A213183 Initialize a(1)=R=1. Repeat: copy the last R preceding terms to current position; increment R; do twice: append the least integer that has not appeared in the sequence yet.

1, 1, 2, 3, 2, 3, 4, 5, 3, 4, 5, 6, 7, 4, 5, 6, 7, 8, 9, 5, 6, 7, 8, 9, 10, 11, 6, 7, 8, 9, 10, 11, 12, 13, 7, 8, 9, 10, 11, 12, 13, 14, 15, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 11, 12

Offset: 1

Author

Alex Ratushnyak, Mar 05 2013

Keywords

• nonn
• easy
• tabf

Comments

Every positive integer k occurs floor((k+3)/2) times: 1 and 2 occur twice, 3 and 4 thrice, 5 and 6 four times, and so on.

Examples

			a(1) = 1  -- initial value
a(2) = 1  -- copied one last term
a(3)=2, a(4)=3  -- appended two terms
a(5)=2, a(6)=3  -- copied two last terms
a(7)=4, a(8)=5  -- appended two terms
a(9)=3, a(10)=4, a(11)=5  -- copied three last terms
a(12)=6, a(13)=7  -- appended two terms
a(14)=4, a(15)=5, a(16)=6, a(17)=7  -- copied four last terms
a(18)=8, a(19)=9  -- appended two terms, and so on.
Comments from _N. J. A. Sloane_, Apr 28 2020, following a suggestion from _Paul Curtz_: (Start)
With an initial -1, 0, this may also be regarded as a triangle read by rows:
  -1;
   0,  1;
   1,  2,  3;
   2,  3,  4,  5;
   3,  4,  5,  6,  7;
   4,  5,  6,  7,  8,  9;
   5,  6,  7,  8,  9, 10, 11;
   6,  7,  8,  9, 10, 11, 12, 13;
  ...
or as an array read by upward antidiagonals:
  -1,  1,  3,  5,  7,  9, 11, ...
   0,  2,  4,  6,  8, 10, ...
   1,  3,  5,  7,  9, ...
   2,  4,  6,  8, ...
   3,  5,  7, ...
   4,  6, ...
   5, ...
  ...
(End)
		

Crossrefs

If we prefix this with -1, 0, and then add 1 to every term, we get A051162.

Programs

Python

a = [1]*992
R = 1
i = 2
while i<900:
        for t in range(R):
                a[i] = a[i-R]
                i += 1
        R += 1
        a[i] = a[i-1] + 1
        i += 1
        a[i] = a[i-1] + 1
        i += 1
for i in range(1,99):
        print(a[i], end=',')