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.

A077192 {Concatenation of n-1 and n+1}/n where n is a member of A069871.

Original entry on oeis.org

8, 90, 92, 98, 992, 998, 9992, 9998, 99992, 99998, 999992, 999994, 999998, 9999992, 9999998, 99999992, 99999998, 999999992, 999999998, 9999999992, 9999999998, 99999999992, 99999999998, 999999999992, 999999999994, 999999999998, 9999999999992
Offset: 1

Views

Author

Amarnath Murthy, Nov 01 2002

Keywords

Examples

			8 and 90 are members since 24/3 = 8, 810/9 = 90.
999994 is in sequence because 999994 = 142856142858/142857 and 142857 is a member of A069871.
		

Crossrefs

Cf. A069871.

Extensions

More terms from Sascha Kurz, Feb 10 2003
Inserted missing terms a(12) and a(25) by Paolo P. Lava, Nov 05 2014
Offset changed by Sean A. Irvine, May 09 2025

A069860 Numbers n that divide the concatenation of n+1 and n+2.

Original entry on oeis.org

1, 2, 3, 4, 6, 17, 34, 51, 167, 334, 501, 1667, 3334, 5001, 14286, 16667, 33334, 50001, 166667, 333334, 500001, 1666667, 3333334, 5000001, 16666667, 33333334, 50000001, 166666667, 333333334, 500000001, 1666666667, 3333333334
Offset: 1

Views

Author

Amarnath Murthy, Apr 18 2002

Keywords

Examples

			17 divides 1819 hence is a member.
		

Crossrefs

Programs

  • Mathematica
    Select[ Range[10^7], Mod[ FromDigits[ Join[ IntegerDigits[ # + 1], IntegerDigits[ # + 2]]], # ] == 0 & ]

Formula

For k > 0, sequence contains (10^k+2)/2, (10^k+2)/3, (10^k+2)/6 and (10^(6k-1)+2)/7. The only other terms are 1 and 3. - David Wasserman, Aug 25 2005

Extensions

Edited and extended by Robert G. Wilson v, Apr 22 2002
More terms from David Wasserman, Aug 25 2005

A088797 Numbers n > 2 such that n divides the concatenation of n-2 and n-1.

Original entry on oeis.org

3, 7, 67, 667, 6667, 66667, 666667, 2857143, 6666667, 66666667, 666666667, 1052631579, 6666666667, 66666666667, 666666666667, 2857142857143, 6666666666667, 11764705882353, 66666666666667, 666666666666667
Offset: 1

Views

Author

Chuck Seggelin, Oct 18 2003

Keywords

Comments

For a 10-digit number, the difference between cat((n+2),(n+1)) and cat((n-2),(n-1)) is 40000000002 (as long as n-2 to n+2 are all numbers with the same number of digits). This difference has only 3 divisors which are ten digits long (1052631579, 2105263158 and 6666666667) of which two belong to the sequence. As 40000000002 has no other 10-digit factors, it is necessary to consider 11-digit numbers to obtain more terms.
From Robert G. Wilson v, Oct 21 2003, Oct 28 2003, Sep 23 2015 & Oct 24 2015: (Start)
All numbers of the forms
2(10^n-1)/3 + 1,
floor(2(10^(6n + 1) - 1)/7 + 1),
floor(2(10^(16n - 2) - 1)/17 + 1), and
floor(2(10^(18n - 8) - 1)/19 + 1), for n > 0 are members.
The only term not one of the above forms so far is 3. But it is included when n=0 for the second form.
(End)
If numbers less than 3 are acceptable, then an argument could be made that 1 is a terms since cat(n-2,n-1) is -10 which is == 0 (mod 1). - Robert G. Wilson v, Sep 29 2015
From Robert Israel, Oct 18 2015: (Start)
Numbers n of the form (2*10^m + 1)/k where k = 3, or k = 7 and m == 1 mod 6, or k = 17 and m == 14 mod 16, or k = 19 and m == 10 mod 18.
This is because n | (n-2)*10^m + (n-1) iff n | 2*10^m + 1.
But since we need 10^m >= n > 10^(m-1), 2*10^m+1 = k*n where 3 <= k <= 20.
The only numbers in that range that ever divide 2*10^m+1 are 3,7,17,19. (End)

Examples

			a(2)=7 because (7-2) concatenated with (7-1) yields 56 and 7 is a divisor of 56.
a(4)=667 because 667 is a divisor of 665666.
.
		

Crossrefs

Programs

  • Maple
    M:= 20: # to get all terms with <= M digits
    select(type, [seq(seq((2*10^m+1)/k, k=[19,17,7,3]),m=1..M)],integer); # Robert Israel, Oct 18 2015
  • Mathematica
    Select[ Range[8250000000], Mod[ FromDigits[ Join[ IntegerDigits[ # - 2], IntegerDigits[ # - 1]]], # ] == 0 &]
    fQ[n_] := Mod[ FromDigits[ Join[ IntegerDigits[n - 2], IntegerDigits[n - 1]]], n] == 0; k = 1; lst = {}; Select[ Flatten@ Table[ Select[ Divisors[4*10^n + 2], 10^(n - 1) < # < 10^n &], {n, 15}], fQ] (* Robert G. Wilson v, Sep 05 2015 *)
  • PARI
    for(n=3, 1e6, if((k=eval(Str(n-2,n-1))) && k % n == 0, print1(n", "))) \\ Altug Alkan, Sep 25 2015

Extensions

Extended by Robert G. Wilson v, Oct 21 2003
Further terms from Chuck Seggelin, Oct 29 2003

A088798 Numbers n that divide the concatenation of n-1, n-2 and n-3.

Original entry on oeis.org

3, 19, 57, 3276457, 9829371, 22997937, 24687460011, 24504559526049, 1152870338086169, 3458611014258507, 19831522709797616449, 54128285729329681609, 59494568129392849347, 61582096835687335289
Offset: 1

Views

Author

Chuck Seggelin, Oct 19 2003

Keywords

Comments

Each member of this sequence also appears to be a divisor of the number formed when concatenating (n+1), (n+2) and (n+3) in that order. Each nonprime member of the terms above appears to be divisible by 3. Further note that apart from 3 itself, if a(n) is a prime, then 3 * a(n) also appears to be a member. 19*3=57, 3276457*3=9829371. More prime members would need to be found to test this.

Examples

			a(2)=19 because 19 is a divisor of 181716. a(4)=3276457 because 3276457 is a divisor of 327645632764553276454.
		

Crossrefs

Programs

  • Maple
    prevcatOld := proc(n,t,o) local i,s; s := ""; for i from 1 to t do if o="a" then s := cat(convert(n-i,string),s) else s := cat(s,convert(n-i,string)) fi; od; parse(s) end; nextdivcat := proc(startAt,endAt,numTerms,catOrder) local i; for i from startAt to endAt while (prevcatOld(i,numTerms,catOrder) mod i > 0) do od; if i<=endAt then i else -1 fi; end; s := NULL; t := 2; for j from 1 to 10 do t := nextdivcat(t+1,23000000,3,"d"); s := s,t od; print(s);
  • Mathematica
    Do[ If[ Mod[ FromDigits[ Join[ IntegerDigits[2n], IntegerDigits[2n - 1], IntegerDigits[2n - 2]]], (2n + 1)] == 0, Print[2n + 1]], {n, 1, 700000000}]

Extensions

Edited by Robert G. Wilson v, Oct 20 2003
More terms from David Wasserman, Aug 25 2005

A088877 Numbers k that are divisors of the number formed by concatenating (k-1), k and (k+1), in that order.

Original entry on oeis.org

1, 3, 9, 11, 33, 111, 117, 143, 189, 231, 259, 273, 297, 333, 351, 407, 429, 481, 693, 777, 819, 1111, 1233, 1507, 2409, 3333, 4521, 7227, 7373, 11111, 26829, 27273, 33333, 81819, 100899, 101101, 108911, 111111, 123321, 128713, 129987, 142857
Offset: 1

Views

Author

Chuck Seggelin, Oct 20 2003

Keywords

Comments

The sequence of numbers that divide the concatenation of k-1 and k+1 (A069871) is a subsequence of this sequence. - David Wasserman, Mar 26 2004

Examples

			a(7)=117 because 117 is a divisor of 116117118.
		

Crossrefs

Cf. A069871.

Programs

  • Mathematica
    Select[Range[143000],MemberQ[Divisors[FromDigits[Flatten[ IntegerDigits/@ {#-1,#,#+1}]]],#]&] (* Harvey P. Dale, Aug 27 2020 *)

A249647 Numbers n that divide the concatenation of n+1 and n-1.

Original entry on oeis.org

1, 3, 9, 11, 33, 99, 111, 333, 999, 1111, 3333, 9999, 11111, 33333, 99999, 111111, 142857, 333333, 999999, 1111111, 3333333, 9999999, 11111111, 33333333, 99999999, 111111111
Offset: 1

Views

Author

Paolo P. Lava, Nov 03 2014

Keywords

Comments

A069871 is a subset of this sequence.
All the numbers of the form (10^k - 1)/9, (10^k - 1)/3 and (10^k-1) are members.
Any concatenation of 142857 with itself is part of the sequence, too.
All integers of the form (10^d-1)/k where 1 <= k <= 9. - Robert Israel, Jan 06 2016
It appears that equivalent definitions are: (A) Numbers which divide the repunit of the same length, n | R(length(n)). (B) Numbers equal to one or more concatenations of one among {1, 3, 9, 142857}. Is there a proof for this? - M. F. Hasler, Jun 10 2016

Examples

			3 belongs to this sequence as 3 divides 42, 11 belongs to this sequence as 11 divides 1210.
9 belongs to this sequence as 9 divides the concatenation of 10 and 8, i.e., 108.
		

Crossrefs

Cf. A069871.

Programs

  • Maple
    with(numtheory): P:=proc(q) local n;
    for n from 1 to q do if type(((n+1)*10^(ilog10(n-1)+1)+n-1)/n,integer) then print(n);
    fi; od; end: P(10^15);
    # alternative:
    seq(op(select(type, [seq((10^d-1)/k, k=[9,7,3,1])],integer)),d=1..20); # Robert Israel, Jan 06 2016
  • Mathematica
    Select[Range[2, 10^7], Divisible[FromDigits[IntegerDigits[# + 1]~Join~IntegerDigits[# - 1]], #] &] (* Michael De Vlieger, Jan 06 2016 *)
  • PARI
    for(n=1,10^8, s=concat(Str(n+1),Str(n-1));if(!(eval(s)%n),print1(n,", "))) \\ Derek Orr, Nov 03 2014

Formula

a(n) = 1000001*a(n-19) + 1000000*a(n-38). - Robert Israel, Jan 06 2016

Extensions

a(23)-a(25) from Derek Orr, Nov 03 2014

A281232 Numbers k such that k+2 divides concat(k, k+1).

Original entry on oeis.org

1, 5, 65, 665, 6665, 66665, 666665, 2857141, 6666665, 66666665, 666666665, 1052631577, 6666666665, 66666666665, 666666666665, 2857142857141, 6666666666665, 11764705882351, 66666666666665, 666666666666665, 6666666666666665, 66666666666666665, 666666666666666665
Offset: 1

Views

Author

Paolo P. Lava, Jan 18 2017

Keywords

Comments

Numbers of the form 60*(10^j - 1)/9 + 5, for j >= 0, belong to the sequence.
The ratios are: 4, 8, 98, 998, 9998, 99998, 999998, 9999994, 9999998, 99999998, 999999998, 9999999982, 9999999998, ...
Numbers of the form t(j) = 20*(10^(6*j) - 1)/7 + 1, for j >= 0, belong to the sequence, because (10^(6*j+1)*t(j) + t(j) + 1)/(t(j) + 2) = 10^(6*j+1) - 6. - Bruno Berselli, Oct 09 2018

Examples

			concat(2857141, 2857142) / 2857143 = 28571412857142 / 2857143 = 9999994.
		

Crossrefs

Programs

  • Maple
    with(numtheory): P:=proc(q) local c,n;
    for n from 1 to q do c:=n*10^(ilog10(n+1)+1)+n+1;
    if type(c/(n+2),integer) then print(n); fi; od; end: P(10^9);
  • Mathematica
    Select[Range[10^7], Divisible[FromDigits@ Flatten@ Map[IntegerDigits, {#, # + 1}], # + 2] &] (* Michael De Vlieger, Jan 19 2017 *)
  • PARI
    isok(n) = !(eval(Str(n, n+1)) % (n+2)); \\ Michel Marcus, Oct 09 2018

Formula

a(n) = A088797(n) - 2. - Alois P. Heinz, Jan 19 2017

Extensions

More terms from Alois P. Heinz, Jan 19 2017

A281233 Numbers k such that k-1 | concat(k, k+1).

Original entry on oeis.org

2, 3, 4, 5, 7, 18, 35, 52, 168, 335, 502, 1668, 3335, 5002, 14287, 16668, 33335, 50002, 166668, 333335, 500002, 1666668, 3333335, 5000002, 16666668, 33333335, 50000002, 166666668, 333333335, 500000002, 1666666668, 3333333335
Offset: 1

Views

Author

Paolo P. Lava, Jan 18 2017

Keywords

Comments

Numbers of the form 10^(j+1) + 60*(10^j-1)/9 + 8, 30*(10^j-1)/9 + 5 and 5*10^j + 2, for j>=0, belong to the sequence.
The ratios are: 23, 17, 15, 14, 13, 107, 104, 103, 1007, 1004, 1003, 10007, 10004, 10003, 100008, 100007, 100004, 100003, 1000007, 1000004, 1000003, 10000007, 10000004, 10000003, 100000007, 100000004, 100000003, 1000000007, 1000000004, 1000000003, ...

Examples

			concat(2,3) / 1 = 23 / 1 = 23; concat(3,4) / 2 = 34 / 2 = 17; etc.
		

Crossrefs

Programs

  • Maple
    with(numtheory): P:=proc(q) local c,n;
    for n from 2 to q do c:=n*10^(ilog10(n+1)+1)+n+1;
    if type(c/(n-1),integer) then print(n); fi; od; end: P(10^9)
  • Mathematica
    Select[Range[2, 10^7], Divisible[FromDigits@ Flatten@ Map[IntegerDigits, {#, # + 1}], # - 1] &] (* Michael De Vlieger, Jan 19 2017 *)

Formula

a(n) = A069860(n) + 1. - Alois P. Heinz, Jan 19 2017

Extensions

Two more terms (using A069860) from Alois P. Heinz, Jan 19 2017

A292885 a(n) is the least number k such that k | concat(k-n,k-n+1,…,k,…,k+n-1,k+n).

Original entry on oeis.org

1, 3, 3, 9, 6, 9, 9, 9, 9, 27, 15, 33, 18, 27, 18, 21, 18, 27, 27, 27, 24, 27, 27, 27, 27, 27, 27, 143, 34, 143, 45, 63, 36, 39, 39, 45, 42, 143, 89, 57, 45, 43, 143, 99, 54, 135, 154, 63, 63, 63, 75, 63, 154, 189, 66, 165, 72, 171, 153, 189, 90, 63, 81, 69, 69
Offset: 0

Views

Author

Paolo P. Lava, Sep 26 2017

Keywords

Examples

			a(4) = 6 because concat(2, 3, 4, 5, 6, 7, 8, 9, 10) = 2345678910 is a multiple of 6 and 6 is the least number to have this property.
		

Crossrefs

Programs

Showing 1-9 of 9 results.