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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A280523 a(n) = Fibonacci(2n + 1) - n.

Original entry on oeis.org

1, 3, 10, 30, 84, 227, 603, 1589, 4172, 10936, 28646, 75013, 196405, 514215, 1346254, 3524562, 9227448, 24157799, 63245967, 165580121, 433494416, 1134903148, 2971215050, 7778742025, 20365011049, 53316291147, 139583862418
Offset: 1

Views

Author

Peter Kagey, Jan 04 2017

Keywords

Comments

Old (and equivalent) definition: these are the indices of records in the Fibonachos sequence A280521: the least k such that A280521(k) = n.
Conjecture: a(n) = A215004(2n - 2). - Peter Kagey. The conjecture is true. See link for analysis of connections with A215004 and A280521. - Nathan Fox, Jan 27 2017
Define the n-th Fibonacci compositions CF(n) by CF(1)={(1)}, CF(2)={(2)}, and CF(n) is 1 adjoined at the end of each composition in CF(n-1) union 2 adjoined to the end of each composition in CF(n-2). The union is disjoint, so |CF(n)| is the n-th Fibonacci number. Define the weight of a composition c by 2^(number of singletons in c). For example, 2122 has 1 singleton and weight 2. Let s(n) be the sum of the weights of CF(n). Conjecture: a(n)= s(2n+4)-s(2n+3). - George Beck, Jan 31 2020

Examples

			a(1) = 1    because A280521(1)    = 1;
a(2) = 3    because A280521(3)    = 2;
a(3) = 10   because A280521(10)   = 3;
a(4) = 30   because A280521(30)   = 4;
a(5) = 84   because A280521(84)   = 5;
a(6) = 227  because A280521(227)  = 6;
a(7) = 603  because A280521(603)  = 7;
a(8) = 1589 because A280521(1589) = 8.

Links

Crossrefs

Cf. A001519, A215004, A280521.

Programs

  • Mathematica
    Table[Fibonacci[2n + 1] - n, {n, 30}] (* Alonso del Arte, Jan 29 2017 *)
LinearRecurrence[{5,-8,5,-1},{1,3,10,30},30] (* Harvey P. Dale, Feb 06 2024 *)
  • PARI
    F=vector(64,n,fibonacci(n+2)-1); \\ Resize as needed
A280521(n)=my(s); while(n, s++; t=setsearch(F,n,1); if(t, n-=F[t-1], return(s))); s
first(n)=my(v=vector(n),k,t,mn=1,gaps=n); while(gaps, t=A280521(k++); if(t>=mn && t<=n && v[t]==0, v[t]=k; while(mn<=n && v[mn], mn++); print("a("t") = "k); gaps--)); v \\ Charles R Greathouse IV, Jan 04 2017

Formula

G.f.: -x*(-1+2*x-3*x^2+x^3) / ( (x^2-3*x+1)*(x-1)^2 ). - R. J. Mathar, Mar 11 2017
a(n) = 5*a(n-1)-8*a(n-2)+5*a(n-3)-a(n-4). - Wesley Ivan Hurt, Apr 26 2021

Extensions

Corrected and extended by Charles R Greathouse IV, Jan 04 2017
a(26) from Charles R Greathouse IV, Jan 09 2017
a(27) from Charles R Greathouse IV, Jan 16 2017
Replaced old definition by simple formula established by Nathan Fox. - N. J. A. Sloane, Jan 30 2017

A281362 a(0) = 1, a(1) = 2; for n>1, a(n) = a(n-1) + a(n-2) + floor(n/2).

Original entry on oeis.org

1, 2, 4, 7, 13, 22, 38, 63, 105, 172, 282, 459, 747, 1212, 1966, 3185, 5159, 8352, 13520, 21881, 35411, 57302, 92724, 150037, 242773, 392822, 635608, 1028443, 1664065, 2692522, 4356602, 7049139, 11405757, 18454912, 29860686, 48315615, 78176319, 126491952
Offset: 0

Views

Author

N. J. A. Sloane, Jan 27 2017, based on a comment in A215004 from Alex Ratushnyak, Jul 31 2012

Keywords

Examples

			G.f. = 1 + 2*x + 4*x^2 + 7*x^3 + 13*x^4 + 22*x^5 + 38*x^6 + 63*x^7 + ...

Links

Crossrefs

Cf. A000045, A215004, A215005.

Programs

  • Magma
    R:=PowerSeriesRing(Integers(), 50); Coefficients(R!((1 - x^2+x^4)/((1-x)^2*(1+x)*(1-x-x^2)))); // G. C. Greubel, Jul 29 2018
  • Mathematica
    CoefficientList[Series[(1-x^2+x^4)/((1-x)^2*(1+x)*(1-x-x^2)), {x,0,50}], x] (* G. C. Greubel, Jul 29 2018 *)
  • PARI
    Vec((1-x^2+x^4)/((1-x)^2*(1+x)*(1-x-x^2)) + O(x^50))
  • PARI
    {a(n) = my(m=n); if( n<0, m=-1-n); polcoeff( (1 - x^2 + x^4) / ((1 - x - x^2 + x^3) * (1 - (-1)^(n<0)*x - x^2)) + x * O(x^m), m)}; /* Michael Somos, Jan 27 2017 */
  • SageMath
    [2*fibonacci(n+2) -(n+2+(n%2))//2 for n in range(51)] # G. C. Greubel, Apr 05 2024

Formula

From Colin Barker, Jan 27 2017: (Start)
a(n) = 2*a(n-1) + a(n-2) - 3*a(n-3) + a(n-5) for n>4.
G.f.: (1 - x^2 + x^4) / ((1 - x)^2*(1 + x)*(1 - x - x^2)). (End)
0 = 3 + a(n)*(-4 - 2*a(n) - a(n+1) + 6*a(n+2) - a(n+3)) + a(n+1)*(-1 + a(n+1) + a(n+2) - 2*a(n+3)) + a(n+2)*(+11 - 5*a(n+2) + 3*a(n+3)) + a(n+3)*(-6) for all n in Z. - Michael Somos, Jan 27 2017
From G. C. Greubel, Apr 05 2024: (Start)
a(n) = 2*Fibonacci(n+2) - (1/4)*(2*n + 5 - (-1)^n).
E.g.f.: (1/2)*(4*exp(x/2)*(cosh(sqrt(5)*x/2) + (3/sqrt(5))*sinh( sqrt(5)*x/2)) - (x+2)*cosh(x) - (x+3)*sinh(x)). (End)
Showing 1-2 of 2 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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