A347113 - OEIS

%I #94 Sep 05 2022 09:55:26

%S 1,4,10,22,46,94,5,2,6,14,3,8,12,26,9,15,18,38,13,7,16,34,20,24,30,62,

%T 21,11,27,32,36,74,25,28,58,118,17,33,40,82,166,334,35,39,42,86,29,44,

%U 48,56,19,45,23,50,54,60,122,41,49,52,106,214,43,55,63,66

%N a(1)=1; for n > 1, a(n) is the smallest unused positive number k such that k != j and gcd(k,j) != 1, where j = a(n-1) + 1.

%C Alternative definition: Lexicographically earliest sequence of distinct positive numbers such that a(n) != a(n-1)+1 and gcd(a(n-1)+1,a(n)) > 1. This makes it a cousin of the EKG sequence A064413, the Yellowstone permutation A098550, the Enots Wolley sequence A336957, and others. - _N. J. A. Sloane_, Sep 01 2021; revised Nov 08 2021.

%C The successive gcd's are listed in A347309.

%H Michel Marcus, <a href="/A347113/b347113.txt">Table of n, a(n) for n = 1..10000</a>

%H Chai Wah Wu, <a href="/A347113/a347113.txt">Table of n, a(n) for n = 1..10^6</a>

%H Chai Wah Wu, <a href="/A347113/a347113.png">Graph of first million terms</a>

%H <a href="/index/Per#IntegerPermutation">Index entries for sequences that are permutations of the integers</a>

%e a(1) = 1, by definition.

%e a(2) = 4; it cannot be 2, because 2 = a(1) + 1, and it cannot be 3, because gcd(a(1) + 1, 3) = 1.

%e a(3) = 10, because gcd(a(3), a(2) + 1) cannot equal 1. a(2) + 1 = 5, so a(3) must be a multiple of 5. It cannot be equal to 5, so it must be 10, the next available multiple of 5.

%e a(4) = 22, because 22 is the smallest positive integer not equal to 11 and not coprime to 11.

%p b:= proc() true end:

%p a:= proc(n) option remember; local j, k; j:= a(n-1)+1;

%p       for k from 2 do if b(k) and k<>j and igcd(k, j)>1

%p         then b(k):= false; return k fi od

%p     end: a(1):= 1:

%p seq(a(n), n=1..100);  # _Alois P. Heinz_, Sep 02 2021

%t Block[{a = {1}, c, k, m = 2}, Do[If[IntegerQ@Log2[i], While[IntegerQ[c[m]], m++]]; Set[k, m]; While[Or[IntegerQ[c[k]], k == # + 1, GCD[k, # + 1] == 1], k++] &[a[[-1]]]; AppendTo[a, k]; Set[c[k], i], {i, 65}]; a] (* _Michael De Vlieger_, Aug 18 2021 *)

%o (PARI) find(va, x) = {my(k=1, s=Set(va)); while ((k==x) || (gcd(k, x) == 1) || setsearch(s, k), k++); k;}

%o lista(nn) = {my(va = vector(nn)); va[1] = 1; for (n=2, nn, va[n] = find(va, va[n-1]+1);); va;} \\ _Michel Marcus_, Aug 21 2021

%o (Python)

%o from math import gcd

%o A347113_list, nset, m = [1], {1}, 2

%o for _ in range(100):

%o     j = A347113_list[-1]+1

%o     k = m

%o     while k == j or gcd(k,j) == 1 or k in nset:

%o         k += 1

%o     A347113_list.append(k)

%o     nset.add(k)

%o     while m in nset:

%o         m += 1 # _Chai Wah Wu_, Sep 01 2021

%Y See A347306 for the inverse, A347307, A347308 for the records, A347309 for the gcd values, A347312 for the parity of a(n), A347314 for the fixed points, and A348780 for partial sums.

%Y For the main diagonal see (A348787(k), A348788(k)).

%Y Cf. A064413, A098550, A336957, A347313, A348779, A348785, A348786, A353712, A353713.

%K nonn

%O 1,2

%A _Grant Olson_, Aug 18 2021

%E Comments edited (including deletion of incorrect comments) by _N. J. A. Sloane_, Sep 05 2021

%E For the moment I am withdrawing my claim that this is a permutation of the positive integers. - _N. J. A. Sloane_, Sep 05 2022