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.

Showing 1-10 of 67 results. Next

A046034 Numbers whose digits are primes.

Original entry on oeis.org

2, 3, 5, 7, 22, 23, 25, 27, 32, 33, 35, 37, 52, 53, 55, 57, 72, 73, 75, 77, 222, 223, 225, 227, 232, 233, 235, 237, 252, 253, 255, 257, 272, 273, 275, 277, 322, 323, 325, 327, 332, 333, 335, 337, 352, 353, 355, 357, 372, 373, 375, 377, 522, 523, 525, 527, 532
Offset: 1

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Author

Keywords

Comments

If n is represented as a zerofree base-4 number (see A084544) according to n=d(m)d(m-1)...d(3)d(2)d(1)d(0) then a(n) = Sum_{j=0..m} c(d(j))*10^j, where c(k)=2,3,5,7 for k=1..4. - Hieronymus Fischer, May 30 2012
According to A153025, it seems that 5, 235 and 72335 are the only terms whose square is also a term, i.e., which are also in the sequence A275971 of square roots of the terms which are squares, listed in A191486. - M. F. Hasler, Sep 16 2016

Examples

			a(100)   = 2277,
a(10^3)  = 55327,
a(9881)  = 3233232,
a(10^4)  = 3235757,
a(10922) = 3333333,
a(10^5)  = 227233257.
		

Crossrefs

Programs

  • Haskell
    a046034 n = a046034_list !! (n-1)
    a046034_list = filter (all (`elem` "2357") . show ) [0..]
    -- Reinhard Zumkeller, Jul 19 2011
    
  • Magma
    [n: n in [2..532] | Set(Intseq(n)) subset [2, 3, 5, 7]];  // Bruno Berselli, Jul 19 2011
    
  • Mathematica
    Table[FromDigits /@ Tuples[{2, 3, 5, 7}, n], {n, 3}] // Flatten (* Michael De Vlieger, Sep 19 2016 *)
  • PARI
    is_A046034(n)=Set(isprime(digits(n)))==[1] \\ M. F. Hasler, Oct 12 2013
    
  • Python
    def A046034(n):
        m = (3*n+1).bit_length()-1>>1
        return int(''.join(('2357'[(3*n+1-(1<<(m<<1)))//(3<<((m-1-j)<<1))&3] for j in range(m)))) # Chai Wah Wu, Feb 08 2023

Formula

A055642(a(n)) = A193238(a(n)). - Reinhard Zumkeller, Jul 19 2011
From Hieronymus Fischer, Apr 20, May 30 and Jun 25 2012: (Start)
a(n) = Sum_{j=0..m-1} ((2*b(j)+1) mod 8 + floor(b(j)/4) - floor((b(j)-1)/4))*10^j, where m = floor(log_4(3*n+1)), b(j) = floor((3*n+1-4^m)/(3*4^j)).
a(n) = Sum_{j=0..m-1} A010877(A005408(b(j)) + A002265(b(j)) - A002265(b(j)-1))*10^j.
Special values:
a(1*(4^n-1)/3) = 2*(10^n-1)/9.
a(2*(4^n-1)/3) = 1*(10^n-1)/3.
a(3*(4^n-1)/3) = 5*(10^n-1)/9.
a(4*(4^n-1)/3) = 7*(10^n-1)/9.
Inequalities:
a(n) <= 2*(10^log_4(3*n+1)-1)/9, equality holds for n = (4^k-1)/3, k>0.
a(n) <= 2*A084544(n), equality holds iff all digits of A084544(n) are 1.
a(n) > A084544(n).
Lower and upper limits:
lim inf a(n)/10^log_4(n) = (7/90)*10^log_4(3) = 0.48232167706987..., for n -> oo.
lim sup a(n)/10^log_4(n) = (2/9)*10^log_4(3) = 1.378061934485343..., for n -> oo.
where 10^log_4(n) = n^1.66096404744...
G.f.: g(x) = (x^(1/3)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(4/3)*(2 + z(j) + 2*z(j)^2 + 2*z(j)^3 - 7*z(j)^4)/(1-z(j)^4), where z(j) = x^4^j.
Also g(x) = (x^(1/3)*(1-x))^(-1) Sum_{j>=0} 10^j*z(j)^(4/3)*(1-z(j))*(2 + 3*z(j) + 5*z(j)^2 + 7*z(j)^3)/(1-z(j)^4), where z(j)=x^4^j.
Also: g(x) = (1/(1-x))*(2*h_(4,0)(x) + h_(4,1)(x) + 2*h_(4,2)(x) + 2*h_(4,3)(x) - 7*h_(4,4)(x)), where h_(4,k)(x) = Sum_{j>=0} 10^j*x^((4^(j+1)-1)/3)*x^(k*4^j)/(1-x^4^(j+1)). (End)
Sum_{n>=1} 1/a(n) = 1.857333779940977502574887651449435985318556794733869779170825138954093657197... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 15 2024

Extensions

More terms from Cino Hilliard, Aug 06 2006
Typo in second formula corrected by Hieronymus Fischer, May 12 2012
Two typos in example section corrected by Hieronymus Fischer, May 30 2012

A007931 Numbers that contain only 1's and 2's. Nonempty binary strings of length n in lexicographic order.

Original entry on oeis.org

1, 2, 11, 12, 21, 22, 111, 112, 121, 122, 211, 212, 221, 222, 1111, 1112, 1121, 1122, 1211, 1212, 1221, 1222, 2111, 2112, 2121, 2122, 2211, 2212, 2221, 2222, 11111, 11112, 11121, 11122, 11211, 11212, 11221, 11222, 12111, 12112, 12121, 12122
Offset: 1

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Author

R. Muller

Keywords

Comments

Numbers written in the dyadic system [Smullyan, Stillwell]. - N. J. A. Sloane, Feb 13 2019
Logic-binary sequence: prefix it by the empty word to have all binary words on the alphabet {1,2}.
The least binary word of length k is a(2^k - 1).
See Mathematica program for logic-binary sequence using (0,1) in place of (1,2); the sequence starts with 0,1,00,01,10. - Clark Kimberling, Feb 09 2012
A007953(a(n)) = A014701(n+1); A007954(a(n)) = A048896(n). - Reinhard Zumkeller, Oct 26 2012
a(n) is n written in base 2 where zeros are not allowed but twos are. The two distinct digits used are 1, 2 instead of 0, 1. To obtain this sequence from the "canonical" base 2 sequence with zeros allowed, just replace any 0 with a 2 and then subtract one from the group of digits situated on the left: (10-->2; 100-->12; 110-->22; 1000-->112; 1010-->122). - Robin Garcia, Jan 31 2014
For numbers made of only two different digits, see also A007088 (digits 0 & 1), A032810 (digits 2 & 3), A032834 (digits 3 & 4), A256290 (digits 4 & 5), A256291 (digits 5 & 6), A256292 (digits 6 & 7), A256340(digits 7 & 8), A256341 (digits 8 & 9), and A032804-A032816 (in other bases). Numbers with exactly two distinct (but unspecified) digits in base 10 are listed in A031955, for other bases in A031948-A031954. - M. F. Hasler, Apr 04 2015
The variant with digits {0, 1} instead of {1, 2} is obtained by deleting all initial digits in sequence A007088 (numbers written in base 2). - M. F. Hasler, Nov 03 2020

Examples

			Positive numbers may not start with 0 in the OEIS, otherwise this sequence would have been written as: 0, 1, 00, 01, 10, 11, 000, 001, 010, 011, 100, 101, 110, 111, 0000, 0001, 0010, 0011, 0100, 0101, 0110, 0111, 1000, 1001, 1010, 1011, 1100, 1101, 1110, 1111, 00000, 00001, 00010, 00011, 00100, 00101, 00110, 00111, 01000, 01001, 01010, 01011, ...
From _Hieronymus Fischer_, Jun 06 2012: (Start)
a(10)   = 122.
a(100)  = 211212.
a(10^3) = 222212112.
a(10^4) = 1122211121112.
a(10^5) = 2111122121211112.
a(10^6) = 2221211112112111112.
a(10^7) = 11221112112122121111112.
a(10^8) = 12222212122221111211111112.
a(10^9) = 22122211221212211212111111112. (End)
		

References

  • J.-P. Allouche and J. Shallit, Automatic Sequences, Cambridge Univ. Press, 2003, p. 2. - From N. J. A. Sloane, Jul 26 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.
  • R. M. Smullyan, Theory of Formal Systems, Princeton, 1961.
  • John Stillwell, Reverse Mathematics, Princeton, 2018. See p. 90.

Crossrefs

Cf. A007932 (digits 1-3), A059893, A045670, A052382 (digits 1-9), A059939, A059941, A059943, A032924, A084544, A084545, A046034 (prime digits 2,3,5,7), A089581, A084984 (no prime digits); A001742, A001743, A001744: loops; A202267 (digits 0, 1 and primes), A202268 (digits 1,4,6,8,9), A014261 (odd digits), A014263 (even digits).
Cf. A007088 (digits 0 & 1), A032810 (digits 2 & 3), A032834 (digits 3 & 4), A256290 (digits 4 & 5), A256291 (digits 5 & 6), A256292 (digits 6 & 7), A256340 (digits 7 & 8), A256341 (digits 8 & 9), and A032804-A032816 (in other bases).
Cf. A020450 (primes).

Programs

  • Haskell
    a007931 n = f (n + 1) where
       f x = if x < 2 then 0 else (10 * f x') + m + 1
         where (x', m) = divMod x 2
    -- Reinhard Zumkeller, Oct 26 2012
    
  • Magma
    [n: n in [1..100000] | Set(Intseq(n)) subset {1,2}]; // Vincenzo Librandi, Aug 19 2016
    
  • Maple
    # Maple program to produce the sequence:
    a:= proc(n) local m, r, d; m, r:= n, 0;
          while m>0 do d:= irem(m, 2, 'm');
            if d=0 then d:=2; m:= m-1 fi;
            r:= d, r
          od; parse(cat(r))/10
        end:
    seq(a(n), n=1..100);  # Alois P. Heinz, Aug 26 2016
    # Maple program to invert this sequence: given a(n), it returns n. - N. J. A. Sloane, Jul 09 2012
    invert7931:=proc(u)
    local t1,t2,i;
    t1:=convert(u,base,10);
    [seq(t1[i]-1,i=1..nops(t1))];
    [op(%),1];
    t2:=convert(%,base,2,10);
    add(t2[i]*10^(i-1),i=1..nops(t2))-1;
    end;
  • Mathematica
    f[n_] := FromDigits[Rest@IntegerDigits[n + 1, 2] + 1]; Array[f, 42] (* Robert G. Wilson v Sep 14 2006 *)
    (* Next, A007931 using (0,1) instead of (1,2) *)
    d[n_] := FromDigits[Rest@IntegerDigits[n + 1, 2] + 1]; Array[FromCharacterCode[ToCharacterCode[ToString[d[#]]] - 1] &, 100] (* Peter J. C. Moses, at request of Clark Kimberling, Feb 09 2012 *)
    Flatten[Table[FromDigits/@Tuples[{1,2},n],{n,5}]] (* Harvey P. Dale, Sep 13 2014 *)
  • PARI
    apply( {A007931(n)=fromdigits([d+1|d<-binary(n+1)[^1]])}, [1..44]) \\ M. F. Hasler, Nov 03 2020, replacing older code from Mar 26 2015
    
  • PARI
    /* inverse function */ apply( {A007931_inv(N)=fromdigits([d-1|d<-digits(N)],2)+2<M. F. Hasler, Nov 09 2020
    
  • Python
    def a(n): return int(bin(n+1)[3:].replace('1', '2').replace('0', '1'))
    print([a(n) for n in range(1, 45)]) # Michael S. Branicky, May 13 2021
    
  • Python
    def A007931(n): return int(s:=bin(n+1)[3:])+(10**(len(s))-1)//9 # Chai Wah Wu, Jun 13 2025

Formula

To get a(n), write n+1 in base 2, remove initial 1, add 1 to all remaining digits: e.g., eleven (11) in base 2 is 1011; remove initial 1 and add 1 to remaining digits: a(10)=122. - Clark Kimberling, Mar 11 2003
Conversely, given a(n), to get n: subtract 1 from all digits, prefix with an initial 1, convert this binary number to base 10, subtract 1. E.g., a(6)=22 -> 11 -> 111 -> 7 -> 6. - N. J. A. Sloane, Jul 09 2012
a(n) = A053645(n+1)+A002275(A000523(n)) = a(n-2^b(n))+10^b(n) where b(n) = A059939(n) = floor(log_2(n+1)-1). - Henry Bottomley, Feb 14 2001
From Hieronymus Fischer, Jun 06 2012 and Jun 08 2012: (Start)
The formulas are designed to calculate base-10 numbers only using the digits 1 and 2.
a(n) = Sum_{j=0..m-1} (1 + b(j) mod 2)*10^j, where m = floor(log_2(n+1)), b(j) = floor((n+1-2^m)/(2^j)).
Special values:
a(k*(2^n-1)) = k*(10^n-1)/9, k= 1,2.
a(3*2^n-2) = (11*10^n-2)/9 = 10^n+2*(10^n-1)/9.
a(2^n-2) = 2*(10^(n-1)-1)/9, n>1.
Inequalities:
a(n) <= (10^log_2(n+1)-1)/9, equality holds for n=2^k-1, k>0.
a(n) > (2/10)*(10^log_2(n+1)-1)/9.
Lower and upper limits:
lim inf a(n)/10^log_2(n) = 1/45, for n --> infinity.
lim sup a(n)/10^log_2(n) = 1/9, for n --> infinity.
G.f.: g(x) = (1/(x(1-x)))*sum_{j=0..infinity} 10^j* x^(2*2^j)*(1 + 2 x^2^j)/(1 + x^2^j).
Also: g(x) = (1/(1-x))*(h_(2,0)(x) + h_(2,1)(x) - 2*h_(2,2)(x)), where h_(2,k)(x) = sum_{j>=0} 10^j*x^(2^(j+1)-1)*x^(k*2^j)/(1-x^2^(j+1)).
Also: g(x) = (1/(1-x)) sum_{j>=0} (1 - 3(x^2^j)^2 + 2(x^2^j)^3)*x^2^j*f_j(x)/(1-x^2^j), where f_j(x) = 10^j*x^(2^j-1)/(1-(x^2^j)^2). The f_j obey the recurrence f_0(x) = 1/(1-x^2), f_(j+1)(x) = 10x*f_j(x^2). (End)

Extensions

Some crossrefs added by Hieronymus Fischer, Jun 06 2012
Edited by M. F. Hasler, Mar 26 2015

A014261 Numbers that contain odd digits only.

Original entry on oeis.org

1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 31, 33, 35, 37, 39, 51, 53, 55, 57, 59, 71, 73, 75, 77, 79, 91, 93, 95, 97, 99, 111, 113, 115, 117, 119, 131, 133, 135, 137, 139, 151, 153, 155, 157, 159, 171, 173, 175, 177, 179, 191, 193, 195, 197, 199, 311, 313, 315, 317, 319
Offset: 1

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Author

Keywords

Comments

Or, numbers whose product of digits is odd.
Complement of A007928; A196563(a(n)) = 0. - Reinhard Zumkeller, Oct 04 2011
If n is represented as a zerofree base-5 number (see A084545) according to n = d(m)d(m-1)...d(3)d(2)d(1)d(0) then a(n) = Sum_{j = 0..m} c(d(j))*10^j, where c(k) = 1, 3, 5, 7, 9 for k = 1..5. - Hieronymus Fischer, Jun 06 2012

Examples

			a(10^3) = 13779.
a(10^4) = 397779.
a(10^5) = 11177779.
a(10^6) = 335777779.
		

Crossrefs

Subsequence of A059708 and of A225985. A066640 and A030096 are subsequences.

Programs

  • Haskell
    a014261 n = a014261_list !! (n-1)
    a014261_list = filter (all (`elem` "13579") . show) [1,3..]
    -- Reinhard Zumkeller, Jul 05 2011
    
  • Magma
    [ n : n in [1..129] | IsOdd(&*Intseq(n,10)) ];
    
  • Mathematica
    Select[Range[400], OddQ[Times@@IntegerDigits[#]] &] (* Alonso del Arte, Feb 21 2014 *)
  • PARI
    is(n)=Set(digits(n)%2)==[1] \\ Charles R Greathouse IV, Jul 06 2017
    
  • PARI
    a(n)={my(k=1); while(n>5^k, n-=5^k; k++); fromdigits([2*d+1 | d<-digits(5^k+n-1, 5)]) - 3*10^k} \\ Andrew Howroyd, Jan 17 2020
    
  • Python
    from itertools import islice, count
    def A014261(): return filter(lambda n: set(str(n)) <= {'1','3','5','7','9'}, count(1,2))
    A014261_list = list(islice(A014261(),20)) # Chai Wah Wu, Nov 22 2021
    
  • Python
    from itertools import count, islice, product
    def agen(): yield from (int("".join(p)) for d in count(1) for p in product("13579", repeat=d))
    print(list(islice(agen(), 60))) # Michael S. Branicky, Jan 13 2022

Formula

A121759(a(n)) = a(n); A000035(A007959(a(n))) = 1. - Reinhard Zumkeller, Nov 30 2007
From Reinhard Zumkeller, Aug 30 2009: (Start)
a(n+1) - a(n) = A164898(n). - Reinhard Zumkeller, Aug 30 2009
a(n+1) = h(a(n)) with h(x) = 1 + (if x mod 10 < 9 then x + x mod 2 else 10*h(floor(x/10)));
a(n) = f(n, 1) where f(n, x) = if n = 1 then x else f(n-1, h(x)). (End)
From Hieronymus Fischer, Jun 06 2012: (Start)
a(n) = Sum_{j = 0..m-1} ((2*b_j(n)+1) mod 10)*10^j, where b_j(n) = floor((4*n+1-5^m)/(4*5^j)), m = floor(log_5(4*n+1)).
a(1*(5^n-1)/4) = 1*(10^n-1)/9.
a(2*(5^n-1)/4) = 1*(10^n-1)/3.
a(3*(5^n-1)/4) = 5*(10^n-1)/9.
a(4*(5^n-1)/4) = 7*(10^n-1)/9.
a(5*(5^n-1)/4) = 10^n - 1.
a((5^n-1)/4 + 5^(n-1)-1) = (10^n-5)/5.
a(n) = (10^log_5(4*n+1)-1)/9 for n = (5^k-1)/4, k > 0.
a(n) < (10^log_5(4*n+1)-1)/9 for (5^k-1)/4 < n < (5^(k+1)-1)/4, k > 0.
a(n) <= 27/(9*2^log_5(9)-1)*(10^log_5(4*n+1)-1)/9 for n > 0, equality holds for n = 2.
a(n) > 0.776*10^log_5(4*n+1)-1)/9 for n > 0.
a(n) >= A001742(n), equality holds for n = (5^k-1)/4, k > 0.
a(n) = A084545(n) if and only if all digits of A084545(n) are 1, else a(n) > A084545(n).
G.f.: g(x)= (x^(1/4)*(1-x))^(-1) Sum_{j >= 0} 10^j*z(j)^(5/4)*(1-z(j))*(1 + 3*z(j) + 5*z(j)^2 + 7*z(j)^3 + 9*z(j)^4)/(1-z(j)^5), where z(j) = x^5^j.
Also: g(x) = (1/(1-x))*(h_(5,0)(x) + 2*h_(5,1)(x) + 2*h_(5,2)(x) + 2*h_(5,3)(x) + 2*h_(5,4)(x) - 9*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)
a(n) = A225985(A226091(n)). - Reinhard Zumkeller, May 26 2013
Sum_{n>=1} 1/a(n) = A194181. - Bernard Schott, Jan 13 2022

Extensions

More terms from Robert G. Wilson v, Oct 18 2002
Examples and crossrefs added by Hieronymus Fischer, Jun 06 2012

A005823 Numbers whose ternary expansion contains no 1's.

Original entry on oeis.org

0, 2, 6, 8, 18, 20, 24, 26, 54, 56, 60, 62, 72, 74, 78, 80, 162, 164, 168, 170, 180, 182, 186, 188, 216, 218, 222, 224, 234, 236, 240, 242, 486, 488, 492, 494, 504, 506, 510, 512, 540, 542, 546, 548, 558, 560, 564, 566, 648, 650, 654, 656, 666, 668, 672, 674
Offset: 1

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Author

Keywords

Comments

The set of real numbers between 0 and 1 that contain no 1's in their ternary expansion is the well-known Cantor set with Hausdorff dimension log 2 / log 3.
Complement of A081606. - Reinhard Zumkeller, Mar 23 2003
Numbers k such that the k-th Apery number is congruent to 1 (mod 3) (cf. A005258). - Benoit Cloitre, Nov 30 2003
Numbers k such that the k-th central Delannoy number is congruent to 1 (mod 3) (cf. A001850). - Benoit Cloitre, Nov 30 2003
Numbers k such that there exists a permutation p_1, ..., p_k of 1, ..., k such that i + p_i is a power of 3 for every i. - Ray Chandler, Aug 03 2004
Subsequence of A125292. - Reinhard Zumkeller, Nov 26 2006
The first 2^n terms of the sequence could be obtained using the Cantor process for the segment [0,3^n-1]. E.g., for n=2 we have [0,{1},2,{3,4,5},6,{7},8]. The numbers outside of braces are the first 4 terms of the sequence. Therefore the terms of the sequence could be called "Cantor's numbers". - Vladimir Shevelev, Jun 13 2008
Mahler proved that positive a(n) is never a square. - Michel Marcus, Nov 12 2012
Define t: Z -> P(R) so that t(k) is the translated Cantor ternary set spanning [k, k+1], and let T be the union of t(a(n)) for all n. T = T * 3 = T / 3 is the closure of the Cantor ternary set under multiplication by 3. - Peter Munn, Oct 30 2019

References

  • K. J. Falconer, The Geometry of Fractal Sets, Cambridge, 1985; p. 14.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Twice A005836.
Cf. A088917 (characteristic function), A306556.

Programs

  • Maple
    a:= proc(n) option remember;
          `if`(n=1, 0, `if`(irem (n, 2, 'q')=0, 3*a(q)+2, 3*a(q+1)))
        end:
    seq(a(n), n=1..100); # Alois P. Heinz, Apr 19 2012
  • Mathematica
    Select[ Range[ 0, 729 ], (Count[ IntegerDigits[ #, 3 ], 1 ]==0)& ]
    Select[Range[0,700],DigitCount[#,3,1]==0&] (* Harvey P. Dale, Mar 12 2016 *)
  • PARI
    is(n)=while(n,if(n%3==1,return(0),n\=3));1 \\ Charles R Greathouse IV, Apr 20 2012
    
  • PARI
    a(n)=n=binary(n-1);sum(i=1,#n,2*n[i]*3^(#n-i)) \\ Charles R Greathouse IV, Apr 20 2012
    
  • PARI
    a(n)=2*fromdigits(binary(n-1),3) \\ Charles R Greathouse IV, Aug 24 2016
    
  • Python
    def A005823(n):
        return 2*int(format(n-1,'b'),3) # Chai Wah Wu, Jan 04 2015

Formula

a(n) = 2 * A005836(n).
a(2n) = 3*a(n)+2, a(2n+1) = 3*a(n+1), a(1) = 0.
a(n) = Sum_{k = 1..n} 1 + 3^A007814(k). - Philippe Deléham, Jul 09 2005
A125291(a(n)) = 1 for n>0. - Reinhard Zumkeller, Nov 26 2006
From Reinhard Zumkeller, Mar 02 2008: (Start)
A062756(a(n)) = 0.
If the offset were changed to zero, then: a(0) = 0, a(n+1) = f(a(n)+1, a(n)+1) where f(x, y) = if x < 3 and x <> 1 then y else if x mod 3 = 1 then f(y+1, y+1) else f(floor(x/3), y). (End)
G.f. g(x) satisfies g(x) = 3*g(x^2)*(1+1/x) + 2*x^2/(1-x^2). - Robert Israel, Jan 04 2015
Sum_{n>=2} 1/a(n) = 1.341426555483087715426958452292349687410838545707857407585878304836140592352... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 12 2022

Extensions

More terms from Sascha Kurz, Mar 24 2002
Offset corrected by N. J. A. Sloane, Mar 02 2008. This may require some of the formulas to be adjusted.

A196564 Number of odd digits in decimal representation of n.

Original entry on oeis.org

0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 0, 1, 0, 1, 0, 1
Offset: 0

Views

Author

Reinhard Zumkeller, Oct 04 2011

Keywords

Crossrefs

Programs

  • Haskell
    a196564 n = length [d | d <- show n, d `elem` "13579"]
    -- Reinhard Zumkeller, Feb 22 2012, Oct 04 2011
    
  • Maple
    A196564 := proc(n)
            if n =0 then
                    0;
            else
                    convert(n,base,10) ;
                    add(d mod 2,d=%) ;
            end if:
    end proc: # R. J. Mathar, Jul 13 2012
  • Mathematica
    Table[Total[Mod[IntegerDigits[n],2]],{n,0,100}] (* Zak Seidov, Oct 13 2015 *)
  • PARI
    a(n) = #select(x->x%2, digits(n)); \\ Michel Marcus, Oct 14 2015
    
  • Python
    def a(n): return sum(1 for d in str(n) if d in "13579")
    print([a(n) for n in range(100)]) # Michael S. Branicky, May 15 2022

Formula

a(n) = A055642(n) - A196563(n);
a(A014263(n)) = 0; a(A007957(n)) > 0.
From Hieronymus Fischer, May 30 2012: (Start)
a(n) = Sum_{j=0..m} (floor(n/(2*10^j) + (1/2)) - floor(n/(2*10^j))), where m=floor(log_10(n)).
a(10*n+k) = a(n) + a(k), 0<=k<10, n>=0.
a(n) = a(floor(n/10)) + a(n mod 10), n>=0.
a(n) = Sum_{j=0..m} a(floor(n/10^j) mod 10), n>=0.
a(A014261(n)) = floor(log_5(4*n+1)), n>0.
G.f.: g(x) = (1/(1-x))*Sum_{j>=0} x^10^j/(1+x^10^j). (End)

A196563 Number of even digits in decimal representation of n.

Original entry on oeis.org

1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 2, 1, 2, 1, 2, 1
Offset: 0

Views

Author

Reinhard Zumkeller, Oct 04 2011

Keywords

Crossrefs

Programs

  • Haskell
    a196563 n = length [d | d <- show n, d `elem` "02468"]
    -- Reinhard Zumkeller, Feb 22 2012, Oct 04 2011
    
  • Maple
    A196563 := proc(n)
            if n =0 then
                    1;
            else
                    convert(n,base,10) ;
                    add(1-(d mod 2),d=%) ;
            end if:
    end proc: # R. J. Mathar, Jul 13 2012
  • Mathematica
    Table[Count[Mod[IntegerDigits[n],2],0][n],{n,0,100}] (* Zak Seidov, Oct 13 2015 *)
    Table[Count[IntegerDigits[n],?EvenQ],{n,0,120}] (* _Harvey P. Dale, Feb 22 2020 *)
  • PARI
    a(n) = #select(x->(!(x%2)), if (n, digits(n), [0])); \\ Michel Marcus, Oct 14 2015
    
  • Python
    def a(n): return sum(1 for d in str(n) if d in "02468")
    print([a(n) for n in range(100)]) # Michael S. Branicky, May 15 2022

Formula

a(n) = A055642(n) - A196564(n);
a(A014261(n)) = 0; a(A007928(n)) > 0.
From Hieronymus Fischer, May 30 2012: (Start)
a(n) = Sum_{j=0..m} (1 + floor(n/(2*10^j)) - floor(n/(2*10^j) + (1/2))), where m=floor(log_10(n)).
a(10*n+k) = a(n) + a(k), 0<=k<10, n>=1.
a(n) = a(floor(n/10))+a(n mod 10), n>=10.
a(n) = Sum_{j=0..m} a(floor(n/10^j) mod 10), n>=0.
a(A014263(n)) = 1 + floor(log_5(n-1)), n>1.
G.f.: g(x) = 1 + (1/(1-x))*Sum_{j>=0} x^(2*10^j)/(1+x^10^j). (End)

A084984 Numbers containing no prime digits.

Original entry on oeis.org

0, 1, 4, 6, 8, 9, 10, 11, 14, 16, 18, 19, 40, 41, 44, 46, 48, 49, 60, 61, 64, 66, 68, 69, 80, 81, 84, 86, 88, 89, 90, 91, 94, 96, 98, 99, 100, 101, 104, 106, 108, 109, 110, 111, 114, 116, 118, 119, 140, 141, 144, 146, 148, 149, 160, 161, 164, 166, 168, 169
Offset: 1

Views

Author

Meenakshi Srikanth (menakan_s(AT)yahoo.com), Jun 27 2003

Keywords

Comments

Complement of A118950. - Reinhard Zumkeller, Jul 19 2011
If n-1 is represented as a base-6 number (see A007092) according to n-1=d(m)d(m-1)...d(3)d(2)d(1)d(0) then a(n)= sum_{j=0..m} c(d(j))*10^j, where c(k)=0,1,4,6,8,9 for k=0..5. - Hieronymus Fischer, May 30 2012

Examples

			166 has digits 1 and 6 and they are nonprime digits.
a(1000) = 8686.
a(10^4) = 118186
a(10^5) = 4090986.
a(10^6) = 66466686.
		

Crossrefs

Programs

  • Haskell
    a084984 n = a084984_list !! (n-1)
    a084984_list = filter (not . any (`elem` "2357") . show ) [0..]
    -- Reinhard Zumkeller, Jul 19 2011
    
  • Magma
    [n: n in [0..169] | forall{d: d in [2,3,5,7] | d notin Set(Intseq(n))}];  // Bruno Berselli, Jul 19 2011
    
  • Mathematica
    npdQ[n_]:=And@@Table[FreeQ[IntegerDigits[n],i],{i,{2,3,5,7}}]; Select[ Range[ 0,200],npdQ] (* Harvey P. Dale, Jul 22 2013 *)
  • PARI
    is(n)=isprime(eval(Vec(Str(n))))==0 \\ Charles R Greathouse IV, Feb 20 2012
    
  • PARI
    my(table=[0,1,4,6,8,9]); \
    a(n) = fromdigits([table[d+1] |d<-digits(n-1,6)]); \\ Kevin Ryde, May 27 2025

Formula

A193238(a(n)) = 0. - Reinhard Zumkeller, Jul 19 2011
a(n) >> n^1.285. - Charles R Greathouse IV, Feb 20 2012
From Hieronymus Fischer, May 30 and Jun 25 2012: (Start)
a(n) = ((2*b_m(n)+1) mod 10 + floor((b_m(n)+4)/5) - floor((b_m(n)+1)/5))*10^m + sum_{j=0..m-1} ((2*b_j(n))) mod 12 + floor(b_j(n)/6) - floor((b_j(n)+1)/6) + floor((b_j(n)+4)/6) - floor((b_j(n)+5)/6)))*10^j, where n>1, b_j(n)) = floor((n-1-6^m)/6^j), m = floor(log_6(n-1)).
Special values:
a(1*6^n+1) = 1*10^n.
a(2*6^n+1) = 4*10^n.
a(3*6^n+1) = 6*10^n.
a(4*6^n+1) = 8*10^n.
a(5*6^n+1) = 9*10^n.
a(2*6^n) = 2*10^n - 1.
a(n) = 10^log_6(n-1) for n=6^k+1, k>0.
Inequalities:
a(n) < 10^log_6(n-1) for 6^k+10.
a(n) > 10^log_6(n-1) for 2*6^k=0.
a(n) <= 4*10^(log_6(n-1)-log_6(2)) = 1.641372618*10^(log_6(n-1)), equality holds for n=2*6^k+1, k>=0.
a(n) > 2*10^(log_6(n-1)-log_6(2)) = 0.820686309*10^(log_6(n-1)).
a(n) = A007092(n-1) iff the digits of A007092(n-1) are 0 or 1, a(n)>A007092(n-1), else.
a(n) >= A202267(n), equality holds if the representation of n-1 as a base-6 number has only digits 0 or 1.
Lower and upper limits:
lim inf a(n)/10^log_6(n) = 2/10^log_6(2) = 0.820686309, for n --> inf.
lim sup a(n)/10^log_6(n) = 4/10^log_6(2) = 1.641372618, for n --> inf.
where 10^log_6(n) = n^1.2850972089...
G.f.: g(x) = (x/(1-x))*sum_{j>=0} 10^j*x^6^j * (1-x^6^j)*((1+x^6^j)^4 + 4(1+2x^6^j) * x^(3*6^j))/(1-x^6^(j+1)).
Also: g(x) = (x/(1-x))*(h_(6,1)(x) + 3*h_(6,2)(x) + 2*h_(6,3)(x) + 2*h_(6,4)(x) + h_(6,5)(x) - 9*h_(6,6)(x)), where h_(6,k)(x) = sum_{j>=0} 10^j*x^(k*6^j)/(1-x^6^(j+1)). (End)
Sum_{n>=2} 1/a(n) = 3.614028405471074989720026361356036456697082276983705341077940360653303099111... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 15 2024

Extensions

0 added by N. J. A. Sloane, Feb 02 2009
100 added by Arkadiusz Wesolowski, Mar 10 2011
Examples for n>=10^3 added by Hieronymus Fischer, May 30 2012

A001744 Numbers n such that every digit contains a loop (version 2).

Original entry on oeis.org

0, 4, 6, 8, 9, 40, 44, 46, 48, 49, 60, 64, 66, 68, 69, 80, 84, 86, 88, 89, 90, 94, 96, 98, 99, 400, 404, 406, 408, 409, 440, 444, 446, 448, 449, 460, 464, 466, 468, 469, 480, 484, 486, 488, 489, 490, 494, 496, 498, 499, 600, 604, 606, 608, 609, 640, 644, 646
Offset: 1

Views

Author

Keywords

Comments

See A001743 for the other version.
If n-1 is represented as a base-5 number (see A007091) according to n-1 = d(m)d(m-1)...d(3)d(2)d(1)d(0) then a(n)= Sum_{j=0..m} c(d(j))*10^j, where c(k)=0,4,6,8,9 for k=0..4. - Hieronymus Fischer, May 30 2012

Examples

			a(1000) = 46999.
a(10^4) = 809999.
a(10^5) = 44499999.
a(10^6) = 668999999.
		

Crossrefs

Programs

  • Mathematica
    FromDigits/@Tuples[{0,4,6,8,9},3] (* Harvey P. Dale, Aug 16 2018 *)
  • PARI
    is(n) = #setintersect(vecsort(digits(n), , 8), [1, 2, 3, 5, 7])==0 \\ Felix Fröhlich, Sep 09 2019

Formula

From Hieronymus Fischer, May 30 2012: (Start)
a(n) = ((2*b_m(n)) mod 8 + 4 + floor(b_m(n)/4) - floor((b_m(n)+1)/4))*10^m + sum_{j=0..m-1} ((2*b_j(n))) mod 10 + 2*floor((b_j(n)+4)/5) - floor((b_j(n)+1)/5) -floor(b_j(n)/5)))*10^j, where n>1, b_j(n)) = floor((n-1-5^m)/5^j), m = floor(log_5(n-1)).
a(1*5^n+1) = 4*10^n.
a(2*5^n+1) = 6*10^n.
a(3*5^n+1) = 8*10^n.
a(4*5^n+1) = 9*10^n.
a(n) = 4*10^log_5(n-1) for n=5^k+1,
a(n) < 4*10^log_5(n-1), otherwise.
a(n) > 10^log_5(n-1) n>1.
a(n) = 4*A007091(n-1), iff the digits of A007091(n-1) are 0 or 1.
G.f.: g(x) = (x/(1-x))*sum_{j>=0} 10^j*x^5^j*(1-x^5^j)*(4 + 6x^5^j + 8(x^2)^5^j + 9(x^3)^5^j)/(1-x^5^(j+1)).
Also: g(x) = (x/(1-x))*(4*h_(5,1)(x) + 2*h_(5,2)(x) + 2*h_(5,3)(x) + h_(5,4)(x) - 9*h_(5,5)(x)), where h_(5,k)(x) = sum_{j>=0} 10^j*(x^5^j)^k/(1-(x^5^j)^5). (End)

Extensions

Ambiguous comment deleted by Zak Seidov, May 25 2010
Examples added by Hieronymus Fischer, May 30 2012

A001743 Numbers in which every digit contains at least one loop (version 1).

Original entry on oeis.org

0, 6, 8, 9, 60, 66, 68, 69, 80, 86, 88, 89, 90, 96, 98, 99, 600, 606, 608, 609, 660, 666, 668, 669, 680, 686, 688, 689, 690, 696, 698, 699, 800, 806, 808, 809, 860, 866, 868, 869, 880, 886, 888, 889, 890, 896, 898, 899, 900, 906, 908, 909, 960, 966, 968, 969
Offset: 1

Views

Author

Keywords

Comments

See A001744 for the other version.
If n-1 is represented as a base-4 number (see A007090) according to n-1 = d(m)d(m-1)...d(3)d(2)d(1)d(0) then a(n) = Sum_{j=0..m} c(d(j))*10^j, where c(k)=0,6,8,9 for k=0..3. - Hieronymus Fischer, May 30 2012

Examples

			a(1000) = 99896.
a(10^4) = 8690099.
a(10^5) = 680688699.
		

Crossrefs

Programs

  • Mathematica
    Union[Flatten[Table[FromDigits/@Tuples[{0,6,8,9},n],{n,3}]]] (* Harvey P. Dale, Sep 04 2013 *)
  • PARI
    is(n) = #setintersect(vecsort(digits(n), , 8), [1, 2, 3, 4, 5, 7])==0 \\ Felix Fröhlich, Sep 09 2019

Formula

From Hieronymus Fischer, May 30 2012: (Start)
a(n) = ((b_m(n)+6) mod 9 + floor((b_m(n)+2)/3) - floor(b_m(n)/3))*10^m + Sum_{j=0..m-1} (b_j(n) mod 4 +5*floor((b_j(n)+3)/4) +floor((b_j(n)+2)/4)- 6*floor(b_j(n)/4)))*10^j, where n>1, b_j(n)) = floor((n-1-4^m)/4^j), m = floor(log_4(n-1)).
a(1*4^n+1) = 6*10^n.
a(2*4^n+1) = 8*10^n.
a(3*4^n+1) = 9*10^n.
a(n) = 6*10^log_4(n-1) for n=4^k+1,
a(n) < 6*10^log_4(n-1), otherwise.
a(n) > 10^log_4(n-1) for n>1.
a(n) = 6*A007090(n-1), iff the digits of A007090(n-1) are 0 or 1.
G.f.: g(x) = (x/(1-x))*Sum_{j>=0} 10^j*x^4^j *(1-x^4^j)* (6 + 8x^4^j + 9(x^2)^4^j)/(1-x^4^(j+1)).
Also: g(x) = (x/(1-x))*(6*h_(4,1)(x) + 2*h_(4,2)(x) + h_(4,3)(x) - 9*h_(4,4)(x)), where h_(4,k)(x) = Sum_{j>=0} 10^j*(x^4^j)^k/(1-(x^4^j)^4). (End)

Extensions

Examples added by Hieronymus Fischer, May 30 2012

A004520 Generalized nim sum n + n in base 10.

Original entry on oeis.org

0, 2, 4, 6, 8, 0, 2, 4, 6, 8, 20, 22, 24, 26, 28, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 40, 42, 44, 46, 48, 60, 62, 64, 66, 68, 60, 62, 64, 66, 68, 80, 82, 84, 86, 88, 80, 82, 84, 86, 88, 0, 2, 4, 6, 8, 0, 2, 4, 6, 8, 20, 22, 24, 26, 28, 20, 22, 24, 26, 28, 40, 42, 44, 46, 48, 40, 42
Offset: 0

Views

Author

Keywords

Comments

a(n) = n + n in carryless arithmetic mod 10. - N. J. A. Sloane, Jul 23 2010.

References

  • E. R. Berlekamp, J. H. Conway and R. K. Guy, Winning Ways, Academic Press, NY, 2 vols., 1982.
  • J. H. Conway, On Numbers and Games. Academic Press, NY, 1976.

Crossrefs

When sorted and duplicates removed, gives A014263. - N. J. A. Sloane, Aug 03 2010

Programs

  • Mathematica
    carrylessAdd[m_, n_, b_] := Block[{lm = IntegerLength[m, b], ln = IntegerLength[n, b]}, mx = Max[lm, ln]; idm = IntegerDigits[m, b, mx]; idn = IntegerDigits[n, b, mx]; FromDigits[ Mod[ idm + idn, b], b]]; Table[ carrylessAdd[n, n, 10], {n, 0, 76}] (* Robert G. Wilson v, Aug 23 2010 *)
  • PARI
    a(n) = fromdigits(digits(n)%5)<<1; \\ Kevin Ryde, Dec 10 2022
  • Python
    def A004520(n):
        return int(''.join(str(2*int(d) % 10) for d in str(n))) # Chai Wah Wu, Jun 29 2020
    

Formula

Generalized nim sum m + n in base q: write m and n in base q and add mod q with no carries, e.g. 5 + 8 in base 3 = "21" + "22" = "10" = 1.

Extensions

More terms from Robert G. Wilson v, Aug 23 2010
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