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.

User: Satya Das

Satya Das's wiki page.

Satya Das has authored 2 sequences.

A342816 Numbers of the form (2^(2*j + 6*k + 10) - 2^(2*j + 2) - 3)/9, with j,k >= 0.

Original entry on oeis.org

113, 453, 1813, 7253, 7281, 29013, 29125, 116053, 116501, 464213, 466005, 466033, 1856853, 1864021, 1864133, 7427413, 7456085, 7456533, 29709653, 29824341, 29826133, 29826161, 118838613, 119297365, 119304533, 119304645, 475354453, 477189461, 477218133
Offset: 1

Views

Author

Satya Das, Mar 22 2021

Keywords

Comments

Sequence is a subsequence of A198584. When any term is iterated using the Collatz function, the last odd integer in the trajectory before 1 is of the form (4^(3*m + 4) - 1)/3.

Links

Crossrefs

Union with A342815 gives A198584.

Programs

  • Mathematica
    Take[Sort[Flatten[Table[(2^(2n1+6n2+10) - 2^(2n1+2) - 3)/9, {n1, 0, 20}, {n2, 0, 20}]]], 50]
  • Python
    seq=[]
for n1 in range(20):
    for n2 in range(20):
        n=(2**(2*n1+6*n2+10) - 2**(2*n1+2) - 3)/9
        seq.append(n)
seq.sort()
print(seq[0:50])

A342815 Numbers of the form (2^(2*j + 6*k + 5) - 2^(2*j + 1) - 3)/9, with j,k >= 0.

Original entry on oeis.org

3, 13, 53, 213, 227, 853, 909, 3413, 3637, 13653, 14549, 14563, 54613, 58197, 58253, 218453, 232789, 233013, 873813, 931157, 932053, 932067, 3495253, 3724629, 3728213, 3728269, 13981013, 14898517, 14912853, 14913077, 55924053, 59594069, 59651413, 59652309
Offset: 1

Views

Author

Satya Das, Mar 22 2021

Keywords

Comments

Sequence is a subsequence of A198584. When any term is iterated using the Collatz function, the last odd integer in the trajectory before 1 is of the form (4^(3*m + 2) - 1)/3.

Links

Crossrefs

Union with A342816 gives A198584.

Programs

  • Mathematica
    Take[Sort[Flatten[Table[(2^(2n1+6n2+5) - 2^(2n1+1) - 3)/9, {n1, 0, 20}, {n2, 0, 20}]]], 50]
  • Python
    seq=[]
for n1 in range(20):
    for n2 in range(20):
        n=(2**(2*n1+6*n2+5) - 2**(2*n1+1) - 3)/9
        seq.append(n)
seq.sort()
print(seq[0:50])

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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