A293430 Persistently squarefree numbers for base-2 shifting: Numbers n such that all terms in finite set [n, floor(n/2), floor(n/4), floor(n/8), ..., 1] are squarefree.
1, 2, 3, 5, 6, 7, 10, 11, 13, 14, 15, 21, 22, 23, 26, 29, 30, 31, 42, 43, 46, 47, 53, 58, 59, 61, 62, 85, 86, 87, 93, 94, 95, 106, 107, 118, 119, 122, 123, 170, 173, 174, 186, 187, 190, 191, 213, 214, 215, 237, 238, 239, 246, 247, 341, 346, 347, 349, 373, 374, 381, 382, 383, 426, 427, 429, 430, 431, 474, 478, 479
Offset: 1
Keywords
Examples
For 479 we see that 479 is prime (thus squarefree, in A005117), [479/2] = 239 is also a prime, [239/2] = 119 = 7*17 (a squarefree composite), [119/2] = 59 (a prime), [59/2] = 29 (a prime), [29/2] = 14 = 2*7 (a squarefree composite), [14/2] = 7 (a prime), [7/2] = 3 (a prime), [3/2] = 1 (the end of halving process 1 is also squarefree), thus all the values obtained by repeated halving were squarefree and 479 is a member of this sequence. Here [ ] stands for taking floor.
Links
- Antti Karttunen, Table of n, a(n) for n = 1..10000
Crossrefs
Programs
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Mathematica
With[{s = Fold[Append[#1, MoebiusMu[#2] #1[[Floor[#2/2]]]] &, {1}, Range[2, 480]]}, Flatten@ Position[s, ?(# != 0 &)]] (* _Michael De Vlieger, Oct 10 2017 *)
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PARI
is_persistently_squarefree(n,base) = { while(n>1, if(!issquarefree(n),return(0)); n \= base); (1); }; isA293430(n) = is_persistently_squarefree(n,2); n=0; k=1; while(k <= 10000, n=n+1; if(isA293430(n),write("b293430.txt", k, " ", n);k=k+1)); \\ Antti Karttunen, Oct 11 2017
Comments