A001232 -id:A001232 - cp's OEIS frontend

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.

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

A124074 Consider the map that sends a number m to m - reverse(m) if that is nonnegative, otherwise to m + reverse(m). Sequence gives those numbers that reach a nonzero cycle under repeated application of this map.

Original entry on oeis.org

119, 129, 169, 179, 188, 198, 200, 210, 218, 220, 228, 230, 240, 249, 250, 260, 268, 270, 278, 280, 287, 290, 297, 299, 300, 301, 310, 311, 317, 320, 321, 327, 330, 331, 340, 341, 348, 349, 350, 351, 360, 361, 367, 370, 371, 377, 380, 381, 386, 389, 390

Offset: 1

Sébastien Dumortier, Nov 05 2006

There are values of n that apparently grow without bounds under repeated application of this map beginning with 10598, 10698, 10798, 10898, 11588, 11688, 11788, 11888, 11999, ... - Ray Chandler, Oct 10 2017

			119 is a member because
119 + 911 = 1030 (since the difference is negative),
1030 - 0301 = 729 (since the difference is positive),
729 + 927 = 1656,
1656 + 6561 = 8217,
8217 - 7128 = 1089,
1089 + 9801 = 10890,
10890 - 09801 = 1089, so the terms are cyclic.

Ray Chandler, Table of n, a(n) for n = 1..5000

Cf. A001232 gives the nonzero fixed points of map.

With[{nn = 400}, Select[Range@ nn, LengthWhile[NestList[If[#1 < 0, #2, #1] & @@ {#1 - #2, #1 + #2} & @@ {#, IntegerReverse@ #} &, #1, #2], # > 0 &] == #2 + 1 & @@ {#, 2^Log2@ #} &]] (* Michael De Vlieger, Oct 10 2017 *)

Definition clarified by Ray Chandler, Oct 10 2017

A222816 Numbers k that divide the base-5 reversal of k.

Original entry on oeis.org

8, 48, 144, 208, 248, 744, 1008, 1248, 3744, 5008, 5208, 6048, 6248, 18744, 25008, 26208, 30048, 31248, 90144, 93744, 125008, 126008, 130208, 131208, 150048, 151048, 155248, 156248, 450144, 468744, 625008, 631008, 650208, 656208, 750048, 756048, 775248, 781248

Offset: 1

N. J. A. Sloane, Mar 13 2013

The base-B reversal of k is obtained by writing k in base B and reversing the order of the digits.

			8 in base 5 is 13_5; its reverse is 31_5, which is 16 in base 10, and 8 divides 16, so 8 is in the sequence.

N. J. A. Sloane, Table of n, a(n) for n = 1..57
Leonard F. Klosinski and Dennis C. Smolarski, On the Reversing of Digits, Math. Mag., 42 (1969), 208-210. [Gives a(1)]
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. A214927, A091079.

Select[Range[10^6],Divisible[s=FromDigits[Reverse@IntegerDigits[#,5],5],#]&&s!=#&] (* Giorgos Kalogeropoulos, Aug 20 2021 *)

isok(n) = (rdn = subst(Polrev(digits(n,  5)), x, 5)) && (rdn != n) && (rdn % n) == 0; \\ Michel Marcus, Oct 10 2014

More terms from Alois P. Heinz, Mar 13 2013

A078273 Smallest multiple of n other than n using only the digits of n (no limit on frequency).

Original entry on oeis.org

11, 22, 33, 44, 55, 66, 77, 88, 99, 100, 1111, 1212, 1131, 1414, 555, 1616, 1717, 1188, 1919, 200, 2121, 2222, 322, 2424, 225, 2262, 2727, 2828, 2929, 300, 1333, 3232, 3333, 3434, 3535, 3636, 333, 3838, 3393, 400, 4141, 4242, 344, 4444, 4455, 644, 4747, 4848

Offset: 1

Amarnath Murthy, Nov 24 2002

a(k) = 10k if k contains a zero. a(n) <= (10^d +1)*n where d is the number of digits in n. There are some patterns in which every digit is used exactly as many times as it occurs in n. (A008918 and A001232). (1) a(2178) = 8712, a(21978) = 87912, a(219978) = 879912, etc... with a(n)/n = 4. A derived pattern is a(21782178) = 87128712, a(217821782178) = 871287128712 etc. (2) a(1089) = 9801, a(10989) = 98901, a(109989)= 989901,... with a(n)/n = 9. More patterns can be derived on similar lines.

			a(30) = 300, a(2178) = 8712, a(1089) = 9801.

Ray Chandler, Table of n, a(n) for n = 1..10000

Cf. A008918, A001232.

smn[n_]:=Module[{k=2},While[!SubsetQ[IntegerDigits[n],IntegerDigits[ k*n]], k++];k*n]; Array[smn,50] (* Harvey P. Dale, Dec 03 2018 *)

Corrected and extended by Sean A. Irvine, Mar 09 2010

A102277 Numbers n such that n = 15*reversal(n).

Original entry on oeis.org

0, 65340, 659340, 6599340, 65999340, 653465340, 659999340, 6534065340, 6599999340, 65340065340, 65934659340, 65999999340, 653400065340, 659340659340, 659999999340, 6534000065340, 6534653465340, 6593400659340, 6599346599340, 6599999999340

Offset: 1

Farideh Firoozbakht, Jan 04 2005

30 divides all terms of the sequence. For all nonnegative integers m and n all numbers of the form f1(m,n) = 660(10^(m + 2) - 1)*(10^((m + 4)*n) - 1)/(10^(m + 4) - 1) are in the sequence, in fact f1(m,n) = (65.(9)(m).34)(n).0 where dot between numbers means concatenation and "(r)(t)" means number of r's is t. With this definition a(1) = 0 = f1(0,0), a(2) = 65340 = f1(0,1), a(3) = 659340 = f1(1,1), a(4) = 6599340 = f1(2,1), a(5) = 65999340 = f1(3,1), a(6) = 653465340 = f1(0,2), a(7) = 659999340 = f1(4,1), a(9) = 6599999340 = f1(5,1), etc. f1(m,1) = 660(10^(m + 2) - 1) = 65.(9)(m).340, f1(m,2) = 65.(9)(m).34.65.(9)(m).340, etc. Let g(s,t,r) = s*(10^((L+t)*(1+r))-1)/(10^(L+t)-1) where L = number of digits of s, in fact g(s,t,r) = (s.(0)(t))(r).s so the function g is the same function that has been defined in the sequence A101704. If s is in the sequence then all numbers of the form g(s,t,r) for nonnegative integers t and r are in the sequence. Next term is greater than 11*10^9. It seems that the eleven next terms are 65340065340, 65934659340, 65999999340, 653400065340, 659340659340 659999999340, 6534000065340, 6534653465340, 6593400659340, 6599346599340 and 6599999999340. Is it true that, all terms of this sequence are of the form g(f1(m,n),r,t)?

			g(65340,0,2)= (65340)(3) = 653406534065340 is in the sequence because reversal(653406534065340) = 43560435604356 = (1/15)*653406534065340.

Ray Chandler, Table of n, a(n) for n = 1..10000

Cf. A001232, A008918, A101704, A101705, A101706.

Do[If[n == 15*FromDigits[Reverse[IntegerDigits[n]]], Print[n]], {n, 0, 11000000000, 30}]

a(n) = 10*A101704(n) = 20*A101706(n). - Ray Chandler, Oct 09 2017

More terms from Ray Chandler, Oct 09 2017

A285040 Numbers n such that three-halves of n equals the reverse of n.

Original entry on oeis.org

4356, 43956, 439956, 4399956, 43564356, 43999956, 435604356, 439999956, 4356004356, 4395643956, 4399999956, 43560004356, 43956043956, 43999999956, 435600004356, 435643564356, 439560043956, 439956439956, 439999999956

Offset: 1

Harvey P. Dale, Apr 08 2017

There are Fibonacci(floor((n-2)/2)) terms with n digits (this is essentially A103609). - Ray Chandler, Oct 12 2017

			439956 times 3/2 equals 659934 which is the reverse of 439956.

David Wells, The Penguin Dictionary of Curious and Interesting Numbers (Rev. ed. 1997), p. 158.

Ray Chandler, Table of n, a(n) for n = 1..10000

Cf. A101704, A101705, A101706, A001232, A008918.

Select[2 Range[10^7], 3(#/2) == FromDigits@ Reverse@ IntegerDigits@ # &] (* Giovanni Resta, Apr 08 2017 *)

isok(n) = 3*n/2 == fromdigits(Vecrev(digits(n))); \\ Michel Marcus, Apr 09 2017

Data corrected by Giovanni Resta, Apr 08 2017

A371034 For n >= 1, a(n) = A004086(n) if A055483(n) = 1, otherwise a(n) = n / A055483(n).

Original entry on oeis.org

1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 4, 31, 41, 5, 61, 71, 2, 91, 10, 7, 1, 32, 4, 52, 13, 3, 14, 92, 10, 13, 23, 1, 43, 53, 4, 73, 83, 13, 10, 14, 7, 34, 1, 5, 23, 74, 4, 94, 10, 17, 25, 35, 6, 1, 65, 19, 85, 95, 10, 16, 31, 7, 32, 56, 1, 76, 34, 23, 10, 17, 8, 37, 47

Offset: 1

Ctibor O. Zizka, Mar 31 2024

Also a(n) = R(n) if (n, R(n)) are coprime, otherwise a(n) = n / GCD(n, R(n)), where R(n) is the digit reversal of n. a(n) = 1 for n from the union of A011557 and A002113 and A001232 and A008918.

			n = 13: A004086(13) = 31, A055483(13) = 1 thus a(13) = 31.
n = 15: A004086(15) = 51, A055483(15) = 3 thus a(15) = 15/3 = 5.

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A001232, A002113, A002275, A004086, A008918, A011557, A055483, A306273.

rev:= proc(n) local L,i;
  L:= convert(n,base,10);
  add(L[-i]*10^(i-1),i=1..nops(L))
end proc:
f:= proc(n) local r,g;
  r:= rev(n);
  g:= igcd(n,r);
  if g = 1 then r else n/g fi
end proc;
map(f, [$1..100]); # Robert Israel, Jul 09 2024

a[n_] := Module[{r = IntegerReverse[n], g}, g = GCD[n, r]; If[g == 1, r, n/g]]; Array[a, 100] (* Amiram Eldar, Mar 31 2024 *)

a(A011557(k)) = 1, k >= 0.
a(A002113(k)) = 1, k >= 2.
a(A001232(k)) = 1, k >= 1.
a(A008918(k)) = 1, k >= 1.

cp's OEIS Frontend

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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A124074 Consider the map that sends a number m to m - reverse(m) if that is nonnegative, otherwise to m + reverse(m). Sequence gives those numbers that reach a nonzero cycle under repeated application of this map.

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A222816 Numbers k that divide the base-5 reversal of k.

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A078273 Smallest multiple of n other than n using only the digits of n (no limit on frequency).

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A102277 Numbers n such that n = 15*reversal(n).

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A285040 Numbers n such that three-halves of n equals the reverse of n.

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A371034 For n >= 1, a(n) = A004086(n) if A055483(n) = 1, otherwise a(n) = n / A055483(n).

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