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.

A266295 2-free tetranacci sequence beginning 1,3,5,7.

Original entry on oeis.org

1, 3, 5, 7, 1, 1, 7, 1, 5, 7, 5, 9, 13, 17, 11, 25, 33, 43, 7, 27, 55, 33, 61, 11, 5, 55, 33, 13, 53, 77, 11, 77, 109, 137, 167, 245, 329, 439, 295, 327, 695, 439, 439, 475, 1, 677, 199, 169, 523, 49, 235, 61, 217, 281, 397, 239, 567, 371, 787
Offset: 1

Views

Author

Jeremy F. Alm, Dec 28 2015

Keywords

Comments

For n>4, a(n) = (a(n-1) + a(n-2) + a(n-3) + a(n-4)) / 2^d, where 2^d is the largest power of 2 dividing a(n-1) + a(n-2) + a(n-3) + a(n-4). In other words, sum the previous four terms, then divide by two until the result is odd.

References

  • Alm, Herald, Miller, and Sexton, 2-Free Tetranacci Sequences, unpublished.

Crossrefs

Programs

  • Mathematica
    nxt[{a_, b_, c_, d_}] := {b, c, d, (a + b + c + d)/2^IntegerExponent[ a + b + c + d, 2]}; NestList[nxt,{1,3,5,7},60][[All,1]] (* Harvey P. Dale, Nov 09 2020 *)
  • PARI
    lista(nn) = {print1(x = 1, ", "); print1(y = 3, ", "); print1(z = 5, ", "); print1(t = 7, ", "); for (n=5, nn, tt = (x+y+z+t); tt /= 2^valuation(tt, 2); print1(tt, ", "); x=y; y=z; z=t; t=tt;);} \\ Michel Marcus, Dec 29 2015
  • Python
    ### CREATES A b-FILE ###
    def main():
        name = "b266295.txt"
        file = open(name, 'w')
        file.write('1' + ' ' + '1\n')
        file.write('2' + ' ' + '3\n')
        file.write('3' + ' ' + '5\n')
        file.write('4' + ' ' + '7\n')
        a, b, c, d = 1, 3, 5, 7
        for i in range(5,10001):
            x=a+b+c+d
            while x%2==0:
                x /= 2
            a, b, c, d = b, c, d, x
            file.write(str(i) + ' ' + str(int(d)) + '\n')
        file.close()
    main()
    

Formula

a(n) = (a(n-1) + a(n-2) + a(n-3) + a(n-4)) / 2^d, where 2^d is the largest power of 2 dividing a(n-1) + a(n-2) + a(n-3) + a(n-4).
a(n) = A000265(a(n-1) + a(n-2) + a(n-3) + a(n-4)). - Michel Marcus, Dec 29 2015