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a(n+1) = a(n-1) + A001414(a(n)) with a(1)=1, a(2)=2.
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%I #18 May 17 2016 16:39:18

%S 1,2,3,5,8,11,19,30,29,59,88,76,111,116,144,130,164,175,181,356,274,

%T 495,296,538,567,557,1124,842,1547,879,1843,995,2047,1107,2097,1346,

%U 2772,1374,3006,1549,4555,2465,4606,2528,4695,2849,4750,2885,5332,2963,8295

%N a(n+1) = a(n-1) + A001414(a(n)) with a(1)=1, a(2)=2.

%C Replace the Fibonacci rule a(n) = a(n-2) + a(n-1) by a(n) = a(n-2) + (sum of prime factors with repetition of a(n-1)).

%H Giovanni Resta, <a href="/A272136/b272136.txt">Table of n, a(n) for n = 1..10000</a>

%e a(3)=a(1)+A001414(2) = 1+2 = 3. a(4)=2+3=5, a(5)=3+5=8, a(6)=5+6=11; etc.

%t a[n_]:= a[n] = If[n < 3, n, a[n-2] + Plus @@ Times @@@ FactorInteger@ a[n-1]]; Array[a, 50] (* _Giovanni Resta_, Apr 24 2016 *)

%Y Cf. A000045, A001414.

%K nonn

%O 1,2

%A _David James Sycamore_, Apr 21 2016

%E a(41)-a(51) from _Giovanni Resta_, Apr 24 2016