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%I #37 Sep 14 2020 12:11:10
%S 1,1,1,1,1,2,1,1,1,2,5,3,2,3,1,1,1,2,8,5,11,18,8,5,3,5,11,7,3,5,1,1,1,
%T 2,13,8,26,42,18,11,26,42,94,58,29,47,13,8,5,8,29,18,36,58,26,16,7,11,
%U 26,16,5,8,1,1,1,2,21,13,60,97,42,26,87,141,317
%N For any number n with k binary digits, a(n) is the number of tilings T of a size k staircase polyomino (as described in A335547) such that the sizes of the polyominoes at the base of T correspond to the lengths of runs of consecutive equal digits in the binary representation of n.
%C a(0) = 1 corresponds to the empty polyomino.
%H Rémy Sigrist, <a href="/A336479/b336479.txt">Table of n, a(n) for n = 0..8192</a>
%H Rémy Sigrist, <a href="/A336479/a336479.gp.txt">PARI program for A336479</a>
%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>
%F A335547(n) = Sum_{k = 2^(n-1)..2^n-1} a(k).
%F a(A000975(n+1)) = A335547(n).
%F a(2^k-1) = 1 for any k >= 0.
%F a(2^k) = 1 for any k >= 0.
%F a(3*2^k) = A000045(k+1) for any k >= 0.
%F a(7*2^k) = A123392(k) for any k >= 0.
%e For n = 13, the binary representation of 13 is "1101", so we count the tilings of a size 4 staircase polyomino whose base has the following shape:
%e .....
%e . .
%e . .....
%e . .
%e +---+ .....
%e | | .
%e | +---+---+---+
%e | 1 1 | 0 | 1 |
%e +-------+---+---+
%e there are 3 such tilings:
%e +---+ +---+ +---+
%e | | | | | |
%e +---+---+ + +---+ +---+---+
%e | | | | | | | |
%e +---+---+---+ +---+---+---+ +---+ +---+
%e | | | | | | | | | | |
%e | +---+---+---+ | +---+---+---+ | +---+---+---+
%e | | | | | | | | | | | |
%e +-------+---+---+, +-------+---+---+, +-------+---+---+
%e so a(13) = 3.
%o (PARI) See Links section.
%Y Cf. A000045, A000975, A101211, A123392, A335547.
%K nonn,base
%O 0,6
%A _Rémy Sigrist_, Sep 13 2020