A173976 -id:A173976

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A174414

a(n) is the smallest natural number k such that the concatenation (n+k)||k is a prime number.

+10
6

3, 1, 1, 3, 1, 1, 3, 3, 1, 9, 19, 1, 3, 1, 7, 3, 1, 1, 3, 1, 13, 17, 1, 1, 3, 1, 1, 3, 7, 1, 9, 1, 23, 3, 3, 19, 17, 7, 1, 3, 1, 1, 3, 21, 1, 11, 3, 1, 3, 7, 1, 9, 1, 7, 21, 1, 7, 3, 1, 7, 3, 1, 1, 3, 1, 17, 9, 1, 1, 3, 3, 7, 9, 1, 1, 9, 13, 7, 3, 1, 1, 3, 3, 11, 3, 7, 1, 27, 7, 1, 9, 3, 1, 9, 3, 1, 9, 1

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,1

COMMENTS

The last digit of a(n) must be 1, 3, 7 or 9.

If n and 10^d+1 are not coprime, then a(n) cannot have d digits. - Robert Israel, Sep 16 2024

REFERENCES

Theo Kempermann, Zahlentheoretische Kostproben, Harri Deutsch, 2. aktualisierte Auflage 2005.

Helmut Kracke, Mathe-musische Knobelisken, Dümmler Bonn, 2. Auflage 1983.

Hugo Steinhaus, Studentenfutter, Urania-Verlag Leipzig-Jena-Berlin, 1991.

LINKS

Robert Israel, Table of n, a(n) for n = 1..10000

FORMULA

a(n) = 1 for n > 1 in A126785.

EXAMPLE

43 = prime(14) = (3+1)||3, a(1) = 3

31 = prime(11) = (1+2)||1, a(2) = 1

41 = prime(13) = (1+3)||1, a(3) = 1

3413 = prime(480) = (13+21)||13, a(21) = 13

11527 = prime(1390) = (27+88)||27, a(88) = 27

Note cases where consecutive values of n give consecutive primes:

n=17: 181 = prime(42) = (1+17)||1, n=18: 191 = prime(43) = (1+18)||1

k=41: 421 = prime(82) = (1+41)||1, n=42: 431 = prime(83) = (1+42)||1

... are there infinitely many of such?

a(11) = 19, 3019 is a resulting "candidate" for n = 301 - 9 = 292, but a(292) = 3 gives 2953 = prime(425)

First twice resulting prime is 5623 = prime(739) = (23+33)||23 = 5623 = (559+3)||3

MAPLE

tcat:= proc(a, b) a*10^(1+ilog10(b))+b end proc:

f:= proc(n) local k, d;

for d from 1 do

if igcd(n, 10^d+1) > 1 then next fi;

for k from 10^(d-1)+`if`(d=1, 0, 1) to 10^d by 2 do

if isprime(tcat(n+k, k)) then return k fi

od od

end proc:

map(f, [$1..100]); # Robert Israel, Sep 16 2024

PROG

(PARI) a(n) = my(k=1); while (!isprime(eval(concat(Str(n+k), Str(k)))), k++); k; \\ Michel Marcus, Sep 17 2024

(Python)

from itertools import count

from math import gcd

from sympy import isprime

def A174414(n):

for l in count(1):

if gcd(n, (m:=10**l)+1)==1:

r = m//10

a = m*(n+r)+r

for k in range(r, m):

if isprime(a):

return k

a += m+1 # Chai Wah Wu, Sep 18 2024

CROSSREFS

Cf. A089712, A126785, A171154, A173976, A174031.

KEYWORD

base,nonn

AUTHOR

Ulrich Krug (leuchtfeuer37(AT)gmx.de), Mar 19 2010

EXTENSIONS

Edited and corrected by Robert Israel, Sep 16 2024

STATUS

approved

A174583

a(k) is the least n such that the concatenation (n - k)"n is a prime number, for k >= 0.

+10
1

1, 3, 3, 7, 7, 17, 7, 9, 9, 11, 13, 17, 13, 19, 17, 19, 21, 21, 23, 27, 27, 23, 43, 33, 41, 27, 27, 29, 31, 33, 31, 33, 39, 47, 37, 39, 37, 39, 39, 41, 51, 47, 47, 61, 47, 49, 49, 53, 49, 51, 51, 59, 57, 57, 61, 57, 57, 61, 63, 63, 71, 63, 63, 67, 67, 77, 67, 69, 77, 71, 73, 77

(list; graph; refs; listen; history; text; internal format)

OFFSET

0,2

COMMENTS

See comments and references for A174414.

10^d*(n - k) + n has to be prime for the least d-digit n > k (k >= 0).

For (n - k)"n to be a prime, n must end in the digit 1, 3, 7, or 9.

Conjecture: a(k) = a(k+1) for an infinite number of k's.

As n > k, the number of a(k) is finite, and can be easily bounded from above.

1, 11, ... appear only once in the sequence; 3, 9, 13, 19, 21, 23, ... appear twice; 7, 17, ... three times; and so on.

Does each n that ends in the digit 1, 3, 7, or 9 appear in this sequence?

Note this interesting observation that first occurs for k = 84: 9291013 = prime(620602) = (1013 - 84)"1013, a(84) = 1013. A second example is: 9381037 = prime(626219) = (1037 - 99)"1037.

Let k be a multiple of 7, 11, or 13, then no 3-digit n exists such that (n - k)"n is prime. Proof: 10^3*(n - k) + n = n * (10^3+1) - k * 10^3 = 7 * 11 * 13 * n - k * 10^3 is not prime, as k is a multiple of 7, 11, or 13.

Similar for k-digit n with given divisors and k > 3: 10^4 + 1 = 73 * 137, 10^5 + 1 = 11 * 9091.

LINKS

Table of n, a(n) for n=0..71.

EXAMPLE

11 = prime(5) = (1 - 0)"1, thus a(0) = 1.

23 = prime(9) = (3 - 1)"3, thus a(1) = 3.

13 = prime(6) = (3 - 2)"3, thus a(2) = 3.

139 = prime(34) = (39 - 38)"39, thus a(38) = 39.

9109 = prime(1130) = (109 - 100)"109, thus a(100) = 109.

MAPLE

mycat := (k, n) -> parse(cat(convert(n - k, string), convert(n, string))):

sol := (k, n) -> isprime(mycat(k, n)):

a := proc(k) local n; for n from k + 1 while not sol(k, n) do od; n end:

seq(a(k), k = 0..71); # Peter Luschny, Sep 20 2024

CROSSREFS

Cf. A174414, A089712, A171154, A173976, A174031.

KEYWORD

base,nonn

AUTHOR

Ulrich Krug (leuchtfeuer37(AT)gmx.de), Mar 23 2010

EXTENSIONS

Edited, offset set to 0 and a(71) corrected by Peter Luschny, Sep 20 2024

STATUS

approved

A177688

Numbers n such that (n+2)//n - (n+1) is prime, where // represents the concatenation of decimals.

+10
0

0, 1, 4, 6, 7, 9, 12, 13, 15, 18, 19, 22, 25, 28, 31, 33, 39, 46, 48, 49, 52, 60, 61, 64, 67, 73, 75, 84, 85, 88, 90, 99, 100, 103, 106, 132, 133, 135, 136, 138, 142, 156, 160, 163, 171, 178, 181, 183, 187, 190, 198, 201, 202, 211, 220, 222, 229, 238, 241, 246, 252

(list; graph; refs; listen; history; text; internal format)

OFFSET

1,3

COMMENTS

If n is a k-digit number, then we demand that p = (n+2) * 10^k + n - (n+1) is a prime number, obviously of the form p = (n+2) * 10^k - 1, so the decimal representation of p is n+1 followed by k times the digit 9.

The sequence is infinite, proof with Dirichlet's prime number (in arithmetic progressions) theorem.

Note that numbers of the form (n+2)//n + (n+1) are multiples of 3 and do not generate primes.

LINKS

Table of n, a(n) for n=1..61.

EXAMPLE

2//0 - 1 = 20 - 1 = 19 = prime(8), 0 is first term;

3//1 - 2 = 31 - 2 = 29 = prime(10), 1 is 2nd term;

6//4 - 5 = 64 - 5 = 59 = prime(17), 4 is 3rd term.

MATHEMATICA

n2ncQ[n_]:=PrimeQ[FromDigits[Join[IntegerDigits[n+2], IntegerDigits[ n]]]- n-1]; Select[Range[0, 300], n2ncQ] (* Harvey P. Dale, Feb 24 2011 *)

CROSSREFS

Cf. A000040, A065582, A095995, A173976, A177435.

KEYWORD

base,easy,nonn

AUTHOR

Ulrich Krug (leuchtfeuer37(AT)gmx.de), May 11 2010

STATUS

approved

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