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%I #25 Oct 29 2017 21:37:34
%S 84,172,348,700,1404,2720,2754,5448,10904,21816,43640,87288
%N Numbers k such that A284644(k) = A284644(k-1) = A284644(k-2) = A284644(k-3).
%C For the first twelve terms of this sequence, corresponding values of A284644(k) are 11*2^2, 11*2^3, 11*2^4, 11*2^5, 11*2^6, 11*31*2^2, 3*5*23*2^2, 11*31*2^3, 11*31*2^4, 11*31*2^5, 11*31*2^6, 11*31*2^7 and k - A284644(k) are 40, 84, 172, 348, 700, 1356, 1374, 2720, 5448, 10904, 21816, 43640.
%C Additionally, a(n) = 2*a(n-1) + 4 for 1 < n < 6 and a(n) = 2*a(n-1) + 8 for 8 < n < 13. In fact, also 5448 = 2720*2 + 8 but there is a(7) = 2754 between 2720 and 5448. In other words, we can partition sequence up to 10^5 as three subsequences: {84, 172, 348, 700, 1404}, {2754}, {2720, 5448, 10904, 21816, 43640, 87288} in order to see curious recursive patterns.
%C If a(13) exists, it must be greater than 7.5*10^9. - _Hans Havermann_, May 27 2017
%H Altug Alkan, Nathan Fox, and Orhan Ozgur Aybar, <a href="https://www.hindawi.com/journals/complexity/aip/2614163/">On Hofstadter Heart Sequences</a>, Complexity, 2017.
%o (PARI) q=vector(10^8); q[1]=q[2]=2;q[3]=1; for(n=4, #q, q[n]=q[n-q[n-1]]+q[n-q[n-2]]); for (k=3, 10^8, if(q[k] == q[k-1] && q[k] == q[k-2] && q[k] == q[k-3], print1(k, ", ")));
%Y Cf. A284644.
%K nonn,more
%O 1,1
%A _Altug Alkan_, following a suggestion from _Nathan Fox_, May 24 2017
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