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 1-9 of 9 results.

A066209 A053041(n)-10^(n-1).

Original entry on oeis.org

0, 0, 2, 0, 0, 2, 6, 0, 8, 0, 10, 8, 12, 4, 5, 0, 16, 8, 18, 0, 5, 12, 22, 8, 0, 16, 8, 8, 28, 20, 30, 0, 32, 24, 10, 8, 36, 28, 17, 0, 40, 2, 42, 12, 35, 36, 46, 32, 41, 0, 2, 40, 52, 8, 10, 32, 14, 48, 58, 20, 60, 52, 26, 0, 10, 56, 66, 20, 38, 50, 70, 8, 72, 64, 50, 64, 10, 74, 78
Offset: 1

Views

Author

Vladeta Jovovic, Dec 17 2001

Keywords

Comments

a(n) = 0 iff n is of form 2^i*5^j (cf. A003592).

Programs

  • Mathematica
    lst = {}; Do[z = n - Mod[10^n/10, n]; If[z == n, z = 0]; AppendTo[lst, z], {n, 79}]; lst (* Arkadiusz Wesolowski, Apr 02 2012 *)

A066557 Largest n-digit multiple of n.

Original entry on oeis.org

9, 98, 999, 9996, 99995, 999996, 9999997, 99999992, 999999999, 9999999990, 99999999990, 999999999996, 9999999999990, 99999999999998, 999999999999990, 9999999999999984, 99999999999999990, 999999999999999990, 9999999999999999990, 99999999999999999980
Offset: 1

Views

Author

Amarnath Murthy, Dec 17 2001

Keywords

Links

Crossrefs

Cf. A053041, A066558.

Programs

Formula

a(n) = floor((10^n-1)/n)*n.

Extensions

More terms and formula from Francisco Salinas (franciscodesalinas(AT)hotmail.com), Dec 24 2001

A066558 a(n) = A066557(n)/n.

Original entry on oeis.org

9, 49, 333, 2499, 19999, 166666, 1428571, 12499999, 111111111, 999999999, 9090909090, 83333333333, 769230769230, 7142857142857, 66666666666666, 624999999999999, 5882352941176470, 55555555555555555, 526315789473684210, 4999999999999999999, 47619047619047619047
Offset: 1

Views

Author

Amarnath Murthy, Dec 17 2001

Keywords

Links

Crossrefs

Cf. A053041, A066557.

Programs

Formula

a(n) = floor((10^n-1)/n).

Extensions

More terms and formula from Francisco Salinas (franciscodesalinas(AT)hotmail.com), Dec 24 2001

A084024 Triangle read by rows in which row n gives n smallest n-digit multiples of n that are palindromes.

Original entry on oeis.org

1, 22, 44, 111, 141, 171, 2112, 2332, 2552, 2772, 50005, 50105, 50205, 50305, 50405, 201102, 204402, 207702, 210012, 213312, 216612, 1002001, 1009001, 1011101, 1018101, 1020201, 1027201, 1036301, 21100112, 21111112, 21122112, 21133112, 21144112, 21155112, 21166112, 21177112
Offset: 1

Views

Author

Amarnath Murthy and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 23 2003

Keywords

Examples

			1
22 44
111 141 171
2112 2332 2552 2772
...

Crossrefs

Cf. A053041, A083123, A084025, A084026, A084027, A084028.

Programs

  • Mathematica
    snm[n_]:=Module[{a=Floor[10^(n-1)],b=Floor[10^n-1]},Select[ Select[ Range[ a,b],Divisible[#,n]&&IntegerLength[#]==n&],PalindromeQ,n]]; Array[ snm,8]//Flatten (* Harvey P. Dale, Aug 23 2020 *)
  • PARI
    isok(k, n) = my(d=digits(k*n)); (#d == n) && (Vecrev(d) == d);
row(n) = {my(v=vector(n), k = ceil(10^(n-1)/n)); for (i=1, n, while(! isok(k, n), k++); v[i] = k*n; k++;); v;} \\ Michel Marcus, Mar 28 2020

Formula

T(n, 1) >= A053041(n); T(n, 1) = A083123(n). - Michel Marcus, Mar 28 2020

Extensions

Corrected and extended by Ray Chandler, Jun 12 2003
More terms from Michel Marcus, Mar 28 2020

A162213 a(n) = the smallest positive multiple of n with exactly n digits when written in binary.

Original entry on oeis.org

1, 2, 6, 8, 20, 36, 70, 128, 261, 520, 1034, 2052, 4108, 8204, 16395, 32768, 65552, 131076, 262162, 524300, 1048593, 2097172, 4194326, 8388624, 16777225, 33554456, 67108878, 134217748, 268435484, 536870940, 1073741854, 2147483648
Offset: 1

Views

Author

Leroy Quet, Jun 28 2009

Keywords

Links

Crossrefs

Cf. A162214, A053041.

Programs

  • Maple
    a := proc (n) local k: for k while nops(convert(k*n, base, 2)) <> n do end do; k*n end proc: seq(a(n), n = 1 .. 24); # Emeric Deutsch, Jul 10 2009
a:= n-> n*ceil(2^(n-1)/n): seq(a(n), n=1..40); # Alois P. Heinz, Jul 11 2009
  • Mathematica
    Array[# Ceiling[2^(# - 1)/#] &, 32] (* Michael De Vlieger, Nov 04 2017 *)

Formula

a(n) = n * ceiling(2^(n-1)/n). - Alois P. Heinz, Jul 11 2009

Extensions

More terms from Emeric Deutsch and Alois P. Heinz, Jul 11 2009

A056968 10^(n-1) modulo n.

Original entry on oeis.org

0, 0, 1, 0, 0, 4, 1, 0, 1, 0, 1, 4, 1, 10, 10, 0, 1, 10, 1, 0, 16, 10, 1, 16, 0, 10, 19, 20, 1, 10, 1, 0, 1, 10, 25, 28, 1, 10, 22, 0, 1, 40, 1, 32, 10, 10, 1, 16, 8, 0, 49, 12, 1, 46, 45, 24, 43, 10, 1, 40, 1, 10, 37, 0, 55, 10, 1, 48, 31, 20, 1, 64, 1, 10, 25, 12, 67, 4, 1, 0, 73, 10, 1
Offset: 1

Views

Author

Henry Bottomley, Jul 20 2000

Keywords

Examples

			a(6)=4 since 100000=6*16666+4

Links

Crossrefs

Cf. A003592, A053041, A056969.

Programs

  • Maple
    0, seq(10&^(n-1) mod n, n=2..100); # Robert Israel, Nov 25 2024
  • Mathematica
    Table[PowerMod[10,n-1,n],{n,100}] (* Harvey P. Dale, Jul 17 2021 *)

Formula

If n is of form 2^i*5^j then a(n)=0, otherwise a(n)=10^(n-1)+n-A053041(n)
From Robert Israel, Nov 25 2024: (Start)
If n is prime other than 2 or 5, then a(n) = 1.
If n = 2^i * 5^j * p where p is a prime > 10^(2^i * 5^j), then a(n) = 10^(2^i * 5^j).
If n = 2^i * 5^j * p where p is a prime and
2^(2^i * 5^j - 1 - i) * 5^(2^i * 5^j -1 - j) > p > 2^(2^i * 5^j-2 - u) * 5^(2^i * 5^j-1-j),
then a(n) = 10^(2^i * 5^j - 1) - 2^i * 5^j * p.
For example, with i = 0 and j = 1 we get a(5*p) = 10^4 - 5*p if p is a prime between 1000 and 2000.
(End)

A066559 a(n) = ceiling(10^(n-1)/n).

Original entry on oeis.org

1, 5, 34, 250, 2000, 16667, 142858, 1250000, 11111112, 100000000, 909090910, 8333333334, 76923076924, 714285714286, 6666666666667, 62500000000000, 588235294117648, 5555555555555556, 52631578947368422, 500000000000000000
Offset: 1

Views

Author

Amarnath Murthy, Dec 17 2001

Keywords

Links

Crossrefs

Cf. A053041, A066209, A066557, A066558.

Programs

Formula

a(n) = A053041(n)/n.

Extensions

Corrected, extended by, and better name from Vladeta Jovovic, Dec 17 2001

A140317 Smallest n-digit number divisible by n^2.

Original entry on oeis.org

1, 12, 108, 1008, 10000, 100008, 1000041, 10000000, 100000008, 1000000000, 10000000109, 100000000080, 1000000000116, 10000000000144, 100000000000125, 1000000000000000, 10000000000000152, 100000000000000008
Offset: 1

Views

Author

Jonathan Vos Post, May 26 2008

Keywords

Examples

			a(7) = 1000041 because 1000041 has 7 digits, 1000041/49 = 20409 = 3 * 6803 and no integer between 1000000 and 1000041 is divisible by 7^2 = 49.
a(9) = 100000008 because 100000008 has 9 digits, 100000008/81 = 1234568 = 23 * 154321 and no integer between 100000000 and 100000008 is divisible by 9^2 = 81.

Links

Crossrefs

Cf. A000290, A053041.

Programs

  • Maple
    A140317 := proc(n) n^2*ceil(10^(n-1)/n^2) ; end: seq(A140317(n),n=1..30) ; # R. J. Mathar, May 31 2008
  • Mathematica
    snd[n_]:=Module[{c=n^2-PowerMod[10,n-1,n^2]},If[Divisible[10^(n-1), n^2], 10^(n-1),10^(n-1)+c]]; Array[snd,20] (* Harvey P. Dale, Dec 14 2012 *)

Formula

a(n) = n^2*ceiling(10^(n-1)/n^2). - Michel Marcus, Aug 25 2020

Extensions

More terms from R. J. Mathar, May 31 2008

A337391 a(n) is the smallest n-digit number divisible by n^3.

Original entry on oeis.org

1, 16, 108, 1024, 10000, 100008, 1000188, 10000384, 100000575, 1000000000, 10000001319, 100000001088, 1000000000792, 10000000000536, 100000000001250, 1000000000000000, 10000000000001886, 100000000000001952, 1000000000000003324, 10000000000000000000, 100000000000000008972, 1000000000000000009208
Offset: 1

Views

Author

Lukas R. Mansour, Aug 25 2020

Keywords

Examples

			a(2) = 16, as 16 is the first 2-digit number divisible by 2^3 = 8.
a(3) = 108, as 108 is the first 3-digit number divisible by 3^3 = 27.
a(4) = 1024, as 1024 is the first 4-digit number divisible by 4^3 = 64.
a(5) = 10000, as 10000 is the first 5-digit number divisible by 5^3 = 125.

Crossrefs

Cf. A000578, A011557, A053041 (divisible by n), A140317 (divisible by n^2).

Programs

  • Mathematica
    Table[n^3 * Ceiling[10^(n - 1)/n^3], {n, 1, 22}] (* Amiram Eldar, Aug 25 2020 *)
  • PARI
    a(n) = n^3 * ceil(10^(n-1) / n^3) \\ David A. Corneth, Aug 25 2020

Formula

a(n) = n^3 * ceiling(10^(n-1) / n^3). - David A. Corneth, Aug 26 2020
Showing 1-9 of 9 results.

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