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.

Previous Showing 11-20 of 24 results. Next

A130246 Partial sums of A130245.

Original entry on oeis.org

0, 1, 3, 6, 10, 14, 18, 23, 28, 33, 38, 44, 50, 56, 62, 68, 74, 80, 87, 94, 101, 108, 115, 122, 129, 136, 143, 150, 157, 165, 173, 181, 189, 197, 205, 213, 221, 229, 237, 245, 253, 261, 269, 277, 285, 293, 301, 310, 319, 328, 337, 346, 355, 364, 373, 382, 391
Offset: 0

Views

Author

Hieronymus Fischer, May 19 2007

Keywords

Crossrefs

Other related sequences: A000032, A130241, A130243, A130244, A130248, A130251, A130252, A130255, A130257, A130261. Fibonacci inverse see A130233 - A130240, A104162.

Programs

  • Magma
    [0] cat [(&+[1+Floor(Log((2*k+1)/2)/Log((1+Sqrt(5))/2)): k in [1..n]]): n in [1..100]]; // G. C. Greubel, Sep 09 2018
  • Mathematica
    Table[Sum[1 + Floor[Log[GoldenRatio, (2*k + 1)/2]], {k, 1, n}], {n, 0, 100}] (* G. C. Greubel, Sep 09 2018 *)
  • PARI
    for(n=0, 100, print1(sum(k=1,n, 1 + floor(log((2*k+1)/2)/log((1+sqrt(5))/2))), ", ")) \\ G. C. Greubel, Sep 09 2018
    

Formula

a(n) = Sum_{k=1..n} A130245(k).
a(n) = 1 +(n+1)*A130245(n) - A000032(A130245(n)+1) for n=0 or n >= 2.
G.f.: 1/(1-x)^2*Sum_{k>=0} x^A000032(k).

A130251 Partial sums of A130249.

Original entry on oeis.org

0, 2, 4, 7, 10, 14, 18, 22, 26, 30, 34, 39, 44, 49, 54, 59, 64, 69, 74, 79, 84, 90, 96, 102, 108, 114, 120, 126, 132, 138, 144, 150, 156, 162, 168, 174, 180, 186, 192, 198, 204, 210, 216, 223, 230, 237, 244, 251, 258, 265, 272, 279, 286, 293, 300, 307, 314, 321
Offset: 0

Views

Author

Hieronymus Fischer, May 20 2007

Keywords

Examples

			G.f. = 2*x + 4*x^2 + 7*x^3 + 10*x^4 + 14*x^5 + 18*x^6 + 22*x^7 + ... - _Michael Somos_, Sep 17 2018
		

Crossrefs

Programs

  • Magma
    [0] cat [(&+[Floor(Log(3*k+1)/Log(2)) : k in [1..n]]): n in [1..100]]; // G. C. Greubel, Sep 09 2018
    
  • Mathematica
    Join[{0}, Table[Sum[Floor[Log[2, 3*k + 1]], {k, 1, n}], {n, 1, 2500}]] (* G. C. Greubel, Sep 09 2018 *)
  • PARI
    for(n=0,100, print1(sum(k=1,n, floor(log(3*k+1)/log(2))), ", ")) \\ G. C. Greubel, Sep 09 2018
    
  • Python
    def A130251(n): return (n+1)*((m:=(3*n+1).bit_length())-1)-(((1<>1) # Chai Wah Wu, Apr 17 2025

Formula

a(n) = Sum_{k=0..n} A130249(k).
a(n) = (n+1)*floor(log_2(3*n+1)) - (1/2)*(A001045(floor(log_2(3*n+1))+2)-1).
G.f.: (1/(1-x)^2)*Sum_{k>=1} x^A001045(k).

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.

A130259 Maximal index k of an even Fibonacci number (A001906) such that A001906(k) = Fib(2k) <= n (the 'lower' even Fibonacci Inverse).

Original entry on oeis.org

0, 1, 1, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5
Offset: 0

Views

Author

Hieronymus Fischer, May 25 2007, Jul 02 2007

Keywords

Comments

Inverse of the even Fibonacci sequence (A001906), since a(A001906(n))=n (see A130260 for another version).
a(n)+1 is the number of even Fibonacci numbers (A001906) <=n.

Examples

			a(10)=3 because A001906(3)=8<=10, but A001906(4)=21>10.
		

Crossrefs

Cf. partial sums A130261. Other related sequences: A000045, A001519, A130233, A130237, A130239, A130255, A130260, A104160. Lucas inverse: A130241 - A130248.

Programs

  • Magma
    [Floor(Log((Sqrt(5)*n+1))/(2*Log((1+Sqrt(5))/2))): n in [0..100]]; // G. C. Greubel, Sep 12 2018
  • Mathematica
    Table[Floor[1/2*Log[GoldenRatio, (Sqrt[5]*n + 1)]], {n, 0, 100}] (* G. C. Greubel, Sep 12 2018 *)
  • PARI
    vector(100, n, n--; floor(log((sqrt(5)*n+1))/(2*log((1+sqrt(5))/2) ))) \\ G. C. Greubel, Sep 12 2018
    

Formula

a(n) = floor(arcsinh(sqrt(5)*n/2)/(2*log(phi))), where phi=(1+sqrt(5))/2.
a(n) = A130260(n+1) - 1.
G.f.: g(x) = 1/(1-x)*Sum_{k>=1} x^Fibonacci(2*k).
a(n) = floor(1/2*log_phi(sqrt(5)*n+1)) for n>=0.

A130261 Partial sums of the 'lower' even Fibonacci Inverse A130259.

Original entry on oeis.org

0, 1, 2, 4, 6, 8, 10, 12, 15, 18, 21, 24, 27, 30, 33, 36, 39, 42, 45, 48, 51, 55, 59, 63, 67, 71, 75, 79, 83, 87, 91, 95, 99, 103, 107, 111, 115, 119, 123, 127, 131, 135, 139, 143, 147, 151, 155, 159, 163, 167, 171, 175, 179, 183, 187, 192, 197, 202, 207, 212, 217
Offset: 0

Views

Author

Hieronymus Fischer, May 25 2007

Keywords

Crossrefs

Programs

  • Magma
    [(&+[Floor(Log((Sqrt(5)*k+1))/(2*Log((1+Sqrt(5))/2))): k in [0..n]]): n in [0..50]]; // G. C. Greubel, Sep 12 2018
  • Mathematica
    Table[Sum[Floor[1/2*Log[GoldenRatio, (Sqrt[5]*k + 1)]], {k, 0, n}], {n, 0, 50}] (* G. C. Greubel, Sep 12 2018 *)
  • PARI
    for(n=0,50, print1(sum(k=0,n, floor(log((sqrt(5)*k+1))/(2*log((1 +sqrt(5))/2)))), ", ")) \\ G. C. Greubel, Sep 12 2018
    

Formula

a(n) = (n+1)*A130259(n) - A001519(A130259(n)+1) + 1.
a(n) = (n+1)*A130259(n) - Fib(2*A130259(n)+1) + 1.
G.f.: g(x) = 1/(1-x)^2*Sum_{k>=1} x^Fib(2*k).

A130244 Partial sums of the 'upper' Lucas Inverse A130242.

Original entry on oeis.org

0, 0, 0, 2, 5, 9, 13, 17, 22, 27, 32, 37, 43, 49, 55, 61, 67, 73, 79, 86, 93, 100, 107, 114, 121, 128, 135, 142, 149, 156, 164, 172, 180, 188, 196, 204, 212, 220, 228, 236, 244, 252, 260, 268, 276, 284, 292, 300, 309, 318, 327, 336, 345, 354, 363, 372, 381, 390
Offset: 0

Views

Author

Hieronymus Fischer, May 19 2007

Keywords

Crossrefs

Other related sequences: A000032, A130241, A130243, A130245, A130246, A130248, A130252, A130258, A130262. Fibonacci inverse see A130233 - A130240, A104162.

Programs

  • Magma
    [0,0] cat [(&+[Ceiling(Log(k + 1/2)/Log((1+Sqrt(5))/2)) : k in [0..n]]): n in [1..50]]; // G. C. Greubel, Sep 12 2018
  • Mathematica
    Join[{0, 0}, Table[Sum[Ceiling[Log[GoldenRatio, k + 1/2]], {k, 0, n}], {n, 1, 50}]] (* G. C. Greubel, Sep 12 2018 *)
  • PARI
    for(n=-1,50, print1(if(n==-1, 0, if(n==0, 0, sum(k=0, n, ceil(log(k + 1/2)/log((1+sqrt(5))/2))))), ", ")) \\ G. C. Greubel, Sep 12 2018
    

Formula

a(n) = Sum_{k=0..n} A130242(k).
a(n) = n*A130242(n) - A000032(A130242(n) +1) for n>=3.
G.f.: x/(1-x)^2*(2*x^2 + Sum{k>=2, x^Lucas(k)}).

A130256 Minimal index k of an odd Fibonacci number A001519 such that A001519(k) = Fibonacci(2*k-1) >= n (the 'upper' odd Fibonacci Inverse).

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, May 24 2007, Jul 02 2007

Keywords

Comments

Inverse of the odd Fibonacci sequence (A001519), nearly, since a(A001519(n))=n except for n=1 (see A130255 for another version).
a(n+1) is the number of odd Fibonacci numbers (A001519) <= n (for n >= 0).

Examples

			a(10)=4 because A001519(4) = 13 >= 10, but A001519(3) = 5 < 10.
		

Crossrefs

Cf. partial sums A130258.
Other related sequences: A000045, A001906, A130234, A130237, A130239, A130255, A130260.
Lucas inverse: A130241 - A130248.

Programs

  • Magma
    [0,0] cat [Ceiling((1/2)*(1 + Log(Sqrt(5)*n-1)/(Log((1+Sqrt(5))/2)))): n in [2..100]]; // G. C. Greubel, Sep 12 2018
  • Mathematica
    Join[{0, 0}, Table[Ceiling[1/2*(1 + Log[GoldenRatio, (Sqrt[5]*n - 1)])], {n, 2, 100}]] (* G. C. Greubel, Sep 12 2018 *)
  • PARI
    for(n=0,100, print1(if(n==0, 0, if(n==1, 0, ceil((1/2)*(1 + log(sqrt(5)*n-1)/(log((1+sqrt(5))/2)))))), ", ")) \\ G. C. Greubel, Sep 12 2018
    

Formula

a(n) = ceiling((1+arccosh(sqrt(5)*n/2)/log(phi))/2), where phi=(1+sqrt(5))/2.
G.f.: (x/(1-x))*Sum_{k>=0} x^Fibonacci(2*k-1).
a(n) = ceiling((1/2)*(1+log_phi(sqrt(5)*n-1))) for n >= 2, where phi=(1+sqrt(5))/2.

A304333 Number of positive integers k such that n - L(k) is a positive squarefree number, where L(k) denotes the k-th Lucas number A000204(k).

Original entry on oeis.org

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

Views

Author

Zhi-Wei Sun, May 11 2018

Keywords

Comments

Conjecture: a(n) > 0 for all n > 1.
This has been verified for n up to 5*10^9.
See also A304331 for a similar conjecture involving Fibonacci numbers.
For all n, a(n) <= A130241(n). - Antti Karttunen, May 13 2018

Examples

			a(2) = 1 with 2 - L(1) = 1 squarefree.
a(3) = 1 with 3 - L(1) = 2 squarefree.
a(67) = 2 with 67 - L(1) = 2*3*11 and 67 - L(7) = 2*19 both squarefree.
		

Crossrefs

Programs

  • Maple
    a := proc(n) local count, lucas, newcas;
    count := 0; lucas := 1; newcas := 2;
    while lucas < n do
        if numtheory:-issqrfree(n - lucas) then count := count + 1 fi;
        lucas, newcas := lucas + newcas, lucas;
    od;
    count end:
    seq(a(n), n=1..90); # Peter Luschny, May 15 2018
  • Mathematica
    f[n_]:=f[n]=LucasL[n];
    tab={};Do[r=0;k=1;Label[bb];If[f[k]>=n,Goto[aa]];If[SquareFreeQ[n-f[k]],r=r+1];k=k+1;Goto[bb];Label[aa];tab=Append[tab,r],{n,1,90}];Print[tab]
  • PARI
    A304333(n) = { my(u1=1,u2=3,old_u1,c=0); if(n<=2,n-1,while(u1Antti Karttunen, May 13 2018

A130260 Minimal index k of an even Fibonacci number A001906 such that A001906(k) = Fib(2k) >= n (the 'upper' even Fibonacci Inverse).

Original entry on oeis.org

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

Views

Author

Hieronymus Fischer, May 25 2007, May 28 2007, Jul 02 2007

Keywords

Comments

Inverse of the even Fibonacci sequence (A001906), since a(A001906(n))=n (see A130259 for another version).
a(n+1) is the number of even Fibonacci numbers (A001906) <=n.

Examples

			a(10)=4 because A001906(4)=21>=10, but A001906(3)=8<10.
		

Crossrefs

Cf. partial sums A130262. Other related sequences: A000045, A001519, A130234, A130237, A130239, A130256, A130259. Lucas inverse: A130241 - A130248.

Programs

  • Magma
    [0] cat [Ceiling(Log(Sqrt(5)*n)/(2*Log((1+ Sqrt(5))/2))): n in [1..100]]; // G. C. Greubel, Sep 12 2018
  • Mathematica
    Join[{0}, Table[Ceiling[Log[GoldenRatio, Sqrt[5]*n]/2], {n, 1, 100}]] (* G. C. Greubel, Sep 12 2018 *)
  • PARI
    for(n=0,100, print1(if(n==0, 0, ceil(log(sqrt(5)*n)/(2*log((1+ sqrt(5))/2)))), ", ")) \\ G. C. Greubel, Sep 12 2018
    

Formula

a(n) = ceiling(arcsinh(sqrt(5)*n/2)/(2*log(phi))) for n>=0.
a(n) = ceiling(arccosh(sqrt(5)*n/2)/(2*log(phi))) for n>=1.
a(n) = ceiling(log_phi(sqrt(5)*n)/2)=ceiling(log_phi(sqrt(5)*n-1)/2) for n>=1, where phi=(1+sqrt(5))/2.
a(n) = A130259(n-1) + 1, for n>=1.
G.f.: g(x)=x/(1-x)*Sum_{k>=0} x^Fib(2*k).

A132664 a(1)=1, a(2)=2, a(n) = a(n-1) + n if the minimal positive integer not yet in the sequence is greater than a(n-1), else a(n) = a(n-1) - 1.

Original entry on oeis.org

1, 2, 5, 4, 3, 9, 8, 7, 6, 16, 15, 14, 13, 12, 11, 10, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 45, 44, 43, 42, 41, 40, 39, 38, 37, 36, 35, 34, 33, 32, 31, 30, 29, 28, 74, 73, 72, 71, 70, 69, 68, 67, 66, 65, 64, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48
Offset: 1

Views

Author

Hieronymus Fischer, Sep 15 2007

Keywords

Comments

Also: a(1)=1, a(2)=2, a(n) = maximal positive number < a(n-1) not yet in the sequence, if it exists, else a(n) = a(n-1) + n.
Also: a(1)=1, a(2)=2, a(n) = a(n-1) - 1, if a(n-1) - 1 > 0 and has not been encountered so far, else a(n) = a(n-1) + n.
A permutation of the positive integers. The sequence is self-inverse, in that a(a(n)) = n.

Crossrefs

For an analog concerning Fibonacci numbers see A132665.
See A132666-A132674 for sequences with a similar recurrence rule.

Formula

G.f.: g(x) = (L'(x) - x^2 - 1/(1-x))/(1-x) where L(x) = Sum_{k>=0} x^Lucas(k) and Lucas(k) = A000032(k). L(x) is the g.f. of the Lucas indicator sequence (see A102460) and L'(x) = derivative of L(x).
a(n) = Lucas(Lucas_inverse(n+1)+2) - n - 3 = A000032(A130241(n+1) + 2) - n - 3 for n > 1.
a(n) = A000032(floor(log_phi(n + 3/2)) + 2) - n - 3 for n > 1, where phi = (1 + sqrt(5))/2 is the golden ratio.
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