%I #26 Sep 22 2023 07:51:02

%S 2,3,8,9,10,11,12,13,18,23,28,33,38,39,40,41,42,43,44,45,46,47,48,49,

%T 50,51,52,53,54,55,56,57,58,59,60,61,62,63,68,73,78,83,88,93,98,103,

%U 108,113,118,123,128,133,138,143,148,153,158,163,168,173,178,183,188

%N Unique monotonic sequence of positive integers satisfying a(a(n)) = k*(n-1) + 3, where k = 5.

%C Numbers m such that the base-5 representation of (2*m-1) starts with 3 or 4 or ends with 0.

%C First differences give a run of 5^i 1's followed by a run of 5^i 5's, for i >= 0.

%H Yifan Xie, <a href="/A353652/b353652.txt">Table of n, a(n) for n = 1..10000</a>

%H <a href="/index/Aa#aan">Index entries for sequences of the a(a(n)) = 2n family</a>

%F For n in the range (5^i + 1)/2 <= n < (3*5^i + 1)/2, for i >= 0:

%F a(n) = n + 5^i.

%F a(n+1) = 1 + a(n).

%F Otherwise, for n in the range (3*5^i + 1)/2 < n <= (5*5^i + 1)/2, for i >= 0:

%F a(n) = 5*(n - 5^i) - 2.

%F a(n+1) = 5 + a(n).

%e a(7) = 12 because (5^1 + 1)/2 <= 7 < (3*5^1 + 1)/2, hence a(7) = 7 + 5^1 = 12;

%e a(11) = 28 because (3*5^1 + 1)/2 <= 11 < (5*5^1 + 1)/2, hence a(11) = 5*(11 - 5^1) - 2 = 28.

%t a[n_] := Module[{n2 = 2n, p}, p = 5^Floor@Log[5, n2]; If[n2 < 3p, n+p, 5(n-p)-2]];

%t Table[a[n], {n, 1, 100}] (* _Jean-François Alcover_, Sep 22 2023, after _Kevin Ryde_ *)

%o (PARI) a(n) = my(n2=n<<1, p=5^logint(n2, 5)); if(n2 < 3*p, n+p, 5*(n-p)-2); \\ _Kevin Ryde_, Apr 18 2022

%o (C++)

%o /* program used to generate the b-file */

%o #include<iostream>

%o using namespace std;

%o int main(){

%o int cnt1=1, flag=0, cnt2=1, a=2;

%o for(int n=1; n<=10000; n++) {

%o cout<<n<<" "<<a<<endl;

%o if(cnt2==cnt1) {

%o flag=1-flag, cnt1=1;

%o if(flag) a+=1;

%o else {

%o a+=5;

%o cnt2*=5;

%o }

%o }

%o else {

%o cnt1++;

%o a+=(flag?5:1);

%o }

%o }

%o return 0;

%o }

%Y For other values of k: A080637 (k=2), A003605 (k=3), A353651 (k=4), this sequence (k=5), A353653 (k=6).

%K nonn,easy

%O 1,1

%A _Yifan Xie_, Jul 15 2022