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 1-10 of 41 results. Next

A130235 Partial sums of the 'lower' Fibonacci Inverse A130233.

Original entry on oeis.org

0, 2, 5, 9, 13, 18, 23, 28, 34, 40, 46, 52, 58, 65, 72, 79, 86, 93, 100, 107, 114, 122, 130, 138, 146, 154, 162, 170, 178, 186, 194, 202, 210, 218, 227, 236, 245, 254, 263, 272, 281, 290, 299, 308, 317, 326, 335, 344, 353, 362, 371, 380, 389, 398, 407, 417, 427
Offset: 0

Views

Author

Hieronymus Fischer, May 17 2007

Keywords

Crossrefs

Programs

  • Magma
    m:=120;
    f:= func< x | (&+[x^Fibonacci(j): j in [1..Floor(3*Log(3*m+1))]])/(1-x)^2 >;
    R:=PowerSeriesRing(Rationals(), m+1);
    [0] cat Coefficients(R!( f(x) )); // G. C. Greubel, Mar 17 2023
    
  • Mathematica
    nmax = 90; CoefficientList[Series[Sum[x^Fibonacci[k], {k, 1, 1 + Log[3/2 + Sqrt[5]*nmax]/Log[GoldenRatio]}]/(1-x)^2, {x, 0, nmax}], x] (* Vaclav Kotesovec, Apr 14 2020 *)
  • SageMath
    m=120
    def f(x): return sum( x^fibonacci(j) for j in range(1, int(3*log(3*m+1))))/(1-x)^2
    def A130235_list(prec):
        P. = PowerSeriesRing(ZZ, prec)
        return P( f(x) ).list()
    A130235_list(m) # G. C. Greubel, Mar 17 2023

Formula

a(n) = Sum_{k=0..n} A130233(k) = (n+1)*A130233(n) - Fib(A130233(n)+2) + 1.
G.f.: 1/(1-x)^2 * Sum_{k>=1} x^Fib(k). [corrected by Joerg Arndt, Apr 14 2020]

A130237 The 'lower' Fibonacci Inverse A130233(n) multiplied by n.

Original entry on oeis.org

0, 2, 6, 12, 16, 25, 30, 35, 48, 54, 60, 66, 72, 91, 98, 105, 112, 119, 126, 133, 140, 168, 176, 184, 192, 200, 208, 216, 224, 232, 240, 248, 256, 264, 306, 315, 324, 333, 342, 351, 360, 369, 378, 387, 396, 405, 414, 423, 432, 441, 450, 459, 468, 477, 486, 550
Offset: 0

Views

Author

Hieronymus Fischer, May 17 2007

Keywords

Crossrefs

Programs

  • Magma
    [n*Floor(Log(3/2 +n*Sqrt(5))/Log((1+Sqrt(5))/2)): n in [0..70]]; // G. C. Greubel, Mar 18 2023
    
  • Mathematica
    Table[n*Floor[Log[GoldenRatio, 3/2 +n*Sqrt[5]]], {n,0,70}] (* G. C. Greubel, Mar 18 2023 *)
  • SageMath
    [n*int(log(3/2 +n*sqrt(5), golden_ratio)) for n in range(71)] # G. C. Greubel, Mar 18 2023

Formula

a(n) = n*A130233(n).
a(n) = n*floor(arcsinh(sqrt(5)*n/2)/log(phi)).
G.f.: (1/(1-x))*Sum_{k>=1} (Fib(k) + x/(1-x))*x^Fib(k).

A385590 Triangle read by rows, based on Fibonacci numbers: Let i > 1 be such that F(i) <= n < F(i+1); i.e., i = A130233(n). Then T(n, k) = F(i-1)^2 + 1 - (i-1) mod 2 + (n - F(i)) * F(i-2) + (k-1) * F(i-1) where F(k) = A000045(k).

Original entry on oeis.org

1, 2, 3, 4, 6, 8, 5, 7, 9, 11, 10, 13, 16, 19, 22, 12, 15, 18, 21, 24, 27, 14, 17, 20, 23, 26, 29, 32, 25, 30, 35, 40, 45, 50, 55, 60, 28, 33, 38, 43, 48, 53, 58, 63, 68, 31, 36, 41, 46, 51, 56, 61, 66, 71, 76, 34, 39, 44, 49, 54, 59, 64, 69, 74, 79, 84, 37, 42, 47, 52, 57, 62, 67, 72, 77, 82, 87, 92, 65, 73, 81, 89, 97
Offset: 1

Views

Author

Werner Schulte, Jul 03 2025

Keywords

Comments

Conjecture: This triangle yields a permutation of the natural numbers.

Examples

			Triangle T(n, k) for 1 <= k <= n starts:
n\ k :   1   2   3   4   5   6   7   8   9  10  11  12  13
==========================================================
   1 :   1
   2 :   2   3
   3 :   4   6   8
   4 :   5   7   9  11
   5 :  10  13  16  19  22
   6 :  12  15  18  21  24  27
   7 :  14  17  20  23  26  29  32
   8 :  25  30  35  40  45  50  55  60
   9 :  28  33  38  43  48  53  58  63  68
  10 :  31  36  41  46  51  56  61  66  71  76
  11 :  34  39  44  49  54  59  64  69  74  79  84
  12 :  37  42  47  52  57  62  67  72  77  82  87  92
  13 :  65  73  81  89  97 105 113 121 129 137 145 153 161
  etc.
		

Crossrefs

Programs

  • PARI
    T(n, k) = i=1; for(j=1,n,if(j==fibonacci(i+1),i=i+1)); (fibonacci(i-1))^2+1-(i-1)%2 + (n-fibonacci(i))*fibonacci(i-2) + (k-1)*fibonacci(i-1)

Formula

Conjecture: Sum_{k=1..n} (-1)^k * binomial(n-1, k-1) * T(n, k) = 0 for n > 2 and (-1)^n for n < 3.

A010056 Characteristic function of Fibonacci numbers: a(n) = 1 if n is a Fibonacci number, otherwise 0.

Original entry on oeis.org

1, 1, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
Offset: 0

Views

Author

Keywords

Comments

Understood as a binary number, Sum_{k>=0} a(k)/2^k, the resulting decimal expansion is 1.910278797207865891... = Fibonacci_binary+0.5 (see A084119) or Fibonacci_binary_constant-0.5 (see A124091), respectively. - Hieronymus Fischer, May 14 2007
a(n)=1 if and only if there is an integer m such that x=n is a root of p(x)=25*x^4-10*m^2*x^2+m^4-16. Also a(n)=1 iff floor(s)<>floor(c) or ceiling(s)<>ceiling(c) where s=arcsinh(sqrt(5)*n/2)/log(phi), c=arccosh(sqrt(5)*n/2)/log(phi) and phi=(1+sqrt(5))/2. - Hieronymus Fischer, May 17 2007
a(A000045(n)) = 1; a(A001690(n)) = 0. - Reinhard Zumkeller, Oct 10 2013
Image, under the map sending a,b,c -> 1, d,e,f -> 0, of the fixed point, starting with a, of the morphism sending a -> ab, b -> c, c -> cd, d -> d, e -> ef, f -> e. - Jeffrey Shallit, May 14 2016

Crossrefs

Decimal expansion of Fibonacci binary is in A084119.
Sequences mentioned in the Allouche et al. "Taxonomy" paper, listed by example number: 1: A003849, 2: A010060, 3: A010056, 4: A020985 and A020987, 5: A191818, 6: A316340 and A273129, 18: A316341, 19: A030302, 20: A063438, 21: A316342, 22: A316343, 23: A003849 minus its first term, 24: A316344, 25: A316345 and A316824, 26: A020985 and A020987, 27: A316825, 28: A159689, 29: A049320, 30: A003849, 31: A316826, 32: A316827, 33: A316828, 34: A316344, 35: A043529, 36: A316829, 37: A010060.
Cf. A079586 (Dirich. g.f. at s=1).

Programs

  • Haskell
    import Data.List (genericIndex)
    a010056 = genericIndex a010056_list
    a010056_list = 1 : 1 : ch [2..] (drop 3 a000045_list) where
       ch (x:xs) fs'@(f:fs) = if x == f then 1 : ch xs fs else 0 : ch xs fs'
    -- Reinhard Zumkeller, Oct 10 2013
    
  • Maple
    a:= n-> (t-> `if`(issqr(t+4) or issqr(t-4), 1, 0))(5*n^2):
    seq(a(n), n=0..144);  # Alois P. Heinz, Dec 06 2020
  • Mathematica
    Join[{1},With[{fibs=Fibonacci[Range[15]]},If[MemberQ[fibs,#],1,0]& /@Range[100]]]  (* Harvey P. Dale, May 02 2011 *)
  • PARI
    a(n)=my(k=n^2);k+=(k+1)<<2; issquare(k) || (n>0 && issquare(k-8)) \\ Charles R Greathouse IV, Jul 30 2012
    
  • Python
    from sympy.ntheory.primetest import is_square
    def A010056(n): return int(is_square(m:=5*n**2-4) or is_square(m+8)) # Chai Wah Wu, Mar 30 2023

Formula

G.f.: (Sum_{k>=0} x^A000045(k)) - x. - Hieronymus Fischer, May 17 2007

A130241 Maximal index k of a Lucas number such that Lucas(k) <= n (the 'lower' Lucas (A000032) Inverse).

Original entry on oeis.org

1, 1, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9
Offset: 1

Views

Author

Hieronymus Fischer, May 19 2007, Jul 02 2007

Keywords

Comments

Inverse of the Lucas sequence (A000032), nearly, since a(Lucas(n))=n for n>=1 (see A130242 and A130247 for other versions). For n>=2, a(n)+1 is equal to the partial sum of the Lucas indicator sequence (see A102460). Identical to A130247 except for n=2.

Examples

			a(10)=4, since Lucas(4)=7<=10 but Lucas(5)=11>10.
		

Crossrefs

For partial sums see A130243. Other related sequences: A000032, A130242, A130245, A130247, A130249, A130255, A130259. Indicator sequence A102460. Fibonacci inverse see A130233 - A130240, A104162.

Programs

  • Magma
    [Floor(Log((2*n+1)/2)/Log((1+Sqrt(5))/2)): n in [2..50]]; // G. C. Greubel, Sep 09 2018
    
  • Mathematica
    Join[{1}, Table[Floor[Log[GoldenRatio, n + 1/2]], {n, 2, 50}]] (* G. C. Greubel, Dec 24 2017 *)
  • PARI
    for(n=1,50, print1(floor(log((2*n+1)/2)/log((1+sqrt(5))/2)), ", ")) \\ G. C. Greubel, Sep 09 2018
    
  • Python
    from itertools import count, islice
    def A130241_gen(): # generator of terms
        a, b = 1, 3
        for i in count(1):
            yield from (i,)*(b-a)
            a, b = b, a+b
    A130241_list = list(islice(A130241_gen(),40)) # Chai Wah Wu, Jun 08 2022

Formula

a(n) = floor(log_phi((n+sqrt(n^2+4))/2)) = floor(arcsinh((n+1)/2)/log(phi)) where phi=(1+sqrt(5))/2.
a(n) = A130242(n+1) - 1 for n>=2.
a(n) = A130247(n) except for n=2.
G.f.: g(x) = 1/(1-x) * Sum{k>=1, x^Lucas(k)}.
a(n) = floor(log_phi(n+1/2)) for n>=2, where phi is the golden ratio.

A130234 Minimal index k of a Fibonacci number such that Fibonacci(k) >= n (the 'upper' Fibonacci Inverse).

Original entry on oeis.org

0, 1, 3, 4, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11
Offset: 0

Views

Author

Hieronymus Fischer, May 17 2007

Keywords

Comments

Inverse of the Fibonacci sequence (A000045), nearly, since a(Fibonacci(n)) = n except for n = 2 (see A130233 for another version). a(n+1) is equal to the partial sum of the Fibonacci indicator sequence (see A104162).

Examples

			a(10) = 7, since Fibonacci(7) = 13 >= 10 but Fibonacci(6) = 8 < 10.
		

Crossrefs

Partial sums: A130236.
Other related sequences: A000045, A130233, A130256, A130260, A104162, A108852.
Lucas inverse: A130241 - A130248.

Programs

  • Maple
    A130234 := proc(n)
        local i;
        for i from 0 do
            if A000045(i) >= n then
                return i;
            end if;
        end do:
    end proc: # R. J. Mathar, Jan 31 2015
  • Mathematica
    a[n_] := For[i = 0, True, i++, If[Fibonacci[i] >= n, Return[i]]];
    a /@ Range[0, 80] (* Jean-François Alcover, Apr 13 2020 *)
  • PARI
    a(n)=my(k);while(fibonacci(k)Charles R Greathouse IV, Feb 03 2014, corrected by M. F. Hasler, Apr 07 2021

Formula

a(n) = ceiling(log_phi((sqrt(5)*n + sqrt(5*n^2-4))/2)) = ceiling(arccosh(sqrt(5)*n/2)/log(phi)) where phi = (1+sqrt(5))/2, the golden ratio, for n > 0.
a(n) = A130233(n-1) + 1 for n > 0.
G.f.: x/(1-x) * Sum_{k >= 0} x^Fibonacci(k).
a(n) = ceiling(log_phi(sqrt(5)*n - 1)) for n > 0, where phi is the golden ratio. - Hieronymus Fischer, Jul 02 2007
a(n) = A108852(n-1). - R. J. Mathar, Jan 31 2015

A130248 Partial sums of the Lucas Inverse A130247.

Original entry on oeis.org

1, 1, 3, 6, 9, 12, 16, 20, 24, 28, 33, 38, 43, 48, 53, 58, 63, 69, 75, 81, 87, 93, 99, 105, 111, 117, 123, 129, 136, 143, 150, 157, 164, 171, 178, 185, 192, 199, 206, 213, 220, 227, 234, 241, 248, 255, 263, 271, 279, 287, 295, 303, 311, 319, 327, 335, 343, 351
Offset: 1

Views

Author

Hieronymus Fischer, May 19 2007

Keywords

Crossrefs

Other related sequences: A000032, A130241, A130242, A130243, A130244, A130245, A130246, A130251, A130252, A130257, A130261. Fibonacci inverse see A130233 - A130240, A104162.

Programs

  • Mathematica
    Join[{1, 1}, Table[Sum[Floor[Log[GoldenRatio, k + 1/2]], {k, 1, n}], {n, 3, 50}]] (* G. C. Greubel, Dec 24 2017 *)

Formula

a(n)=sum{1<=k<=n, A130247(k)}=2+(n+1)*A130247(n)-A000032(A130247(n)+2) for n>=3. G.f.: g(x)=1/(1-x)^2*(sum{k>=1, x^Lucas(k)}-x^2).

A130249 Maximal index k of a Jacobsthal number such that A001045(k)<=n (the 'lower' Jacobsthal inverse).

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, May 20 2007

Keywords

Comments

Inverse of the Jacobsthal sequence (A001045), nearly, since a(A001045(n))=n except for n=1 (see A130250 for another version). a(n)+1 is equal to the partial sum of the Jacobsthal indicator sequence (see A105348).

Examples

			a(12)=5, since A001045(5)=11<=12, but A001045(6)=21>12.
		

Crossrefs

For partial sums see A130251.
Other related sequences A130250, A130253, A105348, A001045, A130233, A130241.
Cf. A000523, A078008 (runlengths).

Programs

  • Magma
    [Floor(Log(3*n+1)/Log(2)): n in [0..30]]; // G. C. Greubel, Jan 08 2018
    
  • Mathematica
    Table[Floor[Log[2, 3*n + 1]], {n, 0, 50}] (* G. C. Greubel, Jan 08 2018 *)
  • PARI
    for(n=0, 30, print1(floor(log(3*n+1)/log(2)), ", ")) \\ G. C. Greubel, Jan 08 2018
    
  • PARI
    a(n) = logint(3*n+1, 2); \\ Ruud H.G. van Tol, May 12 2024
    
  • Python
    def A130249(n): return (3*n+1).bit_length()-1 # Chai Wah Wu, Jun 08 2022

Formula

a(n) = floor(log_2(3n+1)).
a(n) = A130250(n+1) - 1 = A130253(n) - 1.
G.f.: 1/(1-x)*(Sum_{k>=1} x^A001045(k)).
a(n) = A000523(3*n+1). - Ruud H.G. van Tol, May 12 2024

A104162 Indicator sequence for the Fibonacci numbers.

Original entry on oeis.org

1, 2, 1, 1, 0, 1, 0, 0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1
Offset: 0

Views

Author

Paul Barry, Apr 01 2005

Keywords

Comments

Without multiplicities, this is A010056.
The number of nonnegative integer solutions of x^4 - 10*n^2*x^2 + 25*n^4 - 16 = 0. - Hieronymus Fischer, May 17 2007

Examples

			a(1)=2 since F(1)=F(2)=1.
		

Crossrefs

Cf. A000045.
Partial sums are in A108852.
See also A130233 and A130234.

Programs

Formula

G.f.: Sum_{k>=0} x^Fibonacci(k).
From Hieronymus Fischer, May 17 2007: (Start)
a(n) = 1+floor(arcsinh(sqrt(5)*n/2)/log(phi))-ceiling(arccosh(sqrt(5)*n/2)/log(phi)), for n>0, where phi=(1+sqrt(5))/2.
a(n) = A108852(n) - A108852(n-1) for n>0.
a(n) = A130233(n) - A130233(n-1) for n>0.
a(n) = 1 + A130233(n) - A130234(n) for n>0.
a(n) = A130234(n+1) - A130234(n) for n>=0. (End)

A130245 Number of Lucas numbers (A000032) <= n.

Original entry on oeis.org

0, 1, 2, 3, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10
Offset: 0

Views

Author

Hieronymus Fischer, May 19 2007, Jul 02 2007

Keywords

Comments

Partial sums of the Lucas indicator sequence A102460.
For n>=2, we have a(A000032(n)) = n + 1.

Examples

			a(9)=5 because there are 5 Lucas numbers <=9 (2,1,3,4 and 7).
		

Crossrefs

Partial sums of A102460.
For partial sums of this sequence, see A130246. Other related sequences: A000032, A130241, A130242, A130247, A130249, A130253, A130255, A130259.
For Fibonacci inverse, see A130233 - A130240, A104162, A108852.

Programs

  • Magma
    [0] cat [1+Floor(Log((2*n+1)/2)/Log((1+Sqrt(5))/2)): n in [1..100]]; // G. C. Greubel, Sep 09 2018
    
  • Mathematica
    Join[{0}, Table[1+Floor[Log[GoldenRatio, (2*n+1)/2]], {n,1,100}]] (* G. C. Greubel, Sep 09 2018 *)
  • PARI
    A102460(n) = { my(u1=1,u2=3,old_u1); if(n<=2,sign(n),while(n>u2,old_u1=u1;u1=u2;u2=old_u1+u2);(u2==n)); };
    A130245(n) = if(!n,n,A102460(n)+A130245(n-1));
    \\ Or just as:
    c=0; for(n=0,123,c += A102460(n); print1(c,", ")); \\ Antti Karttunen, May 13 2018
    
  • Python
    from itertools import count, islice
    def A130245_gen(): # generator of terms
        yield from (0, 1, 2)
        a, b = 3,4
        for i in count(3):
            yield from (i,)*(b-a)
            a, b = b, a+b
    A130245_list = list(islice(A130245_gen(),40)) # Chai Wah Wu, Jun 08 2022

Formula

a(n) = 1 +floor(log_phi((n+sqrt(n^2+4))/2)) = 1 +floor(arcsinh(n/2)/log(phi)) for n>=2, where phi = (1+sqrt(5))/2.
a(n) = A130241(n)+1 = A130242(n+1) for n>=2.
G.f.: g(x) = 1/(1-x)*sum{k>=0, x^Lucas(k)}.
a(n) = 1 +floor(log_phi(n+1/2)) for n>=1, where phi is the golden ratio.
Sum_{n>=1} (-1)^(n+1)/a(n) = 3/2 - Pi/(6*sqrt(3)) - log(3)/2. - Amiram Eldar, Jul 25 2025
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