proposed

approved

editing

proposed

n-th step is to add a(n) to each previous number a(k) (excluding itself, i.e. , k < n) to produce n more terms of the sequence, starting with a(0)=1, a(1)=1.

Start with (1,1). So after initial step have (1, *1*, 1+1 = 2), then (1, 1, *2*, 1+2 = 3, 1+2 = 3), then (1, 1, 2, *3*, 3, 1+3 = 4, 1+3 = 4, 2+3 = 5), then (1, 1, 2, 3, *3*, 4, 4, 5, 1+3 = 4, 1+3 = 4, 2+3 = 5, 3+3 = 6) , etc.

approved

editing

_Henry Bottomley (se16(AT)btinternet.com), _, Aug 31 2001

n-th step is to add a(n) to each previous number a(k) (excluding itself, i.e. k<n) to produce n more terms of the sequence, starting with a(0)=1, a(1)=1.

1, 1, 2, 3, 3, 4, 4, 5, 4, 4, 5, 6, 5, 5, 6, 7, 7, 5, 5, 6, 7, 7, 8, 6, 6, 7, 8, 8, 9, 9, 5, 5, 6, 7, 7, 8, 8, 9, 5, 5, 6, 7, 7, 8, 8, 9, 8, 6, 6, 7, 8, 8, 9, 9, 10, 9, 9, 7, 7, 8, 9, 9, 10, 10, 11, 10, 10, 11, 6, 6, 7, 8, 8, 9, 9, 10, 9, 9, 10, 11, 6, 6, 7, 8, 8, 9, 9, 10, 9, 9, 10, 11, 10, 7, 7, 8, 9, 9

0,3

Start with (1,1). So after initial step have (1,*1*,1+1=2), then (1,1,*2*,1+2=3,1+2=3), then (1,1,2,*3*,3,1+3=4,1+3=4,2+3=5), then (1,1,2,3,*3*,4,4,5,1+3=4,1+3=4,2+3=5,3+3=6) etc.

Each positive number appears A063894 number of times. Cf. A064064, A064065, A064067.

nonn

Henry Bottomley (se16(AT)btinternet.com), Aug 31 2001

approved