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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A321539 3^n with digits rearranged into nonincreasing order.

Original entry on oeis.org

1, 3, 9, 72, 81, 432, 972, 8721, 6651, 98631, 99540, 777411, 544311, 9543321, 9987642, 98744310, 76443210, 964321110, 988744320, 7666422111, 8876444310, 65433321000, 99865331100, 98877443211, 988654432221, 988876444320, 9888655432221
Offset: 0

Views

Author

N. J. A. Sloane, Nov 19 2018

Keywords

Crossrefs

The following are parallel families: A000079 (2^n), A004094 (2^n reversed), A028909 (2^n sorted up), A028910 (2^n sorted down), A036447 (double and reverse), A057615 (double and sort up), A263451 (double and sort down); A000244 (3^n), A004167 (3^n reversed), A321540 (3^n sorted up), A321539 (3^n sorted down), A163632 (triple and reverse), A321542 (triple and sort up), A321541 (triple and sort down).
Cf. A004186.

Programs

  • Mathematica
    A321539[n_]:=FromDigits[ReverseSort[IntegerDigits[3^n]]];Array[A321539,40,0] (* Paolo Xausa, Aug 10 2023 *)
  • Python
    def A321539(n): return int(''.join(sorted(str(3**n),reverse=True))) # Chai Wah Wu, Nov 10 2022

Formula

a(n) = A004186(A000244(n)). - Michel Marcus, Nov 10 2022

A321540 3^n with digits rearranged into nondecreasing order.

Original entry on oeis.org

1, 3, 9, 27, 18, 234, 279, 1278, 1566, 13689, 4599, 114777, 113445, 1233459, 2467899, 1344789, 1234467, 11123469, 23447889, 1112246667, 134446788, 12333456, 113356899, 11234477889, 122234456889, 23444678889, 1222345568889, 2445567778899
Offset: 0

Views

Author

N. J. A. Sloane, Nov 19 2018

Keywords

Crossrefs

The following are parallel families: A000079 (2^n), A004094 (2^n reversed), A028909 (2^n sorted up), A028910 (2^n sorted down), A036447 (double and reverse), A057615 (double and sort up), A263451 (double and sort down); A000244 (3^n), A004167 (3^n reversed), A321540 (3^n sorted up), A321539 (3^n sorted down), A163632 (triple and reverse), A321542 (triple and sort up), A321541 (triple and sort down).
Cf. A004185.

Programs

  • Magma
    [Seqint(Reverse(Sort(Intseq(3^n)))):n in [0..35]]; // Vincenzo Librandi, Jan 22 2020
    
  • Mathematica
    Table[FromDigits[Sort[IntegerDigits[3^n]]], {n, 0, 40}] (* Vincenzo Librandi, Jan 22 2020 *)
  • Python
    def A321540(n): return int(''.join(sorted(str(3**n)))) # Chai Wah Wu, Nov 10 2022

Formula

a(n) = A004185(A000244(n)). - Michel Marcus, Nov 10 2022

A321541 a(0)=1; thereafter a(n) = 3*a(n-1) with digits rearranged into nonincreasing order.

Original entry on oeis.org

1, 3, 9, 72, 621, 8631, 98532, 996552, 9986652, 99996552, 999986652, 9999996552, 99999986652, 999999996552, 9999999986652, 99999999996552, 999999999986652, 9999999999996552, 99999999999986652, 999999999999996552, 9999999999999986652, 99999999999999996552, 999999999999999986652
Offset: 0

Views

Author

N. J. A. Sloane, Nov 19 2018

Keywords

Comments

In contrast to A321542, this sequence increases forever.
Proof: The terms from a(7) onwards can be described as follows:
3 times the number 9 (2k times) 6552 is 2 9 (2k-1 times) 89656 which becomes 9 (2k times) 86652 when sorted;
then 3 times the number 9 (2k times) 86652 is 2 9 (2k times) 59956 which becomes 9 (2k+2 times) 6552 when sorted. QED

Crossrefs

The following are parallel families: A000079 (2^n), A004094 (2^n reversed), A028909 (2^n sorted up), A028910 (2^n sorted down), A036447 (double and reverse), A057615 (double and sort up), A263451 (double and sort down); A000244 (3^n), A004167 (3^n reversed), A321540 (3^n sorted up), A321539 (3^n sorted down), A163632 (triple and reverse), A321542 (triple and sort up), A321541 (triple and sort down).

Programs

  • Mathematica
    NestList[FromDigits[ReverseSort[IntegerDigits[3*#]]] &, 1, 25] (* Paolo Xausa, Aug 02 2024 *)

Formula

From Chai Wah Wu, Nov 20 2018: (Start)
a(n) = 10*a(n-1) + a(n-2) - 10*a(n-3) for n > 9.
G.f.: (118800*x^9 + 8910*x^8 + 8811*x^7 + 12321*x^6 + 2439*x^5 - 78*x^4 - 11*x^3 - 22*x^2 - 7*x + 1)/((x - 1)*(x + 1)*(10*x - 1)). (End)

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
Previous Showing 11-14 of 14 results.