A084984 -id:A084984 - cp's OEIS frontend

A193238 Number of prime digits in decimal representation of n.

0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 2, 2, 1, 2, 1, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 2, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 2, 2, 1, 2, 1, 2, 1, 1, 0, 0, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 2, 2, 1, 2, 1, 2, 1, 1, 0, 0, 1, 1, 0, 1

Offset: 0

Reinhard Zumkeller, Jul 19 2011

Reinhard Zumkeller, Table of n, a(n) for n = 0..10000

Cf. A010051.

Cf. A211681, A046034, A052382, A055640, A055641, A055642, A102669 - A102685, A122640, A117804, A196563, A196564.

a193238 n = length $ filter (`elem` "2357") $ show n

Count[IntegerDigits[#],?PrimeQ]&/@Range[0,100] (* _Harvey P. Dale, Nov 13 2022 *)

a(n)=n=eval(Vec(Str(n)));sum(i=1,#n,isprime(n[i])) \\ Charles R Greathouse IV, Jul 29 2011

a(A084984(n))=0; a(A118950(n))>0; a(A092620(n))=1; a(A092624(n))=2; a(A092625(n))=3; a(A046034(n))=A055642(A046034(n));
a(A000040(n)) = A109066(n).
From Hieronymus Fischer, May 30 2012: (Start)
a(n) = sum_{j=1..m+1} (floor(n/10^j+0.3) + floor(n/10^j+0.5) + floor(n/10^j+0.8) - floor(n/10^j+0.2) - floor(n/10^j+0.4) - floor(n/10^j+0.6)), where m=floor(log_10(n)), n>0.
a(10n+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(A046034(n)) = floor(log_4(3n+1)), n>0.
a(A211681(n)) = 1 + floor((n-1)/4), n>0.
G.f.: g(x) = (1/(1-x))*sum_{j>=0} (x^(2*10^j) + x^(3*10^j)+ x^(5*10^j) + x^(7*10^j))*(1-x^10^j)/(1-x^10^(j+1)).
Also: g(x) = (1/(1-x))*sum_{j>=0} (x^(2*10^j)- x^(4*10^j)+ x^(5*10^j)- x^(6*10^j)+ x^(7*10^j)- x^(8*10^j))/(1-x^10^(j+1)). (End)

A202267 Numbers in which all digits are noncomposites (1, 2, 3, 5, 7) or 0.

Original entry on oeis.org

0, 1, 2, 3, 5, 7, 10, 11, 12, 13, 15, 17, 20, 21, 22, 23, 25, 27, 30, 31, 32, 33, 35, 37, 50, 51, 52, 53, 55, 57, 70, 71, 72, 73, 75, 77, 100, 101, 102, 103, 105, 107, 110, 111, 112, 113, 115, 117, 120, 121, 122, 123, 125, 127, 130, 131, 132, 133, 135, 137, 150

Offset: 1

Jaroslav Krizek, Dec 25 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)=0,1,2,3,5,7 for k=0..5. - Hieronymus Fischer, May 30 2012

			a(1000) = 5353.
a(10^4) = 115153
a(10^5) = 2070753.
a(10^6) = 33233353.

Hieronymus Fischer, Table of n, a(n) for n = 1..10000
Robert Baillie and Thomas Schmelzer, Summing Kempner's Curious (Slowly-Convergent) Series, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008.
Index entries for 10-automatic sequences.

Supersequence of A001742 and A046034.
Cf. A046034 (numbers in which all digits are primes), A001742 (numbers in which all digits are noncomposites excluding 0), A202268 (numbers in which all digits are nonprimes excluding 0), A084984 (numbers in which all digits are nonprimes), A029581 (numbers in which all digits are composites).
Cf. A007092, A001743, A001744, A193238.

Union[Flatten[FromDigits/@Tuples[{0,1,2,3,5,7},3]]] (* Harvey P. Dale, Mar 11 2015 *)

From Hieronymus Fischer, May 30 2012: (Start)
a(n) = (b_m(n)+1) mod 10 + floor((b_m(n)+2)/5) + floor((b_m(n)+1)/5) - 2*floor(b_m(n)/5))*10^m + sum_{j=0..m-1} (b_j(n) mod 6 + floor((b_j(n)+1)/6) + floor((b_j(n)+2)/6) - 2*floor(b_j(n)/6)))*10^j, where n>1, b_j(n)) = floor((n-1-6^m)/6^j), m = floor(log_6(n-1)).
a(1*6^n+1) = 1*10^n.
a(2*6^n+1) = 2*10^n.
a(3*6^n+1) = 3*10^n.
a(4*6^n+1) = 5*10^n.
a(5*6^n+1) = 7*10^n.
a(n) = 10^log_6(n-1) for n=6^k+1, k>0,
a(n) < 10^log_6(n-1) else.
a(n) = A007092(n-1) iff the digits of A007092(n-1) are <= 3, a(n)>A007092(n-1), else.
a(n) <= A084984(n), equality holds if the representation of n-1 as a base-6 number only has digits 0 or 1.
G.f.: g(x) = (x/(1-x))*sum_{j>=0} 10^j*x^6^j *(1-x^6^j)* (1 + 2x^6^j + 3(x^2)^6^j + 5(x^3)^6^j + 7(x^4)^6^j)/(1-x^6^(j+1)).
Also: g(x) = (x/(1-x))*(h_(6,1)(x) + h_(6,2)(x) + h_(6,3)(x) + 2*h_(6,4)(x) + 2*h_(6,5)(x) - 7*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) = 4.945325883472729555972742252181522711968119529132581193614012706741310832798... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 15 2024

Examples added by Hieronymus Fischer, May 30 2012

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

Original entry on oeis.org

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

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

A118950 Numbers containing at least one prime digit.

Original entry on oeis.org

2, 3, 5, 7, 12, 13, 15, 17, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33, 34, 35, 36, 37, 38, 39, 42, 43, 45, 47, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 62, 63, 65, 67, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 82, 83, 85, 87, 92, 93, 95, 97, 102, 103, 105, 107, 112

Offset: 1

Rick L. Shepherd, May 06 2006

A193238(a(n)) > 0; complement of A084984; A092620, A092624 and A092625 are subsequences. - Reinhard Zumkeller, Jul 19 2011

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Index entries for 10-automatic sequences.

Cf. A118951, A092620, A084984, A011540, A011531, A046034, A179336 (primes).

a118950 n = a118950_list !! (n-1)
a118950_list = filter (any (`elem` "2357") . show ) [0..]
-- Reinhard Zumkeller, Jul 19 2011

Select[Range[150],AnyTrue[IntegerDigits[#],PrimeQ]&] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jul 19 2018 *)

is(n)=!!#select(isprime, digits(n)) \\ Charles R Greathouse IV, Sep 15 2015

a(n) = n + O(n^k) with k = log 6/log 10 = 0.77815.... - Charles R Greathouse IV, Sep 15 2015

A062115 Numbers with no prime substring in their decimal expansion.

Original entry on oeis.org

0, 1, 4, 6, 8, 9, 10, 14, 16, 18, 40, 44, 46, 48, 49, 60, 64, 66, 68, 69, 80, 81, 84, 86, 88, 90, 91, 94, 96, 98, 99, 100, 104, 106, 108, 140, 144, 146, 148, 160, 164, 166, 168, 169, 180, 184, 186, 188, 400, 404, 406, 408, 440, 444, 446, 448, 460, 464, 466

Offset: 1

Erich Friedman, Jun 28 2001

This is a 10-automatic sequence, a consequence of the finitude of A071062. - Charles R Greathouse IV, Sep 27 2011

Subsequence of A202259 (right-truncatable nonprimes). Supersequence of A202262 (composite numbers in which all substrings are composite), A202265 (nonprime numbers in which all substrings and reversal substrings are nonprimes). - Jaroslav Krizek, Jan 28 2012

			25 is not included because 5 is prime.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
Jeffrey Shallit, Minimal primes, Journal of Recreational Mathematics 30:2 (1999-2000), pp. 113-117.
Index entries for 10-automatic sequences.

Subsequence of A084984. [Arkadiusz Wesolowski, Jul 05 2011]
Cf. A071062.
Cf. A163753 (complement).

a062115 n = a062115_list !! (n-1)
a062115_list = filter ((== 0) . a039997) a084984_list
-- Reinhard Zumkeller, Jan 31 2012

from sympy import isprime
def ok(n):
    s = str(n)
    ss = (int(s[i:j]) for i in range(len(s)) for j in range(i+1, len(s)+1))
    return not any(isprime(k) for k in ss)
print([k for k in range(500) if ok(k)]) # Michael S. Branicky, May 02 2023

# faster for initial segment of sequence; uses ok, import above
from itertools import chain, count, islice, product
def agen(): # generator of terms
    yield from chain((0,), (int(t) for t in (f+"".join(r) for d in count(1) for f in "14689" for r in product("014689", repeat=d-1)) if ok(t)))
print(list(islice(agen(), 100))) # Michael S. Branicky, May 02 2023

A039997(a(n)) = 0. - Reinhard Zumkeller, Jul 16 2007
From Charles R Greathouse IV, Mar 23 2010: (Start)
a(n) = O(n^(log_4 10)) = O(n^1.661) because numbers containing only 0,4,6,8 are in this sequence.
a(n) = Omega(n^(log_13637 1000000)) = Omega(n^1.451) for similar reasons; this bound can be increased by considering longer sequences of digits. (End)

Offset corrected by Arkadiusz Wesolowski, Jul 27 2011

A029581 Numbers in which all digits are composite.

Original entry on oeis.org

4, 6, 8, 9, 44, 46, 48, 49, 64, 66, 68, 69, 84, 86, 88, 89, 94, 96, 98, 99, 444, 446, 448, 449, 464, 466, 468, 469, 484, 486, 488, 489, 494, 496, 498, 499, 644, 646, 648, 649, 664, 666, 668, 669, 684, 686, 688, 689, 694, 696, 698, 699, 844, 846, 848

Offset: 1

N. J. A. Sloane

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)=4,6,8,9 for k=1..4. - Hieronymus Fischer, May 30 2012

			From _Hieronymus Fischer_, May 30 2012: (Start)
a(1000) = 88649.
a(10^4) = 6468989
a(10^5) = 449466489. (End)

Hieronymus Fischer, Table of n, a(n) for n = 1..10000
Robert Baillie and Thomas Schmelzer, Summing Kempner's Curious (Slowly-Convergent) Series, Mathematica Notebook kempnerSums.nb, Wolfram Library Archive, 2008.
Index entries for 10-automatic sequences.

Cf. A002808, A001744, A046034, A084544, A084984, A017042, A001743, A014261, A014263, A202267, A202268.

[n: n in [1..1000] | Set(Intseq(n)) subset [4, 6, 8, 9]]; // Vincenzo Librandi, Dec 17 2018

Table[FromDigits/@Tuples[{4, 6, 8, 9}, n], {n, 3}] // Flatten (* Vincenzo Librandi, Dec 17 2018 *)

From Hieronymus Fischer, May 30 and Jun 25 2012: (Start)
a(n) = Sum_{j=0..m-1} (2*b(j) mod 8 + 4 + 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)).
Also: a(n) = Sum_{j=0..m-1} (A010877(2*b(j)) + 4 + A002265(b(j)) - A002265(b(j)+1))*10^j.
Special values:
a(1*(4^n-1)/3) = 4*(10^n-1)/9.
a(2*(4^n-1)/3) = 2*(10^n-1)/3.
a(3*(4^n-1)/3) = 8*(10^n-1)/9.
a(4*(4^n-1)/3) = 10^n-1.
a(n) < 4*(10^log_4(3*n+1)-1)/9, equality holds for n=(4^k-1)/3, k > 0.
a(n) < 4*A084544(n), equality holds iff all digits of A084544(n) are 1.
a(n) > 2*A084544(n).
Lower and upper limits:
lim inf a(n)/10^log_4(n) = 1/10*10^log_4(3) = 0.62127870, for n --> inf.
lim sup a(n)/10^log_4(n) = 4/9*10^log_4(3) = 2.756123868970, for n --> inf.
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)*(1-z(j))*(4 + 6z(j) + 8*z(j)^2 + 9*z(j)^3)/(1-z(j)^4), where z(j) = x^4^j.
Also: g(x) = (1/(1-x))*(4*h_(4,0)(x) + 2*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+1)-1)/3)*(x^(k*4^j)/(1-x^4^(j+1)). (End)
Sum_{n>=1} 1/a(n) = 1.039691381254753739202528087006945643166147087095114911673083135126969046250... (calculated using Baillie and Schmelzer's kempnerSums.nb, see Links). - Amiram Eldar, Feb 15 2024

Offset corrected by Arkadiusz Wesolowski, Oct 03 2011

A092620 Numbers with exactly one prime digit.

Original entry on oeis.org

2, 3, 5, 7, 12, 13, 15, 17, 20, 21, 24, 26, 28, 29, 30, 31, 34, 36, 38, 39, 42, 43, 45, 47, 50, 51, 54, 56, 58, 59, 62, 63, 65, 67, 70, 71, 74, 76, 78, 79, 82, 83, 85, 87, 92, 93, 95, 97, 102, 103, 105, 107, 112, 113, 115, 117, 120, 121, 124, 126, 128, 129, 130, 131, 134

Offset: 1

Jani Melik, Apr 11 2004

A193238(a(n))=1; subsequence of A118950. - Reinhard Zumkeller, Jul 19 2011

			12 has one prime digit, 2;
102 has one prime digit, 2.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

Subsequence of A118950.

Cf. A046034, A084984, A092624, A092625.

import Data.List (elemIndices)
a092620 n = a092620_list !! (n-1)
a092620_list = elemIndices 1 a193238_list
-- Reinhard Zumkeller, Jul 19 2011

stev_sez:=proc(n) local i, tren, st, ans, anstren; ans:=[ ]: anstren:=[ ]: tren:=n: for i while (tren>0) do st:=round( 10*frac(tren/10) ): ans:=[ op(ans), st ]: tren:=trunc(tren/10): end do; for i from nops(ans) to 1 by -1 do anstren:=[ op(anstren), op(i,ans) ]; od; RETURN(anstren); end: ts_stpf:=proc(n) local i, stpf, ans; ans:=stev_sez(n): stpf:=0: for i from 1 to nops(ans) do if (isprime(op(i,ans))='true') then stpf:=stpf+1; # number of prime digits fi od; RETURN(stpf) end: ts_pr_n:=proc(n) local i, stpf, ans, ans1, tren; ans:=[ ]: stpf:=0: tren:=1: for i from 1 to n do if ( isprime(i)='true' and ts_stpf(i) =0) then ans:=[ op(ans), i ]: tren:=tren+1; fi od; RETURN(ans) end: ts_pr_n(300);

Select[Range[150],Count[IntegerDigits[#],?(PrimeQ)]==1&] (* _Harvey P. Dale, Mar 23 2018 *)

There are 6^n*(n-1/6)*2/3 n-digit members of this sequence for n > 1. - Charles R Greathouse IV, Apr 23 2022

A085557 Numbers that have more prime digits than nonprime digits.

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, 122, 123, 125, 127, 132, 133, 135, 137, 152, 153, 155, 157, 172, 173, 175, 177, 202, 203, 205, 207, 212, 213, 215, 217, 220, 221, 222, 223, 224, 225, 226, 227, 228, 229, 230, 231, 232

Offset: 1

Jason Earls, Jul 04 2003

Begins to differ from A046034 at the 21st term (which is the first 3-digit term).

			133 is in the sequence as the prime digits are 3 and 3 (those are two digits; counted with multiplicity) and one nonprime digit 1 and so there are more prime digits than nonprime digits. - _David A. Corneth_, Sep 06 2020

David A. Corneth, Table of n, a(n) for n = 1..10000

Cf. A193238, A046034, A046035, A118950, A019546, A203263, A035232, A039996, A085823, A052382, A084544, A084984, A017042, A001743, A001744, A014261, A014263, A202267, A202268, A211681.

is(n) = my(d = digits(n), c = 0); for(i = 1, #d, if(isprime(d[i]), c++)); c<<1 > #d \\ David A. Corneth, Sep 06 2020

from itertools import count, islice
def A085557_gen(startvalue=1): # generator of terms
    return filter(lambda n:len(s:=str(n))<(sum(1 for d in s if d in {'2','3','5','7'})<<1),count(max(startvalue,1)))
A085557_list = list(islice(A085557_gen(),20)) # Chai Wah Wu, Feb 08 2023

cp's OEIS Frontend

A193238 Number of prime digits in decimal representation of n.

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A202267 Numbers in which all digits are noncomposites (1, 2, 3, 5, 7) or 0.

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

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A084544 Alternate number system in base 4.

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A084545 Alternate number system in base 5.

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A118950 Numbers containing at least one prime digit.

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A062115 Numbers with no prime substring in their decimal expansion.

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