cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 11 results. Next

A036527 Smallest cube containing exactly n 0's.

Original entry on oeis.org

1, 0, 140608, 1000, 4096000, 140608000, 1000000, 4096000000, 140608000000, 1000000000, 4096000000000, 140608000000000, 1000000000000, 4096000000000000, 140608000000000000, 1000000000000000, 4096000000000000000, 140608000000000000000, 1000000000000000000, 4096000000000000000000, 140608000000000000000000, 1000000000000000000000
Offset: 0

Views

Author

Keywords

Comments

a(n)^(1/3) = A048365(n) is the index of the first occurrence of n in A269250. -- For n = 3k, obviously a(n) = 10^n. The first terms for indices n = 3k+1 and n = 3k+2 equals 4096*10^3k resp. 140608*10^3k. Is there an index from where on this is no longer true? - M. F. Hasler, Feb 20 2016

Crossrefs

Cf. A036528 - A036536 for other digits 1 - 9.
Analog for squares: A036507 = A048345^2.

Programs

  • Mathematica
    nsmall = Table[Infinity, 20];
    For[i = 0, i <= 10^6, i++, n0 = Count[IntegerDigits[i^3], 0];
      If[nsmall[[n0 + 1]] > i^3, nsmall[[n0 + 1]] = i^3]];
    Cases[nsmall, ?NumberQ] (* _Robert Price, Mar 20 2020 *)

Formula

a(n) = A048365(n)^3; a(3n) = 10^(3n); a(3n+1) <= 4096*10^(3n) = (16*10^n)^3 for n>0; a(3n+2) <= 140608*10^(3n) = (52*10^n)^3, with equality for all known terms. - M. F. Hasler, Feb 20 2016

Extensions

Extended to a(0) = 1 and three lines of data completed by M. F. Hasler, Feb 20 2016

A037059 Smallest prime containing exactly n 3's.

Original entry on oeis.org

2, 3, 233, 2333, 23333, 313333, 3233333, 31333333, 333233333, 3233333333, 23333333333, 333313333333, 3333333333383, 33133333333333, 323333333333333, 1333333333333333, 23333333333333333, 333333133333333333, 3333313333333333333, 33313333333333333333
Offset: 0

Views

Author

Patrick De Geest, Jan 04 1999

Keywords

Comments

For almost all n >= 0, a(n) equals [10^(n+1)/3] with one of the (first) digits 3 replaced by a digit 1 or 2. We conjecture that in the few other cases (e.g., for n = 12, 119, ...) the statement holds with some digit 3 replaced by a digit among {4, 5, 7, 8}, except for the special case a(1) = 3. - M. F. Hasler, Feb 22 2016

Crossrefs

Programs

  • Mathematica
    f[n_, b_] := Block[{k = 10^(n + 1), p = Permutations[ Join[ Table[b, {i, 1, n}], {x}]], c = Complement[Table[j, {j, 0, 9}], {b}], q = {}}, Do[q = Append[q, Replace[p, x -> c[[i]], 2]], {i, 1, 9}]; r = Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]; If[r ? Infinity, r, p = Permutations[ Join[ Table[ b, {i, 1, n}], {x, y}]]; q = {}; Do[q = Append[q, Replace[p, {x -> c[[i]], y -> c[[j]]}, 2]], {i, 1, 9}, {j, 1, 9}]; Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]]]; Table[ f[n, 3], {n, 1, 18}]
    Table[Sort[Flatten[Table[Select[FromDigits/@Permutations[Join[{n},PadRight[{},i,3]]], PrimeQ],{n,0,9}]]][[1]],{i,20}] (* Harvey P. Dale, Feb 28 2015 *)
  • PARI
    A037059(n)={if(n==1,3,my(t=10^(n+1)\3); forvec(v=[[-1, n], [-2, -1]], ispseudoprime(p=t+10^(n-v[1])*v[2]) && return(p)); forvec(v=[[0, n], [1, 5]], ispseudoprime(p=t+10^v[1]*v[2]) && return(p)))} \\ M. F. Hasler, Feb 22 2016

Formula

a(n) = prime(A037058(n)). - Amiram Eldar, Jul 21 2025

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 23 2003
More terms and a(0) = 2 prefixed by M. F. Hasler, Feb 22 2016

A037071 Smallest prime containing exactly n 9's.

Original entry on oeis.org

2, 19, 199, 1999, 49999, 199999, 2999999, 19999999, 799999999, 9199999999, 59999999999, 959999999999, 9919999999999, 59999999999999, 499999999999999, 9299999999999999, 99919999999999999, 994999999999999999, 9991999999999999999, 29999999999999999999, 989999999999999999999
Offset: 0

Views

Author

Patrick De Geest, Jan 04 1999

Keywords

Comments

We conjecture that for all n >= 0, a(n) equals [10^(n+1)/9]*9 with one of the (first) digits 9 replaced by a digit among {1, 2, 4, 5, 7, 8}. - M. F. Hasler, Feb 22 2016

Crossrefs

Programs

  • Mathematica
    f[n_, b_] := Block[{k = 10^(n + 1), p = Permutations[ Join[ Table[b, {i, 1, n}], {x}]], c = Complement[Table[j, {j, 0, 9}], {b}], q = {}}, Do[q = Append[q, Replace[p, x -> c[[i]], 2]], {i, 1, 9}]; r = Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]; If[r ? Infinity, r, p = Permutations[ Join[ Table[ b, {i, 1, n}], {x, y}]]; q = {}; Do[q = Append[q, Replace[p, {x -> c[[i]], y -> c[[j]]}, 2]], {i, 1, 9}, {j, 1, 9}]; Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]]]; Table[ f[n, 9], {n, 1, 20}]
  • PARI
    A037071(n)={my(t=10^(n+1)\9*9); forvec(v=[[-1, n], [-8, -1]], ispseudoprime(p=t+10^(n-v[1])*v[2]) && return(p));error} \\ M. F. Hasler, Feb 22 2016

Formula

a(n) = prime(A037070(n)). - Amiram Eldar, Jul 21 2025

Extensions

More terms from Vladeta Jovovic, Jan 10 2002
a(0) = 2 prepended by M. F. Hasler, Feb 22 2016

A037067 Smallest prime containing exactly n 7's.

Original entry on oeis.org

2, 7, 277, 1777, 47777, 727777, 7477777, 77767777, 577777777, 1777777777, 67777777777, 377777777777, 7177777777777, 17777777777777, 577777777777777, 2777777777777777, 77777767777777777, 377777777777777777, 2777777777777777777, 71777777777777777777
Offset: 0

Views

Author

Patrick De Geest, Jan 04 1999

Keywords

Comments

We conjecture that for all n >= 2, a(n) equals floor(10^(n+1)/9)*7 with one of the (first) digits 7 replaced by a digit among {0, ..., 6}. - M. F. Hasler, Feb 22 2016
The conjecture is false: a(668) = 7*(10^669-1)/9 + 10^276. - Robert Israel, Jul 13 2016

Crossrefs

Programs

  • Maple
    F:= proc(n) local x0,i,j;
      x0:= 7/9*(10^(n+1)-1);
      for j from 1 to 6 do
        if isprime(x0 + (j-7)*10^n) then
          return x0 + (j-7)*10^n fi od;
      for i from n-1 to 0 by -1 do
        for j from 0 to 6 do
         if isprime(x0 + (j-7)*10^i) then
           return x0 + (j-7)*10^i fi od od;
      for i from 0 to n do
        for j from 8 to 9 do
           if isprime(x0 + (j-7)*10^i) then
             return x0 + (j-7)*10^i fi
      od od:
    end proc:
    F(0):= 2: F(1):= 7:
    map(F, [$0..100]); # Robert Israel, Jul 13 2016
  • Mathematica
    f[n_, b_] := Block[{k = 10^(n + 1), p = Permutations[ Join[ Table[b, {i, 1, n}], {x}]], c = Complement[Table[j, {j, 0, 9}], {b}], q = {}}, Do[q = Append[q, Replace[p, x -> c[[i]], 2]], {i, 1, 9}]; r = Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]; If[r ? Infinity, r, p = Permutations[ Join[ Table[ b, {i, 1, n}], {x, y}]]; q = {}; Do[q = Append[q, Replace[p, {x -> c[[i]], y -> c[[j]]}, 2]], {i, 1, 9}, {j, 1, 9}]; Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]]]; Table[ f[n, 7], {n, 1, 18}]
  • PARI
    A037067(n)={my(t=10^(n+1)\9*7); forvec(v=[[-1, n], [-7, -1]], ispseudoprime(p=t+10^(n-v[1])*v[2]) && return(p)); error} \\ M. F. Hasler, Feb 22 2016

Formula

a(n) = prime(A037066(n)). - Amiram Eldar, Jul 21 2025

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 23 2003
More terms from and a(0) = 2 prepended by M. F. Hasler, Feb 22 2016

A037055 Smallest prime containing exactly n 1's.

Original entry on oeis.org

2, 13, 11, 1117, 10111, 101111, 1111151, 11110111, 101111111, 1111111121, 11111111113, 101111111111, 1111111118111, 11111111111411, 111111111116111, 1111111111111181, 11111111101111111, 101111111111111111, 1111111111111111171, 1111111111111111111, 111111111111111119111
Offset: 0

Views

Author

Patrick De Geest, Jan 04 1999

Keywords

Comments

For n > 1, A037055 is conjectured to be identical to A084673. - Robert G. Wilson v, Jul 04 2003
a(n) = A002275(n) for n in A004023. For all other n < 900, a(n) has n+1 digits. - Robert Israel, Feb 21 2016

Crossrefs

Programs

  • Maple
    f:= proc(n) local m,d,r,x;
       r:= (10^n-1)/9;
       if isprime(r) then return r fi;
       r:= (10^(n+1)-1)/9;
       for m from n-1 to 1 by -1 do
         x:= r - 10^m;
         if isprime(x) then return x fi;
       od;
       for m from 0 to n do
         for d from 1 to 8 do
            x:= r + d*10^m;
            if isprime(x) then return x fi;
         od
       od;
       error("Needs more than n+1 digits")
    end proc:
    map(f, [$0..100]); # Robert Israel, Feb 21 2016
  • Mathematica
    f[n_, b_] := Block[{k = 10^(n + 1), p = Permutations[ Join[ Table[b, {i, 1, n}], {x}]], c = Complement[Table[j, {j, 0, 9}], {b}], q = {}}, Do[q = Append[q, Replace[p, x -> c[[i]], 2]], {i, 1, 9}]; r = Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]; If[r ? Infinity, r, p = Permutations[ Join[ Table[ b, {i, 1, n}], {x, y}]]; q = {}; Do[q = Append[q, Replace[p, {x -> c[[i]], y -> c[[j]]}, 2]], {i, 1, 9}, {j, 1, 9}]; Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]]]; Table[ f[n, 1], {n, 1, 18}]
    Join[{2, 13}, Table[Sort[Flatten[Table[Select[FromDigits/@Permutations[Join[{n}, PadRight[{}, i, 1]]], PrimeQ], {n, 0, 9}]]][[1]], {i, 2, 20}]] (* Vincenzo Librandi, May 11 2017 *)
  • PARI
    A037055(n)={my(p,t=10^(n+1)\9); forstep(k=n+1,1,-1, ispseudoprime(p=t-10^k) && return(p)); forvec(v=[[0, n], [1, 8]], ispseudoprime(p=t+10^v[1]*v[2]) && return(p))} \\ M. F. Hasler, Feb 22 2016

Formula

a(n) = the smallest prime in { R-10^n, R-10^(n-1), ..., R-10; R+a*10^b, a=1, ..., 8, b=0, 1, 2, ..., n }, where R = (10^(n+1)-1)/9 is the (n+1)-digit repunit. - M. F. Hasler, Feb 25 2016
a(n) = prime(A037054(n)). - Amiram Eldar, Jul 21 2025

Extensions

More terms from Sascha Kurz, Feb 10 2003
Edited by Robert G. Wilson v, Jul 04 2003
a(0) = 2 inserted by Robert Israel, Feb 21 2016

A037057 Smallest prime containing exactly n 2's.

Original entry on oeis.org

3, 2, 223, 2221, 22229, 1222229, 20222227, 22222223, 222222227, 20222222221, 22222222223, 2122222222229, 21222222222221, 22222222222229, 222222222222227, 21222222222222221, 202222222222222229, 222222222222222221, 22222202222222222221, 220222222222222222229, 2202222222222222222229
Offset: 0

Views

Author

Patrick De Geest, Jan 04 1999

Keywords

Comments

For n > 1, the last digit cannot be 2, so a(n) has at least n+1 digits. The probability is big that none of [10^n/9]*20 + {1,3,7,9} is prime, in which case a(n) must have at least n+2 digits. This is the most frequent case. We can even conjecture that for all n > 1, a(n) equals [10^(n+1)/9]*20 + b with 1 <= b <= 9 and one of the (first) digits 2 replaced by 0 or 1. - M. F. Hasler, Feb 22 2016

Crossrefs

Programs

  • Mathematica
    f[n_, b_] := Block[{k = 10^(n + 1), p = Permutations[ Join[ Table[b, {i, 1, n}], {x}]], c = Complement[Table[j, {j, 0, 9}], {b}], q = {}}, Do[q = Append[q, Replace[p, x -> c[[i]], 2]], {i, 1, 9}]; r = Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]; If[r ? Infinity, r, p = Permutations[ Join[ Table[ b, {i, 1, n}], {x, y}]]; q = {}; Do[q = Append[q, Replace[p, {x -> c[[i]], y -> c[[j]]}, 2]], {i, 1, 9}, {j, 1, 9}]; Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]]]; Table[ f[n, 2], {n, 1, 18}]
  • PARI
    A037057(n)={my(p,t=10^(n+1)\9*20); n>1 && forvec(v=[[-1,n],[-2,-1]],nextprime(p=t+10^(n-v[1])*v[2])-p<10 && return(nextprime(p)));3-n} \\ M. F. Hasler, Feb 22 2016

Formula

a(n) = prime(A037056(n)). - Amiram Eldar, Jul 21 2025

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 23 2003
More terms and a(0) = 3 from M. F. Hasler, Feb 22 2016

A037063 Smallest prime containing exactly n 5's.

Original entry on oeis.org

2, 5, 557, 5557, 155557, 555557, 15555557, 55555553, 3555555551, 5555555557, 525555555557, 555555555551, 5555555555551, 355555555555559, 555555555555557, 51555555555555551, 545555555555555551, 555555555555555559, 15555555555555555557, 155555555555555555551
Offset: 0

Views

Author

Patrick De Geest, Jan 04 1999

Keywords

Comments

For n > 1, the last digit of n cannot be 5, therefore a(n) must have at least n+1 digits. It is probable that none among [10^n/9]*50 + {1,3,7,9} is prime in which case a(n) must have n+2 digits. We conjecture that for all n > 1, a(n) equals [10^(n+1)/9]*50 + b with 1 <= b <= 9 and one of the (first) digits 5 replaced by a 0, 1, 2, 3 or 4. - M. F. Hasler, Feb 22 2016

Crossrefs

Programs

  • Mathematica
    f[n_, b_] := Block[{k = 10^(n + 1), p = Permutations[ Join[ Table[b, {i, 1, n}], {x}]], c = Complement[Table[j, {j, 0, 9}], {b}], q = {}}, Do[q = Append[q, Replace[p, x -> c[[i]], 2]], {i, 1, 9}]; r = Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]; If[r ? Infinity, r, p = Permutations[ Join[ Table[ b, {i, 1, n}], {x, y}]]; q = {}; Do[q = Append[q, Replace[p, {x -> c[[i]], y -> c[[j]]}, 2]], {i, 1, 9}, {j, 1, 9}]; Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]]]; Table[ f[n, 5], {n, 1, 18}]
  • PARI
    A037063(n)={my(p, t=10^(n+1)\9*50); n>1 && forvec(v=[[-1, n], [-5, -1]], nextprime(p=t+10^(n-v[1])*v[2])-p<10 && return(nextprime(p)));1+4^n} \\ M. F. Hasler, Feb 22 2016

Formula

a(n) = prime(A037062(n)). - Amiram Eldar, Jul 21 2025

Extensions

More terms from Randall L Rathbun, Jan 11 2002
Edited and corrected by Robert G. Wilson v, Jul 04 2003
More terms and a(0) = 2 from M. F. Hasler, Feb 22 2016

A037065 Smallest prime containing exactly n 6's.

Original entry on oeis.org

2, 61, 661, 6661, 166667, 666667, 16666669, 66666667, 666666667, 6666666661, 66666666667, 1666666666661, 36666666666661, 166666666666667, 3666666666666667, 16666666666666661, 616666666666666661, 666666666666666661, 36666666666666666661, 66666666666666666667
Offset: 0

Views

Author

Patrick De Geest, Jan 04 1999

Keywords

Comments

Note that a(11) is palindromic.
The last digit of n cannot be 6, therefore a(n) must have at least n+1 digits. It is probable that none among [10^n/9]*60 + {1,3,7,9} is prime in which case a(n) must have n+2 digits. We conjecture that for all n >= 0, a(n) equals [10^(n+1)/9]*60 + b with 1 <= b <= 9 and one of the (first) digits 6 replaced by a digit among {0, 1, 2, 3, 4, 5}. - M. F. Hasler, Feb 22 2016

Crossrefs

Programs

  • Mathematica
    f[n_, b_] := Block[{k = 10^(n + 1), p = Permutations[ Join[ Table[b, {i, 1, n}], {x}]], c = Complement[Table[j, {j, 0, 9}], {b}], q = {}}, Do[q = Append[q, Replace[p, x -> c[[i]], 2]], {i, 1, 9}]; r = Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]; If[r ? Infinity, r, p = Permutations[ Join[ Table[ b, {i, 1, n}], {x, y}]]; q = {}; Do[q = Append[q, Replace[p, {x -> c[[i]], y -> c[[j]]}, 2]], {i, 1, 9}, {j, 1, 9}]; Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]]]; Table[ f[n, 6], {n, 1, 17}]
  • PARI
    A037065(n)={my(p, t=10^(n+1)\9*60); forvec(v=[[-1, n], [-6, -1]], nextprime(p=t+10^(n-v[1])*v[2])-p<10 && return(nextprime(p)))} \\ M. F. Hasler, Feb 22 2016

Formula

a(n) = prime(A037064(n)). - Amiram Eldar, Jul 21 2025

Extensions

More terms from Hans Havermann, Jun 16 2001
More terms and a(0) = 2 from M. F. Hasler, Feb 22 2016

A037061 Smallest prime containing exactly n 4's.

Original entry on oeis.org

2, 41, 443, 4441, 44449, 444443, 24444443, 424444441, 444444443, 4444444447, 44444444441, 444444444443, 14444444444449, 440444444444441, 2444444444444447, 44044444444444441, 424444444444444447, 4344444444444444449, 42444444444444444443, 44444444444444444447
Offset: 0

Views

Author

Patrick De Geest, Jan 04 1999

Keywords

Comments

The last digit of n cannot be 4, therefore a(n) must have at least n+1 digits. It is probable that none among [10^n/9]*40 + {1,3,7,9} is prime in which case a(n) must have n+2 digits. We conjecture that for all n >= 0, a(n) equals [10^(n+1)/9]*40 + b with 1 <= b <= 9 and one of the (first) digits 4 replaced by a 0, 1, 2 or 3. - M. F. Hasler, Feb 22 2016

Crossrefs

Programs

  • Mathematica
    f[n_, b_] := Block[{k = 10^(n + 1), p = Permutations[ Join[ Table[b, {i, 1, n}], {x}]], c = Complement[Table[j, {j, 0, 9}], {b}], q = {}}, Do[q = Append[q, Replace[p, x -> c[[i]], 2]], {i, 1, 9}]; r = Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]; If[r ? Infinity, r, p = Permutations[ Join[ Table[ b, {i, 1, n}], {x, y}]]; q = {}; Do[q = Append[q, Replace[p, {x -> c[[i]], y -> c[[j]]}, 2]], {i, 1, 9}, {j, 1, 9}]; Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]]]; Table[ f[n, 4], {n, 1, 18}]
  • PARI
    A037061(n)={my(p, t=10^(n+1)\9*40); forvec(v=[[-1, n], [-4, -1]], nextprime(p=t+10^(n-v[1])*v[2])-p<10 && return(nextprime(p)))} \\ M. F. Hasler, Feb 22 2016

Formula

a(n) = prime(A037060(n)). - Amiram Eldar, Jul 21 2025

Extensions

More terms from Antonio G. Astudillo (afg_astudillo(AT)lycos.com), Mar 23 2003
More terms and a(0) = 2 from M. F. Hasler, Feb 22 2016

A037069 Smallest prime containing exactly n 8's.

Original entry on oeis.org

2, 83, 881, 8887, 88883, 888887, 28888883, 88888883, 888888883, 48888888883, 288888888889, 888888888887, 48888888888883, 88888888888889, 888888888888883, 18888888888888883, 88888888888888889, 2888888888888888887, 8888888888888888881, 388888888888888888889
Offset: 0

Views

Author

Patrick De Geest, Jan 04 1999

Keywords

Comments

The last digit of n cannot be 8, therefore a(n) must have at least n+1 digits. It is probable (using [] for floor) that none among [10^n/9]*80 + {1,3,7,9} is prime in which case a(n) must have n+2 digits. We conjecture that for all n >= 0, a(n) equals [10^(n+1)/9]*80 + b with 1 <= b <= 9 and one of the (first) digits 8 replaced by a digit among {0, ..., 7}. - M. F. Hasler, Feb 22 2016

Crossrefs

Programs

  • Mathematica
    f[n_, b_] := Block[{k = 10^(n + 1), p = Permutations[ Join[ Table[b, {i, 1, n}], {x}]], c = Complement[Table[j, {j, 0, 9}], {b}], q = {}}, Do[q = Append[q, Replace[p, x -> c[[i]], 2]], {i, 1, 9}]; r = Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]; If[r ? Infinity, r, p = Permutations[ Join[ Table[ b, {i, 1, n}], {x, y}]]; q = {}; Do[q = Append[q, Replace[p, {x -> c[[i]], y -> c[[j]]}, 2]], {i, 1, 9}, {j, 1, 9}]; Min[ Select[ FromDigits /@ Flatten[q, 1], PrimeQ[ # ] & ]]]]; Table[ f[n, 8], {n, 1, 18}]
  • PARI
    A037069(n)={my(p, t=10^(n+1)\9*80); forvec(v=[[-1, n], [-8, -1]], nextprime(p=t+10^(n-v[1])*v[2])-p<10 && return(nextprime(p)))} \\ ~

Formula

a(n) = prime(A037068(n)). - Amiram Eldar, Jul 21 2025

Extensions

Corrected by Jud McCranie, Jan 04 2001
More terms from Erich Friedman, Jun 03 2001
More terms and a(0) = 2 from M. F. Hasler, Feb 22 2016
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