A069554 -id:A069554 - cp's OEIS frontend

A071686 Smallest solution to gcd(x, Rev(x)) = 2^n.

2, 4, 8, 2192, 21920, 291008, 610688, 2112256, 2131456, 2937856, 25329664, 230465536, 694018048, 2344321024, 4688642048, 2112421888, 65012891648, 650128916480, 4494196736, 63769149440, 637691494400, 23842827272192, 276298064723968, 420127895977984, 4897795987210240

Offset: 1

Labos Elemer, Jun 03 2002

Giovanni Resta, Table of n, a(n) for n = 1..32

Cf. A004086, A055483, A069554, A072005, A072021, A072050.

a[n_] := Block[{k = 2^n}, While[GCD[k, FromDigits@ Reverse@ IntegerDigits@ k] != 2^n, k += 2^n]; k]; Array[a, 17] (* Giovanni Resta, Nov 14 2019 *)

a(n) = A069554(2^n).

a(22)-a(25) from Giovanni Resta, Oct 29 2019

A072005 Smallest solution to gcd(k, reverse(k)) = 3^n.

Original entry on oeis.org

1, 3, 9, 2889, 2899999989, 4899999987, 19899999972, 29898999693, 49989958299, 49999917897, 99884394999, 372797889885, 1989767716659, 2678052898989, 17902896898419, 137530987695297, 189281899170567, 368055404997498, 14048104419899757, 437893473401621955, 218264275944702783

Offset: 0

Labos Elemer, Jun 04 2002

			n=4: 3^4 = 81, a(4) = 2899999989 = 3*3*3*3*35802469, reverse(a(4)) = 2*3*3*3*3*61111111; gcd = 81 = 3^n.

Sean A. Irvine, Table of n, a(n) for n = 0..25

Cf. A004086, A055483, A069554, A071686, A071686, A072021, A072050.

a(n) = A069554(3^n).

a(15)-a(20) from Giovanni Resta, Oct 30 2019

A072021 Smallest solution to gcd(x, reverse(x)) = 5^n.

Original entry on oeis.org

5, 5200, 521000, 5213750, 521875, 5218750, 52130234375, 5734841796875, 57869714843750, 526046650390625, 5265674365234375, 52187008544921875, 526515306396484375, 5213023309008789062500, 5213596736358642578125, 5260466086273193359375, 526041911745452880859375

Offset: 1

Labos Elemer, Jun 06 2002

			For n = 4, gcd(521875, 578125) = 3125 = 5^4.
For n = 8, a(8) = 5734841796875 = 5^9*2936239, reverse(a(8)) = 5786971484375 = 5^8*71*208657.

Hiroaki Yamanouchi, Table of n, a(n) for n = 1..25

Cf. A004086, A055483, A069554, A071686, A072005, A072050, A072016-A072018.

a(n) = {my(k = 1); while (gcd(k, fromdigits(Vecrev(digits(k)))) != 5^n, k++); k;} \\ Michel Marcus, Jul 13 2018

a(n) = A069554(5^n).

a(9)-a(18) from Hiroaki Yamanouchi, Sep 10 2014

A072016 Numbers k such that gcd(k, reverse(k)) = 27 = 3^3, where reverse(x) = A004086(x).

Original entry on oeis.org

2889, 3699, 3888, 3969, 4779, 4887, 5589, 5697, 5778, 5859, 5886, 5967, 6399, 6669, 6777, 6885, 6939, 7398, 7479, 7587, 7668, 7695, 7749, 7776, 7857, 7884, 7938, 7965, 8289, 8397, 8559, 8667, 8775, 8829, 8883, 8937, 9099, 9288, 9369, 9396, 9477, 9558, 9585

Offset: 1

Labos Elemer, Jun 05 2002

Solutions to gcd(k, reverse(k)) = 1,3,9 (lower powers of 3) are trivial (see A072005).

			2889 = 107*3*3*3, 9889 = 3*3*3*3*2*61.

Vincenzo Librandi, Table of n, a(n) for n = 1..1030

Cf. A004086, A055483, A069554, A071686, A072005.

[n: n in [1..10^4] | Gcd(n,Seqint(Reverse(Intseq(n)))) eq 27]; // Vincenzo Librandi, Jul 11 2018

Select[Range[10^4], GCD[#, FromDigits[Reverse[IntegerDigits[#]]]] == 27 &] (* Vincenzo Librandi, Jul 11 2018 *)

isok(n) = gcd(n, fromdigits(Vecrev(digits(n)))) == 27; \\ Michel Marcus, Jul 11 2018

A072050 Smallest solution to GCD(x,A004086(x))=7^n.

Original entry on oeis.org

7, 18718, 343, 125204947, 231012215, 11298657013, 211066659013, 117088913464607, 2846847905744815, 108244538579770418, 2080795357577501075, 18312871825384462928, 26268977180287044053417, 1734582041294009627423816

Offset: 1

Labos Elemer, Jun 10 2002

Cf. A004086, A055483, A069554, A071686, A072005, A072021.

Mathematica
```
(* See A071686. *)
```

a(n) = A069554(7^n).

a(8)-a(9) from Max Alekseyev, Jun 17 2011

a(10)-a(14) from Giovanni Resta, Oct 30 2019

A333666 Smallest k > 0 with gcd(k, rev(k)) = n, where rev(k) is digit reversal of k and with sum of digits of k = n, or 0 if no such k exists.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 209, 48, 4000009, 21182, 5055, 21184, 13328, 288, 12844, 0, 1596, 2398, 13892, 2976, 52675, 45890, 2889, 61768, 178292, 0, 177475, 29984, 42999, 279718, 529865, 29988, 1009009009009, 485678, 1951599, 0, 694499, 655998, 1677688, 658988

Offset: 1

Ruediger Jehn and Kester Habermann, Sep 03 2020

This differs from A069554 in that, additionally, the sum of the digits of a(n) must be equal to n. This is not required in A069554.
As gcd(k, rev(k)) = n, n | k and n | rev(k). - David A. Corneth, Sep 03 2020
Since the sum of the digits of k is n and n | k, all the terms that are not 0 are Niven numbers (A005349). - Amiram Eldar, Sep 03 2020
The first digit of any number of this sequence is less than or equal to the last digit of this number (provided that the last digit is nonzero), because if k satisfies all requirements, also rev(k) does. This means that numbers starting with a "9" are quite rare. So far we have found only 9. But numbers ending with a "1" seem to be even less frequent. Amongst the first 303 terms of this sequence there is none except the trivial solution a(1) = 1. The second term of this sequence ending with a "1", if it exists, is still to be found. - Ruediger Jehn, Sep 20 2020 [Corrected by Pontus von Brömssen, Oct 07 2021]

			a(11) = 209. The sum of the digits is 11 and gcd(209,902) = 11.
a(12) = 48. The sum of the digits is 12 and gcd(48,84) = 12.

Ruediger Jehn, Table of n, a(n) for n = 1..303
Ruediger Jehn, Proofs for difficult terms
Rüdiger Jehn, Porous Numbers, arXiv:2104.02482 [math.GM], 2021.

Cf. A004086, A005349, A007953, A069554.

m = 36; s = Table[0, {m}]; c = 0; n = 1; While[c < m - Quotient[m, 10], g = GCD[n, FromDigits @ Reverse @ (d = IntegerDigits[n])]; If[g <= m && g == Plus @@ d && s[[g]] == 0, c++; s[[g]] = n]; n++]; s (* Amiram Eldar, Sep 03 2020 *)

a(n) = {if ((n % 10) == 0, return(0)); my(k=n); while (! ((sumdigits(k)==n) && (gcd(k, fromdigits(Vecrev(digits(k)))) == n)), k+=n); k;} \\ Michel Marcus, Sep 03 2020

for n in range(11, 20):
    for k in range(n, 1000000000, n):
       s = str(k)
       revk = "" # digit reversal of k
       sum = 0
       for i in range(len(s)):
          revk = revk + s[len(s) - i - 1]
          sum = sum + int(s[i])
       g = gcd(k,int(revk))
       if g == n and sum == n:
          print(n, k, revk, g)
          break

a(10*n) = 0 since all multiples of 10 have a 0 at the end, but their reverse numbers have no 0 at the end and therefore 10*n cannot be their gcd.

A072018 Numbers k for which gcd(k, reverse(k)) = 243 = 3^5.

Original entry on oeis.org

4899999987, 4989999897, 4999889997, 4999997889, 5889998997, 5889999969, 5898989997, 5898998988, 5899899789, 5899979979, 5899987998, 5899989699, 5899996989, 5979999879, 5988899997, 5988998898, 5989889979, 5989897998

Offset: 1

Labos Elemer, Jun 05 2002

			k = 4899999987 = 3*3*3*3*3*157*128437 and reverse(k) = 78999999984 = 2*2*2*2*3*3*3*3*3*3*2031893, gcd = 243. Numerous but not all solutions are obtained by inserting strings of 9's between digits of A071016. Further such regular transformations exist.

Sean A. Irvine, Table of n, a(n) for n = 1..1000

Cf. A004086, A055483, A069554, A071686, A072005, A072016.

A071685 Non-palindromic numbers n, not divisible by 10, such that either n divides R(n) or R(n) divides n, where R(n) is the digit-reversal of n.

Original entry on oeis.org

1089, 2178, 8712, 9801, 10989, 21978, 87912, 98901, 109989, 219978, 879912, 989901, 1099989, 2199978, 8799912, 9899901, 10891089, 10999989, 21782178, 21999978, 87128712, 87999912, 98019801, 98999901, 108901089, 109999989

Offset: 1

Labos Elemer, Jun 03 2002

The quotient R(n)/n or n/R(n) is always 4 or 9.
This is the union of the four sequence A001232, A222814, A008918, A222815. Equivalently, the union of A008919 and A031877.
There are 4*Fibonacci(floor((n-2)/2)) terms with n digits (this is 2*A214927 or essentially 4*A103609). - Ray Chandler, Oct 12 2017
Conjecture: every term mod 100 is equal to 1, 12, 78, or 89. - Harvey P. Dale, Dec 13 2017

			Palindromic solutions like 12021 or also solutions divisible by 10 were filtered out like {8380,838; q=10} or {8400,48; q=175}. In case of m>R(m), q=m/R(m)=4 or 9.

Ray Chandler, Table of n, a(n) for n = 1..10000
N. J. A. Sloane, 2178 And All That, arXiv:1307.0453 [math.NT], 2013; Fib. Quart., 52 (2014), 99-120.
N. J. A. Sloane, 2178 And All That [Local copy]

Cf. A001232, A222814, A008918, A222815, A008919, A031877.

Cf. A004086, A055483, A069554, A103609, A214927.

nd[x_, y_] := 10*x+y tn[x_] := Fold[nd, 0, x] ed[x_] := IntegerDigits[x] red[x_] := Reverse[IntegerDigits[x]] Do[s=Mod[Max[{n, tn[red[n]]}], Min[{n, r=tn[red[n]]}]]; If[Equal[s, 0]&&!Equal[Mod[n, 10], 0] &&!Equal[n, r], Print[{n, r/n}]], {n, 1, 1000000}]
npnQ[n_]:=Module[{r=IntegerReverse[n]},!PalindromeQ[n]&&!Divisible[ n,10] &&(Mod[n,r]==0||Mod[r,n]==0)]; Select[Range[11*10^7],npnQ] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Dec 13 2017 *)

x = q*R(x), q is an integer q<>1, q<>10^j and neither of x or R(x) is divisible by 10.

Corrected and extended by Harvey P. Dale, Jul 01 2013
Edited by N. J. A. Sloane, Jul 02 2013
Missing terms inserted by Ray Chandler, Oct 09 2017
Incorrect comment removed by Ray Chandler, Oct 12 2017

A071687 Non-palindromic numbers such that either x=q1.Rev[x] or Rev[x]=q2.x, where R[x]=A004086[x] and q1 or q2 are integers not divisible by 10.

Original entry on oeis.org

510, 540, 810, 1089, 2100, 2178, 4200, 5200, 5610, 5700, 5940, 6300, 8400, 8712, 8910, 9801, 10989, 21978, 23100, 27000, 46200, 51510, 52200, 52800, 54540, 56610, 57200, 59940, 65340, 69300, 81810, 87912, 89910, 98901, 109989, 212100, 217800

Offset: 1

Labos Elemer, Jun 03 2002

			Includes special cases of A071685. Examples represented by {n, Rev[n], integer-quotient} triples: {1089, 9801, 9}, {87912, 21979, 4}, {5610, 165, 34}, {610000, 16, 38125}, etc.

Cf. A071685, A004046, A055483, A069554.

nd[x_, y_] := 10*x+y tn[x_] := Fold[nd, 0, x] ed[x_] := IntegerDigits[x] red[x_] := Reverse[ed[x]] Do[s=Mod[ma=Max[{n, tn[red[n]]}], mi=Min[{n, r=tn[red[n]]}]]; If[Equal[s, 0]&&!Equal[n, r] &&!Equal[Mod[ma/mi, 10], 0], Print[{n, r, Max[r/n, n/r]}]], {n, 1, 1000000}]

q=Max[n/Rev[n], Rev[n]/n]=10m+r integer, where r>0, q>1.

A072017 Numbers k such that gcd(k, reverse(k)) = 81 = 3^4, where reverse(x) = A004086(x).

Original entry on oeis.org

2899999989, 2989999899, 2999889999, 3799999899, 3898989999, 3899799999, 3899999988, 3979989999, 3988899999, 3989999898, 3989999979, 3998999889, 3999889998, 3999898989, 3999899799, 3999979989, 3999988899, 4699998999

Offset: 1

Labos Elemer, Jun 05 2002

Numerous solutions can be constructed by inserting strings of suitable digits between digits of terms in A071016.

			k = 3*3*3*3*3*449*64157 and reverse(k) = 2*2*3*3*3*3*31*67*14827, GCD = 81.

Sean A. Irvine, Table of n, a(n) for n = 1..1000

Cf. A004086, A055483, A069554, A071686, A072005, A072016, A072018.

cp's OEIS Frontend

A071686 Smallest solution to gcd(x, Rev(x)) = 2^n.

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A072005 Smallest solution to gcd(k, reverse(k)) = 3^n.

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A072021 Smallest solution to gcd(x, reverse(x)) = 5^n.

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A072016 Numbers k such that gcd(k, reverse(k)) = 27 = 3^3, where reverse(x) = A004086(x).

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A072050 Smallest solution to GCD(x,A004086(x))=7^n.

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A333666 Smallest k > 0 with gcd(k, rev(k)) = n, where rev(k) is digit reversal of k and with sum of digits of k = n, or 0 if no such k exists.

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A072018 Numbers k for which gcd(k, reverse(k)) = 243 = 3^5.

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A071685 Non-palindromic numbers n, not divisible by 10, such that either n divides R(n) or R(n) divides n, where R(n) is the digit-reversal of n.

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