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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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

Cf. A000057, A001177, A214030.

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}.

A230359 Prime numbers p such that their Fibonacci entry points are less than p+1.

Original entry on oeis.org

5, 11, 13, 17, 19, 29, 31, 37, 41, 47, 53, 59, 61, 71, 73, 79, 89, 97, 101, 107, 109, 113, 131, 137, 139, 149, 151, 157, 173, 179, 181, 191, 193, 197, 199, 211, 229, 233, 239, 241, 251, 257, 263, 269, 271, 277, 281, 293, 307, 311, 313, 317, 331, 337, 347, 349, 353, 359, 373, 379, 389, 397, 401, 409, 419, 421, 431, 433, 439, 449, 457, 461, 479, 491, 499
Offset: 1

Views

Author

Brandon Avila and Tanya Khovanova, Oct 16 2013

Keywords

Comments

For these primes p there exists a Fibonacci like sequence that doesn't contain multiples of p.
For other primes p the Fibonacci entry points are p+1. These primes are sequence A000057: Primes dividing all Fibonacci sequences.

Links

Crossrefs

A002144 is a subsequence.
Cf. A000057, A001177, A001602.

Programs

  • Maple
    filter:= proc(n) local i,a,b,c;
  if not isprime(n) then return false fi;
  a:= 0; b:= 1;
  for i from 1 to n-1 do
   c:= b;
   b:= a+b mod n; if b = 0 then return true fi;
   a:= c;
  od;
false
end proc:
select(filter, [seq(i,i=3..1000,2)]); # Robert Israel, Sep 01 2020
  • Mathematica
    A001177[n_] := For[k = 1, True, k++, If[Divisible[Fibonacci[k], n], Return[k]]]; A230359 = Reap[For[p = 2, p <= 499, p = NextPrime[p], If[A001177[p] < 1+p, Sow[p]]]][[2, 1]] (* Jean-François Alcover, Oct 21 2013 *)
  • Sage
    def isA230359(p):
    return any(p.divides(fibonacci(k)) for k in (1..p))
print([p for p in primes(1, 500) if isA230359(p)]) # Peter Luschny, Nov 01 2019

Formula

{p in A000040: A001177(p) < 1+p}.

A213893 Fixed points of a sequence h(n) defined by the minimum number of 4's in the relation n*[n,4,4,...,4,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

3, 7, 43, 67, 103, 127, 163, 223, 283, 367, 463, 487, 523, 547, 607, 643, 727, 787, 823, 883, 907, 1063, 1123, 1303, 1327, 1423, 1447, 1543, 1567, 1627, 1663, 1723, 1747, 1783, 1867, 1987, 2083, 2143, 2203, 2287, 2347, 2383, 2467, 2683, 2707, 2767, 2803, 2887
Offset: 1

Views

Author

Art DuPre, Jun 23 2012

Keywords

Comments

In a variant of A213891, multiply n by a number with simple continued fraction [n,4,4,...,4,n] and increase the number of 4's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2*[2,4,2] = [4,2,4],
3*[3,4,4,4,3] = [9,1,2,2,2,1,9],
4*[4,4,4] = [16,1,16],
5*[5,4,4,4,4,5] = [26,5,1,1,5,26].
The number of 4's needed defines the sequence h(n) = 1, 3, 1, 4, 3, 7, 3, 3, 9, ... (n>=2).
The current sequence contains the fixed points of h, i.e., those n where h(n)=n.
We conjecture that this sequence contains prime numbers analogous to the sequence of prime numbers A000057, in the sense that, instead of referring to the Fibonacci sequences(sequences satisfying f(n) = f(n-1) + f(n-2) with arbitrary positive integer values for f(1) and f(2)) it refers to the sequences satisfying f(n) = 4*f(n-1) + f(n-2), A001076, A001077, A015448, etc. This would mean that a prime is in the sequence if and only if it divides some term in each of the sequences satisfying f(n) = 4*f(n-1) + f(n-2).
The above sequence h() is recorded as A262214. - M. F. Hasler, Sep 15 2015

Crossrefs

Cf. A000057, A213891, A213892, A213894 - A213899, A261311; A213358.
Cf. A213648, A262212 - A262220, A213900, A262211.

Programs

  • Mathematica
    f[m_, n_] := Block[{c, k = 1}, c[x_, y_] := ContinuedFraction[x FromContinuedFraction[Join[{x}, Table[m, {y}], {x}]]]; While[First@ c[n, k] != Last@ c[n, k], k++]; k]; Select[Range[2, 1000], f[4, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)
  • PARI
    {a(n) = local(t, m=1); if( n<2, 0, while( 1,
   t = contfracpnqn( concat([n, vector(m,i,4), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

A213894 Fixed points of a sequence h(n) defined by the minimum number of 5's in the relation n*[n,5,5,...,5,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

2, 3, 11, 19, 31, 43, 47, 79, 127, 131, 163, 211, 251, 271, 307, 311, 331, 367, 379, 443, 503, 563, 599, 607, 659, 743, 751, 823, 839, 859, 887, 907, 911, 947, 967, 1063, 1087, 1091, 1123, 1163, 1171, 1187, 1259, 1279, 1291, 1303, 1307, 1319, 1423, 1447, 1471, 1487
Offset: 1

Views

Author

Art DuPre, Jun 23 2012

Keywords

Comments

In a variant of A213891, multiply n by a number with simple continued fraction [n,5,5,...,5,n] and increase the number of 5's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 5, 5, 2] = [4, 2, 1, 1, 2, 4],
3 * [3, 5, 5, 5, 3] = [9, 1, 1, 2, 1, 2, 1, 1, 9],
4 * [4, 5, 5, 5, 5, 5, 4] = [16, 1, 3, 2, 1, 4, 1, 2, 3, 1, 16] ,
5 * [5, 5, 5] = [25, 1, 25].
The number of 5's needed defines the sequence h(n) = 2, 3, 5, 1, 11, 5, 5, 3, 5, 11, 11, ... (n >= 2).
The current sequence contains the fixed points of h, i.e., those n where h(n)=n.
We conjecture that this sequence contains prime numbers analogous to the sequence of prime numbers A000057, in the sense that, instead of referring to the Fibonacci sequences (sequences satisfying f(n) = f(n-1) + f(n-2) with arbitrary positive integer values for f(1) and f(2)) it refers to the generalized Fibonacci sequences satisfying f(n) = 5*f(n-1) + f(n-2), A052918, A015449, A164581, etc. This would mean that a prime is in the sequence if and only if it divides some term in each of the sequences satisfying f(n) = 5*f(n-1) + f(n-2).
The above sequence h() is recorded as A262215. - M. F. Hasler, Sep 15 2015

Crossrefs

Cf. A000057, A213891 - A213893, A213895 - A213899, A261311; A213358.
Cf. A213648, A262212 - A262220, A213900, A262211.

Programs

  • Mathematica
    f[m_, n_] := Block[{c, k = 1}, c[x_, y_] := ContinuedFraction[x FromContinuedFraction[Join[{x}, Table[m, {y}], {x}]]]; While[First@ c[n, k] != Last@ c[n, k], k++]; k]; Select[Range[2, 1000], f[5, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)
  • PARI
    {a(n) = local(t, m=1); if( n<2, 0, while( 1,
   t = contfracpnqn( concat([n, vector(m,i,5), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

A213895 Fixed points of a sequence h(n) defined by the minimum number of 6's in the relation n*[n,6,6,...,6,n] = [x,...,x] between simple continued fractions.

Original entry on oeis.org

7, 11, 23, 47, 127, 139, 211, 223, 251, 331, 367, 379, 383, 463, 487, 499, 607, 619, 691, 727, 739, 743, 811, 823, 863, 887, 967, 971, 983, 1051, 1063, 1087, 1171, 1291, 1303, 1327, 1367, 1423, 1447, 1451, 1459
Offset: 1

Views

Author

Art DuPre, Jun 23 2012

Keywords

Comments

In a variant of A213891, multiply n by a number with simple continued fraction [n,6,6,...,6,n] and increase the number of 6's until the continued fraction of the product has the same first and last entry (called x in the NAME). Examples are
2 * [2, 6, 2] = [4, 3, 4],
3 * [3, 6, 3] = [9, 2, 9],
4 * [4, 6, 6, 6, 4] = [16, 1, 1, 1, 5, 1, 1, 1, 16],
5 * [5, 6, 6, 6, 6, 5] = [25, 1, 4, 3, 3, 4, 1, 25],
6 * [6, 6, 6] = [36, 1, 36],
7 * [7, 6, 6, 6, 6, 6, 6, 6, 7] = [50, 7, 2, 1, 4, 4, 4, 1, 2, 7, 50].
The number of 6's needed defines the sequence h(n) = 1, 1, 3, 4, 1, 7, 7, 5, 9, ... (n>=2).
The current sequence contains the fixed points of h, i.e., those n where h(n)=n.
We conjecture that this sequence contains numbers is analogous to the sequence of prime numbers A000057, in the sense that, instead of referring to the Fibonacci sequences (sequences satisfying f(n) = f(n-1) + f(n-2) with arbitrary positive integer values for f(1) and f(2)) it refers to the generalized Fibonacci sequences satisfying f(n) = 6*f(n-1) + f(n-2), A005668, A015451, A179237, etc. This would mean that a prime is in the sequence if and only if it divides some term in each of the sequences satisfying f(n) = 6*f(n-1) + f(n-2).
The above sequence h() is recorded as A262216. - M. F. Hasler, Sep 15 2015

Crossrefs

Cf. A000057, A213891 - A213894, A213896 - A213899, A261311; A213358.
Cf. A213648, A262212 - A262220, A213900, A262211.

Programs

  • Mathematica
    f[m_, n_] := Block[{c, k = 1}, c[x_, y_] := ContinuedFraction[x FromContinuedFraction[Join[{x}, Table[m, {y}], {x}]]]; While[First@ c[n, k] != Last@ c[n, k], k++]; k]; Select[Range[2, 1000], f[6, #] == # &] (* Michael De Vlieger, Sep 16 2015 *)
  • PARI
    {a(n) = local(t, m=1); if( n<2, 0, while( 1,
   t = contfracpnqn( concat([n, vector(m,i,6), n]));
   t = contfrac(n*t[1,1]/t[2,1]);
   if(t[1]

A262213 Minimum number of 3's such that n*[n; 3, ..., 3, n] = [x; ..., x] for some x, where [...] denotes simple continued fractions.

Original entry on oeis.org

2, 1, 5, 2, 5, 7, 5, 5, 2, 3, 5, 12, 23, 5, 11, 7, 5, 19, 5, 7, 11, 21, 5, 14, 38, 17, 23, 6, 5, 31, 23, 3, 23, 23, 5, 18, 59, 25, 5, 6, 23, 41, 11, 5, 65, 47, 11, 55, 14, 7, 77, 25, 17, 11, 23, 19, 20, 11, 5, 29, 95, 23, 47, 38, 11, 67, 23, 21, 23, 71, 5, 36, 56, 29, 59, 7, 77, 25, 11, 53
Offset: 2

Views

Author

M. F. Hasler, Sep 15 2015

Keywords

Comments

Sequence A213892 lists fixed points of this sequence.

Crossrefs

Cf. A213648, A262212 - A262220, A213900, A262211; A000057, A213891 - A213899, A261311.

Programs

  • PARI
    cf(v)={t=v[#v];forstep(i=#v-1,1,-1,t=v[i]+1/t);t}
A262213(n,d=3)=for(k=1,9e9,(c=contfrac(cf(vector(k+2,i,if(i>1&&i

A270653 Integers k such that A003266(k) is divisible by k.

Original entry on oeis.org

1, 5, 6, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 41, 42, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 64, 65, 66, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 80
Offset: 1

Views

Author

Altug Alkan, Mar 20 2016

Keywords

Comments

Note that this sequence is not the complement of A000057.
See A230359 for the prime terms of this sequence.

Examples

			11 is a term because 1*1*2*3*5*8*13*21*34*55*89 is divisible by 11.

Links

Crossrefs

Complement of A270777.
Cf. A000057, A003266, A230359.

Programs

  • Mathematica
    Select[Range@ 80, Divisible[Fibonorial@ #, #] &] (* Version 10, or *) Select[Range@ 80, Divisible[Product[Fibonacci@ k, {k, #}], #] &] (* Michael De Vlieger, Mar 23 2016 *)
  • PARI
    t(n) = prod(k=1, n, Mod(fibonacci(k), n));
for(n=1, 1e2, if(lift(t(n)) == 0, print1(n, ", ")));

A270777 Integers k such that A003266(k) is not divisible by k.

Original entry on oeis.org

2, 3, 4, 7, 23, 43, 67, 83, 103, 127, 163, 167, 223, 227, 283, 367, 383, 443, 463, 467, 487, 503, 523, 547, 587, 607, 643, 647, 683, 727, 787, 823, 827, 863, 883, 887, 907, 947, 983, 1063, 1123, 1163, 1187, 1283, 1303, 1327, 1367, 1423, 1447, 1487, 1543, 1567, 1583, 1607, 1627, 1663, 1667
Offset: 1

Views

Author

Altug Alkan, Mar 22 2016

Keywords

Examples

			4 is a term because 1*1*2*3 = 6 is not divisible by 4.

Links

Crossrefs

Complement of A270653.
Cf. A000057, A003266.

Programs

  • Mathematica
    Select[Range@ 1680, ! Divisible[Fibonorial@ #, #] &] (* Version 10, or *)
Select[Range@ 1680, ! Divisible[Product[Fibonacci@ k, {k, #}], #] &] (* Michael De Vlieger, Mar 27 2016 *)
  • PARI
    t(n) = prod(k=1, n, Mod(fibonacci(k), n));
for(n=1, 2000, if(lift(t(n)) != 0, print1(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

Cf. A214030, A214031, A000057, A001177.

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,", ")));

A254141 The average of a(n) consecutive Fibonacci numbers is never an integer.

Original entry on oeis.org

8, 16, 21, 28, 32, 40, 52, 55, 56, 64, 65, 68, 69, 80, 84, 85, 87, 88, 92, 93, 99, 104, 105, 112, 117, 119, 128, 132, 133, 136, 140, 141, 145, 148, 152, 153, 155, 156, 160, 161, 164, 165, 171, 172, 176, 184, 187, 188, 196, 200, 203, 204, 205, 207, 208, 209, 212
Offset: 1

Views

Author

Paolo P. Lava, Jan 26 2015

Keywords

Comments

Subset of A033949 and A175594 (essentially the same sequence).
Numbers of the form 2^k, with k>=3, appear to be part of the sequence.
The file "List of indexes and steps (k, x, y)" (see Links) for k = 1, 2, 3, 4, ... consecutive Fibonacci numbers gives the minimum index to start to calculate the average ( x ) and the step to add to get all the other averages ( y ).
E.g.: for k = 7 we have 7, 6, 8. This means that we must start from the 6th Fibonacci number to add 7 consecutive Fibonacci numbers and get an average that is an integer. Fibonacci(6) + Fibonacci(7) + ... + Fibonacci(12) = 8 + 13 + 21 + 34 + 55 + 89 + 144 = 364 and 364 / 7 = 52.
Then 6 + 1*8 = 14, 6 + 2*8 = 22, 6 + 3*8 = 30, etc. are the other indexes:
Fibonacci(14) + Fibonacci (15) + ... + Fibonacci(20) = 377 + 610 + 987 + 1597 + 2584 + 4181 + 6765 = 17101 and 17101 / 7 = 2443;
Fibonacci(22) + Fibonacci(23) + ... + Fibonacci(28) = 17711 + 28657 + 46368 + 75025 + 121393 + 196418 + 317811 = 803383 and 803383 / 7 = 114769;
Fibonacci(30) + Fibonacci(31) + ... + Fibonacci(36) = 832040 + 1346269 + 2178309 + 3524578 + 5702887 + 9227465 + 14930352 = 37741900 and 37741900 / 7 = 5391700; etc.
In particular we note that:
x = 0 is A219612; x = 1 is A124456; x = 0 and y = k - 1 is A106535;
y = 1 is A141767; x = k - 1 and y = k + 1 is A000057;
x = y - 1 or y|k is A023172; y = k is A000351;
x = y - k + 1 appears to give only prime numbers: 3,11,19,31,59,71,79,131,179,191,239,251,271,311,359,379,419,431,439,479,491,499,571,599,631,659,719,739,751,839,971, etc.

Links

Crossrefs

Cf. A000045, A000057, A000071, A023172, A033949, A106535, A124456, A141767, A175594, A219612.

Programs

  • Maple
    with(numtheory); with(combinat):P:=proc(q) local a,b,k,j,n,ok;
for j from 1 to q do b:=0; ok:=1;
for n from 0 to q do a:=add(fibonacci(n+k),k=0..j-1)/j;
if type(a,integer) then ok:=0; break; fi; od;
if ok=1 then print(j); fi; od; end: P(20000);
Previous Showing 21-30 of 43 results. Next

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