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.

A291481 Number of terms of A293630 at stage n.

Original entry on oeis.org

2, 4, 7, 13, 37, 73, 145, 289, 865, 1729, 3457, 10369, 20737, 41473, 82945, 248833, 497665, 995329, 1990657, 3981313, 11943937, 23887873, 47775745, 143327233, 286654465, 573308929, 1146617857, 3439853569, 6879707137, 20639121409, 41278242817, 82556485633
Offset: 0

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Author

Benoit Cloitre, Oct 15 2017

Keywords

Comments

limsup a(n+1)/a(n) = 3, liminf a(n+1)/a(n) = 2 (n->oo). It seems that lim_{n->oo} a(n)^(1/n) = C with C > 2.
Limit_{n->oo} a(n)^(1/n) = 2.236151... (see A297890). - Jon E. Schoenfield, Dec 23 2017
The previous limit is also equal to 2^(2 - d) * 3^(d - 1), where d = 1.275261... (see A296564). - Iain Fox, Dec 24 2017

Examples

			A293630 at stage n:
  n = 0: [1, 2];                                      2 terms
  n = 1: [1, 2, 1, 1];                                4 terms
  n = 2: [1, 2, 1, 1, 1, 2, 1];                       7 terms
  n = 3: [1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, 2];    13 terms
  n = 4: [1, 2, 1, 1, 1, 2, 1, 1, 2, 1, 1, 1, ...];  37 terms
   ...
		

Crossrefs

Programs

  • Mathematica
    Length /@ NestList[Join[#, Join @@ ConstantArray[Most[#], Last[#]]] &, {1, 2}, 24] (* Michael De Vlieger, Jan 21 2018 *)
  • PARI
    v=[1,2];for(n=1,17,l=length(v);w=vector(l-1,i,v[i]);v=concat(v,if(v[l]-1,concat(w,w),w));print1(length(v),","));
    
  • PARI
    lista(nn) = {
    my(S = [1, 2], n = 2, L, nPrev, E);
    print1("2, ");
    for(j = 1, nn, L = S[#S]; n = n*(1+L)-L; nPrev = #S; for(r = 1, L, for(i = 1, nPrev-1, S = concat(S, S[i]))); print1(n, ", "));
    E = S;
    for(j = nn + 1, nn + #E, L = E[#E+1-(j-nn)]; n = n*(1+L)-L; print1(n, ", "))
    } \\ Iain Fox, Jan 21 2018
    
  • Python
    a, z = [1, 2], [2]
    while z[-1]<1000:
        a += a[:-1]*a[-1]
        z.append(len(a))
    for i in range(100):
        z.append((z[-1]-1)*(a.pop()+1)+1)
    print(z)
    # Andrey Zabolotskiy, Oct 15 2017

Formula

From Iain Fox, Jan 21 2018: (Start)
a(n) = (1 + A293630(a(n-1)))*a(n-1) - A293630(a(n-1)).
a(n) ~ c^n, where c = 2.236151... (see comments or A297890).
(End)

Extensions

More terms from Andrey Zabolotskiy, Oct 15 2017