|
%I #31 Jul 30 2020 14:16:07
%S 4,9,18,31,48,69,94,123,156,193,234,279,328,381,438,499,564,633,706,
%T 783,864,949,1038,1131,1228,1329,1434,1543,1656,1773,1894,2019,2148,
%U 2281,2418,2559,2704,2853,3006,3163,3324,3489,3658,3831,4008,4189,4374,4563
%N Positions of occurrences of the natural numbers as a second subsequence in A100035.
%C For n > 1, A100035(a(n)) = n and A100035(m) != n for a(n-1) <= m < a(n);
%C A100036(n) < a(n) < A100038(n) < A100039(n).
%H Guo-Niu Han, <a href="/A196265/a196265.pdf">Enumeration of Standard Puzzles</a>, 2011. [Cached copy]
%H Guo-Niu Han, <a href="https://arxiv.org/abs/2006.14070">Enumeration of Standard Puzzles</a>, arXiv:2006.14070 [math.CO], 2020.
%F a(n) = 2*n^2 - n + 3 (conjectured). - _Ralf Stephan_, May 15 2007
%e First terms (10 = A, 11 = B, 12 = C) of A100035(a(n)):
%e ...1....2........3............4................5......
%e 1231435425165764736271879869584938291A9BA8B7A6B5A4B3A2B;
%e a(1) = A084849(2) = 4, A100035(4) = 1;
%e a(2) = A014107(2) = 9, A100035(9) = 2;
%e a(3) = A033537(3) = 18, A100035(18) = 3;
%e a(4) = A100040(4) = 31, A100035(31) = 4;
%e a(5) = A100041(5) = 48, A100035(48) = 5.
%Y Cf. A014107, A033537, A084849, A100035, A100036, A100038, A100039, A100040, A100041.
%K nonn
%O 1,1
%A _Reinhard Zumkeller_, Oct 31 2004
|
