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.

Showing 1-4 of 4 results.

A214031 Fixed points of A214030.

Original entry on oeis.org

13, 19, 23, 37, 41, 47, 89, 139, 157, 211, 277, 281, 331, 373, 379, 397, 499, 503, 521, 571, 613, 619, 641, 647, 691, 733, 739, 743, 757, 761, 811, 853, 859, 863, 877, 983, 997, 1051, 1093, 1103, 1117, 1171, 1213, 1223, 1237, 1289, 1297, 1409, 1453, 1459, 1481, 1487
Offset: 1

Views

Author

Art DuPre, Jul 12 2012

Keywords

Comments

This sequence is to A214030 as A000057 is to A001177. It would be nice to have an interpretation of this sequence akin to the interpretation of A000057 as the set of primes which divide all Fibonacci sequences, having arbitrary initial values for a(1),a(2). The linearly recursive sequence which seems to be associated to this is 3*f(n) = 6*f(n-1) + 2*f(n-2), but this does not have integral values.
If we use the sequence 3,2,3,2,3,2,... instead of 2,3,2,3,... we end up with the same sequence a(n).

Crossrefs

Programs

  • PARI
    {b23(n)=local(t,m=1,s=[n]); if (n<2,0,while(1,
    if(m%2,s=concat(s,2),s=concat(s,3));
    t=contfracpnqn(concat(s,n));
    t=contfrac(n*t[1,1]/t[2,1]);
    if(t[1]A214031(n) to the screen,
    for(i=1,500,if(b23(i)==i,print1(i,", ")));

Formula

{n: A214030(n)=n}.

A214032 Places n where A214030(n) = n-2.

Original entry on oeis.org

17, 67, 109, 137, 181, 191, 229, 233, 239, 257, 283, 307, 311, 349, 353, 359, 479, 523, 547, 593, 599, 617, 643, 709, 719, 829, 839, 657, 883, 907, 911, 953, 977, 1021, 1031, 1069, 1097, 1123, 1151, 1193, 1217, 1319, 1433, 1439, 1483
Offset: 1

Views

Author

Art DuPre, Jul 12 2012

Keywords

Comments

This sequence is to A214030 as A000057 is to A001177. It would be nice to have an interpretation of this sequence akin to the interpretation of A000057 as the set of primes which divide all Fibonacci sequences, having arbitrary initial values for a(1),a(2). The linearly recursive sequence which seems to be associated to this is 3*f(n)=6*f(n-1)+2*f(n-2), but this does not have integral values.
If we use the sequence 3,2,3,2,3,2.. instead of 2,3,2,3,... we end up with the same sequence a(n).

Crossrefs

Programs

  • PARI
    {b23(n)=local(t,m=1,s=[n]); if (n<2,0,while(1,
    if(m%2,s=concat(s,2),s=concat(s,3));
    t=contfracpnqn(concat(s,n));
    t=contfrac(n*t[1,1]/t[2,1]);
    if(t[1]A214032(n) to the screen, */
    for(i=1,1500,if(b23(i)==i-2,print1(i,", ")));

A214033 Places n where A214030(n) = n or A214030(n) = n-2.

Original entry on oeis.org

13, 17, 19, 23, 37, 41, 47, 67, 89, 109, 137, 139, 157, 181, 191, 211, 229, 233, 239, 257, 277, 281, 283, 307, 311, 331, 349, 353, 359, 373, 379, 397, 479, 499, 503, 521, 523, 547, 571, 593, 599, 613, 617, 619, 641
Offset: 1

Views

Author

Art DuPre, Jul 12 2012

Keywords

Comments

It always has been one of the great mysteries of mathematics, that the superdiagonal sequence of A001177 consists of prime numbers A000057.
Here, regarding A214031 and A214032,there is the further conjecture that these two disjoint sequences are primes and roughly comparable in density. It isn't clear that these two sequences have a density, without appealing to the Riemann Hypothesis, but they are certainly close to one another in growing size.
Since these two sequences are disjoint, it is natural to take their union.

Crossrefs

Programs

  • PARI
    {b23(n)=local(t,m=1,s=[n]); if (n<2,0,while(1,
    if(m%2,s=concat(s,2),s=concat(s,3));
    t=contfracpnqn(concat(s,n));
    t=contfrac(n*t[1,1]/t[2,1]);
    if(t[1]
    				

A214034 The apparition sequence for 2,4 continued fractions.

Original entry on oeis.org

1, 2, 3, 3, 5, 7, 7, 2, 3, 5, 11, 13, 7, 11, 15, 8, 5, 8, 3, 23, 5, 21, 23, 19, 13, 8, 7, 27, 11, 31, 31, 5, 17, 7, 11, 37, 17, 41, 7, 20, 23, 20, 11, 11, 21, 45, 47, 7, 19, 8, 27, 51, 17, 11, 7, 8, 27, 14, 11, 61, 31, 23, 63, 27, 5, 32, 35, 65, 7, 69, 23, 35, 37, 59, 35, 23, 41, 79, 15
Offset: 1

Views

Author

Art DuPre, Jul 21 2012

Keywords

Comments

See A214030 for an explanation. There seems to be no corresponding linear recurrence.

Crossrefs

Cf. A214030.

Programs

  • PARI
    {a24(n)=local(t,m=1,s=[n]);if(n<2,0,while(1,
    if(m%2,s=concat(s,2),s=concat(s,4));
    t=contfracpnqn(concat(s,n));
    t=contfrac(n*t[1,1]/t[2,1]);
    if(t[1]A214034 */
    for(i=2,70,print1(a24(i),", "))

Extensions

More terms from Michel Marcus, Jan 15 2025
Showing 1-4 of 4 results.