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%I #17 Nov 15 2019 23:23:00
%S 1,3,4,5,9,10,14,23,37,38,42,51,65,102,144,209,353,354,358,367,381,
%T 418,460,525,669,1022,1380,1761,2221,2890,4270,6491,10761,10762,10766,
%U 10775,10789,10826,10868,10933,11077,11430,11788
%N a(n) = a(n-1) + a(m) for n >= 3, where m = 2*n - 3 - 2^(p+1) and p is the unique integer such that 2^p < n - 1 <= 2^(p+1), starting with a(1) = 1 and a(2) = 3.
%C The author of the Mathematica program below uses the initial conditions a(1) = 1, a(2) = 3, and a(3) = 4. This is not necessary. We get the same sequence by using the initial conditions a(1) = 1 and a(2) = 3. - _Petros Hadjicostas_, Nov 15 2019
%H Ivan Neretin, <a href="/A050068/b050068.txt">Table of n, a(n) for n = 1..8193</a>
%p a := proc(n) option remember; `if`(n < 3, [1,3][n],
%p a(n - 1) + a(-2^ceil(log[2](n - 1)) + 2*n - 3))
%p end proc:
%p seq(a(n), n = 1..40); # _Petros Hadjicostas_, Nov 15 2019
%t Fold[Append[#1, #1[[-1]] + #1[[#2]]] &, {1, 3, 4}, Flatten@Table[2 k - 1, {n, 5}, {k, 2^n}]] (* _Ivan Neretin_, Sep 07 2015 *)
%Y Same as A050036 and A050052 except for the second term.
%Y Cf. similar sequences with different initial conditions: A050024 (1,1,1), A050028 (1,1,2), A050032 (1,1,3), A050036 (1,1,4), A050040 (1,2,1), A050044 (1,2,2), A050048 (1,2,3), A050052 (1,2,4), A050056 (1,3,1), A050060 (1,3,2), A050064 (1,3,3).
%K nonn
%O 1,2
%A _Clark Kimberling_
%E Name edited by _Petros Hadjicostas_, Nov 15 2019
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