A276349 Numbers consisting of a nonempty string of 1's followed by a nonempty string of 0's.
10, 100, 110, 1000, 1100, 1110, 10000, 11000, 11100, 11110, 100000, 110000, 111000, 111100, 111110, 1000000, 1100000, 1110000, 1111000, 1111100, 1111110, 10000000, 11000000, 11100000, 11110000, 11111000, 11111100, 11111110, 100000000, 110000000, 111000000
Offset: 1
Examples
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
Links
- Robert Israel, Table of n, a(n) for n = 1..10000
- Luboš Pick, Dirichletovy šuplíčky, Pokroky matematiky, fyziky a astronomie, Vol. 61, No. 2 (2016), pp. 106-118. (In Czech; The Dirichlet pigeonhole principle)
Programs
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Magma
[n: n in [1..10^7] | Seqint(Setseq(Set(Sort(Intseq(n))))) eq 10 and Seqint(Sort((Intseq(n)))) eq n];
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Maple
seq(seq(10^(m+1)*(1-10^(-j))/9,j=1..m),m=1..20); # Robert Israel, Sep 02 2016
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Mathematica
Table[FromDigits@ Join[ConstantArray[1, #1], ConstantArray[0, #2]] & @@@ Transpose@ {#, n - #} &@ Range[n - 1], {n, 2, 9}] // Flatten (* Michael De Vlieger, Aug 30 2016 *) Flatten[Table[FromDigits[Join[PadRight[{},n,1],PadRight[{},k,0]]],{n,8},{k,8}]]//Sort (* Harvey P. Dale, Jan 09 2019 *)
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PARI
is(n) = vecmin(digits(n))==0 && vecmax(digits(n))==1 && digits(n)==vecsort(digits(n), , 4) \\ Felix Fröhlich, Aug 30 2016
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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 \\ given an element n, computes the next element of the sequence. nxt(n) = my(d = digits(n), qd=#d, s = vecsum(d)); if(qd-s>1, n+10^(qd-s-1), 10^qd) \\ given an element n of the sequence, computes its place in the sequence. inv(n) = my(d = digits(n)); binomial(#d-1,2) + vecsum(d) \\ David A. Corneth, Aug 31 2016
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Python
from math import isqrt, comb def A276349(n): return 10*(10**(m:=isqrt(n<<3)+1>>1)-10**(comb(m+1,2)-n))//9 # Chai Wah Wu, Jun 16 2025
Formula
A227362(a(n)) = 10.
From Robert Israel, Sep 02 2016: (Start)
a((m^2-m)/2+j) = 10^(m+1)*(1-10^(-j))/9 for m>=1, 1<=j<=m.
a(n) = 10*(10^m - 10^(-n+m*(m+1)/2))/9 where m = A002024(n). (End)
Sum_{n>=1} 1/a(n) = A073668. - Amiram Eldar, Feb 20 2022
a(n) = 10*A309761(n). - Chai Wah Wu, Jun 16 2025
Comments