A346296 a(0) = 1; thereafter a(n) = 2*a(n-1) + 1, with digits rearranged into nondecreasing order.
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
Examples
a(3) = A004185(2*7+1) = A004185(15) = 15. a(4) = A004185(2*15+1) = A004185(31) = 13.
Links
- Harvey P. Dale, Table of n, a(n) for n = 0..1000
- Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,100,-100).
Programs
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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 *)
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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
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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