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 1-8 of 8 results.

A069862 Smallest k such that n divides the concatenation of numbers from (n+1) to (n+k), where (n+1) is on the most significant side.

Original entry on oeis.org

1, 2, 2, 2, 5, 2, 9, 4, 8, 10, 10, 8, 22, 16, 5, 4, 2, 8, 3, 20, 20, 10, 17, 12, 25, 22, 26, 16, 25, 20, 110, 20, 11, 2, 20, 8, 998, 52, 38, 20, 60, 20, 4, 32, 35, 42, 50, 20, 96, 50, 2, 96, 93, 26, 10, 20, 3, 50, 44, 20, 46, 40, 45, 40, 50, 32, 86, 32, 17, 20, 75, 72, 26, 926, 50
Offset: 1

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Author

Amarnath Murthy, Apr 18 2002

Keywords

Comments

Minimum number of consecutive subsequent integers after n that must be concatenated together in ascending order such that n divides the concatenated term.
Concatenation always begins at n+1. Note that multiples of 11 seems to require more terms than any other number. 385 requires 9860. 451 requires 100270 terms be concatenated together into a 495,000 digit number. - Chuck Seggelin, Oct 29 2003; corrected by Chai Wah Wu, Oct 19 2014

Examples

			a(7) = 9 as 7 divides 8910111213141516 the concatenation of numbers from 8(= 7+1) to 16 (= 7+9).
a(5) = 5 because 5 will divide the number formed by concatenating the 5 integers after 5 in ascending order (i.e. 678910). a(385) = 9860 because 385 will divide the concatenation of 386,387,388,...,10245.
		

Crossrefs

Programs

  • Maple
    c[1] := 1:for n from 2 to 172 do k := 1:g := (n+k) mod n:while(true) do k := k+1:b := convert(n+k,base,10):g := (g*10^nops(b)+n+k) mod n: if((g mod n)=0) then c[n] := k:break:fi:od:od:seq(c[l],l=1..172);
  • Mathematica
    f[n_] := Block[{k = n + 1}, d = k; While[ !IntegerQ[d/n], k++; d = d*10^Floor[Log[10, k] + 1] + k]; k - n]; Table[ f[n], {n, 1, 75}] (* Robert G. Wilson v, Nov 04 2003 *)
  • Python
    def A069862(n):
        nk, kr, r = n+1, 1, 1 if n > 1 else 0
        while r:
            nk += 1
            kr = (kr + 1) % n
            r = (r*(10**len(str(nk)) % n)+kr) % n
        return nk-n # Chai Wah Wu, Oct 20 2014

Extensions

More terms from Sascha Kurz, Jan 28 2003

A088800 Numbers n which are divisors of the number formed by concatenating (n-1), (n-2), (n-3) and (n-4) in that order.

Original entry on oeis.org

16, 43, 86, 3923, 7846, 1320532, 14869252, 37031789, 74063578, 1770748607, 3541497214, 7082994428, 50541770557, 1040474831558, 1076026363388, 2080949663116, 2746369613531, 3376243036861, 5492739227062, 6529386313577
Offset: 1

Views

Author

Chuck Seggelin, Oct 20 2003

Keywords

Comments

Each member of this sequence appears to also be a factor of the number formed by concatenating (n+1), (n+2), (n+3) and (n+4) in that order. When evaluating concat((n+1),(n+2),(n+3),(n+4)) - concat((n-1),(n-2),(n-3),(n-4)) for members larger than 86 the difference appears to always be a number of the form 2(0)...4(0)...6(0)...8 with the same number of zeros following the 2, 4 and 6. The member will be a factor of this number. Further terms for the sequence can be produced by factoring numbers of this form. Let z=the number of zeros in one of the segments of a number d of the form 2(0)...4(0)...6(0)...8. Find the divisors of d. All divisors which are not of length z+1 are not members of this sequence and those that are of length z+1 are likely candidates and should be tested (note that apart from 16, candidates which are divisible by 8 appear to never be members). For example let d = 2000000000000000400000000000000060000000000000008. z=15. The divisors of d are numerous, but only one is z+1 (16) digits long: 7547657634163187. Testing this candidate confirms that it is also a member of this sequence.

Examples

			a(3)=86 because 86 is a factor of 85848382.
		

Crossrefs

A088868 Numbers n which are divisors of the number formed by concatenating (n-4), (n-3), (n-2) and (n-1) in that order.

Original entry on oeis.org

49, 109, 307121, 3837881, 415922011, 44886856951, 73071640562111, 11741452251865261, 138599925259848671
Offset: 1

Views

Author

Chuck Seggelin, Oct 20 2003

Keywords

Comments

Each member of this sequence appears to also be a factor of the number formed by concatenating (n+4), (n+3), (n+2) and (n+1) in that order. When evaluating concat((n+4),(n+3),(n+2),(n+1)) - concat((n-4),(n-3),(n-2),(n-1)) for members of this sequence the difference appears to always be a number of the form 8(0)...6(0)...4(0)...2 with the same number of zeros following the 8, 6 and 4. The member will be a factor of this number. Terms for this sequence can be produced by factoring numbers of this form. Let z=the number of zeros in one of the segments of a number d of the form 8(0)...6(0)...4(0)...2. Find the divisors of d. All divisors which are not of length z+1 are not members of this sequence and those that are of length z+1 are possible candidates and should be tested. For example let d = 8000000000000000006000000000000000004000000000000000002. z=17. The divisors of d are numerous, but only two are z+1 (18) digits long: 138599925259848671 and 27719985051 9697342. Testing these candidates confirms that the first one is a member of this sequence.
No more terms < 10^29. - David Wasserman, Aug 26 2005

Examples

			a(2)=109 because 109 is a factor of 105106107108.
		

Crossrefs

A088870 Numbers n which are divisors of the number produced by concatenating (n-5), (n-4), ... (n-1) in that order.

Original entry on oeis.org

13677, 14647, 21291, 29567, 43941, 69031, 88701, 105991, 126507, 317973, 156304482823, 468913448469, 21729950852487, 2212933498428421, 6638800495285263, 12049739358792173, 36149218076376519, 11316117499289108644863
Offset: 1

Views

Author

Chuck Seggelin, Oct 20 2003

Keywords

Examples

			a(1)=13677 because 13677 is a factor of 1367213673136741367513676.
		

Crossrefs

Extensions

More terms from David Wasserman, Aug 26 2005

A088872 Numbers n which are divisors of the number produced by concatenating (n-10), (n-9), (n-8), ... (n-1) in that order.

Original entry on oeis.org

37, 1793, 12247, 13061, 12345679, 17675669, 59454523, 10035074789, 11207086103, 12174047803, 13606288721, 14036302277, 17989804447, 19541987747, 20070727459, 21841045129, 22431989513, 23140930781, 32217871691
Offset: 1

Views

Author

Chuck Seggelin, Oct 20 2003

Keywords

Examples

			a(1)=37 because 37 is a factor of 27282930313233343536.
		

Crossrefs

Extensions

More terms from David Wasserman, Aug 25 2005

A088869 Numbers n which are divisors of the number produced by concatenating (n-1), (n-2), ... (n-5) in that order.

Original entry on oeis.org

5, 15, 197, 591, 599, 985, 23363, 70089, 182415, 32533013, 97599039, 200814277564911, 334690462608185, 100208875079675883, 119690233947031245, 120541110603088381, 152213696825889183, 167014791799459805
Offset: 1

Views

Author

Chuck Seggelin, Oct 20 2003

Keywords

Examples

			a(2)=15 because 15 is a factor of 1413121110.
		

Crossrefs

Extensions

More terms from David Wasserman, Aug 26 2005

A088871 Numbers n which are divisors of the number produced by concatenating (n-1), (n-2), ... (n-10) in that order.

Original entry on oeis.org

10, 22, 55, 181, 190, 362, 905, 2035, 4070, 10490, 11203, 14686, 17833, 22406, 23065, 35666, 36715, 46130, 56015, 73430, 78421, 89165, 100705, 201410, 1004530, 1093165, 1425313, 1480309, 1695710, 1956190, 2186330, 2850626, 2882707
Offset: 1

Views

Author

Chuck Seggelin, Oct 20 2003

Keywords

Examples

			a(2)=22 because 22 is a factor of 21201918171615141312.
		

Crossrefs

Extensions

More terms from David Wasserman, Aug 26 2005

A088886 Minimum number of consecutive previous nonnegative integers to n that must be concatenated together in ascending order such that n divides the concatenated term, or zero if n divides no such concatenation.

Original entry on oeis.org

1, 0, 2, 0, 0, 0, 2, 0, 8, 0, 3, 0, 0, 0, 0, 0, 0, 0, 0, 0, 14, 0, 0, 0, 0, 0, 26, 0, 6, 0, 0, 0, 11, 0, 0, 0, 10, 0, 0, 0, 16, 0, 15, 0, 0, 0, 25, 0, 4, 0, 45, 0, 0, 0, 0, 0, 20, 0, 51, 0, 45, 0, 0, 0, 0, 0, 2, 0, 35, 0, 22, 0, 0, 0, 0, 0, 21, 0, 0, 0, 0, 0, 81, 0, 0, 0, 6, 0, 0, 0, 0, 0, 66, 0, 0, 0, 13
Offset: 1

Views

Author

Chuck Seggelin, Oct 29 2003

Keywords

Comments

Concatenation always end at n-1 and cannot start further than n-n (zero). Hence the maximum value of a(n) is n.

Examples

			a(7) = 2 because 7 will divide the number formed by concatenating the 2 integers prior to 7 in ascending order (i.e. 56). a(6) = 0 because 6 will not divide 5, 45, 345, 2345, 12345, or 012345.
		

Crossrefs

Showing 1-8 of 8 results.