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Revision History for A265181

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A265181

Prime numbers resulting from the concatenation of at least two copies of a cubic number followed by a trailing "1."
(history; published version)

#46 by Alois P. Heinz at Mon Feb 20 14:50:05 EST 2023

STATUS

proposed

approved

#45 by Chai Wah Wu at Mon Feb 20 13:23:28 EST 2023

STATUS

editing

proposed

#44 by Chai Wah Wu at Mon Feb 20 13:23:21 EST 2023

PROG

(Python)

from itertools import count, islice

from sympy import isprime

def A265181_gen(): # generator of terms

return filter(isprime, (int(str(k**3)*2)*10+1 for k in count(1)))

A265181_list = list(islice(A265181_gen(), 20)) # Chai Wah Wu, Feb 20 2023

STATUS

approved

editing

#43 by N. J. A. Sloane at Wed Jan 20 23:50:39 EST 2016

STATUS

editing

approved

#42 by N. J. A. Sloane at Wed Jan 20 23:50:34 EST 2016

NAME

Prime numbers resulting from the concatenation of at least two copies of a cubic number repeated at least twice and followed by a trailing "1."

STATUS

proposed

editing

#41 by Robert Israel at Wed Jan 13 19:30:58 EST 2016

STATUS

editing

proposed

Discussion

Wed Jan 20

23:49

N. J. A. Sloane: "repeated twice" is a classical ambiguity in the English language. If you say something and repeat it twice, you actually say it three times.I will rewrite the definition to avoid this problem.

#40 by Robert Israel at Wed Jan 13 19:30:47 EST 2016

LINKS

Robert Israel, <a href="/A265181/b265181.txt">Table of n, a(n) for n = 1..10000</a>

MAPLE

N:= 20: # to get all terms with at most N digits

M:= floor((N-1)/2):

res:= {}:

for s from 1 to floor(10^(M/3)) do

x:= s^3;

m:= 1+ilog10(x);

for k from 2 to floor((N-1)/m) do

p:= x*add(10^(1+m*i), i=0..k-1)+1;

if isprime(p) then res:= res union {p} fi;

od

od:

sort(convert(res, list)); # Robert Israel, Jan 13 2016

STATUS

proposed

editing

#39 by Michael De Vlieger at Tue Jan 05 16:10:31 EST 2016

STATUS

editing

proposed

#38 by Michael De Vlieger at Tue Jan 05 16:00:40 EST 2016

MATHEMATICA

Take[Sort@ Flatten[Select[#, PrimeQ] & /@ Table[FromDigits@ Append[Flatten@ IntegerDigits@ Table[n^3, {#}], 1] & /@ Range[2, 20], {n, 1, 300}] /. {} -> Nothing], 27] (* Michael De Vlieger, Jan 05 2016 *)

STATUS

proposed

editing

Discussion

Tue Jan 05

16:03

Michael De Vlieger: Got it. Pseudocode: make a number with more than 2 instances of the cube of n, add 1 at the end. I use a range (2, 20) iterations, then find all the instances that are prime. Do this over a range of (1, 300), eliminate all the n that don't produce primes, and sort. This sort of sequence requires quite a lot of range to ensure you have all the qualifying terms over a given range of a(n).

16:10

Michael De Vlieger: Vicenzo, dobbiamo trovare tutti i numeri primi fatto per variare da due a forse 20 volte, per ogni valuta di n tra 1 a circa 300. Qualche valuta di n non fara numeri primi, invece fara un insieme vuoto {}, che dobbiamo eliminare. Poi dobbiamo ordinare i numeri risultanti ad arrivare alla sequenza. (This is a translation of my last pink box for Vicenzo).

#37 by Thomas S. Pedigo at Tue Jan 05 07:56:50 EST 2016

STATUS

editing

proposed

Discussion

Tue Jan 05

08:22

Michel Marcus: We really need a program

15:51

Michael De Vlieger: Vicenzo, ho cambiato la mia programma a fare output per due volte e trovo la stessa data / I changed my program to run two iterations of cubes and reach the same data as you do: {881, 27271, 7297291, 133113311, 337533751, 19683196831, 42875428751, 68921689211, 1038231038231, 1574641574641, 2053792053791, 4218754218751, 7290007290001, 106120810612081, 224809122480911, 274400027440001, 280322128032211, 317652331765231}. Simply add "2 Ceiling@ Log10[n^3] + 1" to the second input for f.

15:51

Michael De Vlieger: Now that I look at it I think I see what Thomas is doing. I think he's finding ALL the primes that derive from the concatenation then sorting them according to magnitude.