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A307213
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Divide the natural numbers into sets of successive sizes 3,4,5,6,7,...,, starting with {1,2,3}. Cycle through each set until you reach a prime; if the prime was the n-th element in its set, jump to the n-th element of the next set.
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2
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1, 2, 5, 9, 10, 11, 16, 17, 23, 30, 31, 39, 40, 41, 50, 51, 52, 43, 53, 64, 65, 66, 67, 79, 92, 93, 94, 95, 96, 97, 111, 112, 113, 128, 129, 130, 131, 147, 148, 149, 166, 167, 185, 186, 187, 169, 170, 171, 172, 173, 192, 193, 213, 214, 215, 216, 217, 218, 219, 220, 221, 222, 223, 244, 245, 246, 247, 248, 249, 250, 229
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OFFSET
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1,2
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COMMENTS
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"Cycle" in the definition, means that if no prime is found, go back to the start of the set.
If a set does not contain a prime, the sequence goes into an infinite loop, but it is conjectured that this does not happen since the sets are of increasing length.
The sets (rather intervals) are I_j = [(j^2 + 3*j - 2)/2, j*(j + 5)/2] =[A034856(j), A095998(j)], for j >= 1. For the number of primes in these intervals see A309121. - Wolfdieter Lang, Jul 13 2019
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LINKS
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EXAMPLE
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The first set is {1,2,3}. We look at 1 then 2. 2 is prime, and it is the second number in the set. The next set is {4,5,6,7}. So we jump to the second element, 5. 5 is also prime, so we jump to the second element of the next set, {8,9,10,11,12}, which is 9, etc. If we reach the end of a set without reaching a prime, we loop back to the first element, which is the only way for a(n) < a(n-1) to happen.
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MATHEMATICA
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Nest[Append[#1, {If[#3 <= Length@ #4, #3, #3 - Length@ #4], If[#2 == #3, {#4[[#3]]}, Join[#4, #4][[#2 ;; #3]]], #4}] & @@ {#1, #2, If[PrimeQ[#3[[#2]] ], #2, #2 + FirstPosition[RotateLeft[#3, #2], _?PrimeQ][[1]] ], #3} & @@ {#1, #2, Range[#3, #3 + #4]} & @@ {#, #[[-1, 1]], 1 + Max@ #[[-1, -1]], Length@ # + 2} &, {{#1, #2[[1 ;; #1]], #2} & @@ {FirstPosition[#, _?PrimeQ][[1]], #}} &@ Range@ 3, 19][[All, 2]] // Flatten (* Michael De Vlieger, Mar 31 2019 *)
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PROG
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(PARI) See Links section.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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