A293194 Primes of the form 2^q * 3^r * 5^s - 1.
2, 3, 5, 7, 11, 17, 19, 23, 29, 31, 47, 53, 59, 71, 79, 89, 107, 127, 149, 179, 191, 199, 239, 269, 359, 383, 431, 449, 479, 499, 599, 647, 719, 809, 863, 971, 1151, 1249, 1279, 1439, 1499, 1619, 1999, 2399, 2591, 2699, 2879, 2999, 4049, 4373, 4799, 4999, 5119, 5399, 6143, 6911
Offset: 1
Keywords
Examples
3 is a member because 3 is a prime number and 2^2 * 3^0 * 5^0 - 1 = 3. 89 is a member because 89 is a prime number and 2^1 * 3^2 * 5^1 - 1 = 89. list of (q, r, s): (0, 1, 0), (2, 0, 0), (1, 1, 0), (3, 0, 0), (2, 1, 0), (1, 2, 0), (2, 0, 1), (3, 1, 0),(1, 1, 1), (5, 0, 0), (4, 1, 0), (1, 3, 0), (2, 1, 1), ...
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
Programs
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GAP
K := 10^5 + 1;; # to get all terms less than or equal to K A := Filtered([1 .. K], IsPrime);; I := [3, 5];; B := List(A, i -> Elements(Factors(i + 1)));; C := List([0 .. Length(I)], j -> List(Combinations(I, j), i -> Concatenation([2], i))); A293194 := Concatenation([2], List(Set(Flat(List([1 .. Length(C)], i -> List([1 .. Length(C[i])], j -> Positions(B, C[i][j]))))), i -> A[i]));
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Maple
N:= 10^6: # to get all terms <= N R:= {}: for c from 0 to floor(log[5]((N+1))) do for b from 0 to floor(log[3]((N+1)/5^c)) do R:= R union select(isprime, {seq(2^a*3^b*5^c-1, a=0..ilog2((N+1)/(3^b*5^c)))}) od od: sort(convert(R,list)); # Robert Israel, Oct 15 2017 -
Mathematica
With[{n = 7000}, Sort@ Select[Flatten@ Table[2^q * 3^r * 5^s - 1, {q, 0, Log[2, n/(1)]}, {r, 0, Log[3, n/(2^q)]}, {s, 0, Log[5, n/(2^q * 3^r)]}], PrimeQ]] (* Michael De Vlieger, Oct 02 2017 *) -
PARI
lista(nn) = {forprime(p=2,nn, if (vecmax(factor(p+1)[,1]) <= 5, print1(p, ", ")););} \\ Michel Marcus, Oct 06 2017 -
Python
from itertools import count, islice from sympy import integer_log, isprime def A293194_gen(): # generator of terms def bisection(f,kmin=0,kmax=1): while f(kmax) > kmax: kmax <<= 1 kmin = kmax >> 1 while kmax-kmin > 1: kmid = kmax+kmin>>1 if f(kmid) <= kmid: kmax = kmid else: kmin = kmid return kmax def f(x): c = x for i in range(integer_log(x,5)[0]+1): for j in range(integer_log(m:=x//5**i,3)[0]+1): c -= (m//3**j).bit_length() return c yield from filter(isprime,(bisection(lambda k:n+f(k),n,n)-1 for n in count(1))) A293194_list = list(islice(A293194_gen(),30)) # Chai Wah Wu, Mar 31 2025
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