A217287 Length of chain of consecutive integers starting with n, where each new integer in the chain has a prime factor which no previous member in the chain has.
3, 2, 3, 4, 3, 2, 5, 4, 3, 5, 5, 4, 3, 2, 3, 8, 7, 6, 5, 4, 3, 5, 4, 3, 5, 6, 5, 4, 3, 2, 5, 4, 3, 6, 5, 9, 8, 7, 6, 5, 7, 6, 5, 4, 3, 8, 7, 6, 5, 4, 3, 8, 7, 6, 5, 7, 7, 6, 5, 4, 3, 2, 7, 8, 7, 6, 5, 4, 3, 5, 9, 8, 7, 6, 5, 5, 4, 3, 11, 10, 9, 8, 7, 6, 5, 10, 9, 8, 7, 6, 5, 4, 3, 6, 5, 9, 8, 7, 9, 8
Offset: 1
Keywords
Examples
Example: a(7)=5 since 7 starts a chain of 5 integers 7-11 with the following property: 7 is divisible by 7, 8 is divisible by 2, 9 is divisible by 3, 10 is divisible by 5, 11 is divisible by 11. And the next integer 12 is divisible by 2 and 3, both of them are prime factors of prior members in the chain.
Links
- Alois P. Heinz, Table of n, a(n) for n = 1..10000
- Lior Manor, First 1000 entries with the associated chains (For n=1, the chain 1,2,3 should be added. - _N. J. A. Sloane_, Apr 25 2020)
- J. S. Myers, R. Schroeppel, S. R. Shannon, N. J. A. Sloane, and P. Zimmermann, Three Cousins of Recaman's Sequence, arXiv:2004:14000 [math.NT], April 2020.
Programs
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Maple
A006530 := n->max(1, op(numtheory[factorset](n))); a:=[]; M:=120; for n from 1 to M do for k from 1 to 3*n do if A006530(n+k) <= k then a:=[op(a),k]; break; fi; od; od: a; # N. J. A. Sloane, Apr 25 2020
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Mathematica
Block[{nn = 111, r}, r = Prime@ Range[PrimePi@ nn]; r = Table[FromDigits[#, 2] &@ Map[Boole[Mod[n, #] == 0] &, r], {n, nn}]; Array[Block[{k = # + 1, s = r[[#]]}, While[UnsameQ[s, Set[s, BitOr[s, r[[k]] ] ] ], k++]; k - #] &, nn - Ceiling@ Sqrt@ nn] ] (* Michael De Vlieger, Apr 30 2020 *)
Extensions
a(1) = 3 added by N. J. A. Sloane, Apr 25 2020
Comments