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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A058058 Generalized Somos-7 sequence: a(n)*a(n+7) = 3*a(n+1)*a(n+6) - 4*a(n+2)* a(n+5) + 4*a(n+3)*a(n+4).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 9, 19, 33, 131, 681, 3921, 23801, 132881, 598121, 7466321, 141401273, 1865484899, 19358862929, 314151573363, 7831607063961, 237725833277411, 8937694547422641, 293153245305595201, 7098035759907924561, 310194702846756799041, 35075042744420641281841
Offset: 1

Views

Author

Robert G. Wilson v, Nov 20 2000

Keywords

References

  • N. Elkies, posting to the NMBRTHRY(AT)LISTSERV.NODAK.EDU newsgroup, Nov. 2000.

Crossrefs

Programs

  • Magma
    I:=[1,1,1,1,1,1,1]; [n le 7 select I[n] else (3*Self(n-1)*Self(n-6) - 4*Self(n-2)*Self(n-5) + 4*Self(n-3)*Self(n-4))/Self(n-7): n in [1..30]]; // G. C. Greubel, Feb 21 2018
  • Mathematica
    a[1] =a[2] =a[3] =a[4] =a[5] =a[6] =a[7] =1; a[n_]:= a[n] = (3*a[n-1]*a[n-6] - 4*a[n-2]*a[n-5] + 4*a[n-3]*a[n-4])/a[n-7]; Table[ a[n], {n, 1, 35}]
  • PARI
    {a(n) = if (n <=7, 1, (3*a(n-1)*a(n-6) - 4*a(n-2)*a(n-5) + 4*a(n-3)*a(n-4))/a(n-7))};
    for(n=1, 30, print1(a(n), ", ")) \\ G. C. Greubel, Feb 21 2018
    

A064268 a(n) = (a(n-1) * a(n-6) + 2 * a(n-3) * a(n-4)) / a(n-7). a(1) = ... = a(7) = 1. Somos-7 variation.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 3, 5, 7, 13, 43, 113, 521, 1241, 3681, 23657, 177721, 679505, 4674203, 27273277, 275517767, 3496390229, 37043734803, 226196947873, 4391322667601, 81041508965617, 1433151398896001, 25397505914206225, 472652420405241521, 9156799134584424289, 499597377081528480243
Offset: 1

Views

Author

Michael Somos, Sep 24 2001

Keywords

Comments

In general, suppose a(n)*a(n-7) = c1*a(n-1)*a(n-6) + c2*a(n-3)*a(n-4) for all n and constants c1,c2. Define u(n) = a(n)*a(n+5)/(a(n+2)*a(n+3)) which satisfies the generalized Lyness recursion u(n) = (c1*u(n-1) + c2)/u(n-2) for all n. For this sequence c1=1, c2=2, u(n) is (1, 1, 3, 5, 7/3, 13/15, 43/35, ...) and satisfies u(n) = (u(n-1) + 2)/u(n-2). See A076839 for Lyness references. - Michael Somos, Sep 26 2022

Crossrefs

Programs

  • Magma
    I:=[1,1,1,1,1,1,1]; [n le 7 select I[n] else (Self(n-1)*Self(n-6) + 2*Self(n-3)*Self(n-4))/Self(n-7): n in [1..30]]; // G. C. Greubel, Feb 21 2018
  • Mathematica
    RecurrenceTable[{a[1]==a[2]==a[3]==a[4]==a[5]==a[6]==a[7]==1,a[n] == (a[n-1]a[n-6]+2a[n-3]a[n-4])/a[n-7]},a,{n,30}] (* Harvey P. Dale, Nov 26 2015 *)
  • PARI
    {a(n) = if( n<1, a(8-n), if( n<8, 1, (a(n-1) * a(n-6) + 2 * a(n-3) * a(n-4)) / a(n-7)))};
    
  • PARI
    { a7=a6=a5=a4=a3=a2=a1=a=1; for (n=1, 100, if (n>7, a=(a1*a6 + 2*a3*a4)/a7; a7=a6; a6=a5; a5=a4; a4=a3; a3=a2; a2=a1; a1=a); write("b064268.txt", n, " ", a) ) } \\ Harry J. Smith, Sep 10 2009
    

Formula

a(8-n) = a(n).

A275695 a(0) = a(1) = a(2) = a(3) = a(4) = a(5) = a(6) = 1; for n>6, a(n) = ( a(n-1)+a(n-3)+a(n-5) )*( a(n-2)+a(n-4)+a(n-6) ) / a(n-7).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 9, 33, 385, 13825, 5474305, 75873853441, 415386585427968001, 3501887406773528570406162401, 44079910680970588907541344275243042224979209, 400942556117903539711475671972145122347091674105174721165559627509313
Offset: 0

Views

Author

Bruno Langlois, Aug 21 2016

Keywords

Crossrefs

Programs

  • Magma
    I:=[1,1,1,1,1,1,1]; [n le 7 select I[n] else (Self(n-1) + Self(n-3) + Self(n-5))*(Self(n-2) + Self(n-4) + Self(n-6))/Self(n-7): n in [1..17]]; // G. C. Greubel, Feb 21 2018
  • Mathematica
    RecurrenceTable[{a[n] == (a[n - 1] + a[n - 3] + a[n - 5]) (a[n - 2] + a[n - 4] + a[n - 6])/a[n - 7], a[0] == a[1] == a[2] == a[3] == a[4] == a[5] == a[6] == 1}, a, {n, 0, 16}] (* Michael De Vlieger, Aug 25 2016 *)
    nxt[{a_,b_,c_,d_,e_,f_,g_}]:={b,c,d,e,f,g,((g+e+c)(f+d+b))/a}; NestList[ nxt,{1,1,1,1,1,1,1},20][[All,1]] (* Harvey P. Dale, May 04 2019 *)
  • PARI
    a(n) = if (n <=6, 1, (a(n-1)+a(n-3)+a(n-5))*(a(n-2)+a(n-4)+a(n-6))/a(n-7)); \\ Michel Marcus, Aug 25 2016
    
  • Ruby
    def A(m, n)
      a = Array.new(2 * m + 1, 1)
      ary = [1]
      while ary.size < n + 1
        i = (1..m).inject(0){|s, i| s + a[2 * i - 1]} * (1..m).inject(0){|s, i| s + a[2 * i]}
        break if i % a[0] > 0
        a = *a[1..-1], i / a[0]
        ary << a[0]
      end
      ary
    end
    def A275695(n)
      A(3, n)
    end # Seiichi Manyama, Aug 27 2016
    

Formula

a(n) = (8-4*(-1)^n)*a(n-1)*a(n-3)*a(n-5) - a(n-2) - a(n-4) - a(n-6).
Previous Showing 11-13 of 13 results.