A019546 -id:A019546 - cp's OEIS frontend

A045336 Palindromic terms from A019546.

2, 3, 5, 7, 353, 373, 727, 757, 32323, 33533, 35353, 35753, 37273, 37573, 72227, 72727, 73237, 75557, 77377, 3222223, 3223223, 3233323, 3252523, 3272723, 3337333, 3353533, 3553553, 3722273, 3732373, 3773773, 7257527, 7327237, 7352537, 7527257, 7722277

Offset: 1

Robert G. Wilson v, Aug 18 2000

a(33) = 7352537 is the smallest palindromic prime using all prime digits (see Prime Curios! link). - Bernard Schott, Nov 10 2020

Michael S. Branicky, Table of n, a(n) for n = 1..12725 (all terms with <= 17 digits; terms 1..330 from Harvey P. Dale)
Chris K. Caldwell and G. L. Honaker, Jr., 7352537, Prime Curios!

Cf. A019546 and A002385.

Select[ Range[ 1, 10^7 ], PrimeQ[ # ] && FreeQ[ RealDigits[ # ][ [ 1 ] ], 0 ] && FreeQ[ RealDigits[ # ][ [ 1 ] ], 1 ] && FreeQ[ RealDigits[ # ][ [ 1 ] ], 4 ] && FreeQ[ RealDigits[ # ][ [ 1 ] ], 6 ] && FreeQ[ RealDigits[ # ][ [ 1 ] ], 8 ] && FreeQ[ RealDigits[ # ][ [ 1 ] ], 9 ] && RealDigits[ # ][ [ 1 ] ] == Reverse[ RealDigits[ # ][ [ 1 ] ] ] & ]
Table[FromDigits/@Select[Tuples[{2,3,5,7},n],#==Reverse[#]&&PrimeQ[ FromDigits[ #]]&],{n,12}]//Flatten (* Harvey P. Dale, Jun 19 2016 *)
Select[Flatten[Table[FromDigits/@Tuples[{2,3,5,7},n],{n,10}]],PrimeQ[#]&&PalindromeQ[#]&] (* Harvey P. Dale, Mar 24 2025 *)
f@n_ := Prime@n;
g@l_ := FromDigits@# & /@ Table[Join[l, {f@i}, Reverse@l], {i, 4}];
Flatten[g@# & /@ (f@# & /@
     Select[Table[IntegerDigits[n, 5], {n, 2000}], FreeQ[#, 0] &])] //
Select[PrimeQ] (* Hans Rudolf Widmer, Dec 18 2021 *)

from sympy import isprime
from itertools import count, product, takewhile
def primedigpals():
    for d in count(1, 2):
        for p in product("2357", repeat=d//2):
            left = "".join(p)
            for mid in "2357":
                yield int(left + mid + left[::-1])
def aupto(N):
    return list(takewhile(lambda x: x<=N, filter(isprime, primedigpals())))
print(aupto(10**7)) # Michael S. Branicky, Dec 18 2021

A055642 Number of digits in the decimal expansion of n.

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3

Offset: 0

Henry Bottomley, Jun 06 2000

From Hieronymus Fischer, Jun 08 2012: (Start)
For n > 0 the first differences of A117804.
The total number of digits necessary to write down all the numbers 0, 1, 2, ..., n is A117804(n+1). (End)
Here a(0) = 1, but a different common convention is to consider that the expansion of 0 in any base b > 0 has 0 terms and digits. - M. F. Hasler, Dec 07 2018

			Examples:
999: 1 + floor(log_10(999)) = 1 + floor(2.x) = 1 + 2 = 3 or
      ceiling(log_10(999+1)) = ceiling(log_10(1000)) = ceiling(3) = 3;
1000: 1 + floor(log_10(1000)) = 1 + floor(3) = 1 + 3 = 4 or
      ceiling(log_10(1000+1)) = ceiling(log_10(1001)) = ceiling(3.x) = 4;
1001: 1 + floor(log_10(1001)) = 1 + floor(3.x) = 1 + 3 = 4 or
      ceiling(log_10(1001+1)) = ceiling(log_10(1002)) = ceiling(3.x) = 4;

Charles R Greathouse IV, Table of n, a(n) for n = 0..10000

Cf. A043537, A178788, A046034, A019546, A054899, A122840, A055640, A055641, A102669-A102685, A117804, A160093, A160094, A196563, A196564, A000120, A000788, A023416, A059015 (for base 2).
Cf. A262190, A262198.
Cf. A007953: sum of digits.

a055642 :: Integer -> Int
a055642 = length . show  -- Reinhard Zumkeller, Feb 19 2012, Apr 26 2011

[ #Intseq(n): n in [0..105] ];   //  Bruno Berselli, Jun 30 2011
(Common Lisp) (defun A055642 (n) (if (zerop n) 1 (floor (log n 10)))) ; James Spahlinger, Oct 13 2012

A055642 := proc(n)
        max(1,ilog10(n)+1) ;
end proc:  # R. J. Mathar, Nov 30 2011

Join[{1}, Array[ Floor[ Log[10, 10# ]] &, 104]] (* Robert G. Wilson v, Jan 04 2006 *)
Join[{1},Table[IntegerLength[n],{n,104}]]
IntegerLength[Range[0,120]] (* Harvey P. Dale, Jul 02 2016 *)

a(n)=#Str(n) \\ M. F. Hasler, Nov 17 2008

A055642(n)=logint(n+!n,10)+1 \\ Increasingly faster than the above, for larger n. (About twice as fast for n ~ 10^7.) - M. F. Hasler, Dec 07 2018

def a(n): return len(str(n))
print([a(n) for n in range(121)]) # Michael S. Branicky, May 10 2022

def A055642(n): # Faster than len(str(n)) from ~ 50 digits on
    L = math.log10(n or 1)
    if L.is_integer() and 10**int(L)>n: return int(L or 1)
    return int(L)+1  # M. F. Hasler, Apr 08 2024

a(A046760(n)) < A050252(A046760(n)); a(A046759(n)) > A050252(A046759(n)). - Reinhard Zumkeller, Jun 21 2011
a(n) = A196563(n) + A196564(n).
a(n) = 1 + floor(log_10(n)) = 1 + A004216(n) = ceiling(log_10(n+1)) = A004218(n+1), if n >= 1. - Daniel Forgues, Mar 27 2014
a(A046758(n)) = A050252(A046758(n)). - Reinhard Zumkeller, Jun 21 2011
a(n) = A117804(n+1) - A117804(n), n > 0. - Hieronymus Fischer, Jun 08 2012
G.f.: g(x) = 1 + (1/(1-x))*Sum_{j>=0} x^(10^j). - Hieronymus Fischer, Jun 08 2012
a(n) = A262190(n) for n < 100; a(A262198(n)) != A262190(A262198(n)). - Reinhard Zumkeller, Sep 14 2015

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

Eric W. Weisstein

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

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

Reinhard Zumkeller, 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.
Eric Weisstein's World of Mathematics, Smarandache Sequences.
Index entries for 10-automatic sequences.

Cf. A046035, A118950, A019546 (primes), A203263, A035232, A039996, A085823, A052382, A084544, A084984, A017042, A001743, A001744, A014261, A014263, A153025, A191486, A193238, A202267, A202268, A211681, A365471 (complement).

a046034 n = a046034_list !! (n-1)
a046034_list = filter (all (`elem` "2357") . show ) [0..]
-- Reinhard Zumkeller, Jul 19 2011

[n: n in [2..532] | Set(Intseq(n)) subset [2, 3, 5, 7]];  // Bruno Berselli, Jul 19 2011

Table[FromDigits /@ Tuples[{2, 3, 5, 7}, n], {n, 3}] // Flatten (* Michael De Vlieger, Sep 19 2016 *)

is_A046034(n)=Set(isprime(digits(n)))==[1] \\ M. F. Hasler, Oct 12 2013

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

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

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

Hieronymus Fischer, Apr 30 2012

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.

			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.

Hieronymus Fischer, Table of n, a(n) for n = 1..250
Index entries for linear recurrences with constant coefficients, signature (1,0,0,9,-9,0,0,10,-10).

Cf. A019546, A035232, A039996, A046034, A085823, A211682, A213299.

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 *)

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)

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

Hieronymus Fischer, Apr 30 2012

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

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

Giovanni Resta, Table of n, a(n) for n = 1..36 (terms < 10^1000)

Cf. A085823, A019546, A046034, A211681.

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

Hieronymus Fischer, Jun 08 2012

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

			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.

Hieronymus Fischer, Table of n, a(n) for n = 1..3000
Index entries for 10-automatic sequences.

Cf. A019546, A035232, A039996, A046034, A069489, A085823, A211681, A211682.

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

Hieronymus Fischer, Jun 08 2012

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.

			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.

Hieronymus Fischer, Table of n, a(n) for n = 1..63 (complete sequence).

Cf. A019546, A035232, A039996, A046034, A069489, A085823, A211681, A211682, A211684.

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

Hieronymus Fischer, Aug 26 2012

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!

			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.

Hieronymus Fischer, Table of n, a(n) for n = 0..32

Cf. A019546, A035232, A039996, A046034, A069489, A085823, A211681, A211682, A211684, A211685.

Cf. A035244, A079307, A213301 - A213321.

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

Hieronymus Fischer, Aug 26 2012

			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)

Hieronymus Fischer, Table of n, a(n) for n = 1..40

Cf. A019546, A035232, A039996, A046034, A069489, A085823, A211681, A211682, A211684, A211685.

Cf. A035244, A079307, A213300 - A213320.

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.

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

Hieronymus Fischer, Nov 22 2012

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.

			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).

Hieronymus Fischer, Table of n, a(n) for n = 0..300

Cf. A019546, A035232, A039996, A046034, A069489, A085823, A211681, A211682, A211684, A211685, A035244, A079397, A213300-A213321, A217303-A217309, A000975.

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

cp's OEIS Frontend

A045336 Palindromic terms from A019546.

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A055642 Number of digits in the decimal expansion of n.

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A046034 Numbers whose digits are primes.

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

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

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

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

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