A004094 -id:A004094 - cp's OEIS frontend

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

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

N. J. A. Sloane, Nov 19 2018

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

Paolo Xausa, Table of n, a(n) for n = 0..900
Index entries for linear recurrences with constant coefficients, signature (10,1,-10).

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).

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

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)

A071586 Powers of 8 written backwards.

Original entry on oeis.org

1, 8, 46, 215, 6904, 86723, 441262, 2517902, 61277761, 827712431, 4281473701, 2954399858, 63767491786, 888318557945, 4011156408934, 23888027348153, 656017679474182, 8425863189971522, 48918490589341081, 278558570881511441

Offset: 0

Benoit Cloitre, Jun 01 2002

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A004094, A004167, A071582, A071583, A071588, A071584, A071587.

for(i=1,50,n=8^i; s=ceil(log(n)/log(10)); print1(sum(i=0,s,10^(s-i-1)*(floor(n/10^i)-10*floor(n/10^(i+1)))),","))

A158625 Lower limit of backward value of 5^n.

Original entry on oeis.org

5, 2, 1, 3, 0, 2, 3, 3, 0, 4, 3, 1, 1, 3, 1, 1, 2, 4, 2, 1, 0, 3, 1, 3, 3, 0, 0, 0, 2, 3, 1, 4, 1, 0, 2, 1, 0, 3, 4, 3, 0, 2, 1, 2, 2, 1, 1, 4, 4, 3, 4, 0, 2, 0, 4, 0, 2, 2, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 3, 3, 2, 0, 1, 1, 4, 4, 2, 0, 1, 4, 2, 4, 1, 2, 0, 4

Offset: 1

Simon Plouffe, Mar 23 2009

Digits are all in {0,1,2,3,4} after the first term.
The upper limit is A158624, 0.5265679578796997657885576975995789586775656...
The sequence is not eventually periodic. Assuming any period results in a condition a(1)=0 mod 10 which contradicts a(1)=5. - Cezary Glowacz, Jul 22 2024

			5^3 = 125 so the backward value is 0.521, 5^10 = 9765625, so backward value is 0.5265679. The lower limit of all values is a constant, which appears to be 0.521302330431131124210313300023141021034302...
From _N. J. A. Sloane_, May 11 2018: (Start)
To describe this another way:  write down the decimal expansion of powers of 5:
  1
  5
  25
  125
  625
  3125
  ...
keep going forever.
Write them backwards:
  1
  5
  52
  526
  5213
  ...
After a while the beginning digits are all the same.
That's the sequence. (End)

Jon E. Schoenfield, Table of n, a(n) for n = 1..3000

Cf. A158624, A004094, A023415, A145679.

D:=87; e:=6; for d in [2..D-1] do t:=Modexp(5,e,10^(d+1)); if t div 10^d ge 5 then e+:=2^(d-2); end if; end for; t:=Modexp(5,e,10^D); IntegerToSequence(t,10); // Jon E. Schoenfield, Feb 05 2018

# lower limit of backward sequence of 5^n
a,i=5,0; x=a
while i < 100:
     i+=1; print(x, end=',')
     x=(-a//pow(5,i)*pow(3,i))%5; a+=x*pow(10,i)
# Cezary Glowacz, Jul 29 2024

a(n) >= 0 and is the minimum satisfying (Sum_{i=1..n} a(i)*10^(i-1)) == 0 (mod 5^n), for n >= 2. - Cezary Glowacz, Jul 24 2024

A034906 Powers of 2 written backwards and sorted.

Original entry on oeis.org

1, 2, 4, 8, 23, 46, 61, 215, 652, 821, 2918, 4201, 6904, 8402, 48361, 63556, 86723, 270131, 441262, 882425, 2517902, 4034914, 6758401, 8068838, 23445533, 46880176, 61277761, 219078635, 654534862

Offset: 1

N. J. A. Sloane

T. D. Noe, Table of n, a(n) for n=1..1000

Cf. A000079, A004094.

Sort[FromDigits[Reverse[IntegerDigits[#]]]&/@(2^Range[0,30])] (* Harvey P. Dale, Mar 06 2012 *)
IntegerReverse[#]&/@(2^Range[0,30])//Sort (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jul 03 2020 *)

A341707 a(n) is the binary representation of n converted to yranib.

Original entry on oeis.org

0, 1, 2, 1, 4, 3, 2, 1, 8, 7, 6, 5, 4, 3, 2, 1, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, -38, -39, -40, -41, -42, -43, -44, -45, -46, -47, -48, -49, -50, -51, -52, -53, 46, 45, 44, 43, 42, 41

Offset: 0

N. J. A. Sloane, Feb 18 2021

If n = Sum_{i=0..k} b_i*2^i, b_i = 0 or 1, b_k = 1, then a(n) = y(k) - Sum_{i=0..k-1} b_i*y(i), where y(j) = A004094(j) = 2^j written backwards in base 10.
If the 2^14 terms from a(16384) to a(32767) were to be considered a packet [call it #1], then the terms from a(32768) to a(49151) [call it #2] are #1 + 38362. #3 = #2 - 48361 (note that 48361 is the reverse of 16384). #4 = #3 + 25194. These successive displacements
     (38362, -48361,   25194, -48361,
     -38362, -48361,  341659, -48361,
     -38362, -48361,   71528, -48361,
     -38362, -48361,  369771, -48361,
     -38362, -48361,   71528, -48361,
     -38362, -48361,  -71491, -48361,
     -38362, -48361,   71528, -48361,
     -38362, -48361,  909934, -48361,
     -38362, -48361,   71528, -48361,
     -38362, -48361,  -71491, -48361,
     -38362, -48361,   71528, -48361,
     -38362, -48361,   27509, -48361,
     -38362, -48361,   71528, -48361,
     -38362, -48361,  -71491, -48361,
     -38362, -48361,   71528, -48361,
     -38362, -48361, 6786009, -48361,
     -38362, -48361,   71528, -48361,
     -38362, -48361,  -71491, -48361,
     -38362, -48361,   71528, -48361,
     -38362, -48361,   27509, -48361,
     -38362, -48361,   71528, -48361,
     -38362, -48361,  -71491, -48361,
     -38362, -48361,   71528, -48361,
     -38362, -48361,   27608, -48361,
     -38362, -48361,   71528, -48361, ...) fully describe the future of the sequence. Can we predict the values of the displacements from first principles? - Hans Havermann, Feb 24 2021

			If n = 48 = 110000_2, b_0 = ... = b_3 = 0, b_4 = b_5 = 1, so a(48) = A004094(5) - A004094(4) = 23 - 61 = -38, which is the first negative term (cf. A341708).

Eric Angelini, Posting to Math Fun Mailing List, Feb 18 2021

Jinyuan Wang, Table of n, a(n) for n = 0..10000

Cf. A004094.
See A341708 for the negative terms.
See A341709 for a different version of a yranib sequence.

{0}~Join~Array[Fold[Subtract, Reverse@ IntegerReverse[2^(-1 + Position[Reverse@ IntegerDigits[#, 2], 1][[All, 1]] )]] &, 69] (* Michael De Vlieger, Feb 25 2021 *)

/* Get decimal value of yranib representation of n written in binary (i.e., write n in binary, e.g., 9[10] = 1001[2], then read this in the yranib system,where the k-th position from the right has value s*R(2^k) where R=reverse(= decimal value read from right to left) and s = -1 except for the largest k. */
y2d(n)=if(n=binary(n),n[1]*=-1);-sum(k=0,#n-1,n[#n-k]*R(2^k))
R(n)=fromdigits(Vecrev(digits(n)))
apply(y2d, [0..99]) \\ M. F. Hasler, Feb 18 2021

def reverse(n):
    s = 0
    while n > 0:
        d, n = n%10, n//10
        s = 10*s+d
    return s
def A341707(n):
    s, t = 0, 1
    while n > 0:
        b, n = n%2, n//2
        if n > 0:
            s, t = reverse(t*b)+s, 2*t
        else:
            s = reverse(t*b)-s
    return s # A.H.M. Smeets, Feb 18 2021

Further terms from M. F. Hasler, Feb 18 2021

A071588 Powers of 6 written backwards.

Original entry on oeis.org

1, 6, 63, 612, 6921, 6777, 65664, 639972, 6169761, 69677001, 67166406, 650797263, 6332876712, 61049606031, 69046146387, 675489481074, 6547099011282, 63744495662961, 614866659955101, 694010047953906, 6792600448516563

Offset: 0

Benoit Cloitre, Jun 01 2002

Seiichi Manyama, Table of n, a(n) for n = 0..1000

Cf. A000400, A004086, A004094, A004167, A071582, A071583, A071584, A071586, A071587.

FromDigits[Reverse[IntegerDigits[#]]]&/@(6^Range[0,30]) (* Harvey P. Dale, Feb 02 2012 *)

for(i=1,50,n=5^i; s=ceil(log(n)/log(10)); print1(sum(i=0,s,10^(s-i-1)*(floor(n/10^i)-10*floor(n/10^(i+1)))),","))

a(n) = A004086(A000400(n)).

A371880 Smallest number reachable starting from 1 and taking n steps either doubling or doubling+reversing.

Original entry on oeis.org

1, 2, 4, 8, 16, 23, 46, 29, 58, 71, 34, 68, 37, 47, 49, 89, 79, 59, 19, 38, 67, 35, 7, 14, 28, 56, 13, 26, 25, 5, 1, 2, 4, 8, 16, 23, 46, 29, 58, 17, 34, 68, 37, 47, 49, 89, 79, 59, 19, 38, 67, 35, 7, 14, 28, 56, 13, 26, 25, 5, 1, 2, 4, 8, 16, 23, 46, 29, 58, 17

Offset: 0

Bryle Morga, Apr 14 2024

At each step you are allowed to either double x -> 2*x or double and reverse x -> R(2*x), where R(x) = A004086(x) is decimal digit reversal.
From Michael S. Branicky, Apr 14 2024: (Start)
Since a(30) = 1 = a(0), a(n) <= a(n-30) for n >= 30. a(39) <= 17 < a(9) = 71 is the first term that strictly lowers the bound. Is it eventually periodic?
Under the map, a term k has preimage k/2 if k is even plus terms of the form R(k)*10^i/2 for i > 1 and for i=0 if R(k) is even. (End)
The condition above implies a(i+30k) is nonincreasing for k >= 0 for all i in 0..29, hence it is periodic (with period being a factor of 30). When does the periodic part of the sequence begin? - Bryle Morga, Apr 15 2024
From David A. Corneth, Apr 15 2024: (Start)
a(n) == 2^n (mod 9).
Because of this, all values 1 <= a(n) <= 9 have a(n + 30*k) = a(n). That is a(30*k) = 1, a(30*k + 1) = 2, a(30*k + 2) = 4, a(30*k + 3) = 8, a(30*k + 22) = 7, a(30*k + 29) = 5, for k >= 0. (End)
Ultimately periodic sequence of period 30 with a(k+30)=a(k) for k != 9. - David Wilson, Apr 19 2024

			a(20) = 67 and here is the 20-move combination that reaches 67: 1, 2, 4, 8, 61, 221, 244, 884, 8671, 17342, 48643, 97286, 275491, 289055, 11875, 23750, 47500, 95000, 190000, 380000, 67.
a(21) = 35 and here is the 21-move combination that reaches 35: 1, 2, 4, 8, 61, 221, 244, 884, 8671, 17342, 48643, 97286, 275491, 289055, 11875, 23750, 47500, 95000, 91, 281, 265, 35.
a(30) = 1 using the path: 1, 2, 4, 8, 61, 122, 442, 488, 976, 2591, 2815, 365, 37, 47, 49, 98, 196, 392, 487, 479, 859, 1718, 3436, 2786, 2755, 155, 13, 26, 25, 5, 1. - _Michael S. Branicky_, Apr 14 2024

David A. Corneth, PARI program.
Bryle Morga et al., Optimal strategy in a game where Doubling and Doubling+Reversing are the allowed moves.
Index entries for linear recurrences with constant coefficients, signature (0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1).

Cf. A000079, A004086, A004094, A036447, A371966.

PARI
```
\\ See PARI link
```

def f(k, d):
      if d == 0:
            return k
      else:
            return min(f(2*k, d-1), f(int(str(2*k)[::-1]), d - 1))
def a(n):
      return f(1, n)
for n in range(25):
      print(a(n))

from itertools import islice
def reverse(n): return int(str(n)[::-1])
def agen(): # generator of terms
    reach = {1}
    while True:
        yield min(reach)
        newreach = set()
        for q in reach: newreach.update([2*q, reverse(2*q)])
        reach = newreach
print(list(islice(agen(), 28))) # Michael S. Branicky, Apr 14 2024

a(n) = f(1, n) where f(k, 0) = k and f(k, n) = min(f(2*k, n-1), f(R(2*k), n-1)).
a(30k) = 1 for k >= 0. - Michael S. Branicky, Apr 14 2024
a(9) = 71. For k != 9, a(k) is the minimum of the positive residues mod 99 of 2^k and 10*2^k. - David Wilson, Apr 19 2024

a(27)-a(34) from Michael S. Branicky, Apr 14 2024

a(35) and beyond from David Wilson, Apr 19 2024

A062018 a(n) = n^n written backwards.

Original entry on oeis.org

1, 4, 72, 652, 5213, 65664, 345328, 61277761, 984024783, 1, 116076113582, 6528440016198, 352295601578203, 61085552860021111, 573958083098398734, 61615590737044764481, 771467633688162042728, 42457573569257080464393

Offset: 1

Amarnath Murthy, Jun 01 2001

			a(5) = 5213, as 5^5 = 3125.

Harry J. Smith, Table of n, a(n) for n = 1..200

Cf. A004086, A004093, A004156, A034906, A004094, A004167, A002108, A002118, A002138, A002140, A002232, A002239, A002241, A002942, A004165, A004087, A004091, A004096, A004098, A004150, A004153.

with(numtheory):for n from 1 to 50 do a := convert(n^n,base,10):b := add(10^(nops(a)- i)*a[i],i=1..nops(a)):printf(`%d,`,b); od:

Table[IntegerReverse[n^n],{n,20}] (* Harvey P. Dale, Jul 31 2022 *)

a(n) = { fromdigits(Vecrev(digits( n^n )))} \\ Harry J. Smith, Jul 29 2009

a(n) = A004086(n^n).

More terms from Jason Earls and Vladeta Jovovic, Jun 01 2001

A103161 GCD of reverse(2^n) and reverse(2^(n+1)), where reverse(k) = A004086(k), the decimal representation of k read backwards.

Original entry on oeis.org

2, 4, 1, 1, 23, 1, 1, 1, 1, 4201, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 1, 1, 1, 7, 1, 1, 1, 1, 4, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 19, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 34, 1, 1, 1, 7, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1

Offset: 1

Labos Elemer, Jan 25 2005

			n=10: GCD of backward written powers of 2 is GCD(4201, 8402) = 4201 = a(10).

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000079, A004094.

rd[x_] :=FromDigits[Reverse[IntegerDigits[x]]] Table[GCD[rd[2^w], rd[2^(w+1)]], {w, 1, 100}]
GCD[IntegerReverse[#[[1]]],IntegerReverse[#[[2]]]]&/@ Partition[ 2^Range[110],2,1] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Aug 24 2017 *)

rev(n) = subst(Polrev(digits(n)), 'x, 10); \\ These two functions from Charles R Greathouse IV, Oct 20 2014
A004094(n) = rev(2^n);
A103161(n) = gcd(A004094(n),A004094(1+n)); \\ Antti Karttunen, Dec 07 2017

a(n) = gcd(A004094(n), A004094(n+1)).

A100825 In decimal representation: minimal number of editing steps (delete, insert, or substitute) to transform 2^n into its reversal.

Original entry on oeis.org

0, 0, 0, 0, 2, 2, 2, 2, 2, 2, 4, 4, 4, 4, 4, 4, 2, 6, 6, 4, 6, 4, 4, 4, 4, 6, 6, 8, 6, 8, 8, 8, 8, 8, 8, 8, 6, 6, 10, 12, 12, 8, 10, 10, 12, 10, 10, 14, 14, 14, 12, 12, 14, 12, 10, 14, 14, 16, 16, 16, 18, 14, 16, 16, 16, 14, 18, 16, 18, 18, 18, 18, 18, 16, 20, 20, 18, 22, 22, 22, 20, 18, 20

Offset: 1

Reinhard Zumkeller, Jan 06 2005

			n=19: 2^19 = 524288=[5]24288 -> 824288=[]824288 ->
8824288=882428[8] -> 882428=88242[8] -> 882425=A004094(19):
a(19) = #{subst[5->8], ins[8], del[8], subst[8->5]} = 4.

Michael Gilleland, Levenshtein Distance [It has been suggested that this algorithm gives incorrect results sometimes. - _N. J. A. Sloane_]

a(n) = LevenshteinDistance(A000079(n), A004094(n)).

cp's OEIS Frontend

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

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A071586 Powers of 8 written backwards.

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A158625 Lower limit of backward value of 5^n.

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A034906 Powers of 2 written backwards and sorted.

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A341707 a(n) is the binary representation of n converted to yranib.

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A071588 Powers of 6 written backwards.

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Mathematica

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A371880 Smallest number reachable starting from 1 and taking n steps either doubling or doubling+reversing.

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PARI

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A062018 a(n) = n^n written backwards.

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