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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A095087 Fib010 primes, i.e., primes p whose Zeckendorf-expansion A014417(p) ends with zero, one and zero.

Original entry on oeis.org

7, 23, 31, 41, 83, 109, 151, 167, 193, 227, 269, 277, 311, 337, 353, 379, 397, 421, 431, 439, 463, 523, 541, 557, 599, 607, 617, 641, 659, 701, 709, 719, 727, 743, 761, 769, 811, 829, 853, 863, 887, 929, 947, 997, 1031, 1049, 1091, 1117, 1151
Offset: 1

Views

Author

Antti Karttunen, Jun 01 2004

Keywords

Crossrefs

Intersection of A000040 and A035336. Cf. A095067.

Programs

  • Python
    from sympy import fibonacci, primerange
    def a(n):
        k=0
        x=0
        while n>0:
            k=0
            while fibonacci(k)<=n: k+=1
            x+=10**(k - 3)
            n-=fibonacci(k - 1)
        return x
    def ok(n): return str(a(n))[-3:]=="010"
    print([n for n in primerange(1, 1201) if ok(n)]) # Indranil Ghosh, Jun 08 2017

A095088 Fib100 primes, i.e., primes p whose Zeckendorf-expansion A014417(p) ends with one and two final zeros.

Original entry on oeis.org

3, 11, 37, 71, 79, 113, 139, 181, 223, 257, 283, 359, 367, 401, 409, 443, 503, 571, 587, 613, 647, 757, 859, 977, 1019, 1087, 1163, 1181, 1223, 1231, 1291, 1307, 1367, 1409, 1451, 1511, 1553, 1579, 1613, 1621, 1663, 1697, 1723, 1867, 1901
Offset: 1

Views

Author

Antti Karttunen, Jun 01 2004

Keywords

Crossrefs

Intersection of A000040 and A035337. Cf. A095068.

Programs

  • Python
    from sympy import fibonacci, primerange
    def a(n):
        k=0
        x=0
        while n>0:
            k=0
            while fibonacci(k)<=n: k+=1
            x+=10**(k - 3)
            n-=fibonacci(k - 1)
        return x
    def ok(n): return str(a(n)).endswith("100")
    print([n for n in primerange(1, 2001) if ok(n)]) # Indranil Ghosh, Jun 08 2017

A095734 Asymmetricity-index for Zeckendorf-expansion A014417(n) of n.

Original entry on oeis.org

0, 0, 1, 1, 0, 1, 0, 2, 1, 0, 2, 1, 0, 1, 0, 2, 2, 1, 2, 1, 3, 1, 0, 2, 2, 1, 1, 0, 2, 2, 1, 3, 1, 0, 1, 0, 2, 2, 1, 2, 1, 3, 2, 1, 3, 3, 2, 2, 1, 3, 1, 0, 3, 2, 4, 1, 0, 2, 2, 1, 2, 1, 3, 1, 0, 2, 2, 1, 2, 1, 3, 3, 2, 1, 0, 2, 2, 1, 3, 1, 0, 3, 2, 4, 2, 1, 3, 1, 0, 1, 0, 2, 2, 1, 2, 1, 3, 2, 1, 3, 3, 2, 2, 1, 3
Offset: 0

Views

Author

Antti Karttunen, Jun 05 2004

Keywords

Comments

Least number of flips of "fibits" (changing either 0 to 1 or 1 to 0 in Zeckendorf-expansion A014417(n)) so that a palindrome is produced.

Examples

			The integers 0 and 1 look as '0' and '1' also in Fibonacci-representation,
and being palindromes, a(0) and a(1) = 0.
2 has Fibonacci-representation '10', which needs a flip of other 'fibit',
that it would become a palindrome, thus a(2) = 1. Similarly 3 has representation
'100', so flipping for example the least significant fibit, we get '101',
thus a(3)=1 as well. 7 (= F(3)+F(5)) has representation '1010', which needs
two flips to produce a palindrome, thus a(7)=2. Here F(n) = A000045(n).
		

Crossrefs

a(n) = A037888(A003714(n)). A094202 gives the positions of zeros. Cf. also A095732.

A319432 The first differences (A129761) of the tribonacci representation numbers (A003714 or A014417) consists of runs of 1's separated by the terms of the present sequence.

Original entry on oeis.org

2, 3, 6, 2, 11, 2, 3, 22, 2, 3, 6, 2, 43, 2, 3, 6, 2, 11, 2, 3, 86, 2, 3, 6, 2, 11, 2, 3, 22, 2, 3, 6, 2, 171, 2, 3, 6, 2, 11, 2, 3, 22, 2, 3, 6, 2, 43, 2, 3, 6, 2, 11, 2, 3, 342, 2, 3, 6, 2, 11, 2, 3, 22, 2, 3, 6, 2, 43, 2, 3, 6, 2, 11, 2, 3, 86, 2, 3, 6, 2, 11, 2, 3, 22, 2, 3, 6, 2, 683, 2
Offset: 1

Views

Author

N. J. A. Sloane, Sep 30 2018

Keywords

Comments

The runs of 1's in A129761 have lengths that apparently are given by the Fibonacci word A005614 (with a slight change at the start). The present sequence shows the terms greater than 1. The structure of this sequence is (partially) analyzed in the comments in A129761.

Crossrefs

Programs

  • PARI
    p=0; for (n=1, 2052, if (bitand(n,2*n)==0, if (n-p>1, print1 (n-p ", ")); p=n)) \\ Rémy Sigrist, Oct 07 2018

A319433 Take Zeckendorf representation of n (A014417(n)), drop least significant bit, take inverse Zeckendorf representation.

Original entry on oeis.org

0, 0, 1, 2, 2, 3, 3, 4, 5, 5, 6, 7, 7, 8, 8, 9, 10, 10, 11, 11, 12, 13, 13, 14, 15, 15, 16, 16, 17, 18, 18, 19, 20, 20, 21, 21, 22, 23, 23, 24, 24, 25, 26, 26, 27, 28, 28, 29, 29, 30, 31, 31, 32, 32, 33, 34, 34, 35, 36, 36, 37, 37, 38, 39, 39, 40, 41, 41, 42, 42, 43
Offset: 0

Views

Author

N. J. A. Sloane, Sep 30 2018

Keywords

Comments

In other words, the Zeckendorf representation of a(n) is obtained from the Zeckendorf representation of n by deleting the least significant bit.
Theorem: The first differences (1,1,0,1,0,1,1,0,...) form the Fibonacci word A005614. (The proof is straightforward.)
The difference sequence agrees with A005614 after the first two terms of A005614. - Clark Kimberling, Dec 29 2020

Examples

			n = 19 has Zeckendorf representation [1, 0, 1, 0, 0, 1], dropping last bit we get [1, 0, 1, 0, 0], which is the Zeckendorf representation of 11, so a(19) = 11.
		

Crossrefs

Programs

  • Mathematica
    r = (1 + Sqrt[5])/2; t = Table[Floor[(n - 1)/r] + 2, {n, 0, 150}] (* A319433 *)
    Differences[t]  (* A005614 after the 1st 2 terms *)
    (* Clark Kimberling, Dec 29 2020 *)
  • PARI
    a(n) = my (f=2, v=0); while (fibonacci(f) < n, f++); while (n > 1, if (n >= fibonacci(f), v += fibonacci(f-1); n -= fibonacci(f); f--); f--); return (v) \\ Rémy Sigrist, Oct 04 2018

Formula

a(n) = floor((n+2)/tau)-1, where tau = golden ratio. - Clark Kimberling, Dec 29 2020; corrected by Harry Altman, Jun 06 2024

Extensions

More terms from Rémy Sigrist, Oct 04 2018

A095336 Sum of 1-fibits in Zeckendorf-expansion A014417(p) summed for all primes p in range ]2^n,2^(n+1)].

Original entry on oeis.org

1, 3, 3, 13, 20, 41, 76, 176, 325, 638, 1353, 2533, 5223, 10186, 20504, 40775, 80661, 163765, 318602, 649948, 1268922, 2571531, 5082895, 10217300, 20327307, 40399966, 82164918, 160343669, 324931245, 640501167, 1290990369, 2567150515
Offset: 1

Views

Author

Antti Karttunen, Jun 04 2004

Keywords

Examples

			a(1)=1, as only prime in range ]2,4] is 3, whose Fibonacci-representation is 100. In the next range we have primes 5 and 7, whose Fibonacci-representations are 1000 and 1010 respectively, thus a(2)=3.
		

Crossrefs

A095733 Zeckendorf-expansion A014417(p) of primes p for which this expansion is palindromic.

Original entry on oeis.org

1010000101, 10010101001, 1000101001010001, 1001010000101001, 10010001010001001, 10100100000100101, 1000001001001000001, 1000100101010010001, 1010010100010100101, 10001010000001010001
Offset: 1

Views

Author

Antti Karttunen, Jun 05 2004

Keywords

Examples

			The first such prime is 127 = F(2)+F(4)+F(9)+F(11) (which thus has a symmetric Zeckendorf-expansion as 1010000101).
		

Crossrefs

a(n) = A014417(A095730(n)).

A212278 Number of adjacent pairs of zeros (possibly overlapping) in the representation of n in base of Fibonacci numbers (A014417).

Original entry on oeis.org

0, 0, 0, 1, 0, 2, 1, 0, 3, 2, 1, 1, 0, 4, 3, 2, 2, 1, 2, 1, 0, 5, 4, 3, 3, 2, 3, 2, 1, 3, 2, 1, 1, 0, 6, 5, 4, 4, 3, 4, 3, 2, 4, 3, 2, 2, 1, 4, 3, 2, 2, 1, 2, 1, 0, 7, 6, 5, 5, 4, 5, 4, 3, 5, 4, 3, 3, 2, 5, 4, 3, 3, 2, 3, 2, 1, 5, 4, 3, 3, 2, 3, 2, 1, 3, 2, 1, 1, 0, 8
Offset: 0

Views

Author

Alex Ratushnyak, May 13 2012

Keywords

Comments

a(n) = 0 only if n = Fibonacci(k)-1.

Examples

			A014417(5) = 1000, two pairs of adjacent zeros, so a(5) = 2.
		

Crossrefs

Programs

  • Maple
    F:= combinat[fibonacci]:
    b:= proc(n) option remember; local j;
          if n=0 then 0
        else for j from 2 while F(j+1)<=n do od;
             b(n-F(j))+2^(j-2)
          fi
        end:
    a:= proc(n) local c, h, m, t;
          c, t, m:= 0, 1, b(n);
          while m>0 do
            h:= irem(m, 2, 'm');
            if h=t and h=0 then c:=c+1 fi;
            t:=h
          od; c
        end:
    seq(a(n), n=0..150);  # Alois P. Heinz, May 18 2012

A372302 Numbers k for which the Zeckendorf representation A014417(k) ends with "1001".

Original entry on oeis.org

6, 19, 27, 40, 53, 61, 74, 82, 95, 108, 116, 129, 142, 150, 163, 171, 184, 197, 205, 218, 226, 239, 252, 260, 273, 286, 294, 307, 315, 328, 341, 349, 362, 375, 383, 396, 404, 417, 430, 438, 451, 459, 472, 485, 493, 506, 519, 527, 540, 548, 561, 574, 582, 595, 603
Offset: 1

Views

Author

A.H.M. Smeets, Apr 25 2024

Keywords

Crossrefs

Tree of Zeckendorf subsequences of positive integers partitioned by their suffix part S (except initial term or offset in some cases). $ is the empty string. length(S) =
0 1 2 3 4 5 6 7
----------------------------------------------------------------------
$: 0: 00: 000: 0000: 00000: 000000:
100000: 0100000:
A035340 <------
10000:
1000: 01000:
A035338 <------
10: 010: 0010:
A035336 <------ A134861
1010: 01010:
A134863 <------
100: 0100:
A035337 <------
1: 01: 001: 0001:
1001: 01001:
A372302 <------
101: 0101:
A134860 <------
Suffixes 10^n, where ^ means n times repeated concatenation, are the (n+1)-th columns in the Wythoff array A083412 and A035513 (n >= 0).

Formula

Equals {A134859}\{A151915}.
a(n) = A134863(n) - 1 = A035338(n) + 1.
a(n) = a(n-1) + 8 + 5*A005614(n-2) = a(n-1) + F(6) + F(5)*A005614(n-2), n > 1, where F(k) is the k-th Fibonacci number (A000045).

A182575 Primes with equal number of 1's and 0's in their representation in base of Fibonacci numbers (A014417).

Original entry on oeis.org

2, 7, 17, 19, 53, 373, 983, 6709, 6761, 6763, 17333, 17707, 43783, 103681, 317789, 785671, 2177321, 5581493, 38770357, 39070457, 1836305137, 10749957121, 12586147631, 12586251313, 32951279111, 86252640919, 86267571127, 86267571263, 213265164691, 225845730829
Offset: 1

Views

Author

Alex Ratushnyak, May 05 2012

Keywords

Examples

			Representation of 19 is 101001, three 1's and three 0's, so 19 is in the sequence.
Representation of 23 is 1000010, two 1's and five 0's, so 23 is not in the sequence.
		

Crossrefs

Programs

  • Mathematica
    nn = 30; f = Join[{0}, Accumulate[Fibonacci[Range[2, 2*nn, 2] - 1]]]; t = Table[hi = f[[n+1]] - 1; Reverse[Table[hi - f[[i]], {i, n - 1}]], {n, 2, nn}]; Select[Flatten[t], PrimeQ] (* T. D. Noe, May 07 2012 *)

Extensions

Extended by T. D. Noe, May 07 2012
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