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%I #29 Feb 13 2024 08:15:33
%S 0,1,1,0,2,2,0,2,1,2,1,0,3,3,0,3,1,3,1,0,3,2,3,2,0,3,2,1,3,2,1,0,4,4,
%T 0,4,1,4,1,0,4,2,4,2,0,4,2,1,4,2,1,0,4,3,4,3,0,4,3,1,4,3,1,0,4,3,2,4,
%U 3,2,0,4,3,2,1,4,3,2,1,0,5,5,0,5,1,5,1
%N Irregular triangle read by rows: strictly decreasing sequences of nonnegative numbers given in lexicographic order.
%C Length of n-th row given by A000120(n);
%C Maximum of n-th row given by A000523(n);
%C Minimum of n-th row given by A007814(n);
%C GCD of n-th row given by A064894(n);
%C Sum of n-th row given by A073642(n + 1).
%C n-th row begins at index A000788(n - 1) for n > 0.
%C The first appearance of n is at A001787(n).
%C a(A001787(n) + 1) = a(A001787(n)) for all n > 0.
%C a(A001787(n) + 2) = 0 for all n > 0.
%C a(A001787(n) + 3) = a(A001787(n)) for all n > 1.
%C a(A001787(n) + 4) = 1 for all n > 1.
%C a(A001787(n) + 5) = a(A001787(n)) for all n > 1.
%C Row n < 1024 lists the digits of A262557(n). - _M. F. Hasler_, Dec 11 2019
%H Peter Kagey, <a href="/A272011/b272011.txt">Table of n, a(n) for n = 0..10000</a>
%e Row n is given by the exponents in the binary expansion of n. For example, row 5 = [2, 0] because 5 = 2^2 + 2^0.
%e Row 0: []
%e Row 1: [0]
%e Row 2: [1]
%e Row 3: [1, 0]
%e Row 4: [2]
%e Row 5: [2, 0]
%e Row 6: [2, 1]
%e Row 7: [2, 1, 0]
%t Map[Length[#] - Flatten[Position[#, 1]] &, IntegerDigits[Range[50], 2]] (* _Paolo Xausa_, Feb 13 2024 *)
%o (PARI) apply( A272011_row(n)=Vecrev(vecextract([0..exponent(n+!n)],n)), [0..39]) \\ For n < 2^10: row(n)=digits(A262557[n]). There are 2^k rows starting with k, they start at row 2^k. - _M. F. Hasler_, Dec 11 2019
%Y Cf. A000120, A000523, A001787, A007814, A064894, A073642.
%Y Cf. A133457 gives the rows in reverse order.
%Y Cf. A272020, A262557.
%K nonn,tabf
%O 0,5
%A _Peter Kagey_, Apr 17 2016
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