%I #58 Jul 02 2024 02:16:38

%S 12,7,4,7,4,7,12,3,12,7,4,7,4,7,12,3,12,7,4,7,4,7,12,3,12,7,4,7,4,7,

%T 12,3,12,7,4,7,4,7,12,3,12,7,4,7,4,7,12,3,12,7,4,7,4,7,12,3,12,7,4,7,

%U 4,7,12,3,12,7,4,7,4,7,12,3,12,7,4,7,4,7,12,3

%N Sequence of gaps between deletions of multiples of 7 in step 4 of the sieve of Eratosthenes.

%C This sequence is a repeating cycle 12, 7, 4, 7, 4, 7, 12, 3 of length A005867(4) = 8 = (prime(1)-1)*(prime(2)-1)*(prime(3)-1).

%C The mean of the cycle is prime(4) = 7.

%C The cycle is constructed from the sieve of Eratosthenes as follows.

%C In the first 2 steps of the sieve, the gaps between the deleted numbers are constant: gaps of 2 in step 1 when we delete multiples of 2, and gaps of 3 in step 2 when we delete multiples of 3.

%C In step 3, when we delete all multiples of 5, the gaps are alternately 7 and 3 (i.e., cycle [7,3]).

%C For this sequence, we look at the interesting cycle from step 4 (multiples of 7).

%C Excluding the final 3, the cycle has reflective symmetry: 12, 7, 4, 7, 4, 7, 12. This is true for every subsequent step of the sieve too.

%C The central element is 7 (BUT not all steps have their active prime number as the central element).

%C a(1) is A054272(4).

%C a(8) = 3, the first appearance of the last element of the cycle, corresponds to deletion of 217 = A002110(4)+7.

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,1).

%F a(n) = A236175(n)+1. - _Peter Munn_, Jan 21 2023

%e After sieve step 3, multiples of 2,3,5 have been eliminated leaving

%e 7,11,13,17,19,23,29,31,37,41,43,47,49,53, ...

%e ^ ^

%e The first two multiples of 7 are 7 itself and 49 and they are distance 12 apart in the list so that a(1) = 12.

%e For n = 2, a(n) = 7, because the third multiple of 7 that is not a multiple of 2, 3 or 5 is 77 = 7 * 11, which is located 7 numbers after 49 = 7*7 in the list of numbers without the multiples of 2, 3 and 5.

%t PadRight[{}, 100, {12, 7, 4, 7, 4, 7, 12, 3}] (* _Paolo Xausa_, Jul 01 2024 *)

%o (Python)

%o numbers = []

%o for i in range(2,880):

%o numbers.append(i)

%o gaps = []

%o step = 4

%o current_step = 1

%o while current_step <= step:

%o prime = numbers[0]

%o new_numbers = []

%o gaps = []

%o gap = 0

%o for i in range(1,len(numbers)):

%o gap += 1

%o if numbers[i] % prime != 0:

%o new_numbers.append(numbers[i])

%o else:

%o gaps.append(gap)

%o gap = 0

%o current_step += 1

%o numbers = new_numbers

%o print(gaps)

%Y Cf. A002110, A005867, A054272, A236175.

%Y Equivalent sequences for steps 1..3: A007395, A010701, A010705 (without the initial 3).

%K nonn,easy

%O 1,1

%A _Alexandre Herrera_, Jan 08 2023