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%I #68 Oct 02 2021 11:06:41
%S 1,1,1,2,1,2,1,1,1,2,1,3,1,3,1,2,1,1,1,1,1,1,1,2,2,2,2,1,1,3,1,4,1,4,
%T 1,3,1,1,2,1,1,1,2,2,1,1,1,2,1,1,1,1,1,1,1,2,1,1,1,3,2,3,2,2,1,2,1,1,
%U 1,2,1,2,3,2,3,1,1,4,1,5,1,5,1,4,1,1,3,1,1,1,3,2,1,2,1,2,1,2,1,1,1,1,2,2,1
%N Triangle read by rows: n-th row is length of run of leftmost 1's, followed by length of run of 0's, followed by length of run of 1's, etc., in the binary representation of n.
%C Row n has A005811(n) elements. In rows 2^(k-1)..2^k-1 we have all the compositions (ordered partitions) of k. Other orderings of compositions: A066099, A108244, and A124734. - _Jason Kimberley_, Feb 09 2013
%C A043276(n) = largest term in n-th row. - _Reinhard Zumkeller_, Dec 16 2013
%C From the first comment it follows that we have a bijection between the positive integers and the set of all compositions. - _Emeric Deutsch_, Jul 11 2017
%C From _Robert Israel_, Jan 23 2018: (Start)
%C If n is even, row 2*n is row n with its last element incremented by 1, and row 2*n+1 is row n with 1 appended.
%C If n is odd, row 2*n+1 is row n with its last element incremented by 1, and row 2*n is row n with 1 appended. (End)
%H Antti Karttunen, <a href="/A101211/b101211.txt">The rows 1..1023 of the table, flattened</a>
%F a(n) = A227736(A227741(n)) = A227186(A056539(A227737(n)),A227740(n)) - _Antti Karttunen_, Jul 27 2013
%e Since 9 is 1001 in binary, the 9th row is 1,2,1.
%e Since 11 is 1011 in binary, the 11th row is 1,1,2.
%e Triangle begins:
%e 1;
%e 1,1;
%e 2;
%e 1,2;
%e 1,1,1;
%e 2,1;
%e 3;
%e 1,3;
%p # Maple program due to W. Edwin Clark:
%p Runs := proc (L) local j, r, i, k; j := 1: r[j] := L[1]: for i from 2 to nops(L) do if L[i] = L[i-1] then r[j] := r[j], L[i] else j := j+1: r[j] := L[i] end if end do: [seq([r[k]], k = 1 .. j)] end proc: RunLengths := proc (L) map(nops, Runs(L)) end proc: c := proc (n) ListTools:-Reverse(convert(n, base, 2)): RunLengths(%) end proc: # Row n is obtained with the command c(n). - _Emeric Deutsch_, Jul 03 2017
%p # Maple program due to W. Edwin Clark, yielding the integer ind corresponding to a given composition (the index of the composition):
%p ind := proc (x) local X, j, i: X := NULL: for j to nops(x) do if type(j, odd) then X := X, seq(1, i = 1 .. x[j]) end if: if type(j, even) then X := X, seq(0, i = 1 .. x[j]) end if end do: X := [X]: add(X[i]*2^(nops(X)-i), i = 1 .. nops(X)) end proc; # Clearly, ind(c(n))= n. - _Emeric Deutsch_, Jan 23 2018
%t Table[Length /@ Split@ IntegerDigits[n, 2], {n, 38}] // Flatten (* _Michael De Vlieger_, Jul 11 2017 *)
%o (Scheme, two variants)
%o (define (A101211 n) (A227736 (A227741 n)))
%o (define (A101211v2 n) (A227186bi (A056539 (A227737 n)) (A227740 n)))
%o ;; Scheme-implementation for A227186bi can be found under A227186. - _Antti Karttunen_, Jul 27 2013
%o (Haskell)
%o import Data.List (group)
%o a101211 n k = a101211_tabf !! (n-1) !! (k-1)
%o a101211_row n = a101211_tabf !! (n-1)
%o a101211_tabf = map (reverse . map length . group) $ tail a030308_tabf
%o -- _Reinhard Zumkeller_, Dec 16 2013
%o (Python)
%o from itertools import groupby
%o def arow(n): return [len(list(g)) for k, g in groupby(bin(n)[2:])]
%o def auptorow(rows):
%o alst = []
%o for i in range(1, rows+1): alst.extend(arow(i))
%o return alst
%o print(auptorow(38)) # _Michael S. Branicky_, Oct 02 2021
%Y A070939(n) gives the sum of terms in row n, while A167489(n) gives the product of its terms. A090996 gives the first column. A227736 lists the terms of each row in reverse order.
%Y Cf. also A227186.
%Y Cf. A030308, A175911.
%K nonn,base,tabf
%O 1,4
%A _Leroy Quet_, Dec 13 2004
%E More terms from _Emeric Deutsch_, Apr 12 2005
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