A185121 -id:A185121 - cp's OEIS frontend

2, 1, 11, 10, 103, 101, 100, 1017, 1011, 1002, 1000, 10037, 10023, 10007, 10002, 10000, 100137, 100073, 100023, 100003, 100002, 100000, 1000313, 1000037, 1000033, 1000023, 1000003, 1000002, 1000000, 10000337, 10000223, 10000137, 10000037, 10000023, 10000013, 10000002, 10000000, 100001733

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 not empty. Proof: Define m(n)=2*sum_{j=i..k} 10^j, where k=floor((sqrt(8*n+1)-1)/2), i:= n-A000217(k). 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, which proves the statement.

The 3 versions according to A213302 - A213304 are quite different. Example: 1002 has 9 nonprime substrings in version 1 (0, 0, 00, 02, 002, 1, 10 100, 1002), in version 2 there are 6 nonprime substrings (02, 002, 1, 10, 100, 1002) and there are 4 nonprime substrings in version 3 (1, 10, 100, 1002).

			a(0)=2, since 2 is the least number with zero nonprime substrings.
a(1)=1, since 1 has 1 nonprime substrings.
a(2)=11, since 11 is the least number with 2 nonprime substrings.
a(3)=10, since 10 is the least number with 3 nonprime substrings, these are 1, 0 and 10 (‘0’ will be counted).

Hieronymus Fischer, Table of n, a(n) for n = 0..200

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

Cf. A035244, A079307, A213300 - A213321.

a(n) >= 10^floor((sqrt(8*n-7)-1)/2) for n>0, equality holds if n is a triangular number > 0 (cf. A000217).

a(A000217(n)) = 10^(n-1), n>0.

a(A000217(n)-k) >= 10^(n-1)+k, n>0, 0<=k

a(A000217(n)-1) = 10^(n-1)+2, n>3, provided 10^(n-1)+1 is not a prime (which is proved to be true for all n-1 <= 50000 (cf. A185121) except n-1=16384 and is generally true for n-1 unequal to a power of 2).

a(A000217(n)-k) = 10^(n-1)+p, where p is the minimal number such that 10^(n-1) + p, has k prime substrings, n>0, 0<=k

Min(a(A000217(n)-k-i), 0<=i<=m) <= 10^(n-1)+p, where p is the minimal number with k prime substrings and m is the number of digits of p, and k+m

Min(a(A000217(n)-k-i), 0<=i<=A055642(A035244(k)) <= 10^(n-1)+A035244(k).

a(A000217(n)-k) <= 10^(n-1)+max(p(i), k<=i<=k+m), where p(i) is the minimal number with i prime substrings and m is the number of digits of p(i), and k+m

a(A000217(n)-k) <= 10^(n-1)+max(A035244(i), k<=i<=k+ A055642(i).

a(n) <= A213305(n).

A038371 Smallest prime factor of 10^n + 1.

Original entry on oeis.org

2, 11, 101, 7, 73, 11, 101, 11, 17, 7, 101, 11, 73, 11, 29, 7, 353, 11, 101, 11, 73, 7, 89, 11, 17, 11, 101, 7, 73, 11, 61, 11, 19841, 7, 101, 11, 73, 11, 101, 7, 17, 11, 29, 11, 73, 7, 101, 11, 97, 11, 101, 7, 73, 11, 101, 11, 17, 7, 101, 11, 73, 11, 101, 7, 1265011073

Offset: 0

Miklos SZABO (mike(AT)ludens.elte.hu)

nonn

a(n) >= 7 for all n >= 1 since 10^n + 1 is then not divisible by 2, 3 or 5.
Record values are a({0, 1, 2, 16, 32, 64, ...}). - M. F. Hasler, Apr 04 2008
The record values (2, 11, 101, 353, 19841, 1265011073, ...) are also found in A185121 and A102050 (smallest prime factor of 10^2^n+1). - M. F. Hasler, Jun 28 2024

			a(12) = 73 as 10^12+1 = 1000000000001 = 73*137*99990001.

Ehrhard Behrends, Five-Minute Mathematics, translated by David Kramer. American Mathematical Society (2008) p. 7

Max Alekseyev, Table of n, a(n) for n = 0..1023 (terms 0..500 from M. F. Hasler)
Makoto Kamada, Factorizations of 100...001.

Cf. A020639 (least prime factor), A062397 (10^n + 1), A003021 (largest prime factor of 10^n + 1), A057934 (number of prime factors of 10^n + 1, with multiplicity), A119704 (as before, without multiplicity), A185121 (smallest prime factor of 10^2^n+1), A102050 (as before, but 1 if 10^2^n+1 is prime).

[Min(PrimeFactors(10^n+1)):n in[0..70]]; // Vincenzo Librandi, Nov 08 2018

Table[FactorInteger[10^n + 1][[1, 1]], {n, 0, 49}] (* Alonso del Arte, Oct 21 2011 *)

A038371(n)=A020639(10^n+1) \\ Much more efficient than the naive {factor(10^n+1)[1,1]}. - M. F. Hasler, Apr 04 2008, edited Jun 29 2024

a(n) = A020639(A062397(n)).
For odd n, a(n) <= 11 since every (base 10) palindrome of even length is divisible by 11. - M. F. Hasler, Apr 04 2008 [See below for more precise formula.]
More generally, for k >= 0 and n == 2^k (mod 2^(k+1)), a(n) <= A185121(k) = (11, 101, 73, 17, 353, ...). This follows from  x^{2q+1} + 1 = (x+1) Sum_{m=0..2q} (-x)^m, with x=10^2^k. - M. F. Hasler, Jul 30 2019
From M. F. Hasler, Jun 28 2024: (Start)
a(2k+1) = 7 iff k == 1 (mod 3), else 11. [Making the 2008 formula more precise.]
a(4k+2) = 29 iff k == 3 (mod 7), else = 61 if k == 7 (mod 15), else = 89 if k == 5 (mod 11), else 101.
a(8k+4) = 73 for all k >= 0.
a(16k+8) = 17 for all k >= 0.
a(32k+16) = 97 iff k==1 (mod 3), else 353.
a(64k+32) = 193 iff k==1 (mod 3), else 1217 if k==9 (mod 19), else 2753 if k==21 (mod 43), else 3137 if k==24 (mod 49), else 3329 if k==6 (mod 13), else 4481 if k==17 (mod 35), else 4673 if k==36 (mod 73), else 5953 if k==15 (mod 31), else 6529 if k==8 (mod 17), else 13633 if k==35 (mod 71), else 15937 if k==41 (mod 83), else 19841. (End)

More terms from Reinhard Zumkeller, Mar 12 2002

A075493 Numbers k such that (sum of digits of k) > (number of divisors of k).

Original entry on oeis.org

3, 4, 5, 6, 7, 8, 9, 13, 14, 15, 16, 17, 18, 19, 23, 25, 26, 27, 28, 29, 31, 33, 34, 35, 37, 38, 39, 41, 43, 44, 45, 46, 47, 48, 49, 51, 52, 53, 54, 55, 56, 57, 58, 59, 61, 62, 63, 64, 65, 66, 67, 68, 69, 71, 73, 74, 75, 76, 77, 78, 79, 81, 82, 83, 85, 86, 87, 88, 89, 91, 92

Offset: 1

Labos Elemer, Sep 26 2002

			Sequence includes all primes with sum of digits > 2, i.e., all primes not of the form 10^d + 1 for nonnegative integers d (the only known primes of this form are 2, 11, and 101; see A185121).
Sequence also includes all squared primes with sum of digits > 3.

Cf. A007953, A000005, A057531, A075492, A075491, A185121.

sud[x_] := Apply[Plus, IntegerDigits[x]] Do[s=sud[n]-DivisorSigma[0, n]; If[s>0, Print[n]], {n, 1, 256}]
Select[Range[100],Total[IntegerDigits[#]]>DivisorSigma[0,#]&] (* Harvey P. Dale, Mar 07 2020 *)

Solutions to A007953(k) > A000005(k).

Edited by Jon E. Schoenfield, Sep 23 2018

A102050 a(n) = 1 if 10^(2^n)+1 is prime, otherwise smallest prime factor of 10^(2^n)+1.

Original entry on oeis.org

1, 1, 73, 17, 353, 19841, 1265011073, 257, 10753, 1514497, 1856104284667693057, 106907803649, 458924033, 3635898263938497962802538435084289

Offset: 0

Klaus Brockhaus and Walter Oberschelp (oberschelp(AT)informatik.rwth-aachen.de), Dec 28 2004

The smallest known prime factors of 10^(2^15)+1 to 10^(2^18)+1 are 65537, 8257537, 175636481, 639631361. - Jeppe Stig Nielsen, Nov 04 2010
Above values for a(15)-a(18) are confirmed. a(19) = 70254593, a(20) = 167772161. - Chai Wah Wu, Oct 16 2019
a(14) <= 1702047085242613845984907230501142529. - Max Alekseyev, Feb 26 2023

			10^(2^4)+1 = 10000000000000001 = 353*449*641*1409*69857, hence a(4) = 353.

factordb.com, Status of 10^(2^n)+1.
Wilfrid Keller, Prime factors of generalized Fermat numbers Fm(10) and complete factoring status

Cf. A000533, A002275, A062397.

spf[n_]:=Module[{c=10^2^n+1},If[PrimeQ[c],1,FactorInteger[c][[1,1]]]]; Array[spf,15,0] (* Harvey P. Dale, Apr 09 2019 *)

for(k=0,8,fac=factor(10^(2^k)+1);print1(if(matsize(fac)[1]==1,1,fac[1,1]),","))

If 10^(2^n)+1 is composite, a(n) = A185121(n).

a(13) from the Keller link, added by Jeppe Stig Nielsen, Nov 04 2010

A262083 Smallest possible prime factor of 10^k+n for any k.

Original entry on oeis.org

2, 7, 2, 7, 2, 3, 2, 17, 2, 7, 2, 3, 2, 7, 2, 5, 2, 3, 2, 7, 2, 11, 2, 3, 2, 5, 2, 7, 2, 3, 2, 7, 2, 7, 2, 3, 2, 7, 2, 7, 2, 3, 2, 7, 2, 5, 2, 3, 2, 13, 2, 7, 2, 3, 2, 5, 2, 7, 2, 3, 2, 7, 2, 17, 2, 3, 2, 7, 2, 7, 2, 3, 2, 7, 2, 5, 2, 3, 2, 7, 2, 7, 2, 3, 2, 5, 2, 7, 2, 3, 2, 17, 2, 7, 2, 3, 2, 7, 2, 7, 2

Offset: 0

Sergio Pimentel, Sep 10 2015

nonn

Is this sequence bounded?  What are the records for a(n)?
From Robert G. Wilson v, Sep 13 2015: (Start)
First occurrence of the i-th prime: 0, 5, 15, 1, 21, 49, 7, 357, 24871, 364021, ..., .
a(n) = 2 when n == 0 (mod 2),
a(n) = 3 when n == 5 (mod 6),
a(n) = 5 when n == 15 or 25 (mod 30),
a(n) = 7 when n == 1, 3, 9, 13, 19, 27, 31, 33, 37, 39, 43, 51, 57, 61, 67, 69, 73, 79, 81, 87, 93, 97, 99, 103, 109, 111, 117, 121, 123, 127, 129, 139, 141, 151, 153, 157, 159, 163, 169, 171, 177, 181, 183, 187, 193, 199, 201 or 207 (mod 210),
a(n) = 11 when n = 21, 133, 441, 483, 637, 903, 1057, 1099, 1407, 1519, 1561, 1827, 1869, 1981, 2023 or 2289 (mod 2310),
a(n) = 13 when n = 49, 147, 217, 231, 259, 399, 469, 511, 651, 679, 693, 763, 777, 861, 987, 1141, 1197, (413 terms missing), 29883 or 29953, ... (mod 30030),
a(n) = 17 when n = 7, 63, 91, 189, 273, 301, 343, 427, 553, 567, 609, 721, 819, 847, 889, 931, 973, 1029, (8044 terms missing), 510349 or 510447 (mod 510510),
a(n) = 19 when n = 357, 1071, 2737, 3451, 6069, 6307, 8211, 9163, 9639, 10353, 12019, 12733, 13447, 13923, 15351, 15589, 17017, 17493, 18207, ... (mod 9699690),
a(n) = 23 when n = 24871, 47481, 74613, 88179, 92701, 106267, 133399, 142443, 160531, 187663, 201229, 210273, 223839, 250971, 264537, 309757, ... (mod 223092870),
a(n) = 29 when n = 364021, 988057, ... (mod 6469693230), etc.
To the question if this sequence is 'bounded', I would answer no.
(End)
For complete lists of when a(n) < 19, see Wilson's Congruencies a-file. - Danny Rorabaugh, Oct 08 2015

			a(1) = 7 since 10^k+1 is not divisible by 2,3 or 5 for all k but is divisible by 7 when k = 3 (i.e., 1001 = 7*11*13).

Robert G. Wilson v, Table of n, a(n) for n = 0..10000
Robert G. Wilson v, Congruencies for A262083

Cf. A000533, A003617, A038371, A185121.

p = Prime@ Range@ 25; f[n_] := Block[{k = 1, lst = {}}, While[k < 25, AppendTo[lst, Position[ Mod[ PowerMod[10, k, p] + n, p] 0, 1, 1][[1, 1]]]; k++]; lst = Union@ lst; Prime@ lst[[1]]]; Array[f, 101, 0] (* Robert G. Wilson v, Sep 13 2015 *)

More terms from Robert G. Wilson v, Sep 13 2015

A309358 Numbers k such that 10^k + 1 is a semiprime.

Original entry on oeis.org

4, 5, 6, 7, 8, 19, 31, 53, 67, 293, 586, 641, 922, 2137, 3011

Offset: 1

Hugo Pfoertner, Jul 29 2019

a(16) > 12000.
10^k + 1 is composite unless k is a power of 2, and it can be conjectured that it is composite for all k > 2, cf. A038371 and A185121. - M. F. Hasler, Jul 30 2019
Suppose k is odd. Then k is a term if and only if (10^k+1)/11 is prime. - Chai Wah Wu, Jul 31 2019

			a(1) = 4 because 10^4 + 1 = 10001 = 73*137.

Cf. A003021, A038371, A046413, A057934, A062397, A092559.

Odd terms in sequence: A001562.

IsSemiprime:=func; [n: n in [2..200] | IsSemiprime(s) where s is 10^n+1]; // Vincenzo Librandi, Jul 31 2019

Mathematica

Select[Range[200], Plus@@Last/@FactorInteger[10^# + 1] == 2 &] (* Vincenzo Librandi, Jul 31 2019 *)

Showing 1-7 of 7 results.

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cp's OEIS Frontend

A062397 a(n) = 10^n + 1.

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A213302 Smallest number with n nonprime substrings (Version 1: substrings with leading zeros are considered to be nonprime).

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A038371 Smallest prime factor of 10^n + 1.

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A075493 Numbers k such that (sum of digits of k) > (number of divisors of k).

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A102050 a(n) = 1 if 10^(2^n)+1 is prime, otherwise smallest prime factor of 10^(2^n)+1.

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A262083 Smallest possible prime factor of 10^k+n for any k.

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A309358 Numbers k such that 10^k + 1 is a semiprime.

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