cp's OEIS Frontend

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.

%I A329130 #35 Oct 08 2020 03:57:43
%S A329130 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,
%T A329130 32,4,23,43,12,33,55,21,44,8,32,57,18,44,3,30,58,14,43,73,26,57,8,40,
%U A329130 73,21,55,1,36,72,15,52,90
%N 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.
%C A329130 Values where a(n) = n appear to be the values of A027941.
%H A329130 James Marjamaa, <a href="/A329130/b329130.txt">Table of n, a(n) for n = 0..10000</a>
%H A329130 Rémy Sigrist, <a href="/A329130/a329130.png">Logarithmic scatterplot of the first 1000000 terms of the ordinal transform of the ordinal transform of the sequence</a>
%F A329130 a(n+1) = a(n) + I, if n >= a(n) ('I' is the next consecutive integer not yet added);
%F A329130        = a(n) - n, if n < a(n).
%F A329130 From _Yan Sheng Ang_, May 20 2020: (Start)
%F A329130 a(A004957(n)) = a(n). Hence it follows that (writing F(n) = A000045(n)):
%F A329130   a(F(2*n)-1) = F(2*n) for n > 1;
%F A329130   a(F(2*n)) = 1;
%F A329130   a(F(2*n+1)-1) = F(2*n+1)-1 (as noted above);
%F A329130   a(F(2*n+1)) = F(2*n+2);
%F A329130   a(F(2*n+1)+1) = F(2*n) for n > 1.
%F A329130 (End)
%e A329130 For n = 0,   a(0) =            0    0 >= a(0)
%e A329130     n = 1,   a(1) = a(0) + 1 = 1,   1 >= a(1), I = 1
%e A329130     n = 2,   a(2) = a(1) + 2 = 3,   2 <  a(2), I = 2
%e A329130     n = 3,   a(3) = a(2) - 2 = 1,   3 >= a(3)
%e A329130     n = 4,   a(4) = a(3) + 3 = 4,   4 >= a(4), I = 3
%e A329130     n = 5,   a(5) = a(4) + 4 = 8,   5 <  a(5), I = 4
%e A329130     n = 6,   a(6) = a(5) - 5 = 3,   6 >= a(6)
%e A329130     n = 7,   a(7) = a(6) + 5 = 8,   7 <  a(7), I = 5
%e A329130     n = 8,   a(8) = a(7) - 7 = 1,   8 >= a(8)
%e A329130     n = 9,   a(9) = a(8) + 6 = 7,   9 >= a(9), I = 6
%e A329130     n = 10,  a(10)= a(9) + 7 = 14,  10<  a(10), I = 7
%e A329130     n = 11,  a(11)= a(10)- 10= 4,   11>= a(11)
%e A329130     n = 12,  a(12)= a(11)+ 8 = 12,  12>= a(12), I = 8
%e A329130     .
%e A329130     .
%e A329130     .
%o A329130 (C)
%o A329130 #include<stdio.h>
%o A329130 void seq(int terms)
%o A329130 {int n = 0; int i = 0; int a = 0; int c = 0;
%o A329130 while (n <= terms)
%o A329130 {
%o A329130     if (c)
%o A329130        {int N = n - 1;
%o A329130         a -= N;
%o A329130         printf("%d\n", a);}
%o A329130     else
%o A329130        {a += i;
%o A329130         printf("%d\n", a);
%o A329130         i++;}
%o A329130     if (a > n)
%o A329130        {c = 1;}
%o A329130     else
%o A329130        {c = 0;}
%o A329130     n++;
%o A329130 }
%o A329130 }
%o A329130 int main(void)
%o A329130 {
%o A329130 seq(1000);
%o A329130 }
%o A329130 (PARI) v=k=0; for (n=0, 69, print1 (v, ", "); v=if (v>n, v-n, v+k++)) \\ _Rémy Sigrist_, Nov 06 2019
%Y A329130 Cf. A027941, A008344.
%K A329130 look,easy,nonn
%O A329130 0,3
%A A329130 _James Marjamaa_, Nov 05 2019