

Kronecker symbols {(d/n)} where d is a fundamental discriminant with |d| <= 24: A109017 (d=-24), A011586 (d=-23), A289741 (d=-20), A011585 (d=-19), A316569 (d=-15), A011582 (d=-11), A188510 (d=-8), A175629 (d=-7), A101455 (d=-4), A102283 (d=-3), A080891 (d=5), this sequence (d=8), A110161 (d=12), A011583 (d=13), A011584 (d=17), A322829 (d=21), A322796 (d=24).

[(n mod 2) * (-1)^((n+1) div 4)  : n in [1..100]]; // Vincenzo Librandi, Oct 31 2014

A091337:= n -> [0, 1, 0, -1, 0, -1, 0, 1][(n mod 8)+1]: seq(A091337(n), n=1..100); # Wesley Ivan Hurt, Sep 07 2015

KroneckerSymbol[Range[100], 2] (* Alonso del Arte, Oct 30 2014 *)

{a(n) = (n%2) * (-1)^((n+1)\4)}; /* Michael Somos, Sep 10 2005 */

{a(n) = kronecker( 2, n)}; /* Michael Somos, Sep 10 2005 */

{a(n) = [0, 1, 0, -1, 0, -1, 0, 1][n%8 + 1]}; /* Michael Somos, Jul 17 2009 */

Euler transform of length 8 sequence [0, -1, 0, -1, 0, 0, 0, 1]. - Michael Somos, Jul 17 2009
a(n) is multiplicative with a(2^e) = 0^e, a(p^e) = 1 if p == 1, 7 (mod 8), a(p^e) = (-1)^e if p == 3, 5 (mod 8). - Michael Somos, Jul 17 2009
G.f.: x*(1 - x^2)/(1 + x^4). a(n) = -a(n + 4) = a(-n) for all n in Z. a(2*n) = 0. a(2*n + 1) = A087960(n). - Michael Somos, Apr 10 2011
Transform of Pell numbers A000129 by the Riordan array A102587. - Paul Barry, Jul 14 2005
a(n) = (2/n) = (n/2), Charles R Greathouse IV explained. - Alonso del Arte, Oct 31 2014
a(n) = (1 - (-1)^n)*(-1)^(n/4 - 1/8 - (-1)^n/8 + (-1)^((2*n + 1 - (-1)^n)/4)/4)/2. - Wesley Ivan Hurt, Sep 07 2015
From Jianing Song, Nov 14 2018: (Start)
a(n) = sqrt(2)*sin(Pi*n/2)*sin(Pi*n/4).
E.g.f.: sqrt(2)*cos(x/sqrt(2))*sinh(x/sqrt(2)).
Moebius transform of A035185.
a(n) = A101455(n)*A188510(n). (End)
a(n) = Sum_{i=1..n} (-1)^(i + floor((i-3)/4)). - Wesley Ivan Hurt, Apr 27 2020
Sum_{n>=1} a(n)/n = A196525. Sum_{n>=1} a(n)/n^2 = A328895. Sum_{n>=1} a(n)/n^3 = A329715. Sum_{n>=1} a(n)/n^4 = A346728. - R. J. Mathar, Dec 17 2024

a(0) prepended by Jianing Song, Nov 14 2024

A175629 Legendre symbol (n,7).

Original entry on oeis.org

0, 1, 1, -1, 1, -1, -1, 0, 1, 1, -1, 1, -1, -1, 0, 1, 1, -1, 1, -1, -1, 0, 1, 1, -1, 1, -1, -1, 0, 1, 1, -1, 1, -1, -1, 0, 1, 1, -1, 1, -1, -1, 0, 1, 1, -1, 1, -1, -1, 0, 1, 1, -1, 1, -1, -1, 0, 1, 1, -1, 1, -1, -1, 0, 1, 1, -1, 1, -1, -1, 0, 1, 1, -1, 1, -1, -1, 0, 1, 1, -1, 1, -1, -1, 0, 1, 1

Offset: 0

R. J. Mathar, Jul 29 2010

This represents a non-principal Dirichlet character modulo 7.

T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1986, page 139, k=7, Chi_2(n).

Index entries for linear recurrences with constant coefficients, signature (-1,-1,-1,-1,-1,-1).

Cf. A089509, A097343.
The Legendre symbols (n,p): A091337 (p = 2, Kronecker symbol), A102283 (p = 3), A080891 (p = 5), this sequence (p = 7), A011582 (p = 11), A011583 (p = 13), ..., A011631 (p = 251), A165573 (p = 257), A165574 (p = 263). Also, many other sequences for p > 263 are in the OEIS.
Moebius transform of A035182.

&cat [[0, 1, 1, -1, 1, -1, -1]^^20]; // Vincenzo Librandi, Jun 30 2018

A := proc(n) numtheory[jacobi](n,7) ; end proc: seq(A(n),n=0..120) ;

LinearRecurrence[{-1,-1,-1,-1,-1,-1},{0,1,1,-1,1,-1},100] (* or *) PadRight[ {},100,{0,1,1,-1,1,-1,-1}] (* Harvey P. Dale, Aug 02 2013 *)
Table[JacobiSymbol[n, 7], {n, 0, 100}] (* Vincenzo Librandi, Jun 30 2018 *)

a(n) = kronecker(n, 7); \\ Michel Marcus, Jan 28 2019

a(n) = a(n+7).
|a(n)| = A109720(n).
a(n) = -a(n-1) - a(n-2) - a(n-3) - a(n-4) - a(n-5) - a(n-6).
G.f.: x*(1 + 2*x + x^2 + 2*x^3 + x^4)/(1 + x + x^2 + x^3 + x^4 + x^5 + x^6).
a(n) == n^3 (mod 7). - Jianing Song, Jun 29 2018

A327653 Composite numbers k coprime to 13 such that k divides A006190(k-Kronecker(13,k)).

Original entry on oeis.org

10, 119, 649, 1189, 1763, 3599, 4187, 5559, 6681, 12095, 12403, 12685, 12871, 12970, 14041, 14279, 15051, 16109, 19043, 22847, 23479, 24769, 26795, 28421, 30743, 30889, 31631, 31647, 33919, 34997, 37949, 38503, 39203, 41441, 46079, 48577, 49141, 50523, 50545, 53301, 56279, 58081, 58589

Offset: 1

Jianing Song, Sep 20 2019

nonn

Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n) = m*x(n-1) + x(n-2) for k >= 2. For primes p, we have (a) p divides x(p-((m^2+4)/p)); (b) x(p) == ((m^2+4)/p) (mod p), where (D/p) is the Kronecker symbol. This sequence gives composite numbers k such that gcd(k, m^2+4) = 1 and that a condition similar to (a) holds for k, where m = 3.
If k is not required to be coprime to m^2 + 4 (= 13), then there are 360 such k <= 10^5 and 1506 such k <= 10^6, while there are only 62 terms <= 10^5 and 197 terms <= 10^6 in this sequence.
Also composite numbers k coprime to 13 such that A322907(k) divides k - Kronecker(13,k).

			A006190(9) = 12970 which is divisible by 10, so 10 is a term.

Amiram Eldar, Table of n, a(n) for n = 1..10000

             m                      m=1            m=2      m=3
k | x(k-Kronecker(m^2+4,k))*  A081264 U A141137  A327651  this seq
k | x(k)-Kronecker(m^2+4,k)        A049062       A099011  A327654
            both                   A212424       A327652  A327655
* k is composite and coprime to m^2 + 4.
Cf. A006190, A322907, A011583 ({Kronecker(13,n)}).

seqmod(n, m)=((Mod([3, 1; 1, 0], m))^n)[1, 2]
isA327653(n)=!isprime(n) && !seqmod(n-Kronecker(13,n), n) && gcd(n,13)==1 && n>1

A289741 a(n) = Kronecker symbol (-20/n).

Original entry on oeis.org

0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, -1, 0, -1, 0, 0, 0, -1, 0, -1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, -1, 0, -1, 0, 0, 0, -1, 0, -1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, -1, 0, -1, 0, 0, 0, -1, 0, -1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, -1, 0, -1, 0, 0, 0, -1, 0, -1, 0

Offset: 0

Jianing Song, Dec 27 2018

Period 20: repeat [0, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, -1, 0, -1, 0, 0, 0, -1, 0, -1].
This sequence is one of the three non-principal real Dirichlet characters modulo 20. The other two are Jacobi or Kronecker symbols {(20/n)} (or {(n/20)}) and {((-100)/n)} (A185276).
Note that (Sum_{i=0..19} i*a(i))/(-20) = 2 gives the class number of the imaginary quadratic field Q(sqrt(-5)). The fact Q(sqrt(-5)) has class number 2 implies that Q(sqrt(-5)) is not a unique factorization domain.

Antti Karttunen, Table of n, a(n) for n = 0..10000
Eric Weisstein's World of Mathematics, Kronecker Symbol
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,-1).

Cf. A035170 (inverse Moebius transform).
Kronecker symbols {(d/n)} where d is a fundamental discriminant with |d| <= 24: A109017 (d=-24), A011586 (d=-23), this sequence (d=-20), A011585 (d=-19), A316569 (d=-15), A011582 (d=-11), A188510 (d=-8), A175629 (d=-7), A101455 (d=-4), A102283 (d=-3), A080891 (d=5), A091337 (d=8), A110161 (d=12), A011583 (d=13), A011584 (d=17), A322829 (d=21), A322796 (d=24).
Cf. A045797, A045798, A185276.

Array[KroneckerSymbol[-20, #]&, 100, 0] (* Amiram Eldar, Jan 10 2019 *)

PARI
```
a(n) = kronecker(-20, n)
```

a(n) = 1 for n in A045797; -1 for n in A045798; 0 for n that are not coprime with 20.
Completely multiplicative with a(p) = a(p mod 20) for primes p.
a(n) = A080891(n)*A101455(n).
a(n) = -a(n+10) = -a(-n) for all n in Z.
Multiplicative with a(2) = a(5) = 0, a(p) = (-1)^floor(p/10) otherwise; equivalently: a(n) = (-1)^floor(n/10) if n is coprime to 2*5, 0 otherwise. - M. F. Hasler, Feb 28 2022

A322796 a(n) = Kronecker symbol (6/n).

Original entry on oeis.org

0, 1, 0, 0, 0, 1, 0, -1, 0, 0, 0, -1, 0, -1, 0, 0, 0, -1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, -1, 0, 0, 0, -1, 0, -1, 0, 0, 0, -1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, -1, 0, 0, 0, -1, 0, -1, 0, 0, 0, -1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 0, 0, 1, 0, -1, 0, 0, 0, -1, 0

Offset: 0

Jianing Song, Dec 26 2018

Period 24: repeat [0, 1, 0, 0, 0, 1, 0, -1, 0, 0, 0, -1, 0, -1, 0, 0, 0, -1, 0, 1, 0, 0, 0, 1].
Also a(n) = Kronecker symbol (24/n).
This sequence is one of the seven non-principal real Dirichlet characters modulo 24. The other six are Jacobi or Kronecker symbols {(-6/n)} (or {(n/6)}, {(-24/n)}, {(n/24)}, A109017), {(-12/n)} (or {(n/12)}, A134667), {(12/n)} (A110161), {(-18/n)} (or {(-72/n)}), {(18/n)} (or {(72/n)}, {(n/72)}) and {(-36/n)}. These sequences all become the same after taking absolute values.

Antti Karttunen, Table of n, a(n) for n = 0..65543
Eric Weisstein's World of Mathematics, Kronecker Symbol (contains this sequence)
Index entries for linear recurrences with constant coefficients, signature (0,0,0,1,0,0,0,-1).

Cf. A035188 (inverse Moebius transform).

Kronecker symbols {(d/n)} where d is a fundamental discriminant with |d| <= 24: A109017 (d=-24), A011586 (d=-23), A289741 (d=-20), A011585 (d=-19), A316569 (d=-15), A011582 (d=-11), A188510 (d=-8), A175629 (d=-7), A101455 (d=-4), A102283 (d=-3), A080891 (d=5), A091337 (d=8), A110161 (d=12), A011583 (d=13), A011584 (d=17), A322829 (d=21), this sequence (d=24).

[KroneckerSymbol(6, n): n in [0..100]]; // Vincenzo Librandi, Jan 01 2019

Array[KroneckerSymbol[6, #] &, 105, 0] (* Michael De Vlieger, Dec 31 2018 *)
Table[KroneckerSymbol[6, n], {n, 0, 100}] (* Vincenzo Librandi, Jan 01 2019 *)

a(n) = kronecker(6, n); \\ --- Argument order corrected by Antti Karttunen, Sep 27 2019

a(n) = 1 for n == 1, 5, 19, 23 (mod 24); -1 for n == 7, 11, 13, 17 (mod 24); 0 for n that are not coprime with 21.
Completely multiplicative with a(p) = a(p mod 24) for primes p.
a(n) = A091337(n)*A102283(n).
a(n) = A109017(n+12) = A109017(n-12).
a(n) = a(-n) = a(n+24) for all n in Z.

Definition corrected by Antti Karttunen, Sep 28 2019

A316569 a(n) = Jacobi (or Kronecker) symbol (n, 15).

Original entry on oeis.org

0, 1, 1, 0, 1, 0, 0, -1, 1, 0, 0, -1, 0, -1, -1, 0, 1, 1, 0, 1, 0, 0, -1, 1, 0, 0, -1, 0, -1, -1, 0, 1, 1, 0, 1, 0, 0, -1, 1, 0, 0, -1, 0, -1, -1, 0, 1, 1, 0, 1, 0, 0, -1, 1, 0, 0, -1, 0, -1, -1, 0, 1, 1, 0, 1, 0, 0, -1, 1, 0, 0, -1, 0, -1, -1, 0

Offset: 0

Jianing Song, Aug 05 2018

Period 15: repeat [0, 1, 1, 0, 1, 0, 0, -1, 1, 0, 0, -1, 0, -1, -1].
Also a(n) = Kronecker(-15, n).
This sequence is one of the three non-principal real Dirichlet characters modulo 15. The other two are Jacobi or Kronecker symbols (n, 45) (or (45, n)) and (n, 75) (or (-75, n)).
Note that (Sum_{i=0..14} i*a(i))/(-15) = 2 gives the class number of the imaginary quadratic field Q(sqrt(-15)).

Eric Weisstein's World of Mathematics, Kronecker Symbol
Index entries for linear recurrences with constant coefficients, signature (1,0,-1,1,-1,0,1,-1).

Cf. A035175 (inverse Moebius transform).

Kronecker symbols: A063524 ((n, 0) or (0, n)), A000012 ((n, 1) or (1, n)), A091337 ((n, 2) or (2, n) or (n, 8) or (8, n)), A102283 ((n, 3) or (-3, n)), A000035 ((n, 4) or (4, n) or (n, 16) or (16, n)), A080891 ((n, 5) or (5, n)), A109017 ((n, 6) or (-6, n)), A175629 ((n, 7) or (-7, n)), A011655 ((n, 9) or (9, n)), A011582 ((n, 11) or (-11, n)), A134667 ((n, 12) or (-12, n)), A011583 ((n, 13) or (13, n)), this sequence ((n, 15) or (-15, n)).

[KroneckerSymbol(-15, n): n in [0..100]]; // Vincenzo Librandi, Aug 28 2018

Array[ JacobiSymbol[#, 15] &, 90, 0] (* Robert G. Wilson v, Aug 06 2018 *)
PadRight[{},100,{0,1,1,0,1,0,0,-1,1,0,0,-1,0,-1,-1}] (* Harvey P. Dale, Feb 20 2023 *)

PARI
```
a(n) = kronecker(n, 15)
```

a(n) = 1 for n == 1, 2, 4, 8 (mod 15); -1 for n == 7, 11, 13, 14 (mod 15); 0 for n that are not coprime with 15.
Completely multiplicative with a(p) = a(p mod 15) for primes p.
a(n) = A102283(n)*A080891(n).
a(n) = a(n+15) = -a(-n) for all n in Z.
From Chai Wah Wu, Feb 16 2021: (Start)
a(n) = a(n-1) - a(n-3) + a(n-4) - a(n-5) + a(n-7) - a(n-8) for n > 7.
G.f.: (x^7 - x^5 + 2*x^4 - x^3 + x)/(x^8 - x^7 + x^5 - x^4 + x^3 - x + 1). (End)

A055088 Triangle of generalized Legendre symbols L(a/b) read by rows, with 1's for quadratic residues and 0's for quadratic non-residues.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0

Offset: 1

Antti Karttunen, Apr 18 2000

L(a/b) is 1 if an integer c exists such that c^2 is congruent to a (mod b) and 0 otherwise.

For every prime of the form 4k+1 (A002144) the row is symmetric and for every prime of the form 4k+3 (A002145) the row is "complementarily symmetric".

			The tenth row gives the quadratic residues and non-residues of 11 (see A011582) and the twelfth row gives the same information for 13 (A011583), with -1's replaced by zeros.
.
Triangle starts:
  [ 1] [1]
  [ 2] [1, 0]
  [ 3] [1, 0, 0]
  [ 4] [1, 0, 0, 1]
  [ 5] [1, 0, 1, 1, 0]
  [ 6] [1, 1, 0, 1, 0, 0]
  [ 7] [1, 0, 0, 1, 0, 0, 0]
  [ 8] [1, 0, 0, 1, 0, 0, 1, 0]
  [ 9] [1, 0, 0, 1, 1, 1, 0, 0, 1]
  [10] [1, 0, 1, 1, 1, 0, 0, 0, 1, 0]
  [11] [1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0]
  [12] [1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 0, 1]

Cf. A002144, A002145, A011582, A011583.

Each row interpreted as a binary number: A055094.

# See A054431 for one_or_zero and trinv.
with(numtheory,quadres); quadres_0_1_array := (n) -> one_or_zero(quadres((n-((trinv(n-1)*(trinv(n-1)-1))/2)), (trinv(n-1)+1)));

row[n_] := With[{rr = Table[Mod[k^2, n + 1], {k, 1, n}] // Union}, Boole[ MemberQ[rr, #]]& /@ Range[n]];
Array[row, 14] // Flatten (* Jean-François Alcover, Mar 05 2016 *)

def A055088_row(n) :
    Q = quadratic_residues(n+1)
    return [int(i in Q) for i in (1..n)]
for n in (1..14) : print(A055088_row(n))  # Peter Luschny, Aug 08 2012

A327654 Composite numbers k coprime to 13 such that k divides A006190(k) - Kronecker(13,k).

Original entry on oeis.org

4, 8, 9, 119, 399, 649, 1023, 1179, 1189, 1199, 1881, 2703, 3519, 4081, 4187, 5151, 7055, 7361, 10349, 12871, 13833, 14041, 15519, 16109, 18639, 22593, 23479, 24769, 26937, 28421, 29007, 31631, 34111, 34997, 38503, 41441, 44671, 48577, 50545, 51711, 53823, 56279, 57407, 58081, 59081

Offset: 1

Jianing Song, Sep 20 2019

nonn

Let {x(n)} be a sequence defined by x(0) = 0, x(1) = 1, x(n) = m*x(n-1) + x(n-2) for k >= 2. For primes p, we have (a) p divides x(p-((m^2+4)/p)); (b) x(p) == ((m^2+4)/p) (mod p), where (D/p) is the Kronecker symbol. This sequence gives composite numbers k such that gcd(k, m^2+4) = 1 and that a condition similar to (b) holds for k, where m = 3.

If k is not required to be coprime to m^2 + 4 (= 13), then there are 352 such k <= 10^5, and 1457 such k <= 10^6, while there are only 54 terms <= 10^5 and 148 terms <= 10^6 in this sequence.

			A006190(8) = 3927 == Kronecker(13,8) (mod 8), so 8 is a term.

Amiram Eldar, Table of n, a(n) for n = 1..2000

             m                      m=1            m=2      m=3
k | x(k-Kronecker(m^2+4,k))*  A081264 U A141137  A327651  A327653
k | x(k)-Kronecker(m^2+4,k)        A049062       A099011  this seq
            both                   A212424       A327652  A327655
* k is composite and coprime to m^2 + 4.
Cf. A006190, A011583 ({Kronecker(13,n)}).

seqmod(n, m)=((Mod([3, 1; 1, 0], m))^n)[1, 2]
isA327654(n)=!isprime(n) && seqmod(n, n)==kronecker(13,n) && gcd(n,13)==1 && n>1

A296937 Rational primes that decompose in the field Q(sqrt(13)).

Original entry on oeis.org

3, 17, 23, 29, 43, 53, 61, 79, 101, 103, 107, 113, 127, 131, 139, 157, 173, 179, 181, 191, 199, 211, 233, 251, 257, 263, 269, 277, 283, 311, 313, 337, 347, 367, 373, 389, 419, 433, 439, 443, 467, 491, 503, 521, 523, 547, 563, 569, 571, 599, 601, 607, 641

Offset: 1

N. J. A. Sloane, Dec 26 2017

Is this the same sequence as A141188 or A038883? - R. J. Mathar, Jan 02 2018
From Jianing Song, Apr 21 2022: (Start)
Primes p such that Kronecker(13, p) = Kronecker(p, 13) = 1, where Kronecker() is the Kronecker symbol. That is to say, primes p that are quadratic residues modulo 13.
Primes p such that p^6 == 1 (mod 13).
Primes p == 1, 3, 4, 9, 10, 12 (mod 13). (End)

Cf. A011583 (kronecker symbol modulo 13), A038883.
Rational primes that decompose in the quadratic field with discriminant D: A139513 (D=-20), A191019 (D=-19), A191018 (D=-15), A296920 (D=-11), A033200 (D=-8), A045386 (D=-7), A002144 (D=-4), A002476 (D=-3), A045468 (D=5), A001132 (D=8), A097933 (D=12), this sequence (D=13), A296938 (D=17).
Cf. A038884 (inert rational primes in the field Q(sqrt(13))).

Load the Maple program HH given in A296920. Then run HH(13, 200); This produces A296937, A038884, A038883.

isA296937(p) = isprime(p) && kronecker(p,13) == 1 \\ Jianing Song, Apr 21 2022

Equals A038883 \ {13}. - Jianing Song, Apr 21 2022

cp's OEIS Frontend

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A091337 a(n) = (2/n), where (k/n) is the Kronecker symbol.

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A175629 Legendre symbol (n,7).

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A327653 Composite numbers k coprime to 13 such that k divides A006190(k-Kronecker(13,k)).

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