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.

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A211682 Prime numbers such that all the substrings of length <= 2 are primes.

Original entry on oeis.org

2, 3, 5, 7, 23, 37, 53, 73, 373, 237373, 537373, 5373737, 53737373, 53737373737, 237373737373, 737373737373737, 53737373737373737, 373737373737373737373, 53737373737373737373737, 53737373737373737373737373, 373737373737373737373737373, 5373737373737373737373737373737
Offset: 1

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Author

Hieronymus Fischer, Apr 30 2012

Keywords

Comments

A211681 restricted to primes. Contains A085823. In contrast to A085823 this sequence seems to be not finite (conjecture).

Examples

			a(10)=237373, since all substrings of length <= 2 are primes (2, 3, 7, 23, 37, 73) and 237373 is prime.
		

Crossrefs

Extensions

The originally submitted terms with index 20 and 21 are composite and have been erased therefore. - Hieronymus Fischer, Sep 24 2018
a(20)-a(22) from Giovanni Resta, Oct 08 2018

A211684 Numbers > 1000 such that all the substrings of length = 3 are primes (substrings with leading '0' are considered to be nonprime).

Original entry on oeis.org

1131, 1137, 1139, 1271, 1277, 1311, 1313, 1317, 1373, 1379, 1397, 1491, 1499, 1571, 1577, 1631, 1673, 1677, 1733, 1739, 1797, 1811, 1911, 1919, 1937, 1971, 1977, 1991, 1997, 2113, 2233, 2239, 2271, 2277, 2293, 2331, 2337, 2397, 2419, 2571
Offset: 1

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Author

Hieronymus Fischer, Jun 08 2012

Keywords

Comments

Only numbers > 1000 are considered, since all 3-digit primes are trivial members. See A069489 for the sequence with prime terms > 1000.
The sequence is infinite (for example, consider the continued concatenation of '19' or of '337': 1919, 19191, 191919, ..., 3373, 33733, 337337, ... are members).
Infinitely many terms are palindromic.
A 10-automatic sequence realized by a linear recurrence relation. - Charles R Greathouse IV, Jan 04 2013

Examples

			a(1) = 1131, since all substrings of length = 3 (113 and 131) are primes.
a(33) = 2271, since all substrings of length = 3 (227, 271) are primes.
		

Crossrefs

A211685 Prime numbers > 1000 such that all the substrings of length >= 3 are primes (substrings with leading '0' are considered to be nonprime).

Original entry on oeis.org

1277, 1373, 1499, 1571, 1733, 1811, 1997, 2113, 2239, 2293, 2719, 3137, 3313, 3373, 3491, 3499, 3593, 3673, 3677, 3733, 3739, 3797, 4211, 4337, 4397, 4673, 4877, 4919, 5233, 5419, 5479, 6131, 6173, 6197, 6199, 6311, 6317, 6599, 6619, 6733
Offset: 1

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Author

Hieronymus Fischer, Jun 08 2012

Keywords

Comments

Only numbers > 1000 are considered, since all 3-digit primes are trivial members.
By definition, each term of the sequence with more than 4 digits is built up by an overlapped union of previous terms, i.e., a(59)=33739 has the two embedded previous terms a(14)=3373 and a(21)=3739.
The sequence is finite, the last term is 349199 (n=63). Proof of finiteness: Let p be a number with more than 6 digits. By the argument above, each 6-digit substring of p must be a previous term. The only 6-digit term is 349199. Thus, there is no number p with the desired property.

Examples

			a(1)=1277, since all substrings of length >= 3 are primes (127, 277, and 1277).
a(63)=349199, all substrings of length >= 3 (349, 491, 919, 199, 3491, 4919, 9199, 34919, 49199 and 349199) are primes.
		

Crossrefs

A213300 Largest number with n nonprime substrings (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

373, 3797, 37337, 73373, 373379, 831373, 3733797, 3733739, 8313733, 9973331, 37337397, 82337397, 99733313, 99733317, 99793373, 733133733, 831373379, 997333137, 997337397, 997933739, 7331337337, 8313733797, 9733733797, 9973331373, 9979337397, 9982337397
Offset: 0

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Author

Hieronymus Fischer, Aug 26 2012

Keywords

Comments

The sequence is well-defined in that for each n the set of numbers with n nonprime substrings is nonempty and finite. Proof of existence: Define m(n):=2*sum_{j=i..k} 10^j, where k:=floor((sqrt(8n+1)-1)/2), i:= n - k(k+1)/2. For n=0,1,2,3,... the m(n) are 2, 22, 20, 222, 220, 200, 2222, 2220, 2200, 2000, 22222, 22220, ... . m(n) has k+1 digits and (k-i+1) 2’s. Thus the number of nonprime substrings of m(n) is ((k+1)(k+2)/2)-k-1+i=(k(k+1)/2)+i=n. This proves existence. Proof of finiteness: Each 4-digit number has at least 1 nonprime substring. Hence each 4*(n+1)-digit number has at least n+1 nonprime substrings. Consequently, there is a boundary b < 10^(4n+3) such that all numbers > b have more than n nonprime substrings. It follows that the set of numbers with n nonprime substrings is finite.
The following statements hold true:
For all n>=0 there are minimal numbers with n nonprime substrings (cf. A213302 - A213304).
For all n>=0 there are maximal numbers with n nonprime substrings (= A213300 = this sequence).
For all n>=0 there are minimal numbers with n prime substrings (cf. A035244).
The greatest number with n prime substrings does not exist. Proof: If p is a number with n prime substrings, than 10*p is a greater number with n prime substrings.
Comment from N. J. A. Sloane, Sep 01 2012: it is a surprise that any number greater than 373 has a nonprime substring!

Examples

			a(0)=373, since 373 is the greatest number such that all substrings are primes, hence it is the maximal number with 0 nonprime substrings.
a(1)=3797, since the only nonprime substring of 3797 is 9 and all greater numbers have more than 1 nonprime substrings.
a(2)=37337, since the nonprime substrings of 37337 are 33 and 7337 and all greater numbers have > 2 nonprime substrings.
		

Crossrefs

Formula

a(n) >= A035244(A000217(A055642(a(n)))-n).

A213321 Minimal prime with n prime substrings (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

2, 13, 23, 113, 137, 373, 1973, 1733, 1373, 11317, 17333, 31379, 37337, 113173, 211373, 313739, 337397, 1113173, 1137337, 2313797, 2337397, 11131733, 12337397, 11373379, 33133733, 111733373, 113137337, 123733739, 291733373, 113733797, 1173313373, 1137333137, 1237337393, 1137337973
Offset: 1

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Author

Hieronymus Fischer, Aug 26 2012

Keywords

Examples

			a(1)=2, since 2 is a prime has 1 prime substring (2).
a(2)=13, since 13 is prime and has 2 prime substrings (3 and 13)
		

Crossrefs

Formula

a(n) > 10^floor((sqrt(8*n+1)-1)/2).
min(a(k), k>=n-1) <= A079397(n-1), n>0.
a(n) >= A035244(n), n>0.

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

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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

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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

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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

A217302 Minimal natural number (in decimal representation) with n prime substrings in binary representation (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

1, 2, 5, 7, 11, 15, 27, 23, 31, 55, 47, 63, 111, 95, 187, 127, 223, 191, 381, 255, 447, 503, 383, 511, 1015, 895, 767, 1023, 1533, 1791, 1535, 1919, 3039, 3069, 3067, 3839, 3967, 6079, 6139, 6135, 7679, 8063, 8159, 12159, 12271, 15359, 16127
Offset: 0

Views

Author

Hieronymus Fischer, Nov 22 2012

Keywords

Comments

The sequence is well-defined in that for each n the set of numbers with n prime substrings in binary representation is not empty. Proof: A000975(n+1) has exactly n prime substrings in binary representation (see A000975).
All terms with n > 1 are odd.

Examples

			a(1) = 2 = 10_2, since 2 is the least number with 1 prime substring (=10_2) in binary representation.
a(2) = 5 = 101_2, since 5 is the least number with 2 prime substrings in binary representation (10_2 and 101_2).
a(4) = 11 = 1011_2, since 11 is the least number with 4 prime substrings in binary representation (10_2, 11_2, 101_2 and 1011_2).
a(8) = 31 = 11111_2, since 31 is the least number with 8 prime substrings in binary representation (4 times 11_2, 3 times 111_2, and 11111_2).
a(9) = 47 = 101111_2, since 47 is the least number with 9 prime substrings in binary representation (10_2, 3 times 11_2, 101_2, 2 times 111_2, 1011_2, and 10111_2).
		

Crossrefs

Formula

a(n) >= 2^ceiling(sqrt(8*n+1)-1)/2).
a(n) <= A000975(n+1).
a(n+1) <= 2*a(n)+1.

A217309 Minimal natural number (in decimal representation) with n prime substrings in base-9 representation (substrings with leading zeros are considered to be nonprime).

Original entry on oeis.org

1, 2, 11, 23, 101, 173, 902, 1562, 1559, 8120, 14032, 14033, 73082, 126290, 604523, 657743, 723269, 1136684, 5918933, 5972147, 10227787, 25051529, 53276231, 54333278, 92071913, 441753767, 479669051, 483743986, 828662228, 3971590751, 4315446629
Offset: 0

Views

Author

Hieronymus Fischer, Nov 22 2012

Keywords

Comments

The sequence is well-defined in that for each n the set of numbers with n prime substrings is not empty. Proof: Define m(0):=1, m(1):=2 and m(n+1):=9*m(n)+2 for n>0. This results in m(n)=2*sum_{j=0..n-1} 9^j = (9^n - 1)/4 or m(n)=1, 2, 22, 222, 2222, 22222, …, (in base-9) for n=0,1,2,3,…. Evidently, for n>0 m(n) has n 2’s and these are the only prime substrings in base-9 representation. This is why every substring of m(n) with more than one digit is a product of two integers > 1 (by definition) and can therefore not be a prime number.
No term is divisible by 9.

Examples

			a(1) = 2 = 2_9, since 2 is the least number with 1 prime substring in base-9 representation.
a(2) = 11 = 12_9, since 11 is the least number with 2 prime substrings in base-9 representation (2_9 and 12_9).
a(3) = 23 = 25_9, since 23 is the least number with 3 prime substrings in base-9 representation (2_9, 3_9, and 23_9).
a(4) = 101 = 122_9, since 101 is the least number with 4 prime substrings in base-9 representation (2 times 2_9, 12_9=11, and 122_9=101).
a(7) = 1562 = 2125_9, since 1562 is the least number with 7 prime substrings in base-9 representation (2 times 2_9, 5_9, 12_9=11, 21_9=19, 25_9=23, and 212_9=173).
		

Crossrefs

Formula

a(n) > 9^floor(sqrt(8*n-7)-1)/2), for n>0.
a(n) <= (9^n - 1)/4, n>0.
a(n+1) <= 9*a(n)+3.
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