A048379 -id:A048379 - cp's OEIS frontend

A169930 Duplicate of A048379.

1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 21, 22, 23, 24, 25, 26, 27, 28, 29, 20, 31, 32, 33, 34, 35, 36

Offset: 0

dead

A002283 a(n) = 10^n - 1.

0, 9, 99, 999, 9999, 99999, 999999, 9999999, 99999999, 999999999, 9999999999, 99999999999, 999999999999, 9999999999999, 99999999999999, 999999999999999, 9999999999999999, 99999999999999999, 999999999999999999, 9999999999999999999, 99999999999999999999, 999999999999999999999, 9999999999999999999999

Offset: 0

N. J. A. Sloane

A friend from Germany remarks that the sequence 9, 99, 999, 9999, 99999, 999999, ... might be called the grumpy German sequence: nein!, nein! nein!, nein! nein! nein!, ...
The Regan link shows that integers of the form 10^n -1 have binary representations with exactly n trailing 1 bits. Also those integers have quinary expressions with exactly n trailing 4's. For example, 10^4 -1 = (304444)5. The first digits in quinary correspond to the number 2^n -1, in our example (30)5 = 2^4 -1. A similar pattern occurs in the binary case. Consider 9 = (1001)2. - Washington Bomfim Dec 23 2010
a(n) is the number of positive integers with less than n+1 digits. - Bui Quang Tuan, Mar 09 2015
From Peter Bala, Sep 27 2015: (Start)
For n >= 1, the simple continued fraction expansion of sqrt(a(2*n)) = [10^n - 1; 1, 2*(10^n - 1), 1, 2*(10^n - 1), ...] has period 2. The simple continued fraction expansion of sqrt(a(2*n))/a(n) = [1; 10^n - 1, 2,  10^n - 1, 2, ...] also has period 2. Note the occurrence of large partial quotients in both expansions.
A theorem of Kuzmin in the measure theory of continued fractions says that large partial quotients are the exception in continued fraction expansions.
Empirically, we also see the presence of unexpectedly large partial quotients early in the continued fraction expansions of the m-th roots of the numbers a(m*n) for m >= 3. Some typical examples are given below. (End)
For n > 0, numbers whose smallest decimal digit is 9. - Stefano Spezia, Nov 16 2023

			From _Peter Bala_, Sep 27 2015: (Start)
Continued fraction expansions showing large partial quotients:
a(12)^(1/3) = [9999; 1, 299999998, 1, 9998, 1, 449999998, 1, 7998, 1, 535714284, 1, 2, 2, 142, 2, 2, 1, 599999999, 3, 1, 1,...].
Compare with a(30)^(1/3) = [9999999999; 1, 299999999999999999998, 1, 9999999998, 1, 449999999999999999998, 1, 7999999998, 1, 535714285714285714284, 1, 2, 2, 142857142, 2, 2, 1, 599999999999999999999, 3, 1, 1,...].
a(24)^(1/4) = [999999; 1, 3999999999999999998, 1, 666665, 1, 1, 1, 799999999999999999, 3, 476190, 7, 190476190476190476, 21, 43289, 1, 229, 1, 1864801864801863, 1, 4, 6,...].
Compare with a(48)^(1/4) = [999999999999; 1, 3999999999999999999999999999999999998, 1, 666666666665, 1, 1, 1, 799999999999999999999999999999999999, 3, 476190476190, 7, 190476190476190476190476190476190476, 21, 43290043289, 1, 229, 1, 1864801864801864801864801864801863, 1, 4, 6,...].
a(25)^(1/5) = [99999, 1, 499999999999999999998, 1, 49998, 1, 999999999999999999998, 1, 33332, 3, 151515151515151515151, 5, 1, 1, 1947, 1, 1, 38, 3787878787878787878, 1, 3, 5,...].
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..100
Peter Bala, A002283 and some continued fraction expansions.
Rick Regan, Nines in quinary, September 8th, 2009.
Amelia Carolina Sparavigna, The groupoids of Mersenne, Fermat, Cullen, Woodall and other Numbers and their representations by means of integer sequences, Politecnico di Torino, Italy (2019), [math.NT].
Amelia Carolina Sparavigna, Some Groupoids and their Representations by Means of Integer Sequences, International Journal of Sciences (2019) Vol. 8, No. 10.
Index entries for linear recurrences with constant coefficients, signature (11,-10).

Cf. A000533, A003020, A007138, A007953, A008591, A010888, A048379, A066138, A073668, A168624, A276352.
Cf. A002275, A002276, A002277, A002278, A002279, A002280, A002281, A002282.
Cf. A075412, A075415, A178630, A178631, A178632, A178633, A178634, A178635.
Cf. A010785.

a002283 = subtract 1 . (10 ^)  -- Reinhard Zumkeller, Feb 21 2014

[(10^n-1): n in [0..20]]; // Vincenzo Librandi, Apr 26 2011

Table[10^n - 1, {n, 0, 22}] (* Michael De Vlieger, Sep 27 2015 *)

A002283(n):=10^n-1$
makelist(A002283(n),n,0,20); /* Martin Ettl, Nov 08 2012 */

a(n)=10^n-1; \\ Charles R Greathouse IV, Jan 30 2012

def a(n): return 10**n-1 # Michael S. Branicky, Feb 25 2023

From Mohammad K. Azarian, Jan 14 2009: (Start)
G.f.: 1/(1-10*x)-1/(1-x).
E.g.f.: e^(10*x)-e^x. (End)
a(n) = A075412(n)/A002275(n) = A178630(n)/A002276(n) = A178631(n)/A002277(n) = A075415(n)/A002278(n) = A178632(n)/A002279(n) = A178633(n)/A002280(n) = A178634(n)/A002281(n) = A178635(n)/A002282(n). - Reinhard Zumkeller, May 31 2010
a(n) = a(n-1) + 9*10^(n-1) with a(0)=0; Also: a(n) = 11*a(n-1) - 10*a(n-2) with a(0)=0, a(1)=9. - Vincenzo Librandi, Jul 22 2010
For n>0, A007953(a(n)) = A008591(n) and A010888(a(n)) = 9. - Reinhard Zumkeller, Aug 06 2010
A048379(a(n)) = 0. - Reinhard Zumkeller, Feb 21 2014
a(n) = Sum_{k=1..n} 9*10^k. - Carauleanu Marc, Sep 03 2016
Sum_{n>=1} 1/a(n) = A073668. - Amiram Eldar, Nov 13 2020
From Elmo R. Oliveira, Jul 19 2025: (Start)
a(n) = 9*A002275(n).
a(n) = A010785(A008591(n)). (End)

More terms from Michael De Vlieger, Sep 27 2015

A035327 Write n in binary, interchange 0's and 1's, convert back to decimal.

Original entry on oeis.org

1, 0, 1, 0, 3, 2, 1, 0, 7, 6, 5, 4, 3, 2, 1, 0, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 31, 30, 29, 28, 27, 26, 25, 24, 23, 22, 21, 20, 19, 18, 17, 16, 15, 14, 13, 12, 11, 10, 9, 8, 7, 6, 5, 4, 3, 2, 1, 0, 63, 62, 61, 60, 59, 58, 57, 56, 55, 54, 53, 52, 51, 50, 49, 48, 47, 46

Offset: 0

N. J. A. Sloane

For n>0: largest m<=n such that no carry occurs when adding m to n in binary arithmetic: A003817(n+1) = a(n) + n = a(n) XOR n. - Reinhard Zumkeller, Nov 14 2009
a(0) could be considered to be 0 (it was set so from 2004 to 2008) if the binary representation of zero was chosen to be the empty string. - Jason Kimberley, Sep 19 2011
For n > 0: A240857(n,a(n)) = 0. - Reinhard Zumkeller, Apr 14 2014
This is a base-2 analog of A048379. Another variant, without converting back to decimal, is given in A256078. - M. F. Hasler, Mar 22 2015
For n >= 2, a(n) is the least nonnegative k that must be added to n+1 to make a power of 2. Hence in a single-elimination tennis tournament with n entrants, a(n-1) is the number of players given a bye in round one, so that the number of players remaining at the start of round two is a power of 2. For example, if 39 players register, a(38)=25 players receive a round-one bye leaving 14 to play, so that round two will have 25+(14/2)=32 players. - Mathew Englander, Jan 20 2024

			8 = 1000 -> 0111 = 111 = 7.

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000
J.-P. Allouche and J. Shallit, The ring of k-regular sequences, II, Theoret. Computer Sci., 307 (2003), 3-29.
Ralf Stephan, Divide-and-conquer generating functions. I. Elementary sequences, arXiv:math/0307027 [math.CO], 2003.
Ralf Stephan, Some divide-and-conquer sequences ...
Ralf Stephan, Table of generating functions
Index entries for sequences related to binary expansion of n
Index entries for sequences related to the Josephus Problem

a(n) = A003817(n) - n, for n>0.
Cf. A080079, A087734, A167831, A167877, A007088, A061601, A171960, A010078, A000225, A006257 (Josephus problem).
Cf. A240857.
Cf. A048379, A256078.

a035327 n = if n <= 1 then 1 - n else 2 * a035327 n' + 1 - b
            where (n',b) = divMod n 2
-- Reinhard Zumkeller, Feb 21 2014

using IntegerSequences
A035327List(len) = [Bits("NAND", n, n) for n in 0:len]
println(A035327List(100))  # Peter Luschny, Sep 25 2021

A035327:=func; // Jason Kimberley, Sep 19 2011

seq(2^(1 + ilog2(max(n, 1))) - 1 - n, n = 0..81); # Emeric Deutsch, Oct 19 2008
A035327 := n -> `if`(n=0, 1, Bits:-Nand(n, n)):
seq(A035327(n), n=0..81); # Peter Luschny, Sep 23 2019

Table[BaseForm[FromDigits[(IntegerDigits[i, 2]/.{0->1, 1->0}), 2], 10], {i, 0, 90}]
Table[BitXor[n, 2^IntegerPart[Log[2, n] + 1] - 1], {n, 100}] (* Alonso del Arte, Jan 14 2006 *)
Join[{1},Table[2^BitLength[n]-n-1,{n,100}]] (* Paolo Xausa, Oct 13 2023 *)
Table[FromDigits[IntegerDigits[n,2]/.{0->1,1->0},2],{n,0,90}] (* Harvey P. Dale, May 03 2025 *)

a(n)=sum(k=1,n,if(bitxor(n,k)>n,1,0)) \\ Paul D. Hanna, Jan 21 2006

a(n) = bitxor(n, 2^(1+logint(max(n,1), 2))-1) \\ Rémy Sigrist, Jan 04 2019

a(n)=if(n, bitneg(n, exponent(n)+1), 1) \\ Charles R Greathouse IV, Apr 13 2020

def a(n): return int(''.join('1' if i == '0' else '0' for i in bin(n)[2:]), 2) # Indranil Ghosh, Apr 29 2017

def a(n): return 1 if n == 0 else n^((1 << n.bit_length()) - 1)
print([a(n) for n in range(100)]) # Michael S. Branicky, Sep 28 2021

def A035327(n): return (~n)^(-1<Chai Wah Wu, Dec 20 2022

def a(n):
    if n == 0:
        return 1
    return sum([(1 - b) << s for (s, b) in enumerate(n.bits())])
[a(n) for n in srange(82)]  # Peter Luschny, Aug 31 2019

a(n) = 2^k - n - 1, where 2^(k-1) <= n < 2^k.
a(n+1) = (a(n)+n) mod (n+1); a(0) = 1. - Reinhard Zumkeller, Jul 22 2002
G.f.: 1 + 1/(1-x)*Sum_{k>=0} 2^k*x^2^(k+1)/(1+x^2^k). - Ralf Stephan, May 06 2003
a(0) = 0, a(2n+1) = 2*a(n), a(2n) = 2*a(n) + 1. - Philippe Deléham, Feb 29 2004
a(n) = number of positive integers k < n such that n XOR k > n. a(n) = n - A006257(n). - Paul D. Hanna, Jan 21 2006
a(n) = 2^{1+floor(log[2](n))}-n-1 for n>=1; a(0)=1. - Emeric Deutsch, Oct 19 2008
a(n) = if n<2 then 1 - n else 2*a(floor(n/2)) + 1 - n mod 2. - Reinhard Zumkeller, Jan 20 2010
a(n) = abs(2*A053644(n) - n - 1). - Mathew Englander, Jan 22 2024

More terms from Vit Planocka (planocka(AT)mistral.cz), Feb 01 2003
a(0) corrected by Paolo P. Lava, Oct 22 2007
Definition completed by M. F. Hasler, Mar 22 2015

A032810 Numbers using only digits 2 and 3.

Original entry on oeis.org

2, 3, 22, 23, 32, 33, 222, 223, 232, 233, 322, 323, 332, 333, 2222, 2223, 2232, 2233, 2322, 2323, 2332, 2333, 3222, 3223, 3232, 3233, 3322, 3323, 3332, 3333, 22222, 22223, 22232, 22233, 22322, 22323, 22332, 22333, 23222, 23223

Offset: 1

Clark Kimberling

Identical to A007931 with substitution of digits 2 -> 3, 1 -> 2, i.e., application of the function A048379 or A256079(n) = n + A002275(A055642(n)). - M. F. Hasler, Mar 21 2015

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for 10-automatic sequences.

Cf. A020458, A143967, A248907 (permutation).

Cf. A032804-A032816 (in other bases), A007088 (digits 0 & 1), A007931 (digits 1 & 2), A032834 (digits 3 & 4), A256290 (digits 4 & 5), A256291 (digits 5 & 6), A256292 (digits 6 & 7), A256340 (digits 7 & 8), A256341 (digits 8 & 9).

a032810 = f 0 . (+ 1) where
   f y 1 = a004086 y
   f y x = f (10 * y + m + 2) x' where (x', m) = divMod x 2
-- Reinhard Zumkeller, Mar 18 2015

[n: n in [1..24000] | Set(Intseq(n)) subset {2, 3}]; // Vincenzo Librandi, May 27 2012

[n eq 1 select 2 else IsOdd(n) select 10*Self(Floor(n/2))+2 else Self(n-1)+1: n in [1..40]]; // Bruno Berselli, May 27 2012

Flatten[Table[FromDigits[#,10]&/@Tuples[{2,3},n],{n,5}]] (* Vincenzo Librandi, May 27 2012 *)

A032810(n)=vector(#n=binary(n+1)[2..-1],i,10^(#n-i))*n~+10^#n\9*2 \\ M. F. Hasler, Mar 26 2015

def A032810(n): return int(bin(n+1)[3:])+(10**((n+1).bit_length()-1)-1<<1)//9 # Chai Wah Wu, Jul 15 2023

a(n) = f(n+1, 0) with f(n, x) = if n=1 then A004086(x) else f(floor(n/2), 10*x + 2 + n mod 2). - Reinhard Zumkeller, Sep 06 2008
a(n) is Theta(n^(log_2 10)); there are about n^(log_10 2) members of this sequence up to n. - Charles R Greathouse IV, Mar 18 2010
a(n) = A007931(n) + A002275(A000523(n+1)). A055642(a(n)) = A000523(n+1). - M. F. Hasler, Mar 21 2015

A061844 Squares that remain squares if you decrease every digit by 1.

Original entry on oeis.org

1, 36, 3136, 24336, 5973136, 71526293136, 318723477136, 264779654424693136, 24987377153764853136, 31872399155963477136, 58396845218255516736, 517177921565478376336, 252815272791521979771662766736, 518364744896318875336864648336, 554692513628187865132829886736

Offset: 1

Erich Friedman, Jun 23 2001

The terms may be calculated efficiently by solving x^2 - y^2 = 111...1; this is done by factoring 111..1 = (x + y)(x - y).
Note that some solutions will produce a square containing a zero digit so the solution is impermissible; for example, 460^2 - 317^2 = 111111 but 460^2 = 211600. - Wendy Appleby, Sep 20 2015
Except for a(1) = 1, we don't allow decreasing the digits to create a leading 0. Thus 126736 = 356^2 is not included, even though 126736 - 111111 = 15625 = 125^2. - Robert Israel, Dec 30 2015
If it exists, a(79) > 10^262. - Max Alekseyev, Sep 05 2023
From Robert Israel, Jan 04 2016: (Start)
The sequence may well be finite.
It is known that A000005(n) = O(n^epsilon) for all epsilon>0.
Therefore if 1 < c < 10/9, for d sufficiently large (10^d-1)/9 has fewer than c^d divisors, and thus fewer than c^d possible candidates for x^2 having d digits.
Heuristically, x^2 has probability ~ (9/10)^d of having no digits 0.
Thus we expect fewer than (9c/10)^d terms having d digits.
Since Sum_d (9c/10)^d converges, we expect only finitely many terms.
Of course, this is only a heuristic argument, but it seems to fit well with the data. (End)

			13225 = 115^2 and 24336 = 156^2.

JungHwan Min, Table of n, a(n) for n = 1..78

Cf. A048379, A052382, A061843, A117755.

A:= {1}:
for d from 1 to 96 do
  r:= (10^d-1)/9;
  f:= subs(X=10,factors((X^d-1)/(X-1))[2]);
  q:= map(t -> op(map(s -> [s[1],t[2]*s[2]], ifactors(t[1])[2])),f);
  divs:= {1};
for t in q do
    divs:= map(x -> seq(x*t[1]^j,j=0..t[2]),divs)
  od;
  for t in select(s -> s^2 > r, divs) do
    x:= (t + r/t)/2;
    if ilog10(x^2) = d-1 and x^2 > 2*10^(d-1) and not has(convert(x^2,base,10),0) then
      A:= A union {x^2};
    fi
  od
od:
sort(convert(A,list)); # Robert Israel, Dec 30 2015

For[digits = 1, digits <= 30, digits++, n = (10^digits - 1)/9; divList = Select[Divisors[n], (#1 >= Sqrt[n])&]; For[j = 1, j <= Length[divList], j++, x = (divList[[j]] + n/divList[[j]])/2; y = (divList[[j]] - n/divList[[j]])/2; dx = IntegerDigits[x^2]; dy = IntegerDigits[y^2]; If[(Length[dx] == digits) && (Length[dy] == digits) && (Select[dx, (#1 == 0)&] == {}), Print[x^2]]]]
Flatten@Prepend[Table[Select[#[[Ceiling[(Length[#] + 1)/2] ;;]] &@(# + Reverse@#)/2 &@Divisors[(10^n - 1)/9], IntegerLength[#^2] == n && (#[[1]] != 1 && FreeQ[#, 0]&[IntegerDigits[#^2]])&]^2, {n, 30}], 1] (* JungHwan Min, Dec 29 2015 *)
Join[{1},Select[Select[Flatten[Table[#^2&/@(x/.Solve[{x^2-y^2 == FromDigits[ PadRight[{},n,1]],x>0,y>0},{x,y},Integers]),{n,2,30}]], DigitCount[ #,10,0]==0&&IntegerDigits[#][[1]]>1&]// Union,IntegerQ[ Sqrt[ FromDigits[IntegerDigits[#]-1]]]&]] (* Harvey P. Dale, Apr 16 2016 *)

a(n) = A048379(A061843(n)). - Max Alekseyev, Jul 26 2023

More terms and program from Jonathan Cross (jcross(AT)wcox.com), Oct 08 2001

A117755 Squares which remain squares when each digit is replaced by the next digit.

Original entry on oeis.org

0, 9, 25, 2025, 13225, 1974025, 4862025, 6943225, 60415182025, 207612366025, 916408817340025, 9960302475729225, 153668543313582025, 1978088677245614025, 13876266042653742025, 20761288044852366025, 47285734107144405625, 406066810454367265225

Offset: 1

Luc Stevens (lms022(AT)yahoo.com), Apr 14 2006

Replace 1 with 2, 2 with 3, ..., 8 with 9 and 9 with 0.

			13225 is in the sequence because (1) it is a square and (2) if we transform it we get 24336 and this is also a square.

Max Alekseyev, Table of n, a(n) for n = 1..22

Cf. A048379, A061843, A061844, A127856.

Select[Range[0, 500000]^2, IntegerQ[Sqrt[FromDigits[(1 + IntegerDigits[ # ]) /. 10-> 0]]] &] (* Harvey P. Dale, Jan 21 2007 *)

More terms from Harvey P. Dale, Jan 21 2007
3 more terms from Donovan Johnson, Apr 03 2008
a(14)-a(21) from Max Alekseyev, Oct 22 2008
a(1) = 0 prepended by Max Alekseyev, Jul 26 2023

A256078 Write n in binary, exchange digits '0' <-> '1'.

Original entry on oeis.org

1, 0, 1, 0, 11, 10, 1, 0, 111, 110, 101, 100, 11, 10, 1, 0, 1111, 1110, 1101, 1100, 1011, 1010, 1001, 1000, 111, 110, 101, 100, 11, 10, 1, 0, 11111, 11110, 11101, 11100, 11011, 11010, 11001, 11000, 10111, 10110, 10101, 10100, 10011, 10010, 10001, 10000

Offset: 0

M. F. Hasler, Mar 22 2015

Binary representation of A035327.

A base-2 analog of A048379.

Robert Israel, Table of n, a(n) for n = 0..10000

Cf. A035327, A048379.

f:= proc(n) local L,i;
  L:= convert(n,base,2);
  add((1-L[i])*10^(i-1),i=1..nops(L))
end proc:
map(f, [$0..100]); # Robert Israel, Sep 17 2024

Table[FromDigits[IntegerDigits[n, 2] /. {0 -> 1, 1 -> 0}], {n, 0, 47}] (* or *)
Table[FromDigits@ IntegerDigits[BitXor[n, 2^IntegerPart[Log[2, n] + 1] - 1], 2], {n, 0, 47}] (* Michael De Vlieger, Mar 22 2015, the latter based on Alonso del Arte at A035327 *)

A256078(n)=!n+eval(Strchr(apply(d->49-d,binary(n))))

def a(n): return int(bin(1 if n==0 else n^((1 << n.bit_length())-1))[2:])
print([a(n) for n in range(48)]) # Michael S. Branicky, Dec 21 2022

A169918 Squares in carryless arithmetic mod 10 with addition of digits defined to be multiplication mod 10 and multiplication of digits defined to be addition mod 10.

Original entry on oeis.org

0, 2, 4, 6, 8, 0, 2, 4, 6, 8, 210, 242, 294, 266, 258, 260, 292, 244, 216, 208, 440, 492, 464, 456, 468, 490, 442, 414, 406, 418, 690, 662, 654, 666, 698, 640, 612, 604, 616, 648, 860, 852, 864, 896, 848, 810, 802, 814, 846, 898, 50, 62, 94, 46, 18, 0, 12, 44, 96, 68, 260, 292

Offset: 0

David Applegate, Marc LeBrun and N. J. A. Sloane, Jul 20 2010

The rules of arithmetic used in A169916, A169917, A169918 have very strange consequences. Many of the familiar laws fail. For instance, the arithmetic in A169916 is not associative: 10*(9*2) = 10*1 = 21 != (10*9)*2 = 9*2 = 1.

			a(17) = 17*17 = 244:
...17
...17
-----
...84 (7*7 = 7+7 mod 10 = 4, 7*1 = 7+1 mod 10 = 8)
..28.
-----
..244
(The rule for "adding" the columns is to multiply mod 10: 8+8 = 8 * 8 mod 10 = 4. Blanks are ignored)

Index entries for sequences related to carryless arithmetic

See A048379, A169931-A169933, A169935 for other examples of calculations in this version of arithmetic.

The four versions are A059729, A169916, A169917, A169918.

A169918(n)={u=vector(#n=digits(n),i,1);n=apply(d->n+d*u,n)%10;sum(i=0,2*#n-2,prod(j=max(1,#n-i),min(2*#n-1-i,#n),n[2*#n-i-j][j])%10*10^i)} \\ M. F. Hasler, Mar 26 2015

Thanks to Rick L. Shepherd for pointing out a typo in the example. - N. J. A. Sloane, Nov 08 2014

A256289 Apply the transformation 0 -> 1 -> 2 -> 3 -> 4 -> 5 -> 6 -> 7 -> 8 -> 0 to the digits of n written in base 9; do not convert back to base 10.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 0, 21, 22, 23, 24, 25, 26, 27, 28, 20, 31, 32, 33, 34, 35, 36, 37, 38, 30, 41, 42, 43, 44, 45, 46, 47, 48, 40, 51, 52, 53, 54, 55, 56, 57, 58, 50, 61, 62, 63, 64, 65, 66, 67, 68, 60, 71, 72, 73, 74, 75, 76

Offset: 0

M. F. Hasler, Mar 22 2015

Base 9 variant of A256078 (base 2) and A048379 (base 10). See A256303 - A256308 for bases 3 through 8, and A256299 for the variant where the result is converted back to base 10.

			a(9) = 21 because 9 = "10" in base 9 becomes "21".
a(80) = 0 because 80 = "88" in base 9 becomes "00".

A256289(n,b=9)=!n+eval(Strchr(apply(d->(d+1)%b+48, digits(n,b))))

A256293 Apply the transformation 0 -> 1 -> 2 -> 0 to the digits of n written in base 3, then convert back to base 10.

Original entry on oeis.org

1, 2, 0, 7, 8, 6, 1, 2, 0, 22, 23, 21, 25, 26, 24, 19, 20, 18, 4, 5, 3, 7, 8, 6, 1, 2, 0, 67, 68, 66, 70, 71, 69, 64, 65, 63, 76, 77, 75, 79, 80, 78, 73, 74, 72, 58, 59, 57, 61, 62, 60, 55, 56, 54, 13, 14, 12, 16, 17, 15

Offset: 0

M. F. Hasler, Mar 22 2015

Base 3 variant of A035327 (base 2) and A048379 (base 10).

See A256294 - A256299 for bases 4 through 9, and A256303 for the variant where the result is not converted back to base 10.

			a(3) = 7 because 3 = 10[3] becomes 21[3] = 7.
a(8) = 0 because 8 = 22[3] becomes 00[3] = 0.

Table[FromDigits[IntegerDigits[n,3]/.{0->1,1->2,2->0},3],{n,0,100}] (* Harvey P. Dale, Jan 21 2019 *)

A256293(n,b=3)=!n+apply(t->(t+1)%b,n=digits(n,b))*vector(#n,i,b^(#n-i))~

cp's OEIS Frontend

A169930 Duplicate of A048379.

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A002283 a(n) = 10^n - 1.

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A035327 Write n in binary, interchange 0's and 1's, convert back to decimal.

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Julia

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A032810 Numbers using only digits 2 and 3.

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A061844 Squares that remain squares if you decrease every digit by 1.

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A117755 Squares which remain squares when each digit is replaced by the next digit.

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