A007931 -id:A007931 - cp's OEIS frontend

A007932 Numbers that contain only 1's, 2's and 3's.

1, 2, 3, 11, 12, 13, 21, 22, 23, 31, 32, 33, 111, 112, 113, 121, 122, 123, 131, 132, 133, 211, 212, 213, 221, 222, 223, 231, 232, 233, 311, 312, 313, 321, 322, 323, 331, 332, 333, 1111, 1112, 1113, 1121, 1122, 1123, 1131, 1132, 1133, 1211, 1212, 1213, 1221

Offset: 1

R. Muller

This sequence is the alternate number system in base 3. - Robert R. Forslund (forslund(AT)tbaytel.net), Jun 27 2003
a(n) is the "bijective base-k numeration" or "k-adic notation" for k=3. - Chris Gaconnet (gaconnet(AT)gmail.com), May 27 2009
a(n) = n written in base 3 where zeros are not allowed but threes are. The three distinct digits used are 1, 2 and 3 instead of 0, 1 and 2. To obtain this sequence from the "canonical" base 3 sequence with zeros allowed, just replace any 0 with a 3 and then subtract one from the group of digits situated on the left: (20-->13; 100-->23; 110-->33; 1000-->223; 1010-->233). This can be done in any integer positive base b, replacing zeros with positive b's and subtracting one from the group of digits situated on the left. And zero is the only digit that can be replaced, since there is always a more significant digit greater than 0, on the left, from which to subtract one. - Robin Garcia, Jan 07 2014

			a(100)  = 3131.
a(10^3) = 323231.
a(10^4) = 111123331.
a(10^5) = 11231311131.
a(10^6) = 1212133131231.
a(10^7) = 123133223331331.
a(10^8) = 13221311111312131.
a(10^9) = 2113123122313232231.
- _Hieronymus Fischer_, Jun 06 2012

K. Atanassov, On the 97th, 98th and the 99th Smarandache Problems, Notes on Number Theory and Discrete Mathematics, Sophia, Bulgaria, Vol. 5 (1999), No. 3, 89-93.
A. Salomaa, Formal Languages, Academic Press, 1973. pages 90-91. [From Chris Gaconnet (gaconnet(AT)gmail.com), May 27 2009]

Hieronymus Fischer, Table of n, a(n) for n = 1..10000
K. Atanassov, On Some of Smarandache's Problems
EMIS, Mirror site for Southwest Journal of Pure and Applied Mathematics
Robert R. Forslund, A Logical Alternative to the Existing Positional Number System, Southwest Journal of Pure and Applied Mathematics. Vol. 1 1995 pp. 27-29.
James E. Foster, A Number System without a Zero-Symbol, Mathematics Magazine, Vol. 21, No. 1. (1947), pp. 39-41.
F. Smarandache, Only Problems, Not Solutions!.
Wikipedia, Bijective numeration
Index entries for 10-automatic sequences.

Cf. A007931, A052382, A084544, A084545, A046034, A089581, A084984, A001742, A001743, A001744, A202267, A202268, A014261, A014263.

NextNbr[n_] := Block[{d = IntegerDigits[n + 1], l}, l = Length[d]; While[l != 1, If[ d[[l]] > 3, d[[l - 1]]++; d[[l]] = 1]; l-- ]; If[ d[[1]] > 3, d[[1]] = 11]; FromDigits[d]]; NestList[ NextNbr, 1, 51]
Table[FromDigits/@Tuples[{1,2,3},n],{n,4}]//Flatten (* Harvey P. Dale, Mar 29 2018 *)

a(n) = my (w=3); while (n>w, n -= w; w *= 3); my (d=digits(w+n-1, 3)); d[1] = 0; fromdigits(d) + (10^(#d-1)-1)/9 \\ Rémy Sigrist, Aug 28 2018

From Hieronymus Fischer, May 30 2012 and Jun 08 2012: (Start)
The formulas are designed to calculate base-10 numbers only using the digits 1, 2, 3.
a(n) = Sum_{j=0..m-1} (1 + b(j) mod 3)*10^j,
where m = floor(log_3(2*n+1)), b(j) = floor((2*n+1-3^m)/(2*3^j)).
Special values:
a(k*(3^n-1)/2) = k*(10^n-1)/9, k=1,2,3.
a((5*3^n-3)/2) = (4*10^n-1)/3 = 10^n + (10^n-1)/3.
a((3^n-1)/2 - 1) = (10^(n-1)-1)/3, n>1.
Inequalities:
a(n) <= (10^log_3(2*n+1)-1)/9, equality holds for n=(3^k-1)/2, k>0.
a(n) > (3/10)*(10^log_3(2*n+1)-1)/9, n>0.
Lower and upper limits:
lim inf a(n)/10^log_3(2*n) = 1/30, for n --> infinity.
lim sup a(n)/10^log_3(2*n) = 1/9, for n --> infinity.
G.f.: g(x) = (x^(1/2)*(1-x))^(-1) Sum_{j=>0} 10^j*(x^3^j)^(3/2) * (1-x^3^j)*(1 + 2x^3^j + 3x^(2*3^j))/(1 - x^3^(j+1)).
Also: g(x) = (1/(1-x)) Sum_{j>=0} (1 - 4(x^3^j)^3 + 3(x^3^j)^4)*x^3^j*f_j(x)/(1-x^3^j), where f_j(x) = 10^j*x^((3^j-1)/2)/(1-(x^3^j)^3). The f_j obey the recurrence f_0(x) = 1/(1-x^3), f_(j+1)(x) = 10x*f_j(x^3).
Also: g(x) = (1/(1-x))*(h_(3,0)(x) + h_(3,1)(x) + h_(3,2)(x) - 3*h_(3,3)(x)), where h_(3,k)(x) = Sum_{j>=0} 10^j*x^((3^(j+1)-1)/2) * (x^3^j)^k/(1-(x^3^j)^3).
(End)

Edited and extended by Robert G. Wilson v, Dec 14 2002

Crossrefs added by Hieronymus Fischer, Jun 06 2012

A256292 Numbers which have only digits 6 and 7 in base 10.

Original entry on oeis.org

6, 7, 66, 67, 76, 77, 666, 667, 676, 677, 766, 767, 776, 777, 6666, 6667, 6676, 6677, 6766, 6767, 6776, 6777, 7666, 7667, 7676, 7677, 7766, 7767, 7776, 7777, 66666, 66667, 66676, 66677, 66766, 66767, 66776, 66777, 67666, 67667

Offset: 1

M. F. Hasler, Mar 27 2015

Vincenzo Librandi, Table of n, a(n) for n = 1..8190
Index entries for 10-automatic sequences.

Cf. A007088 (digits 0 & 1), A007931 (digits 1 & 2), A032810 (digits 2 & 3), A032834 (digits 3 & 4), A256290 (digits 4 & 5), A256291 (digits 5 & 6), A256340 (digits 7 & 8), A256341 (digits 8 & 9).

[n: n in [1..35000] | Set(IntegerToSequence(n, 10)) subset {7, 6}];

[n: n in [1..100000] | Set(Intseq(n)) subset {6,7}]; // Vincenzo Librandi, Aug 19 2016

Flatten[Table[FromDigits[#,10]&/@Tuples[{6,7},n],{n,5}]]

A256292(n)=vector(#n=binary(n+1)[2..-1],i,10^(#n-i))*n~+10^#n\9*6

a(n) = A007931(n) + A002279(A000523(n+1)) = A256291(n) + A256077(n) etc.

A014701 Number of multiplications to compute n-th power by the Chandah-sutra method.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 4, 3, 4, 4, 5, 4, 5, 5, 6, 4, 5, 5, 6, 5, 6, 6, 7, 5, 6, 6, 7, 6, 7, 7, 8, 5, 6, 6, 7, 6, 7, 7, 8, 6, 7, 7, 8, 7, 8, 8, 9, 6, 7, 7, 8, 7, 8, 8, 9, 7, 8, 8, 9, 8, 9, 9, 10, 6, 7, 7, 8, 7, 8, 8, 9, 7, 8, 8, 9, 8, 9, 9, 10, 7, 8, 8, 9, 8, 9, 9

Offset: 1

James Kilfiger (jamesk(AT)maths.warwick.ac.uk)

In other words, number of steps to reach 1 starting from n and using the process: x -> x-1 if n is odd and x -> x/2 otherwise.
a(n) = number of 0's + twice number of 1's (disregarding the leading digit 1) in the binary expansion of n, i.e., A007088(n). - Lekraj Beedassy, May 28 2010
From Daniel Forgues, Jul 31 2012: (Start)
For the binary Fibonacci rabbits sequence (A036299) (cf. OEIS Wiki link below) we have the substitution/concatenation rule: a(n), n >= 3, may be obtained by the concatenation of a(n-1) and a(n-2), with a(1) = 0, a(2) = 1. Thus, using . (dot) as the concatenation operator, we have the recursive substitution/concatenation
    a(n) = a(n-0)
    a(n) = a(n-1).a(n-2)
    a(n) = a(n-2).a(n-3).a(n-3).a(n-4)
    a(n) = a(n-3).a(n-4).a(n-4).a(n-5).a(n-4).a(n-5).a(n-5).a(n-6)
  which suggests the sequence
    {0}
    {1, 2}
    {2, 3, 3, 4}
    {3, 4, 4, 5, 4, 5, 5, 6}
  whose concatenation gives A014701 (this sequence).
Number of multiplications to compute n-th power by the Chandah-sutra method, also called left-to-right binary exponentiation:
  x^1 = x^(   1_2) =                                 (x)   (0 prod)
  x^2 = x^(  10_2) =                         (x^2)         (1 prod)
  x^3 = x^(  11_2) =                         (x^2) * (x)   (2 prod)
  x^4 = x^( 100_2) =               (x^2)^2                 (2 prod)
  x^5 = x^( 101_2) =               (x^2)^2         * (x)   (3 prod)
  x^6 = x^( 110_2) =               (x^2)^2 * (x^2)         (3 prod)
  x^7 = x^( 111_2) =               (x^2)^2 * (x^2) * (x)   (4 prod)
  x^8 = x^(1000_2) = ((x^2)^2)^2                           (3 prod) (End)
From Ya-Ping Lu, Mar 03 2021: (Start)
Index at which record m occurs is A052955(m).
First appearance of m in the sequence (or the record value m) is at n = 2^(m/2 + 1) - 1 for even m, and at n = 3*2^((m - 1)/2) - 1 for odd m.
The last appearance of m in the sequence is at n = 2^m. (End)
a(n) is the digit sum of n-1 in bijective base-2. Since the Fibonacci number F(m) can be defined as the number of ways to compose m as the sum of 1s and 2s, we get that m appears F(m) times in the sequence. - Oscar Cunningham, Apr 14 2024
Conjecture: a(n+1) is the minimal number of steps to go from 0 to n, by choosing before each step, after the first step, whether to keep the same step length or double it. The initial step length is 1. - Jean-Marc Rebert, May 15 2025

			5 -> 4 -> 2 -> 1 so 3 steps are needed to reach 1 hence a(5)=3; 9 -> 8 -> 4 -> 2 -> 1 hence a(9)=4.

Alois P. Heinz, Table of n, a(n) for n = 1..16384 (first 1000 terms from T. D. Noe)
Wawrzyniec Bieniawski, Piotr Masierak, Andrzej Tomski, and Szymon Łukaszyk, Assembly Theory - Formalizing Assembly Spaces and Discovering Patterns and Bounds, Preprints (2025).
C. K. Caldwell, The Prime Glossary, binary exponentiation
Hermann Gruber and Markus Holzer, Optimal Regular Expressions for Palindromes of Given Length, Proceedings of the 46th International Symposium on Mathematical Foundations of Computer Science, Article No. 53, pp. 53:1-53:15, 2021.
J. Jordan and R. Southwell, Further Properties of Reproducing Graphs, Applied Mathematics, Vol. 1 No. 5, 2010, pp. 344-350. - From _N. J. A. Sloane_, Feb 03 2013
Szymon Łukaszyk and Wawrzyniec Bieniawski, Assembly Theory of Binary Messages (How to Assemble a Black Hole and Use it to Assemble New Binary Information?), Preprints (2024).
OEIS Wiki, Binary Fibonacci rabbits sequence
Index to sequences related to the complexity of n
Index entries for sequences related to binary expansion of n

Cf. A003313, A056791, A056792, A115964, A052955.

a014701 1 = 0
a014701 n = a007953 $ a007931 (n - 1)
-- Reinhard Zumkeller, Oct 26 2012

A014701 := proc(n) local j,k; j := n; k := 0; while(j>1) do if j mod 2=1 then j := j-1 else j := j/2 fi; k := k+1 od end;
# second Maple program:
a:= n-> add(i+1, i=Bits[Split](n))-2:
seq(a(n), n=1..128);  # Alois P. Heinz, Aug 30 2021

a[n_] := DigitCount[n, 2] /. {x_, y_} -> 2x + y - 2; Array[a, 100] (* Robert G. Wilson v, Jul 31 2012 *)

a(n)=hammingweight(n)+logint(n,2)-1 \\ Charles R Greathouse IV, Dec 29 2016

def a(n):
    if n==1:
        return 0
    return a(n//2)+1+n%2
for i in range(1,60):
    print(a(i), end=", ")
# Pablo Hueso Merino, Oct 28 2020

a(n) = A056792(n) - 1 = A056791(n) - 2.
a(n) = floor(log_2(n)) + (number of 1's in binary representation of n) - 1. - Corrected (- 1 at end) by Daniel Forgues, Aug 01 2012
a(2^n) = n, a(2^n-1) = 2*(n-1), and for n >= 2, log_2(n) <= a(n) <= 2*log_2(n) - 1. - Robert FERREOL, Oct 01 2014
Let u(1) = 1, u(2*n) = u(n)+1, u(2*n+1) = u(2*n)+1; then a(1) = 0 and a(n) = u(n-1). - Benoit Cloitre, Dec 19 2002
G.f.: -2/(1-x) + (1/(1-x)) * Sum_{k>=0} (2*x^2^k + x^2^(k+1))/(1+x^2^k). - Ralf Stephan, Aug 15 2003
From {0}, apply the substitution rule (n -> n+1, n+2) repeatedly, giving {{0}, {1, 2}, {2, 3, 3, 4}, {3, 4, 4, 5, 4, 5, 5, 6}, ...} and concatenate. - Daniel Forgues, Jul 31 2012
For n > 1: a(n) = A007953(A007931(n-1)). - Reinhard Zumkeller, Oct 26 2012
a(n) >= A003313(n). - Charles R Greathouse IV, Jan 03 2018
a(n) = a(floor(n/2)) + 1 + (n mod 2) for n > 1. - Pablo Hueso Merino, Oct 28 2020
a(n+1) = max_{1<=i<=n} (H(i) + H(n-i)) where H(n) denotes the Hamming weight of n (A000120(n)). See Lemma 8 in Gruber/Holzer 2021 article. - Hermann Gruber, Jun 26 2024

A020450 Primes that contain digits 1 and 2 only.

Original entry on oeis.org

2, 11, 211, 2111, 2221, 12211, 21121, 21211, 21221, 22111, 111121, 111211, 112111, 112121, 1111211, 1121221, 1212121, 1212221, 1221221, 2121121, 2211211, 2221111, 11221211, 12111221, 12121121, 12121211, 12122111, 12122221, 12212111, 12222121

Offset: 1

David W. Wilson

Jason Bard, Table of n, a(n) for n = 1..10000 (first 1000 terms from Vincenzo Librandi)

Cf. A020449 (digits 0 & 1), ..., A020472 (digits 8 & 9). [From M. F. Hasler, Mar 18 2010]

Subsequence of A007931.

[p: p in PrimesUpTo(12222121) | Set(Intseq(p)) subset [1, 2]]; // Vincenzo Librandi, Jul 28 2012

Flatten[Table[Select[FromDigits/@Tuples[{1,2},n],PrimeQ],{n,8}]] (* Vincenzo Librandi, Jul 28 2012 *)

for(nd=1,9, forvec(v=vector(nd,i,[49,50-(i==nd && i>1)]), isprime(t=eval(Strchr(Vecsmall(v)))) && print1(t","))) \\ M. F. Hasler, Mar 18 2010

from sympy import primerange
def checkd(a, c):
    b =  set(int(i) for i in set(str(a)))
    return b.issubset(c)
for n in primerange(2, 2000000):
    if checkd(n, [1, 2]):
        print(n)
# Abhiram R Devesh, May 08 2015

A214676 A(n,k) is n represented in bijective base-k numeration; square array A(n,k), n>=1, k>=1, read by antidiagonals.

Original entry on oeis.org

1, 1, 11, 1, 2, 111, 1, 2, 11, 1111, 1, 2, 3, 12, 11111, 1, 2, 3, 11, 21, 111111, 1, 2, 3, 4, 12, 22, 1111111, 1, 2, 3, 4, 11, 13, 111, 11111111, 1, 2, 3, 4, 5, 12, 21, 112, 111111111, 1, 2, 3, 4, 5, 11, 13, 22, 121, 1111111111

Offset: 1

Alois P. Heinz, Jul 25 2012

The digit set for bijective base-k numeration is {1, 2, ..., k}.

			Square array A(n,k) begins:
:         1,   1,  1,  1,  1,  1,  1,  1, ...
:        11,   2,  2,  2,  2,  2,  2,  2, ...
:       111,  11,  3,  3,  3,  3,  3,  3, ...
:      1111,  12, 11,  4,  4,  4,  4,  4, ...
:     11111,  21, 12, 11,  5,  5,  5,  5, ...
:    111111,  22, 13, 12, 11,  6,  6,  6, ...
:   1111111, 111, 21, 13, 12, 11,  7,  7, ...
:  11111111, 112, 22, 14, 13, 12, 11,  8, ...

Alois P. Heinz, Antidiagonals n = 1..18, flattened
R. R. Forslund, A logical alternative to the existing positional number system, Southwest Journal of Pure and Applied Mathematics, Vol. 1, 1995, 27-29.
Eric Weisstein's World of Mathematics, Zerofree
Wikipedia, Bijective numeration

Columns k=1-9 give: A000042, A007931, A007932, A084544, A084545, A057436, A214677, A214678, A052382.

A(n+1,n) gives A010850.

A:= proc(n, b) local d, l, m; m:= n; l:= NULL;
      while m>0 do  d:= irem(m, b, 'm');
        if d=0 then d:=b; m:=m-1 fi;
        l:= d, l
      od; parse(cat(l))
    end:
seq(seq(A(n, 1+d-n), n=1..d), d=1..12);

A[n_, b_] := Module[{d, l, m}, m = n; l = Nothing; While[m > 0, {m, d} = QuotientRemainder[m, b]; If[d == 0, d = b; m--]; l = {d, l}]; FromDigits @ Flatten @ l];
Table[A[n, d-n+1], {d, 1, 12}, {n, 1, d}] // Flatten (* Jean-François Alcover, May 28 2019, from Maple *)

A084544 Alternate number system in base 4.

Original entry on oeis.org

1, 2, 3, 4, 11, 12, 13, 14, 21, 22, 23, 24, 31, 32, 33, 34, 41, 42, 43, 44, 111, 112, 113, 114, 121, 122, 123, 124, 131, 132, 133, 134, 141, 142, 143, 144, 211, 212, 213, 214, 221, 222, 223, 224, 231, 232, 233, 234, 241, 242, 243, 244, 311, 312, 313, 314, 321

Offset: 1

Robert R. Forslund (forslund(AT)tbaytel.net), Jun 27 2003

			From _Hieronymus Fischer_, Jun 06 2012: (Start)
a(100)  = 1144.
a(10^3) = 33214.
a(10^4) = 2123434.
a(10^5) = 114122134.
a(10^6) = 3243414334.
a(10^7) = 211421121334.
a(10^8) = 11331131343334.
a(10^9) = 323212224213334. (End)

Hieronymus Fischer, Table of n, a(n) for n = 1..10000
EMIS, Mirror site for Southwest Journal of Pure and Applied Mathematics
R. R. Forslund, A logical alternative to the existing positional number system, Southwest Journal of Pure and Applied Mathematics, Vol. 1 1995 pp. 27-29.
R. R. Forslund, Positive Integer Pages [Broken link]
James E. Foster, A Number System without a Zero-Symbol, Mathematics Magazine, Vol. 21, No. 1. (1947), pp. 39-41.
Index entries for 10-automatic sequences.

Cf. A007931, A007932, A052382, A084545, A046034, A089581, A084984, A001742, A001743, A001744, A202267, A202268, A014261, A014263.

def A084544(n):
    m = (3*n+1).bit_length()-1>>1
    return int(''.join((str(((3*n+1-(1<<(m<<1)))//(3<<((m-1-j)<<1))&3)+1) for j in range(m)))) # Chai Wah Wu, Feb 08 2023

From Hieronymus Fischer, Jun 06 and Jun 08 2012: (Start)
The formulas are designed to calculate base-10 numbers only using the digits 1..4.
a(n) = Sum_{j=0..m-1} (1 + b(j) mod 4)*10^j,
where m = floor(log_4(3*n+1)), b(j) = floor((3*n+1-4^m)/(3*4^j)).
Special values:
a(k*(4^n-1)/3) = k*(10^n-1)/9, k = 1,2,3,4.
a((7*4^n-4)/3) = (13*10^n-4)/9 = 10^n + 4*(10^n-1)/9.
a((4^n-1)/3 - 1) = 4*(10^(n-1)-1)/9, n > 1.
Inequalities:
a(n) <= (10^log_4(3*n+1)-1)/9, equality holds for n=(4^k-1)/3, k>0.
a(n) > (4/10)*(10^log_4(3*n+1)-1)/9, n > 0.
Lower and upper limits:
lim inf a(n)/10^log_4(3*n) = 2/45, for n --> infinity.
lim sup a(n)/10^log_4(3*n) = 1/9, for n --> infinity.
G.f.: g(x) = (x^(1/3)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(4/3)*(1 - 5z(j)^4 + 4z(j)^5)/((1-z(j))(1-z(j)^4)), where z(j) = x^4^j.
Also: g(x) = (1/(1-x)) Sum_{j>=0} (1-5(x^4^j)^4 + 4(x^4^j)^5)*x^4^j*f_j(x)/(1-x^4^j), where f_j(x) = 10^j*x^((4^j-1)/3)/(1-(x^4^j)^4). The f_j obey the recurrence f_0(x) = 1/(1-x^4), f_(j+1)(x) = 10x*f_j(x^4).
Also: g(x) = (1/(1-x))* (h_(4,0)(x) + h_(4,1)(x) + h_(4,2)(x) + h_(4,3)(x) - 4*h_(4,4)(x)), where h_(4,k)(x) = Sum_{j>=0} 10^j*x^((4^(j+1)-1)/3) * (x^4^j)^k/(1-(x^4^j)^4).
(End)
a(n) = A045926(n) / 2. - Reinhard Zumkeller, Jan 01 2013

Offset set to 1 according to A007931, A007932 by Hieronymus Fischer, Jun 06 2012

A084545 Alternate number system in base 5.

Original entry on oeis.org

1, 2, 3, 4, 5, 11, 12, 13, 14, 15, 21, 22, 23, 24, 25, 31, 32, 33, 34, 35, 41, 42, 43, 44, 45, 51, 52, 53, 54, 55, 111, 112, 113, 114, 115, 121, 122, 123, 124, 125, 131, 132, 133, 134, 135, 141, 142, 143, 144, 145, 151, 152, 153, 154, 155, 211, 212, 213, 214, 215, 221, 222

Offset: 1

Robert R. Forslund (forslund(AT)tbaytel.net), Jun 27 2003

			From _Hieronymus Fischer_, Jun 06 2012: (Start)
a(100)  = 345.
a(10^3) = 12445.
a(10^4) = 254445.
a(10^5) = 11144445.
a(10^6) = 223444445.
a(10^7) = 4524444445.
a(10^8) = 145544444445.
a(10^9) = 3521444444445. (End)

Hieronymus Fischer, Table of n, a(n) for n = 1..10000
EMIS, Mirror site for Southwest Journal of Pure and Applied Mathematics
R. R. Forslund, A logical alternative to the existing positional number system, Southwest Journal of Pure and Applied Mathematics, Vol. 1 1995 pp. 27-29.
R. R. Forslund, Positive Integer Pages [Wayback Machine link]
James E. Foster, A Number System without a Zero-Symbol, Mathematics Magazine, Vol. 21, No. 1. (1947), pp. 39-41.
Index entries for 10-automatic sequences.

Cf. A007931, A007932, A052382, A084544, A046034, A089581, A084984, A001742, A001743, A001744, A202267, A202268, A014261, A014263.

a(n) = my (w=5); while (n>w, n -= w; w *= 5); my (d=digits(w+n-1, 5)); d[1] = 0; fromdigits(d) + (10^(#d-1)-1)/9 \\ Rémy Sigrist, Dec 04 2019

From Hieronymus Fischer, Jun 06 and Jun 08 2012: (Start)
The formulas are designed to calculate base-10 numbers only using the digits 1..5.
a(n) = Sum_{j=0..m-1} (1 + b(j) mod 5)*10^j, where m = floor(log_5(4*n+1)), b(j) = floor((4*n+1-5^m)/(4*5^j)).
a(k*(5^n-1)/4) = k*(10^n-1)/9, for k = 1,2,3,4,5.
a((9*5^n-5)/4) = (14*10^n-5)/9 = 10^n + 5*(10^n-1)/9.
a((5^n-1)/4 - 1) = 5*(10^(n-1)-1)/9, n>1.
a(n) <= (10^log_5(4*n+1)-1)/9, equality holds for n=(5^k-1)/4, k>0.
a(n) > (5/10)*(10^log_5(4*n+1)-1)/9, n>0.
lim inf a(n)/10^log_5(4*n) = 1/18, for n --> infinity.
lim sup a(n)/10^log_5(4*n) = 1/9, for n --> infinity.
G.f.: g(x) = (x^(1/4)*(1-x))^(-1) sum_{j>=0} 10^j*z(j)^(5/4)*(1 - 6z(j)^5 + 5z(j)^6)/((1-z(j))(1-z(j)^5)), where z(j) = x^5^j.
Also: g(x) = (1/(1-x)) sum_{j>=0} (1-6(x^5^j)^5+5(x^5^j)^6)*x^5^j*f_j(x)/(1-x^5^j), where f_j(x) = 10^j*x^((5^j-1)/4)/(1-(x^5^j)^5). The f_j obey the recurrence f_0(x) = 1/(1-x^5), f_(j+1)(x) = 10x*f_j(x^5).
Also: g(x) = 1/(1-x))*(h_(5,0)(x) + h_(5,1)(x) + h_(5,2)(x) + h_(4,1)(x) + h_(5,4)(x) - 5*h_(5,5)(x)), where h_(5,k)(x) = sum_{j>=0} 10^j*x^((5^(j+1)-1)/4) * (x^5^j)^k/(1-(x^5^j)^5).
(End)

Offset set to 1 according to A007931, A007932 and more terms added by Hieronymus Fischer, Jun 06 2012

A256340 Numbers which have only digits 7 and 8 in base 10.

Original entry on oeis.org

7, 8, 77, 78, 87, 88, 777, 778, 787, 788, 877, 878, 887, 888, 7777, 7778, 7787, 7788, 7877, 7878, 7887, 7888, 8777, 8778, 8787, 8788, 8877, 8878, 8887, 8888, 77777, 77778, 77787, 77788, 77877, 77878, 77887, 77888, 78777, 78778, 78787, 78788, 78877, 78878

Offset: 1

M. F. Hasler, Mar 27 2015

Vincenzo Librandi, Table of n, a(n) for n = 1..8190
Index entries for 10-automatic sequences.

Cf. A007088 (digits 0 & 1), A007931 (digits 1 & 2), A032810 (digits 2 & 3), A032834 (digits 3 & 4), A256290 (digits 4 & 5), A256291 (digits 5 & 6), A256292 (digits 6 & 7), A256341 (digits 8 & 9).

[n: n in [1..35000] | Set(IntegerToSequence(n, 10)) subset {7, 8}];

[n: n in [1..100000] | Set(Intseq(n)) subset {7,8}]; // Vincenzo Librandi, Aug 19 2016

Flatten[Table[FromDigits[#,10]&/@Tuples[{7,8},n],{n,5}]]

A256340(n)=vector(#n=binary(n+1)[2..-1],i,10^(#n-i))*n~+10^#n\9*7

def a(n): return int(bin(n+1)[3:].replace('0', '7').replace('1', '8'))
print([a(n) for n in range(1, 45)]) # Michael S. Branicky, Jul 08 2021

a(n) = A007931(n) + A002280(A000523(n+1)) = A256292(n) + A256077(n) etc.

A214218 List of words over {1,2} with equal numbers of 1's and 2's.

Original entry on oeis.org

12, 21, 1122, 1212, 1221, 2112, 2121, 2211, 111222, 112122, 112212, 112221, 121122, 121212, 121221, 122112, 122121, 122211, 211122, 211212, 211221, 212112, 212121, 212211, 221112, 221121, 221211, 222111, 11112222, 11121222, 11122122, 11122212, 11122221

Offset: 1

N. J. A. Sloane, Jul 18 2012

Of course the empty word also has this property.
All of these, interpreted as decimal integers are divisible by 3, because each pair of "1" and "2" contributes a digital sum of 3, hence the total is divisible by 3. Is there a semiprime in the sequence after 21? - Jonathan Vos Post, Jul 18 2012
The semiprime subsequence contains 21, 11222121, 12122211, 21221121, 22211121, 22212111, and continues with 14 10-digit entries etc. - R. J. Mathar, Jul 19 2012

J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 2.

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Subsequence of A007931, A111066.

sort([seq(seq((10^(2*d)-1)/9+add(10^i,i=s),s=combinat:-choose([$0..(2*d-1)],d)),d=1..4)]); # Robert Israel, Jan 02 2018

Sort[FromDigits/@Flatten[Table[Permutations[PadRight[{},2n,{1,2}]],{n,3}],1]] (* Harvey P. Dale, Aug 30 2016 *)

from itertools import count, islice
from sympy.utilities.iterables import multiset_permutations as mp
def agen():
    for d in count(2, 2):
        for s in mp("1"*(d//2) + "2"*(d//2), d):
            yield int("".join(s))
print(list(islice(agen(), 33))) # Michael S. Branicky, Dec 21 2021

A256341 Numbers which have only digits 8 and 9 in base 10.

Original entry on oeis.org

8, 9, 88, 89, 98, 99, 888, 889, 898, 899, 988, 989, 998, 999, 8888, 8889, 8898, 8899, 8988, 8989, 8998, 8999, 9888, 9889, 9898, 9899, 9988, 9989, 9998, 9999, 88888, 88889, 88898, 88899, 88988, 88989, 88998, 88999, 89888, 89889

Offset: 1

M. F. Hasler, Mar 27 2015

Vincenzo Librandi, Table of n, a(n) for n = 1..8190
Index entries for 10-automatic sequences.

Cf. A007088 (digits 0 & 1), A007931 (digits 1 & 2), A032810 (digits 2 & 3), A032834 (digits 3 & 4), A256290 (digits 4 & 5) - A256292 (digits 6 & 7), A256340 (digits 7 & 8).

[n: n in [1..35000] | Set(IntegerToSequence(n, 10)) subset {8, 9}];

[n: n in [1..100000] | Set(Intseq(n)) subset {8,9}]; // Vincenzo Librandi, Aug 19 2016

Flatten[Table[FromDigits[#,10]&/@Tuples[{8,9},n],{n,5}]]

A256341(n)=vector(#n=binary(n+1)[2..-1],i,10^(#n-i))*n~+10^#n\9*8

def a(n): return int(bin(n+1)[3:].replace('0', '8').replace('1', '9'))
print([a(n) for n in range(1, 45)]) # Michael S. Branicky, Aug 09 2021

a(n) = A007931(n) + A002281(A000523(n+1)) = A256341(n) + A256077(n) etc.

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