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.

A346296 a(0) = 1; thereafter a(n) = 2*a(n-1) + 1, with digits rearranged into nondecreasing order.

Original entry on oeis.org

1, 3, 7, 15, 13, 27, 55, 111, 223, 447, 589, 1179, 2359, 1479, 2599, 1599, 1399, 2799, 5599, 11199, 22399, 44799, 58999, 117999, 235999, 147999, 259999, 159999, 139999, 279999, 559999, 1119999, 2239999, 4479999, 5899999, 11799999, 23599999, 14799999, 25999999
Offset: 0

Views

Author

Ctibor O. Zizka, Jul 13 2021

Keywords

Examples

			a(3) = A004185(2*7+1) = A004185(15) = 15.
a(4) = A004185(2*15+1) = A004185(31) = 13.
		

Crossrefs

Programs

  • Mathematica
    a[0] = 1; a[n_] := a[n] = FromDigits @ Sort @ IntegerDigits[2*a[n - 1] + 1]; Array[a, 45, 0] (* Amiram Eldar, Jul 13 2021 *)
    NestList[FromDigits[Sort[IntegerDigits[2#+1]]]&,1,40] (* Harvey P. Dale, Oct 01 2023 *)
  • PARI
    lista(nn) = {my(va = vector(nn)); va[1] = 1; for (n=2, nn, va[n] = fromdigits(vecsort(digits(2*va[n-1]+1)));); va;} \\ Michel Marcus, Aug 31 2021
    
  • Python
    from itertools import accumulate
    def atis(anm1, _): return int("".join(sorted(str(2*anm1+1))))
    print(list(accumulate([1]*39, atis))) # Michael S. Branicky, Aug 31 2021

Formula

a(n) = A004185(2*a(n-1)+1).
For k >= 1;
a(12*k-9) = 100^(k-1) * 16 - 1;
a(12*k-8) = 100^(k-1) * 14 - 1;
a(12*k-7) = 100^(k-1) * 28 - 1;
a(12*k-6) = 100^(k-1) * 56 - 1;
a(12*k-5) = 100^(k-1) * 112 - 1;
a(12*k-4) = 100^(k-1) * 224 - 1;
a(12*k-3) = 100^(k-1) * 448 - 1;
a(12*k-2) = 100^(k-1) * 590 - 1;
a(12*k-1) = 100^(k-1) * 1180 - 1;
a(12*k) = 100^(k-1) * 2360 - 1;
a(12*k+1) = 100^(k-1) * 1480 - 1;
a(12*k+2) = 100^(k-1) * 2600 - 1.
G.f.: -(1800*x^15 -720*x^14 +1080*x^13 -1080*x^12 -590*x^11 -142*x^10 -224*x^9 -112*x^8 -56*x^7 -28*x^6 -14*x^5 +2*x^4 -8*x^3 -4*x^2 -2*x -1) / ((x-1)*(10*x^6-1)*(10*x^6+1)). - Alois P. Heinz, Aug 02 2021
a(n) = 100*a(n-12) + 99 for n >= 15. - Pontus von Brömssen, Sep 01 2021