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A359632
Sequence of gaps between deletions of multiples of 7 in step 4 of the sieve of Eratosthenes.
1
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, 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, 4, 7, 12, 3, 12, 7, 4, 7, 4, 7, 12, 3, 12, 7, 4, 7, 4, 7, 12, 3
OFFSET
1,1
COMMENTS
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).
The mean of the cycle is prime(4) = 7.
The cycle is constructed from the sieve of Eratosthenes as follows.
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.
In step 3, when we delete all multiples of 5, the gaps are alternately 7 and 3 (i.e., cycle [7,3]).
For this sequence, we look at the interesting cycle from step 4 (multiples of 7).
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.
The central element is 7 (BUT not all steps have their active prime number as the central element).
a(1) is A054272(4).
a(8) = 3, the first appearance of the last element of the cycle, corresponds to deletion of 217 = A002110(4)+7.
FORMULA
a(n) = A236175(n)+1. - Peter Munn, Jan 21 2023
EXAMPLE
After sieve step 3, multiples of 2,3,5 have been eliminated leaving
7,11,13,17,19,23,29,31,37,41,43,47,49,53, ...
^ ^
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.
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.
MATHEMATICA
PadRight[{}, 100, {12, 7, 4, 7, 4, 7, 12, 3}] (* Paolo Xausa, Jul 01 2024 *)
PROG
(Python)
numbers = []
for i in range(2, 880):
numbers.append(i)
gaps = []
step = 4
current_step = 1
while current_step <= step:
prime = numbers[0]
new_numbers = []
gaps = []
gap = 0
for i in range(1, len(numbers)):
gap += 1
if numbers[i] % prime != 0:
new_numbers.append(numbers[i])
else:
gaps.append(gap)
gap = 0
current_step += 1
numbers = new_numbers
print(gaps)
CROSSREFS
Equivalent sequences for steps 1..3: A007395, A010701, A010705 (without the initial 3).
Sequence in context: A093763 A002548 A363151 * A364135 A305939 A206423
KEYWORD
nonn,easy
AUTHOR
Alexandre Herrera, Jan 08 2023
STATUS
approved