A168219 -id:A168219 - cp's OEIS frontend

A168147 Primes of the form 10*n^3 + 1.

11, 271, 641, 2161, 33751, 40961, 58321, 138241, 196831, 270001, 297911, 466561, 506531, 795071, 1326511, 1406081, 1851931, 2160001, 3890171, 4218751, 5314411, 5513681, 6585031, 7290001, 8043571, 11910161, 12597121, 12950291, 14815441

Offset: 1

Eva-Maria Zschorn (e-m.zschorn(AT)zaschendorf.km3.de), Nov 19 2009

(1) These primes all with end digit 1=1^3 are concatenations of two CUBIC numbers: "n^3 1".
(2) It is conjectured that the sequence is infinite.
(3) It is an open problem if 3 consecutive naturals n exist which give such a prime.
No three such integers exist, as every n = 2 (mod 3) yields 10n^3 + 1 = 0 (mod 3). - Charles R Greathouse IV, Apr 24 2010

Harold Davenport, Multiplicative Number Theory, Springer-Verlag New-York 1980
Leonard E. Dickson: History of the Theory of numbers, vol. I, Dover Publications 2005

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Cf. A030430 (primes of the form 10*n+1).
Cf. A167535 (concatenation of two square numbers which give a prime).
See A168219 for the numbers n.

[ a: n in [1..150] | IsPrime(a) where a is 10*n^3+1 ]; // Vincenzo Librandi, Jul 25 2011

Select[Table[10*n^3+1,{n,1000}],PrimeQ] (* Vincenzo Librandi, Aug 01 2012 *)

for(n=1,2e2, isprime(n^3*10+1) && print1(n^3*10+1", "))  \\ M. F. Hasler, Jul 24 2011

a(n) = 10*A168219(n)^3 + 1. \\ M. F. Hasler, Jul 24 2011

A173836 Natural numbers n such that the concatenation 1331//n^3 is a prime number.

Original entry on oeis.org

21, 27, 29, 41, 101, 119, 141, 171, 173, 177, 191, 197, 219, 243, 267, 291, 309, 327, 333, 369, 371, 383, 411, 417, 1019, 1049, 1059, 1091, 1157, 1163, 1211, 1311, 1337, 1343, 1359, 1371, 1379, 1409, 1461, 1473, 1481, 1503, 1521, 1593, 1599, 1613, 1637

Offset: 1

Eva-Maria Zschorn (e-m.zschorn(AT)zaschendorf.km3.de), Feb 26 2010

Given the cube n^3 with k = A111393(n) decimal digits, we have to check whether the concatenation, 11^3 * 10^k + n^3, is a prime.
The number k of digits that 1331=11^3 is shifted is not a multiple of 3,
because the form a^3+b^3 = (a^2+a*b+b^2) * (a - b) cannot construct a prime.

			21 is in the sequence because 21^3=9261, and the concatenation is 13319261=prime(868687).
27 is in the sequence because 27^3=19683, and the concatenation is 133119683=prime(7545064).

K. Haase, P. Mauksch: Spass mit Mathe, Urania-Verlag Leipzig, Verlag Dausien Hanau, 2. Auflage 1985

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A102006, A167535, A168147, A168219, A168274, A173579, A173733

Select[Range[2000],PrimeQ[FromDigits[Join[{1,3,3,1}, IntegerDigits[ #^3]]]]&] (* Harvey P. Dale, Oct 14 2011 *)

Comments sligthly rephrased - R. J. Mathar, Mar 05 2010

A173579 Natural numbers n which give primes when 1331 = 11^3 is prefixed.

Original entry on oeis.org

3, 17, 21, 53, 57, 69, 83, 87, 107, 119, 123, 153, 207, 227, 243, 249, 251, 261, 269, 279, 293, 299, 327, 329, 333, 339, 347, 377, 381, 383, 399, 411, 431, 437, 443, 471, 489, 497, 513, 521, 527, 549, 567, 573, 579, 587, 591, 597, 599, 611, 633, 641, 647, 657

Offset: 1

Eva-Maria Zschorn (e-m.zschorn(AT)zaschendorf.km3.de), Feb 22 2010

Concatenation of N = 1331 = 11^3 = palindrome(113) and natural n is a prime. No zeros "between" N and n.
13 = emirp(1) = prime(6), R(13) = 31 = emirp(3) = prime(11).
Necessarily n = 3 * k or n = 3 * k + 2, but not n = 3 * k + 1, because sod(1331) = 8. So no prime twins are terms of the sequence.

			13313 = prime(1581) => a(1) = 3.
133117 = prime(12425) => a(2) = 17.
133103, 133109 are prime, but "0" included: "03" resp. "09" are no terms of the sequence.

Leonard E. Dickson: History of the Theory of numbers, vol. I, Dover Publications 2005
K. Haase, P. Mauksch: Spass mit Mathe, Urania-Verlag Leipzig, Verlag Dausien Hanau, 2. Auflage 1985
Theo Kempermann: Zahlentheoretische Kostproben, Harri Deutsch, 2. aktualisierte Auflage 2005

Harvey P. Dale, Table of n, a(n) for n = 1..1000

A102006, A167535, A168147, A168219, A168274

Select[Range[700],PrimeQ[1331*10^IntegerLength[#]+#]&] (* Harvey P. Dale, Jun 25 2020 *)

isok(n) = isprime(n + 1331*10^(length(Str(n)))); \\ Michel Marcus, Aug 27 2013

A168274 Primes p such that 10*p^3 + 1 is prime.

Original entry on oeis.org

3, 31, 37, 43, 73, 109, 163, 211, 241, 313, 337, 409, 499, 673, 739, 757, 907, 1033, 1039, 1069, 1237, 1327, 1447, 1483, 1489, 1597, 1663, 1741, 1777, 1933, 2083, 2143, 2251, 2437, 2683, 2797, 3001, 3181, 3307, 3319, 3463, 3739, 3793, 4051

Offset: 1

Eva-Maria Zschorn (e-m.zschorn(AT)zaschendorf.km3.de), Nov 22 2009

nonn

Subsequence of A168219.

			10*3^3+1 = 271 is prime, so prime 3 is in the sequence.
10*313^3+1 = 306642971 is prime, so prime 313 is in the sequence.

G. C. Greubel, Table of n, a(n) for n = 1..1000

Cf. A168219, A168147.

[ p: p in PrimesUpTo(5000) | IsPrime(10*p^3+1) ]; // Vincenzo Librandi, Jan 29 2011

Select[Prime[Range[100]], PrimeQ[10*#^3 + 1] &] (* G. C. Greubel, Jul 16 2016 *)

select(p->isprime(10*p^3+1), primes(1000)) \\ Charles R Greathouse IV, Jul 16 2016

Edited by Charles R Greathouse IV, Apr 24 2010

A173733 Primes p which give primes when 1331 = 11^3 is prefixed (see A173579).

Original entry on oeis.org

3, 17, 53, 83, 107, 227, 251, 269, 293, 347, 383, 431, 443, 521, 587, 599, 641, 647, 683, 719, 761, 773, 821, 857, 929, 1031, 1097, 1217, 1223, 1301, 1367, 1409, 1433, 1451, 1619, 1637, 1709, 1787, 1973, 2081, 2087, 2129, 2399, 2477, 2591, 2633, 2657, 2693

Offset: 1

Eva-Maria Zschorn (e-m.zschorn(AT)zaschendorf.km3.de), Feb 23 2010

N = 1331 = 11^3, p k-digit prime, to check if q = N * 10^k + p is prime

With exception of 3 necessarily p of form 3k+2, as sod(1331 = 8)

			13313 = prime(1581) => a(1) = prime(2) = 3
133117 = prime(12425) => a(2) = prime(7) = 17
133153 = prime(12427) => a(3) = prime(16) = 53
13311217 = prime(868166) => a(28) = prime(199) = 1217
13311223 = prime(868167) => a(29) = prime(200) = 1223
Note: two consecutive primes P = prime(n), Q = prime(n+1) yield consecutive prime concatenations "N P" = prime(m) and "N Q" = prime(m+1)

K. Haase, P. Mauksch: Spass mit Mathe, Urania-Verlag Leipzig, Verlag Dausien Hanau, 2. Auflage 1985
Helmut Kracke, Mathe-musische Knobelisken, Duemmler Bonn, 2. Auflage 1983

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A102006, A167535, A168147, A168219, A168274, A173579

Select[Prime[Range[400]],PrimeQ[FromDigits[Join[{1,3,3,1}, IntegerDigits[ #]]]]&] (* Harvey P. Dale, Jun 09 2015 *)

Edited and extended by Charles R Greathouse IV, Apr 24 2010

A176722 Primes of the form k^3 + 13, k >= 0.

Original entry on oeis.org

13, 229, 1013, 1741, 39317, 64013, 74101, 157477, 438989, 551381, 830597, 1906637, 2000389, 4096013, 7077901, 9261013, 10941061, 15625013, 16003021, 21024589, 24897101, 27000013, 69934541, 74088013, 79507013, 93576677, 122023949

Offset: 1

Eva-Maria Zschorn (e-m.zschorn(AT)zaschendorf.km3.de), Apr 25 2010

nonn

Necessarily, k = 6 * j or k = 6 * j + 4.

Values of k corresponding to terms of the sequence: 0, 6, 10, 12, 34, 40, 42, 54, 76, 82, 94, 124, 126, 160, 192, 210, 222, 250, 252, 276, 292, 300, 412, 420, 430, 454, 496, 502, 570, 586, 612, 622, 640, 670, 684, 712, 720, 724, 726, 756, 784, 822, 826, 874, 882, 894, 934, 952, 964, 1006, 1056.

			0^3 + 13 = 13 = prime(6) = a(1);
6^3 + 13 = 229 = prime(50) = a(2);
300^3 + 13 = 27000013 = prime(1683067) = a(22).

H. Rademacher, Topics in Analytic Number Theory, Springer-Verlag Berlin, 1973.

Robert Israel, Table of n, a(n) for n = 1..10000
D. R. Heath-Brown and B. Z. Moroz, Primes Represented by Binary Cubic Forms, Proceedings of the London Math. Soc. vol. 84(2), pp. 257-288, 2002; see also.

Cf. A000040, A000578, A037396, A138375, A144953, A154648, A157772, A168147, A168219, A168327.

[a: n in [0..500]|IsPrime(a) where a is n^3+13] // Vincenzo Librandi, Dec 22 2010

select(isprime,[seq(seq((6*j+m)^3+13,m=[0,4]),j=0..1000)]); # Robert Israel, Jun 28 2018

Select[Range[0,1000]^3+13,PrimeQ]  (* Harvey P. Dale, Mar 12 2011 *)

A173874 Primes in A173836.

Original entry on oeis.org

29, 41, 101, 173, 191, 197, 383, 1019, 1049, 1091, 1163, 1409, 1481, 1613, 1637, 1721, 1823, 1913, 1973, 2027, 2099, 2243, 2339, 2351, 2447, 2729, 2837, 2897, 2999, 3023, 3089, 3137, 3167, 3203, 3251, 3407, 3881, 4019, 4349, 4397, 4451, 4457

Offset: 1

Eva-Maria Zschorn (e-m.zschorn(AT)zaschendorf.km3.de), Mar 01 2010

For a prime p and its k-digit cube p^3 we need to check if q = 11^3 * 10^k + p^3 is a prime.
11^3*10^k is congruent to 2 (mod 3), so p^3 must be congruent to 2 (mod 3) because otherwise the sum q cannot become a prime.
In turn, all p in the sequence are also congruent to 2 (mod 3) (see A003627).

			The prime 29 is in the sequence because 29^3=24389, and the concatenation 133124389=prime(7545294) is a prime number.

K. Haase and P. Mauksch: Spass mit Mathe, Urania-Verlag Leipzig, Verlag Dausien Hanau, 2. Auflage 1985

Harvey P. Dale, Table of n, a(n) for n = 1..1000

Cf. A000578, A168147, A168219, A173579, A173733, A173836

cat2 := proc(a,b) ndgs := max(1, ilog10(b)+1) ; a*10^ndgs+b ; end proc:
for i from 1 to 800 do p := ithprime(i) ; if isprime(cat2(1331,p^3)) then printf("%d,",p) ; end if; end do: # R. J. Mathar, Mar 26 2010

Select[Prime[Range[2000]],PrimeQ[FromDigits[Join[{1,3,3,1}, IntegerDigits[ #^3]]]]&] (* Harvey P. Dale, Oct 14 2011 *)

Definition simplified, missing numbers 2243, 2339 etc. inserted, numbers like 2621, 2693 removed - R. J. Mathar, Mar 26 2010

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A168147 Primes of the form 10*n^3 + 1.

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A173836 Natural numbers n such that the concatenation 1331//n^3 is a prime number.

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A173579 Natural numbers n which give primes when 1331 = 11^3 is prefixed.

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A168274 Primes p such that 10*p^3 + 1 is prime.

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A173733 Primes p which give primes when 1331 = 11^3 is prefixed (see A173579).

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A176722 Primes of the form k^3 + 13, k >= 0.

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A173874 Primes in A173836.

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