cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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.

Original entry on oeis.org

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

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Author

James Marjamaa, Nov 05 2019

Keywords

Comments

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

Examples

			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
    .
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Crossrefs

Programs

  • C
    #include
    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

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)