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Start with a(1)=1; for n >= 1, a(n+1)=a(n)+a(k) with k=[n - n-th digit of sqrt(2)]. If k<0 or k=0, then a(k)=0.
1

%I #9 Oct 23 2014 09:34:19

%S 1,1,1,2,2,3,5,7,8,9,16,23,25,34,68,71,87,174,208,224,247,494,741,828,

%T 899,986,1194,2093,2921,4115,8230,8724,9710,10538,11524,23048,25969,

%U 28890,38600,48310,56540,68064,79588,147652,170700,199590,256130,403782

%N Start with a(1)=1; for n >= 1, a(n+1)=a(n)+a(k) with k=[n - n-th digit of sqrt(2)]. If k<0 or k=0, then a(k)=0.

%C Terms computed by Gilles Sadowski.

%H Gilles Sadowski, <a href="/A133393/b133393.txt">Table of n, a(n) for n = 1..101</a>

%e For n=7 we have a(8)=a(7)+a(k) with k=(7-3) because "3" is the 7th digit of sqrt(2): 1,4,1,4,2,1,(3),5,6,2,3,... So a(8)=a(7)+a(4)=5+2=7.

%K base,easy,nonn

%O 1,4

%A _Eric Angelini_, Nov 23 2007