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A100037
Positions of occurrences of the natural numbers as a second subsequence in A100035.
21
4, 9, 18, 31, 48, 69, 94, 123, 156, 193, 234, 279, 328, 381, 438, 499, 564, 633, 706, 783, 864, 949, 1038, 1131, 1228, 1329, 1434, 1543, 1656, 1773, 1894, 2019, 2148, 2281, 2418, 2559, 2704, 2853, 3006, 3163, 3324, 3489, 3658, 3831, 4008, 4189, 4374, 4563
OFFSET
1,1
COMMENTS
For n > 1, A100035(a(n)) = n and A100035(m) != n for a(n-1) <= m < a(n);
A100036(n) < a(n) < A100038(n) < A100039(n).
LINKS
Guo-Niu Han, Enumeration of Standard Puzzles, 2011. [Cached copy]
Guo-Niu Han, Enumeration of Standard Puzzles, arXiv:2006.14070 [math.CO], 2020.
FORMULA
a(n) = 2*n^2 - n + 3 (conjectured). - Ralf Stephan, May 15 2007
EXAMPLE
First terms (10 = A, 11 = B, 12 = C) of A100035(a(n)):
...1....2........3............4................5......
1231435425165764736271879869584938291A9BA8B7A6B5A4B3A2B;
a(1) = A084849(2) = 4, A100035(4) = 1;
a(2) = A014107(2) = 9, A100035(9) = 2;
a(3) = A033537(3) = 18, A100035(18) = 3;
a(4) = A100040(4) = 31, A100035(31) = 4;
a(5) = A100041(5) = 48, A100035(48) = 5.
KEYWORD
nonn
AUTHOR
Reinhard Zumkeller, Oct 31 2004
STATUS
approved