Juan Martin Rodriguez - cp's OEIS frontend

User: Juan Martin Rodriguez

Juan Martin Rodriguez has authored 1 sequences.

A381677 a(n) = floor(a(n-1)*3/2) with a(1) = 5.

5, 7, 10, 15, 22, 33, 49, 73, 109, 163, 244, 366, 549, 823, 1234, 1851, 2776, 4164, 6246, 9369, 14053, 21079, 31618, 47427, 71140, 106710, 160065, 240097, 360145, 540217, 810325, 1215487, 1823230, 2734845, 4102267, 6153400, 9230100, 13845150, 20767725, 31151587, 46727380, 70091070, 105136605, 157704907

Offset: 1

Juan Martin Rodriguez, Mar 03 2025

The final number of animals if at each step two animals are bred and produce one offspring, starting with 5 animals (as in the game Minecraft, and waiting for all animals to be fully grown before the next round of breeding).

			a(3) = floor(a(2)/2)+a(2) = floor(7/2)+7 = 3+7 = 10.

Cf. A032766, A061418.

Cf. A112088 (first differences).

class Main {
    public static void main(String[] args) {
        long a=5; // if you need more than 103 terms, you will need to use BigInteger
        System.out.print(a);
        for(int n = 0; n<104;n++){
            a += a/2;
            System.out.print(","+a);
        }
    }
}

a[1]=5;a[n_]:=Floor[(a[n-1]*3/2)];Array[a,44] (* James C. McMahon, Mar 08 2025 *)

lista(n)= my(t=10/3); vector(n, i, t=t*3\2) \\ Ruud H.G. van Tol, Mar 09 2025

cp's OEIS Frontend

User: Juan Martin Rodriguez

A381677 a(n) = floor(a(n-1)*3/2) with a(1) = 5.

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