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Start with a(1)=1; now a(n+1)=a(n)+a(k) with k=[n-n-th digit of Pi]. If k<0 or k=0, then a(k)=0.
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%I #5 Jul 12 2012 00:39:49

%S 1,1,2,2,4,4,4,8,9,11,15,19,21,23,27,31,52,79,106,121,152,179,300,352,

%T 473,652,952,1058,1531,2483,2783,2962,3914,7828,10790,11742,12800,

%U 16714,29514,31997,35911,67908,79650,87478,123389,135131,147931,235409,271320

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

%C Terms of this "Pibonacci sequence" computed by Gilles Sadowski.

%H Gilles Sadowski, <a href="/A133389/b133389.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=(7-2) [because "2" is the 7th digit of Pi: 3,1,4,1,5,9,(2),6...] So a(8)=a(7)+a(5)=4+4=8.

%K base,easy,nonn

%O 1,3

%A _Eric Angelini_, Nov 23 2007