A111393 -id:A111393 - cp's OEIS frontend

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

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

A276621 After a(0)=0, each n occurs A261234(n-1) times.

Original entry on oeis.org

0, 1, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6

Offset: 0

Antti Karttunen, Sep 11 2016

Auxiliary function for computing A276622 & A276623.

Antti Karttunen, Table of n, a(n) for n = 0..6250

Cf. A261233, A276622, A276623.

After a(0), a(n) differs from A111393(n+1) for the first time at n=46, where a(46)=5, while A111393(47)=6.

(define (A276621 n) (let loop ((k 0)) (if (>= (A261233 k) n) k (loop (+ 1 k)))))

A338433 Values of n for which A070939(n^3) differs from A004221(n).

Original entry on oeis.org

1, 20, 40, 80, 101, 126, 127, 159, 160, 161, 200, 201, 202, 203, 252, 253, 254, 255, 317, 318, 319, 320, 321, 322, 399, 400, 401, 402, 403, 404, 405, 406, 502, 503, 504, 505, 506, 507, 508, 509, 510, 511, 631, 632, 633, 634, 635, 636, 637, 638, 639, 640, 641, 642, 643, 644, 645

Offset: 1

Jeremy Gardiner, Oct 27 2020

nonn

Sequence gives the values of n for which the length of the binary representation of n^3 differs from ceiling(10*log_10(n)) rounded up.

The largest number not in the sequence is 158489319246111348520210137339 = floor(10^29.2). - Robert Israel, Oct 27 2020

Jeremy Gardiner, Table of n, a(n) for n = 1..1083
Index entries for linear recurrences with constant coefficients, signature (2,-1).

Cf. A004221, A070939, A111393, A329193.

filter:= n -> evalb(ilog2(n^3)+1 <> ceil(10*log[10](n))):
select(filter, [$1..1000]); # Robert Israel, Oct 27 2020

Select[Range[1000], IntegerLength[#^3, 2] != Ceiling[10*Log10[#]] &] (* Amiram Eldar, Oct 27 2020 *)

cp's OEIS Frontend

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

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A276621 After a(0)=0, each n occurs A261234(n-1) times.

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A338433 Values of n for which A070939(n^3) differs from A004221(n).

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