A329130 a(0)=0; for any n >= 0, if a(n) > n then a(n+1) = a(n) - n, otherwise a(n+1) = a(n) + k, where k is the total number of terms a(m) <= m with m <= n.

0, 1, 3, 1, 4, 8, 3, 8, 1, 7, 14, 4, 12, 21, 8, 18, 3, 14, 26, 8, 21, 1, 15, 30, 7, 23, 40, 14, 32, 4, 23, 43, 12, 33, 55, 21, 44, 8, 32, 57, 18, 44, 3, 30, 58, 14, 43, 73, 26, 57, 8, 40, 73, 21, 55, 1, 36, 72, 15, 52, 90

Offset

0,3

Comments

Values where a(n) = n appear to be the values of A027941.

Links

James Marjamaa, Table of n, a(n) for n = 0..10000

Rémy Sigrist, Logarithmic scatterplot of the first 1000000 terms of the ordinal transform of the ordinal transform of the sequence

Formula

a(n+1) = a(n) + I, if n >= a(n) ('I' is the next consecutive integer not yet added);

= a(n) - n, if n < a(n).

From Yan Sheng Ang, May 20 2020: (Start)

a(A004957(n)) = a(n). Hence it follows that (writing F(n) = A000045(n)):

a(F(2*n)-1) = F(2*n) for n > 1;

a(F(2*n)) = 1;

a(F(2*n+1)-1) = F(2*n+1)-1 (as noted above);

a(F(2*n+1)) = F(2*n+2);

a(F(2*n+1)+1) = F(2*n) for n > 1.

(End)

Example

For n = 0,   a(0) =            0    0 >= a(0)
    n = 1,   a(1) = a(0) + 1 = 1,   1 >= a(1), I = 1
    n = 2,   a(2) = a(1) + 2 = 3,   2 <  a(2), I = 2
    n = 3,   a(3) = a(2) - 2 = 1,   3 >= a(3)
    n = 4,   a(4) = a(3) + 3 = 4,   4 >= a(4), I = 3
    n = 5,   a(5) = a(4) + 4 = 8,   5 <  a(5), I = 4
    n = 6,   a(6) = a(5) - 5 = 3,   6 >= a(6)
    n = 7,   a(7) = a(6) + 5 = 8,   7 <  a(7), I = 5
    n = 8,   a(8) = a(7) - 7 = 1,   8 >= a(8)
    n = 9,   a(9) = a(8) + 6 = 7,   9 >= a(9), I = 6
    n = 10,  a(10)= a(9) + 7 = 14,  10<  a(10), I = 7
    n = 11,  a(11)= a(10)- 10= 4,   11>= a(11)
    n = 12,  a(12)= a(11)+ 8 = 12,  12>= a(12), I = 8
    .
    .
    .

Prog

(C)
#include<stdio.h>
void seq(int terms)
{int n = 0; int i = 0; int a = 0; int c = 0;
while (n <= terms)
{
    if (c)
       {int N = n - 1;
        a -= N;
        printf("%d\n", a); }
    else
       {a += i;
        printf("%d\n", a);
        i++; }
    if (a > n)
       {c = 1; }
    else
       {c = 0; }
    n++;
}
}
int main(void)
{
seq(1000);
}
(PARI) v=k=0; for (n=0, 69, print1 (v, ", "); v=if (v>n, v-n, v+k++)) \\ Rémy Sigrist, Nov 06 2019

Crossrefs

Cf. A027941, A008344.

Sequence in context: A193605 A193667 A205878 * A057049 A245672 A303029

Adjacent sequences: A329127 A329128 A329129 * A329131 A329132 A329133

Keyword

look,easy,nonn

Author

James Marjamaa, Nov 05 2019

Status

approved