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 61 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

A211681 Numbers such that all the substrings of length <= 2 are primes.

Original entry on oeis.org

2, 3, 5, 7, 23, 37, 53, 73, 237, 373, 537, 737, 2373, 3737, 5373, 7373, 23737, 37373, 53737, 73737, 237373, 373737, 537373, 737373, 2373737, 3737373, 5373737, 7373737, 23737373, 37373737, 53737373, 73737373, 237373737, 373737373, 537373737, 737373737
Offset: 1

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Author

Hieronymus Fischer, Apr 30 2012

Keywords

Comments

The terms are primes for n = 1, 2, 3, 4, 5, 6, 7, 8, 10, 21, 23, 27, 31, 43, 45, 60, 67, 82, 91, .... The further terms until index 102 are composite. For the subsequence with prime terms see A211682. - [updated by Hieronymus Fischer, Oct 02 2018]
From Hieronymus Fischer, Oct 02 2018: (Start)
For indices n > 8, prime terms satisfy n mod 24 = 1, 3, 5, 7, 10, 12, 19, 21, 23. However, this condition is not sufficient. Indeed, for n <= 200 most of those terms are proven composite unless the terms with n = 103, 106, 123, 156, 165, 175, 178, 191, 193 and 195 which are potentially prime.
The terms are composite for n > 10 and n mod 24 = 0, 2, 4, 6, 8, 9, 11, 13, 14, 15, 16, 17, 18, 20, 22 (see formula section for the details).
(End)
Cf. A213299 for the partial sums.

Examples

			a(11)=537, since all substrings of length <= 2 are primes (5, 3, 7, 53 and 37).
a(21)=237373, the substrings of length <= 2 are 2, 3, 7, 23, 37, 73.
		

Crossrefs

Programs

  • Mathematica
    Table[FromDigits/@Select[Tuples[{2,3,5,7},n],AllTrue[FromDigits/@ Partition[ #,2,1],PrimeQ]&],{n,9}]//Flatten (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Jun 13 2020 *)

Formula

a(1+8*k) = 2*10^(2k) + 37*(10^(2k)-1)/99,
a(2+8*k) = 3*10^(2k) + 73*(10^(2k)-1)/99,
a(3+8*k) = 5*10^(2k) + 37*(10^(2k)-1)/99,
a(4+8*k) = 7*10^(2k) + 37*(10^(2k)-1)/99,
a(5+8*k) = 23*10^(2k) + 73*(10^(2k)-1)/99,
a(6+8*k) = 37*10^(2k) + 37*(10^(2k)-1)/99,
a(7+8*k) = 53*10^(2k) + 73*(10^(2k)-1)/99,
a(8+8*k) = 73*10^(2k) + 73*(10^(2k)-1)/99, for k >= 0.
a(n) = ((2*n+7) mod 8 + dn3 - dn2)*10^dn_1 + floor((37+36*(dn2-dn1))*10^dn_1/99), where dn1 = floor((n+1)/4), dn2 = floor((n+2)/4), dn3 = floor((n+3)/4), dn_1 = floor((n-1)/4). [updated by Hieronymus Fischer, Oct 02 2018]
From Hieronymus Fischer, Oct 02 2018: (Start)
a(24k + 0) = 73*(10^(6k-2) + (10^(6k-2)-1)/99), for k > 0.
a(24k + 2) = 3*(1245790*(10^(6k)-1)/999999 + 1),
a(24k + 4) = 7*(1053390*(10^(6k)-1)/999999 + 1),
a(24k + 6) = 37*(10^(6k) + (10^(6k)-1)/99),
a(24k + 8) = 73*(10^(6k) + (10^(6k)-1)/99),
a(24k + 9) = 3*(79124500*(10^(6k)-1)/999999 + 79),
a(24k + 11) = 3*(79124500*(10^(6k)-1)/999999 + 79 + 10^(6k+2)),
a(24k + 13) = 3*(791245000*(10^(6k)-1)/999999 + 791),
a(24k + 14) = 37*(10^(6k+2) + (10^(6k+2)-1)/99),
a(24k + 15) = 3*(791245000*(10^(6k)-1)/999999 + 791 + 10^(6k+3)),
a(24k + 16) = 73*(10^(6k+2) + (10^(6k+2)-1)/99),
a(24k + 17) = 7*(3391050000*(10^(6k)-1)/999999 + 3391),
a(24k + 18) = 7*(5339100000*(10^(6k)-1)/999999 + 5339),
a(24k + 20) = 3*(24579100000*(10^(6k)-1)/999999 + 24579),
a(24k + 22) = 37*(10^(6k+4) + (10^(6k+4)-1)/99), for k >= 0.
(End)
Recursion for n>8:
a(n) = 10*(1+a(n-4)) - a(n-4) mod 10.
G.f.: (2*x*(1+x^10) + 3*x^2*(1 + x^3 + x^5 + x^6) + 5*x^3*(1+x^6) + 7*x^4*(1+x^2))/((1-10*x^4)*(1-x^8)). [corrected by Hieronymus Fischer, Sep 03 2012]
From Chai Wah Wu, Feb 08 2023: (Start)
a(n) = a(n-1) + 9*a(n-4) - 9*a(n-5) + 10*a(n-8) - 10*a(n-9) for n > 9.
G.f.: x*(2*x^7 - 2*x^6 + 5*x^5 - 2*x^4 + 2*x^3 + 2*x^2 + x + 2)/((x - 1)*(x^4 + 1)*(10*x^4 - 1)). (End)

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

A035244 Smallest number with exactly n prime substrings.

Original entry on oeis.org

1, 2, 13, 23, 113, 137, 373, 1137, 1733, 1373, 11317, 11373, 13733, 31373, 113173, 131373, 137337, 337397, 1113173, 1137337, 1373373, 2337397, 3733797, 11373137, 11373379, 13733797, 37337397, 111373379, 123733739
Offset: 0

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Author

Keywords

Comments

No leading 0's allowed in substrings.
The sequence is well-defined in that for each n the set of numbers with n prime substrings is not empty. Proof by induction: '1' has 0 prime substrings and '2' has 1 prime substring. Let m be a number with n prime substrings. Then 10m+2 is a number with n+1 prime substrings (since m and 10m have identical prime substrings, and '2' is one additional prime substring, but 10m+2 cannot be prime). - Hieronymus Fischer, Aug 26 2012

Examples

			a(4)=113 since 3, 11, 13 and 113 are prime and no smaller number works.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Block[{s = IntegerDigits[n], c = 0, d = {}}, l = Length[s]; t = Flatten[ Table[ Take[s, {i, j}], {i, 1, l}, {j, i, l}], 1]; k = l(l + 1)/2; While[k > 0, If[ t[[k]][[1]] != 0, d = Append[d, FromDigits[ t[[k]] ]]]; k-- ]; Count[ PrimeQ[d], True]]; a = Table[0, {25}]; Do[ b = f[n]; If[ a[[b + 1]] == 0, a[[b + 1]] = n], {n, 1, 15000000}]; a

Formula

a(n) > 10^floor((sqrt(8*n-7)-1)/2) for n > 0. - Hieronymus Fischer, Jun 25 2012
Min_{k>=n} a(k) <= A079397(n-1), n > 0. - Hieronymus Fischer, Aug 26 2012
a(n+1) <= 10*a(n) + 2. - Hieronymus Fischer, Aug 26 2012

Extensions

Edited by Robert G. Wilson v, Feb 25 2003
a(25)-a(40) from Hieronymus Fischer, Jun 25 2012 and Aug 25 2012

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

Views

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.

A079397 Smallest prime with memory = n.

Original entry on oeis.org

2, 13, 23, 113, 137, 1237, 1733, 1373, 12373, 11317, 23719, 111317, 113171, 211373, 1131379, 1113173, 1317971, 2313797, 11131733, 11373379, 23931379, 113193797, 52313797, 129733313, 113733797, 523137971, 1113179719, 1317971939
Offset: 0

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Author

Joseph L. Pe, Feb 16 2003

Keywords

Comments

The memory of a prime p is the number of previous primes contained as substrings in (the decimal representation of) p.
Also the minimal prime such that the number of different prime substrings is n+1 (substrings with leading zeros are considered to be nonprime). - Hieronymus Fischer, Aug 26 2012

Examples

			113 is the smallest prime with memory = 3. (The smaller primes 3, 11, 13 are substrings of 113.) Hence a(3) = 113.
		

Crossrefs

Programs

  • Mathematica
    f[n_] := Block[{id = IntegerDigits@n}, len = Length@id - 1; Count[ PrimeQ@ Union[ FromDigits@# & /@ Flatten[ Table[ Partition[id, k, 1], {k, len}], 1]], True] + 1]; t = Table[0, {30}]; p = 2; While[p < 11500000000, a = f@p; If[t[[a]] == 0, pp = PrimePi@p; t[[a]] = pp; Print[{a, p, pp}]]; p = NextPrime@p]; t (* Robert G. Wilson v, Aug 03 2010 *)

Formula

a(n) > 10^floor((sqrt(8*n+1)-1)/2). - Hieronymus Fischer, Aug 26 2012
a(n) >= min(A035244(k+1), k >= n). - Hieronymus Fischer, Aug 26 2012

Extensions

Edited and extended by Robert G. Wilson v, Feb 25 2003
a(24)-a(27) from Robert G. Wilson v, Aug 03 2010

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.
Showing 1-10 of 61 results. Next