A353652 Unique monotonic sequence of positive integers satisfying a(a(n)) = k*(n-1) + 3, where k = 5.

2, 3, 8, 9, 10, 11, 12, 13, 18, 23, 28, 33, 38, 39, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 50, 51, 52, 53, 54, 55, 56, 57, 58, 59, 60, 61, 62, 63, 68, 73, 78, 83, 88, 93, 98, 103, 108, 113, 118, 123, 128, 133, 138, 143, 148, 153, 158, 163, 168, 173, 178, 183, 188

Offset: 1

Author

Yifan Xie, Jul 15 2022

Keywords

• nonn
• easy

Comments

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

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

Examples

			a(7) = 12 because (5^1 + 1)/2 <= 7 < (3*5^1 + 1)/2, hence a(7) = 7 + 5^1 = 12;
a(11) = 28 because (3*5^1 + 1)/2 <= 11 < (5*5^1 + 1)/2, hence a(11) = 5*(11 - 5^1) - 2 = 28.
		

Links

• Yifan Xie, Table of n, a(n) for n = 1..10000
• Index entries for sequences of the a(a(n)) = 2n family

Crossrefs

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

Programs

Mathematica

a[n_] := Module[{n2 = 2n, p}, p = 5^Floor@Log[5, n2]; If[n2 < 3p, n+p, 5(n-p)-2]];
Table[a[n], {n, 1, 100}] (* Jean-François Alcover, Sep 22 2023, after Kevin Ryde *)

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
(C++)
/* program used to generate the b-file */
#include
using namespace std;
int main(){
    int cnt1=1, flag=0, cnt2=1, a=2;
    for(int n=1; n<=10000; n++) {
        cout<
				

Formula

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

a(n) = n + 5^i.

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

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

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

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