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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A077058 Minimal positive solution a(n) of Diophantine equation b(n)^2 - b(n)*a(n) - G(n)*a(n)^2 = +1 or -1 with G(n) := A078358(n). The companion sequence is b(n)=A077057(n).

Original entry on oeis.org

1, 1, 2, 1, 1, 8, 2, 10, 1, 1, 40, 5, 2, 3, 250, 1, 1, 106, 3, 1138, 2, 8, 25, 146, 1, 1, 2968, 15, 298, 16, 2, 5, 352, 17, 1856, 1, 1, 9384, 97, 10, 8, 253970, 2, 72664, 3, 6440, 5, 521904, 1, 1, 3034, 5, 9148450, 3, 1084152, 117, 2, 45, 746, 10, 88, 157, 126890, 1, 1
Offset: 1

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Author

Wolfdieter Lang, Nov 29 2002

Keywords

Comments

This equation can also be written as (2*b(n)-a(n))^2 - D(n)*a(n)^2 = +4 or -4 with D(n) := A077425(n)=1+4*G(n).
This is from Perron's table (see reference p. 108, for n = 1..28) which gives the minimal x,y values which solve the above mentioned Diophantine equations.
For Pell equation x^2 - D*y^2 = +4, see A077428 and A078355. For Pell equation x^2 - D*y^2 = -4, see A078356 and A078357.

References

  • O. Perron, "Die Lehre von den Kettenbruechen, Bd.I", Teubner, 1954, 1957 (Sec. 30, Satz 3.35, p. 109 and table p. 108).

Programs

  • Mathematica
    g[n_] := Ceiling[ Sqrt[n] ] + n - 1; r[n_] := Reduce[an > 0 && (bn^2 - bn *an - g[n]*an^2 == 1 || bn^2 - bn *an - g[n]*an^2 == - 1), {an, bn}, Integers] /. C -> c; ab[n_] := DeleteCases[ Flatten[ Table[{an, bn} /. {ToRules[r[n]]} // Simplify, {c[1], 0, 1}] , 1] , an | bn]; a[n_] := a[n] = Min[ ab[n][[All, 1]] ]; Table[ Print[{n, a[n]}]; a[n], {n, 1, 65}] (* Jean-François Alcover, Oct 03 2012 *)
  • PARI
    forstep(D=1,1000,4, if(issquare(D),next); u=bnfinit(x^2-D).fu[1]; k=1; while( denominator(t=polcoeff(lift(u^k),1)*2)>1, k++); print1(abs(t),", "); ) \\ Max Alekseyev, Feb 06 2010

Extensions

More terms from Max Alekseyev, Feb 06 2010

A053373 Write fundamental unit for real quadratic field of discriminant n as x + y*omega; sequence gives values of y for n == 1 (mod 4).

Original entry on oeis.org

1, 1, 2, 1, 1, 8, 2, 10, 1, 40, 5, 2, 3, 250, 1, 1, 106, 3, 1138, 2, 8, 25, 146, 2968, 15, 298, 16, 2, 5, 17, 1856, 1, 1, 9384, 97, 10, 253970, 2, 72664, 3, 6440, 5, 521904, 1, 1, 3034, 5, 9148450, 1084152, 117, 2, 746, 10, 88, 157, 126890, 1, 1, 1311, 56, 287
Offset: 1

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Author

N. J. A. Sloane, Jan 06 2000

Keywords

Comments

Entries are indexed by values of n from A039955.
Subsequence of A077058 excluding terms for which A077425(n) is not squarefree. - Max Alekseyev, Dec 12 2012

References

  • R. A. Mollin, Quadratics, CRC Press, 1996, Tables B1-B3.

Crossrefs

Programs

  • Mathematica
    2*NumberFieldFundamentalUnits[ Sqrt[#] ][[1, 2, 2]] & /@ Select[ Range[5, 309, 4], SquareFreeQ ]  (* Jean-François Alcover, Jul 09 2013 *)
  • PARI
    forstep(n=5,1000,4, if(!issquarefree(n),next); print1( 2*polcoeff(lift(bnfinit(x^2-n).fu[1]),1), ", " )) /* Max Alekseyev */

A166409 Odd numbers corresponding to the positions of zeros in A166406.

Original entry on oeis.org

5, 13, 17, 21, 29, 33, 37, 41, 45, 53, 57, 61, 65, 69, 73, 77, 85, 89, 93, 97, 99, 101, 105, 109, 113, 117, 125, 129, 133, 137, 141, 145, 147, 149, 153, 157, 161, 165, 173, 177, 181, 185, 189, 193, 197, 201, 205, 207, 209, 213, 217, 221, 229, 233, 237, 241
Offset: 1

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Author

Antti Karttunen, Oct 21 2009, Oct 22 2009

Keywords

Comments

Those odd numbers 2n+1 for which the sum of i in [1,2n+1] with J(i,2n+1)=-1 is equal to the sum of i in [1,2n+1] with J(i,2n+1)=+1. Here J(i,k) is the Jacobi symbol.
Probably a union of A077425 & A165603: It is clear that A077425 is a subsequence of this sequence. For the remaining terms to be equal to A165603, it is at least required that the intersection of A165603 and A095100 be empty.

Crossrefs

Programs

  • Python
    from sympy import jacobi_symbol as J
    def a(n):
        l=0
        m=0
        for i in range(1, 2*n + 2):
            if J(i, 2*n + 1)==-1: l+=i
            elif J(i, 2*n + 1)==1: m+=i
        return l - m
    print([2*n + 1 for n in range(201) if a(n)==0]) # Indranil Ghosh, Jun 12 2017
Previous Showing 11-13 of 13 results.