%I #47 Feb 20 2022 22:38:24
%S 10,100,110,1000,1100,1110,10000,11000,11100,11110,100000,110000,
%T 111000,111100,111110,1000000,1100000,1110000,1111000,1111100,1111110,
%U 10000000,11000000,11100000,11110000,11111000,11111100,11111110,100000000,110000000,111000000
%N Numbers consisting of a nonempty string of 1's followed by a nonempty string of 0's.
%C Intersection of A037415 and A009996 except for 1 [Corrected by _David A. Corneth_, Aug 30 2016].
%C Set of terms from sequence A052983.
%C a(n) is the binary expansion of A043569(n). - _Michel Marcus_, Sep 04 2016
%H Robert Israel, <a href="/A276349/b276349.txt">Table of n, a(n) for n = 1..10000</a>
%H Luboš Pick, <a href="https://eudml.org/doc/280143">Dirichletovy šuplíčky</a>, Pokroky matematiky, fyziky a astronomie, Vol. 61, No. 2 (2016), pp. 106-118. (In Czech; The Dirichlet pigeonhole principle)
%F A227362(a(n)) = 10.
%F From _Robert Israel_, Sep 02 2016: (Start)
%F a((m^2-m)/2+j) = 10^(m+1)*(1-10^(-j))/9 for m>=1, 1<=j<=m.
%F a(n) = 10*(10^m - 10^(-n+m*(m+1)/2))/9 where m = A002024(n). (End)
%F A002275(A002260(n)) * 10^A004736(n) - _Peter Kagey_, Sep 02 2016
%F Sum_{n>=1} 1/a(n) = A073668. - _Amiram Eldar_, Feb 20 2022
%e 60 is of the form binomial(a, 2) + b where 0 < b <= a and a = 11, b = 5. So a(60) has (11 + 1) digits and 5 leading ones. The other digits are 0. Giving a(60) = 111110000000. It has 7 (more than 1) trailing zeros so the next one, a(61) is a(60) + 10^(7 - 1). - _David A. Corneth_, Aug 30 2016
%p seq(seq(10^(m+1)*(1-10^(-j))/9,j=1..m),m=1..20); # _Robert Israel_, Sep 02 2016
%t Table[FromDigits@ Join[ConstantArray[1, #1], ConstantArray[0, #2]] & @@@ Transpose@ {#, n - #} &@ Range[n - 1], {n, 2, 9}] // Flatten (* _Michael De Vlieger_, Aug 30 2016 *)
%t Flatten[Table[FromDigits[Join[PadRight[{},n,1],PadRight[{},k,0]]],{n,8},{k,8}]]//Sort (* _Harvey P. Dale_, Jan 09 2019 *)
%o (Magma) [n: n in [1..10^7] | Seqint(Setseq(Set(Sort(Intseq(n))))) eq 10 and Seqint(Sort((Intseq(n)))) eq n]
%o (PARI) is(n) = vecmin(digits(n))==0 && vecmax(digits(n))==1 && digits(n)==vecsort(digits(n), , 4) \\ _Felix Fröhlich_, Aug 30 2016
%o (PARI) a(n) = my(r = ceil((sqrt(1+8*n)+1)/2), k = n - binomial(r-1, 2));10^(r-k)*(10^(k)-1)/9
%o \\ given an element n, computes the next element of the sequence.
%o nxt(n) = my(d = digits(n), qd=#d, s = vecsum(d)); if(qd-s>1, n+10^(qd-s-1), 10^qd)
%o \\ given an element n of the sequence, computes its place in the sequence.
%o inv(n) = my(d = digits(n)); binomial(#d-1,2) + vecsum(d) \\ _David A. Corneth_, Aug 31 2016
%Y Cf. A002024, A009996, A037415, A043569, A052983, A073668, A227362, A276348.
%K nonn,base
%O 1,1
%A _Jaroslav Krizek_, Aug 30 2016