cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A076352 Squarefree numbers k such that A076341(k) = 0.

Original entry on oeis.org

1, 2, 15, 30, 143, 286, 2145, 3599, 4290, 5183, 7198, 10366, 11663, 23326, 32399, 36863, 51983, 53985, 57599, 64798, 73726, 77745, 97343, 103966, 107970, 115198, 121103, 155490, 174945, 176399, 186623, 194686, 242206, 349890, 352798, 359999, 373246, 435599, 485985
Offset: 1

Views

Author

Reinhard Zumkeller, Oct 08 2002

Keywords

Examples

			Applying the map as defined in A076340, A076341:
A005117(19) = 30 = 5*3*2 = (4+1)*(4-1)*2 -> (4,1)*(4,-1)*(2,0) = (4*4+1,4-4)*(2,0) = (34,0), therefore A076340(30) = 34 and A076341(30) = 0, hence 30 is a term.
A005117(28) = 42 = 7*3*2 = (8-1)*(4-1)*2 -> (8,-1)*(4,-1)*(2,0) = (8*4-1,-8-4)*(2,0) = (62,-24), therefore A076340(42) = 62 and A076341(42) = -24, hence 42 is not a term.
		

Crossrefs

Programs

  • Mathematica
    f[p_, e_] := 4*(Floor[p/4] + Floor[Mod[p, 4]/2]) + (2 - Mod[p, 4])*I; f[2, e_] := 2; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; Select[Range[500000], SquareFreeQ[#] && Im[s[#]] == 0 &] (* Amiram Eldar, Feb 24 2024 *)

Extensions

More terms from Amiram Eldar, Feb 24 2024

A070750 0 if n-th prime is even, 1 if n-th prime is == 1 (mod 4), and -1 if n-th prime is == 3 (mod 4).

Original entry on oeis.org

0, -1, 1, -1, -1, 1, 1, -1, -1, 1, -1, 1, 1, -1, -1, 1, -1, 1, -1, -1, 1, -1, -1, 1, 1, 1, -1, -1, 1, 1, -1, -1, 1, -1, 1, -1, 1, -1, -1, 1, -1, 1, -1, 1, 1, -1, -1, -1, -1, 1, 1, -1, 1, -1, 1, -1, 1, -1, 1, 1, -1, 1, -1, -1, 1, 1, -1, 1, -1, 1, 1, -1
Offset: 1

Views

Author

Reinhard Zumkeller, May 04 2002

Keywords

Comments

Also, sin(prime(n)*Pi/2), where prime(n) = A000040(n), Pi=3.1415... (original definition).
Also imaginary part of primes mapped as defined in A076340, A076341: a(n) = A076341(A000040(n)), real part = A076342.
Legendre symbol (-1/prime(n)) for n > 1. - T. D. Noe, Nov 05 2003
For n > 1, let p = prime(n) and m = (p-1)/2. Then c(m) - a(n) == 0 (mod p), where c(m) = (2*m)!/(m!)^2 = A000984(m) is the central binomial coefficient. [Proof: By definition, c(m)*(m!)^2 - (p-1)! = 0 and therefore c(m)*(m!)^2*(-1)^(m+1) - (p-1)!*(-1)^(m+1) = 0. Now apply Wilson's theorem, (p-1)! == 1 (mod p), and its corollary, (m!)^2 == (-1)^(m+1) (mod p), and finally use the formula by T. D. Noe listed below to replace (-1)^m with a(n).] Similarly, C_m - 2*a(n) == 0 (mod p), with C_m = A000108(m) being the m-th Catalan number. [Proof: By definition, C_m*(p+1)*(m!)^2 - 2*(p-1)! = 0. The result follows proceeding as in the first proof.] - Stanislav Sykora, Aug 11 2014

Examples

			p = 4*k+1 (see A002144): a(p) = sin((4*k+1)*Pi/2) = sin(2*k*Pi + Pi/2) = sin(Pi/2) = 1.
p = 4*k+3 (see A002145): a(p) = sin((4*k+3)*Pi/2) = sin(2*k*Pi + 3*Pi/2) = sin(3*Pi/2) = -1.
		

Crossrefs

Programs

Formula

a(n) = 2 - prime(n) mod 4 = 2 - A039702(n).
a(n) = (-1)^((prime(n)-1)/2) for n > 1. - T. D. Noe, Nov 05 2003
From Amiram Eldar, Dec 24 2022: (Start)
Product_{n>=1} (1 - a(n)/prime(n)) = 4/Pi (A088538).
Product_{n>=1} (1 + a(n)/prime(n)) = 2/Pi (A060294). (End)

Extensions

Wording of definition changed by N. J. A. Sloane, Jun 21 2015

A076340 Real part of the function defined multiplicatively on the complex numbers by 2->(2,0) and p->((floor(p/4)+floor((p mod 4)/2))*4,2-(p mod 4)) for odd primes p.

Original entry on oeis.org

1, 2, 4, 4, 4, 8, 8, 8, 15, 8, 12, 16, 12, 16, 17, 16, 16, 30, 20, 16, 31, 24, 24, 32, 15, 24, 52, 32, 28, 34, 32, 32, 47, 32, 33, 60, 36, 40, 49, 32, 40, 62, 44, 48, 68, 48, 48, 64, 63, 30, 65, 48, 52, 104, 49, 64, 79, 56, 60, 68, 60, 64, 112, 64, 47, 94, 68, 64, 95, 66, 72
Offset: 1

Views

Author

Reinhard Zumkeller, Oct 08 2002

Keywords

Comments

a(n)>0 for n<2187=3^7, a(2187)=-5816, A076341(2187)=-20047.

Examples

			n=21: 21 = 3*7 = (4-1)*(8-1) = (4,-1)*(8,-1) -> (32-(-1)*(-1),-4+(-8)) = (31,-12), therefore a(21)=31, A076341(21)=-12;
n=35: 35 = 5*7 = (4+1)*(8-1) = (4,1)*(8,-1) -> (32-1*(-1),-4+8) = (33,4), therefore a(35)=33, A076341(35)=4.
		

Crossrefs

Programs

  • Mathematica
    b[n_] := If[n == 1, 1, Product[{p, e} = pe; If[p == 2, 2, ((Floor[p/4] + Floor[Mod[p, 4]/2])*4 + (2 - Mod[p, 4]) I)]^e, {pe, FactorInteger[n]}]];
    a[n_] := Re[b[n]];
    Array[a, 100] (* Jean-François Alcover, Dec 12 2021 *)

Formula

a(A000040(n)) = A076342(n).
a(A001358(n)) = A076343(n).
a(A000961(n)) = A076345(n).
a(A005117(n)) = A076347(n).
a(A000290(n)) = A076349(n).
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