A214927 -id:A214927 - cp's OEIS frontend

A222819 a(n) = number of nontrivial reverse multipliers for base n.

0, 1, 1, 2, 2, 2, 4, 3, 2, 5, 4, 3, 6, 6, 3, 7, 5, 7, 7, 6, 5, 10, 10, 5, 7, 8, 5, 12, 11, 9, 12, 5, 8, 13, 9, 8, 12, 16, 8, 14, 11, 12, 16, 12, 10, 19, 15, 11, 11, 9, 10, 19, 18, 17, 18, 13, 9, 23, 14, 15, 21, 19, 14, 19, 12, 18, 16, 19, 17, 26, 17, 11, 20, 16, 15, 21, 13, 26, 24, 13

Offset: 2

N. J. A. Sloane, Mar 13 2013

If there is a number m such that the reversal of m in base n is c times m, then c is called a reverse multiplier for n. For example, 2 is a reverse multiplier for base n=5, since 8 (base 10) = 13 (base 5), and 2*8 = 16 (base 10) = 31 (base 5).

The trivial reverse multiplier 1 is excluded.

N. J. A. Sloane, Table of n, a(n) for n = 2..100
N. J. A. Sloane, 2178 And All That, Fib. Quart., 52 (2014), 99-120.
N. J. A. Sloane, 2178 And All That [Local copy]
Anne Ludington Young, k-Reverse multiples, Fib. Q., 30 (1992), 126-132.

Cf. A214927, A222817, A222818, A222820.

See A214927 for other cross-references.

A222820 a(n) is the number of reverse multipliers for base n.

Original entry on oeis.org

1, 2, 2, 3, 3, 3, 5, 4, 3, 6, 5, 4, 7, 7, 4, 8, 6, 8, 8, 7, 6, 11, 11, 6, 8, 9, 6, 13, 12, 10, 13, 6, 9, 14, 10, 9, 13, 17, 9, 15, 12, 13, 17, 13, 11, 20, 16, 12, 12

Offset: 2

N. J. A. Sloane, Mar 13 2013

If there is a number m such that the reversal of m in base n is c times m, then c is called a reverse multiplier for n. For example, 2 is a reverse multiplier for base n=5, since 8 (base 10) = 13 (base 5), and 2*8 = 16 (base 10) = 31 (base 5).
The trivial reverse multiplier 1 is included.
a(n)-1 is the length of row n of A222817. - Michel Marcus, Apr 12 2020

For a complete list of references and links related to this problem see A214927.

N. J. A. Sloane, 2178 And All That, arXiv:1307.0453 [math.NT], 2013; see also, Fib. Quart., 52 (2014), 99-120.
N. J. A. Sloane, 2178 And All That [Local copy]
Anne Ludington Young, k-Reverse multiples, Fib. Q., 30 (1992), 126-132.

Cf. A214927, A222817, A222818, A222819.

See A214927 for other cross-references.

A173951 Positive integers with the property that if the base-3 representation is reversed the result is twice the original number.

Original entry on oeis.org

32, 104, 320, 968, 2624, 2912, 7808, 8744, 23360, 25376, 26240, 70016, 75920, 78728, 209984, 212576, 227552, 233600, 236192, 629888, 638312, 682448, 700160, 708584, 1889600, 1897376, 1915520, 2047136, 2054912, 2099840, 2117984, 2125760

Offset: 1

John W. Layman, Mar 03 2010

The number of terms of this sequence containing n ternary digits is given by {d(n)}={0,0,0,1,1,1,1,2,2,3,3,5,5,8,8,13,13,21,...} for n=1,2,3,... and thus appears to be essentially the doubling-up of the Fibonacci numbers A103609. For example, 2624 = 10121012(base-3) and 2912 = 10222212(base-3) are the only two terms that have 8 digits when written in base 3, so d(8)=2.
(This conjecture is correct - see A223077. - N. J. A. Sloane, Mar 19 2013)
All terms of sequence A173952, defined by b(1)=32 and, for n>1, b(n)=9*b(n-1)+32, appear to be terms of the above sequence {a(n)}; in fact each term b(n) appears to be the largest term of {a(k)} that has 2n+2 digits when written in base 3.

Cf. A103609, A173952, A223077, A214927.

A222809 Number of n-digit numbers N such that the reversal of N divides N but is different from N.

Original entry on oeis.org

0, 9, 21, 122, 228, 1167, 2123, 11270, 20440, 110971, 201475, 1103592, 2005388

Offset: 1

N. J. A. Sloane, Mar 10 2013

Suggested by A214927.
Conjecture: a(n) = A222811(n) - 9*10^floor((n-1)/2). - Lars Blomberg, Jul 03 2014
Proof of this conjecture: A number N with n digits which equals its own reversal has the first and identical last digit from {1, 2, ..., 9}. If n is even the other n-2 digits come in n/2 - 1 pairs of equal numbers from {0, 1, ..., 9}. If n is odd with n >= 3 then the other n-2 numbers come in (n-3)/2 pairs of equal numbers from {0, 1, ..., 9} and an additional middle digit also from {0, 1, ..., 9}. Therefore there are 9*10^(n/2-1) such numbers N for even n, and 9*10^((n-1)/2) for odd n, fitting 9*10^floor((n-1)/2). - Wolfdieter Lang, Jul 13 2014

			Some of the smallest solutions are:
[10, 20, 30, 40, 50, 60, 70, 80, 90] (so a(2) = 9),
[100, 110, 200, 220, 300, 330, 400, 440, 500, 510, 540, 550, 600, 660, 700, 770, 800, 810, 880, 900, 990] (so a(3) = 21),
[1000, 1010, 1100, 1110, 1210, 1310, 1410, ...].

Cf. A214927, A222810, A222811, A222812.

a(n) = sum(i=10^(n-1), 10^n-1, (irev=eval(concat(Vecrev(Str(i))))) && irev!=i && !(i % irev)); \\ Michel Marcus, Jul 03 2014

a(7)-a(12) from Lars Blomberg, Jul 03 2014

a(13) from Giovanni Resta, Aug 15 2019

A222812 Number of n-digit numbers N such that the number formed by some nontrivial permutation of the digits of N divides N.

Original entry on oeis.org

0, 18, 329, 5000, 65931, 779504, 8517616, 88555255, 897147508, 8997325290, 90000000000, 900000000000, 9000000000000, 90000000000000, 900000000000000, 9000000000000000, 90000000000000000, 900000000000000000, 9000000000000000000, 90000000000000000000

Offset: 1

N. J. A. Sloane, Mar 10 2013

Suggested by A214927.

			Some of the smallest solutions are:
[10, 11, 20, 22, 30, 33, 40, 44, 50, 55, 60, 66, 70, 77, 80, 88, 90, 99] (so a(2) = 18),
[100, 101, 105, 108, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120, 121, 122, 130, 131, 133, 140, 141, 144, 150, 151, 155, 160, 161, 166, 170, 171, 177, 180, 181, 188, 190, 191, 199, 200, 202, 210, 211, 212, 220, 221, 222, 223, 224, ...]
Note that 11 is in the sequence because permuting the two digits gives 11, and 11 divides 11.

Index entries for linear recurrences with constant coefficients, signature (10).

Cf. A214927, A222809, A222810, A222811.

a(n) = 9 * 10^(n-1) for n >= 11. - Hiroaki Yamanouchi, Sep 03 2014.

G.f.: x^2*(26747100*x^9 +25850210*x^8 +11594958*x^7 +3379095*x^6 +722576*x^5 +120194*x^4 +15931*x^3 +1710*x^2 +149*x +18)/(1-10*x). - Robert Israel, Sep 03 2014

a(7)-a(20) from Hiroaki Yamanouchi, Sep 03 2014

A222818 Irregular triangle read by rows: row n gives list of reverse multipliers for base n.

Original entry on oeis.org

1, 1, 2, 1, 3, 1, 2, 4, 1, 2, 5, 1, 3, 6, 1, 2, 3, 5, 7, 1, 2, 4, 8, 1, 4, 9, 1, 2, 3, 5, 7, 10, 1, 2, 3, 5, 11, 1, 5, 6, 12, 1, 2, 3, 4, 6, 9, 13, 1, 2, 3, 4, 7, 11, 14, 1, 3, 7, 15, 1, 2, 4, 5, 8, 10, 11, 16, 1, 2, 5, 7, 8, 17, 1, 3, 4, 6, 7, 9, 14, 18, 1, 2, 3, 4, 6, 9, 13, 19

Offset: 2

N. J. A. Sloane, Mar 13 2013

If there is a number m such that the reversal of m in base n is c times m, then c is called a reverse multiplier for n. For example, 2 is a reverse multiplier for base n=5, since 8 (base 10) = 13 (base 5), and 2*8 = 16 (base 10) = 31 (base 5).
The trivial reverse multiplier 1 is included.
The last entry in each row is n-1; the number of terms in row n is A222820(n).

			Triangle begins:
  1,
  1,2,
  1,3,
  1,2,4,
  1,2,5,
  1,3,6,
  1,2,3,5,7,
  1,2,4,8,
  1,4,9,
  1,2,3,5,7,10,
  1,2,3,5,11,
  1,5,6,12,
  1,2,3,4,6,9,13,
  1,2,3,4,7,11,14,
  1,3,7,15
 ...

For a complete list of references and links related to this problem see A214927.

N. J. A. Sloane, Table giving n, number of nontrivial reverse multipliers, list of nontrivial reverse multipliers, for n = 3..50
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]
Anne Ludington Young, k-Reverse multiples, Fib. Q., 30 (1992), 126-132.

Cf. A214927, A222817, A222819, A222820.

See A214927 for other cross-references.

A226516 Number of (18,7)-reverse multiples with n digits.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 3, 3, 4, 4, 6, 5, 8, 6, 10, 8, 13, 11, 17, 15, 23, 20, 31, 26, 41, 34, 54, 45, 71, 60, 94, 80, 125, 106, 166, 140, 220, 185, 291, 245, 385, 325, 510, 431, 676, 571, 896, 756, 1187, 1001, 1572, 1326, 2082, 1757, 2758, 2328, 3654, 3084, 4841, 4085, 6413

Offset: 0

N. J. A. Sloane, Jun 16 2013

Comment from Emeric Deutsch, Aug 21 2016 (Start):
Given an increasing sequence of positive integers S = {a0, a1, a2, ... }, let
          F(x) = x^{a0} + x^{a1} + x^{a2} + ... .
Then the g. f. for the number of palindromic compositions of n with parts in S is (see Hoggatt and Bicknell, Fibonacci Quarterly, 13(4), 1975):
      (1 + F(x))/(1 - F(x^2))
Playing with this, I have found easily that
1. number of palindromic compositions of n into {3,4,5,...} = A226916(n+4);
2. number of palindromic compositions of n into {1,4,7,10,13,...} = A226916(n+6);
3. number of palindromic compositions of n into {1,4} = A226517(n+10);
4. number of palindromic compositions of n into {1,5} = A226516(n+11).
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
V. E. Hogatt, M. Bicknell, Palindromic Compositions, Fib. Quart. 13(4) (1975) 350-356
N. J. A. Sloane, 2178 And All That, Fib. Quart., 52 (2014), 99-120.
N. J. A. Sloane, 2178 And All That [Local copy]
Index entries for linear recurrences with constant coefficients, signature (0,1,0,0,0,0,0,0,0,1).

Cf. A214927, A226517, A226916.

f:=proc(n) option remember;
if
n <= 5 then 0
elif n=6 then 1
elif n <= 10 then 0
elif n <= 12 then 1
else f(n-2)+f(n-10)
fi;
end;
[seq(f(n),n=0..100)]

CoefficientList[Series[x^6 (1 - x^2 + x^5 + x^6) / (1 - x^2 - x^10), {x, 0, 80}], x] (* Vincenzo Librandi, Jun 18 2013 *)
LinearRecurrence[{0,1,0,0,0,0,0,0,0,1},{0,0,0,0,0,0,1,0,0,0,0,1,1},80] (* Harvey P. Dale, Jun 17 2015 *)

G.f.: x^6*(1+x)*(1-x+x^5)/(1-x^2-x^10).
a(n) = a(n-2) + a(n-10) for n>12, with initial values a(0)-a(12) equal to 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1. [Bruno Berselli, Jun 17 2013]
a(2n+1) = A003520(n-5). a(2n) = A098523(n-6). - R. J. Mathar, Dec 13 2022

A226517 Number of (19,14)-reverse multiples with n digits.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 2, 1, 2, 1, 3, 2, 4, 3, 6, 4, 8, 5, 11, 7, 15, 10, 21, 14, 29, 19, 40, 26, 55, 36, 76, 50, 105, 69, 145, 95, 200, 131, 276, 181, 381, 250, 526, 345, 726, 476, 1002, 657, 1383, 907, 1909, 1252, 2635, 1728, 3637, 2385, 5020, 3292, 6929, 4544, 9564, 6272, 13201, 8657, 18221

Offset: 0

N. J. A. Sloane, Jun 16 2013

Comment from Emeric Deutsch, Aug 21 2016 (Start):
Given an increasing sequence of positive integers S = {a0, a1, a2, ... }, let
          F(x) = x^{a0} + x^{a1} + x^{a2} + ... .
Then the g. f. for the number of palindromic compositions of n with parts in S is (see Hoggatt and Bicknell, Fibonacci Quarterly, 13(4), 1975, 350 - 356):
      (1 + F(x))/(1 - F(x^2))
Playing with this, I have found easily that
1. number of palindromic compositions of n into {3,4,5,...} = A226916(n+4);
2. number of palindromic compositions of n into {1,4,7,10,13,...} = A226916(n+6);
3. number of palindromic compositions of n into {1,4} = A226517(n+10);
4. number of palindromic compositions of n into {1,5} = A226516(n+11).
(End)

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
V. E. Hogatt, M. Bicknell, Palindromic Compositions, Fib. Quart. 13(4) (1975) 350-356
N. J. A. Sloane, 2178 And All That, Fib. Quart., 52 (2014), 99-120.
N. J. A. Sloane, 2178 And All That [Local copy]
Index entries for linear recurrences with constant coefficients, signature (0,1,0,0,0,0,0,1).

Cf. A214927, A226516, A226916.

f:=proc(n) option remember;
if
n <= 5 then 0
elif n=6 then 1
elif n <= 9 then 0
elif n <= 11 then 1
else f(n-2)+f(n-8)
fi;
end;
[seq(f(n),n=0..120)];

CoefficientList[Series[x^6 (1 - x^2 + x^4 + x^5) / (1 - x^2 - x^8), {x, 0, 80}], x] (* Vincenzo Librandi, Jun 18 2013 *)
LinearRecurrence[{0,1,0,0,0,0,0,1},{0,0,0,0,0,0,1,0,0,0,1,1},80] (* Harvey P. Dale, Aug 23 2019 *)

G.f.: x^6*(1+x)*(1-x+x^4)/(1-x^2-x^8).
a(n) = a(n-2) + a(n-8) for n>11, with initial values a(0)-a(11) =  0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1. [Bruno Berselli, Jun 17 2013]
a(2n+1)=A003269(n-4). a(2n)=A103632(n-3). - R. J. Mathar, Dec 13 2022

A223077 Positive integers, written in base 3, with the property that if the base-3 representation is reversed the result is twice the original number.

Original entry on oeis.org

1012, 10212, 102212, 1022212, 10121012, 10222212, 101201012, 102222212, 1012001012, 1021210212, 1022222212, 10120001012, 10212010212, 10222222212, 101200001012, 101210121012, 102120010212, 102212102212, 102222222212, 1012000001012, 1012102121012, 1021200010212, 1022120102212, 1022222222212

Offset: 1

N. J. A. Sloane, Mar 14 2013

For the decimal representations of these same numbers see A173951.

Theorem: The number of terms of length n is equal to A103609(n-2).

N. J. A. Sloane, Table of n, a(n) for n = 1..108

Cf. A173951, A214927.

b3rQ[n_]:=FromDigits[Reverse[IntegerDigits[n]],3]/FromDigits[ IntegerDigits[ n],3] ==2; Select[FromDigits/@Tuples[{0,1,2},13],b3rQ]//Quiet (* Harvey P. Dale, Jun 10 2018 *)

A223080 Numbers n with distinct digits such that the reversal of n divides n.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 30, 40, 50, 60, 70, 80, 90, 510, 540, 810, 5610, 5940, 8712, 8910, 9801, 65340, 87120, 87912, 659340, 879120

Offset: 1

N. J. A. Sloane, Mar 20 2013

Suggested by A214927.
There are no terms with 7 digits.
Sequence is complete. - Giovanni Resta, Mar 20 2013

			8712 reversed is 2178, which divides 8712.

Cf. A222810, A214927, A223081, A223082.

Select[Range[10^6],Max[DigitCount[#]]==1&&Divisible[#,IntegerReverse[#]]&] (* Harvey P. Dale, Jun 21 2022 *)

cp's OEIS Frontend

A222819 a(n) = number of nontrivial reverse multipliers for base n.

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A222820 a(n) is the number of reverse multipliers for base n.

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A173951 Positive integers with the property that if the base-3 representation is reversed the result is twice the original number.

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A222809 Number of n-digit numbers N such that the reversal of N divides N but is different from N.

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A222812 Number of n-digit numbers N such that the number formed by some nontrivial permutation of the digits of N divides N.

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A222818 Irregular triangle read by rows: row n gives list of reverse multipliers for base n.

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A226516 Number of (18,7)-reverse multiples with n digits.

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A226517 Number of (19,14)-reverse multiples with n digits.

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Maple

Mathematica

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A223077 Positive integers, written in base 3, with the property that if the base-3 representation is reversed the result is twice the original number.

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