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 1-5 of 5 results.

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).

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.

A102460 a(n) = 1 if n is a Lucas number, else a(n) = 0.

Original entry on oeis.org

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

Views

Author

Casey Mongoven, Apr 18 2005

Keywords

Comments

From Hieronymus Fischer, Jul 02 2007: (Start)
a(n) is the number of nonnegative integer solutions to 25*x^4-10*n^2*x^2+n^4-16=0.
a(n)=1 if and only if there is an integer m such that x=n is a root of p(x)=x^4-10*m^2*x^2+25*m^4-16.
For n>=3: a(n)=1 iff floor(log_phi(n+1/2))=ceiling(log_phi(n-1/2)). (End)

Crossrefs

Programs

  • Mathematica
    {0}~Join~ReplacePart[ConstantArray[0, Last@ #], Map[# -> 1 &, #]] &@ Array[LucasL, 11, 0] (* Michael De Vlieger, Nov 22 2017 *)
    With[{nn=130,lc=LucasL[Range[0,20]]},Table[If[MemberQ[lc,n],1,0],{n,0,nn}]] (* Harvey P. Dale, Jul 03 2022 *)
  • PARI
    a(n)=my(f=factor(25*'x^4-10*n^2*'x^2+n^4-16)[,1]); sum(i=1,#f, poldegree(f[i])==1 && polcoeff(f[i],0)<=0) \\ Charles R Greathouse IV, Nov 06 2014
    
  • 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)); }; \\ Antti Karttunen, Nov 22 2017
    
  • Python
    from sympy.ntheory.primetest import is_square
    def A102460(n): return int(is_square(m:=5*(n**2-4)) or is_square(m+40)) # Chai Wah Wu, Jun 13 2024

Formula

From Hieronymus Fischer, Jul 02 2007: (Start)
G.f.: g(x) = Sum_{k>=0} x^A000032(k).
a(n) = 1+floor(arcsinh(n/2)/log(phi))-ceiling(arccosh(n/2)/log(phi)) for n>=3, where phi=(1+sqrt(5))/2.
a(n) = 1+A130241(n)-A130242(n) for n>=3.
a(n) = 1+A130247(n)-A130242(n) for n=>2.
a(n) = A130245(n)-A130245(n-1) for n>=1.
For n>=3: a(n)=1 iff A130241(n)=A130242(n). (End)

Extensions

Data section extended up to a(123) by Antti Karttunen, Nov 22 2017

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

A130242 Minimal index k of a Lucas number such that Lucas(k)>=n (the 'upper' Lucas (A000032) Inverse).

Original entry on oeis.org

0, 0, 0, 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

Inverse of the Lucas sequence (A000032), nearly, since a(Lucas(n))=n except for n=1 (see A130241 and A130247 for other versions). For n>=2, a(n+1) is equal to the partial sum of the Lucas indicator sequence (see A102460).

Examples

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

Crossrefs

For partial sums see A130244.
Other related sequences: A000032, A130241, A130245, A130247, A130250, A130256, A130260.
Indicator sequence A102460.
Fibonacci inverse see A130233 - A130240, A104162.

Programs

  • Mathematica
    Join[{0, 0, 0}, Table[Ceiling[Log[GoldenRatio, n + 1/2]], {n, 2, 50}]] (* G. C. Greubel, Dec 24 2017 *)
  • Python
    from itertools import islice, count
    def A130242_gen(): # generator of terms
        yield from (0,0,0,2)
        a, b = 3, 4
        for i in count(3):
            yield from (i,)*(b-a)
            a, b = b, a+b
    A130242_list = list(islice(A130242_gen(),40)) # Chai Wah Wu, Jun 08 2022

Formula

a(n) = ceiling(log_phi((n+sqrt(n^2-4))/2))=ceiling(arccosh(n/2)/log(phi)) where phi=(1+sqrt(5))/2.
a(n) = A130241(n-1) + 1 = A130245(n-1) for n>=3.
G.f.: x/(1-x)*(2x^2+sum{k>=2, x^Lucas(k)}).
a(n) = ceiling(log_phi(n-1/2)) for n>=3, where phi is the golden ratio.
Showing 1-5 of 5 results.