A201221 The smallest number of previous terms of sequence A008829 which are required to sum to A008829(n).

4, 6, 8, 10, 12, 14, 16, 16, 19, 16, 18, 17, 19, 18, 28, 34, 72, 49, 57, 35, 51, 36, 97, 65, 56, 56, 57, 69, 64, 65, 219, 136, 96, 72, 79, 75, 78, 74, 78, 70, 72, 73

Offset: 2

Author

Alastair Stanley, Nov 30 2011

Keywords

• nonn
• base
• more

Examples

			For n=2, the corresponding term of A008829 is 4, and the only preceding term is 1, therefore four 1s are needed and term a(2)=4.
Likewise, for n=3, the corresponding term is 15, which is three 4s and three 1s, so term a(3)=6.
		

Crossrefs

Cf. A008829.

Extensions

More terms from Alastair Stanley, Nov 29 2018