A366836 a(1) = 1 and a(n) = prime(a(n-1)+n) mod (a(n-1)+n).

1, 2, 1, 1, 1, 3, 9, 8, 8, 7, 7, 10, 14, 23, 11, 22, 11, 22, 15, 9, 23, 17, 13, 9, 3, 22, 31, 41, 69, 28, 41, 2, 9, 19, 35, 69, 47, 14, 29, 2, 19, 39, 11, 37, 11, 41, 17, 53, 47, 24, 4, 39, 19, 2, 41, 24, 14, 71, 83, 108, 164, 73, 89, 118, 178, 85, 121, 184, 89, 142, 25, 24, 24

Offset

1,2

Comments

In each step we take the (a(n-1)+n)th prime and we find the remainder when we divide it with a(n-1)+n.

If we examine the plot, we notice some rectangles and we get the same fractal pattern every time we scale ~2.4 times. Why does this happen? (See Angelini's link).

Links

Table of n, a(n) for n=1..73.

Eric Angelini, A fractal sequence by GK, Personal blog "Cinquante signes", Oct 2023.

Formula

a(1) = 1, a(n) = prime(a(n-1)+n) mod (a(n-1)+n).

Example

a(2) = 2 because a(1) = 1 and prime(1+2) mod (1+2) is 5 mod 3.
a(7) = 9 because a(6) = 3 and prime(3+7) mod (3+7) is 29 mod 10.

Mathematica

a[1] = 1;
 a[n_] := a[n] = Mod[Prime[a[n - 1] + n], a[n - 1] + n]; Array[a, 100]

Prog

(PARI) a(n) = if (n==1, 1, my(x=a(n-1)+n); prime(x) % x); \\ Michel Marcus, Oct 29 2023

Crossrefs

Cf. A004648.

Sequence in context: A279453 A054252 A240472 * A007442 A362483 A054772

Adjacent sequences: A366833 A366834 A366835 * A366837 A366838 A366839

Keyword

nonn

Author

Giorgos Kalogeropoulos and Eric Angelini, Oct 29 2023

Status

approved