A001232 -id:A001232 - cp's OEIS frontend

A222817 Irregular triangle read by rows: row n gives list of nontrivial reverse multipliers for base n.

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

Offset: 3

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.
The last entry in each row is n-1; the number of terms in row n is A222819(n).

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

N. J. A. Sloane, Table giving n, list of nontrivial reverse multipliers, for n = 3..100
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, A222818, A222819, A222820.

See A214927 for other cross-references.

A061851 Digital representation of n contains only 1's and 0's, is palindromic and contains no singleton 1's or 0's.

Original entry on oeis.org

11, 111, 1111, 11111, 110011, 111111, 1100011, 1111111, 11000011, 11100111, 11111111, 110000011, 111000111, 111111111, 1100000011, 1100110011, 1110000111, 1111001111, 1111111111, 11000000011, 11001110011, 11100000111, 11110001111, 11111111111, 110000000011, 110001100011

Offset: 1

Henry Bottomley, May 10 2001

The terms can be constructed by gluing together terms from A355280 with their reversal as follows: The terms with odd length L = 2k-1 are given from the k-digit terms of A355280 by replacing the last digit with the reversal of the term. (Equivalently, concatenate with the reversal and delete one of the middle digits.) Terms with an even number L = 2k of digits are given as concatenation(m, reverse(m)) = m*10^L(m) + A004086(m) where m runs over the k-digit terms from A355280, and the (k-1)-digit terms with the 1's complement of the last digit appended. This explains the formula given in CROSSREFS for the number of terms of given length. - M. F. Hasler, Oct 17 2022

Ray Chandler, Table of n, a(n) for n = 1..10000 (first 608 terms from N. J. A. Sloane)
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]

Number of terms with k digits is Fibonacci(floor(k/2)) = A000045(A004526(k)).
Union of this sequence and twice this sequence is A061852 and 99 times that is A008919.
Cf. A214927, A001232, A222814, A008918, A008919, A222813 (terms converted to decimal).
Cf. A355280 (palindromic binary numbers with no digit run of length < 2).

concat(apply( {A061851_row(n)=[fromdigits(binary(m))|m<-A222813_row(n)]}, [1..11])) \\ M. F. Hasler, Oct 17 2022

def A061851_row(n): return [] if n < 2 else [10**n//9] if n < 6 else [
    m*10**(n//2) + A004086(m//10) for m in A355280_row(n//2+1)] if n&1 else [
    m*10**(n//2) + A004086(m) for m in sorted(A355280_row(n//2)+
                        [x*10+1-x%10 for x in A355280_row(n//2-1)])]
# M. F. Hasler, Oct 17 2022

a(n) = A001232(n)/99 = A008918(n)/198.

a(n) = A007088(A222813(n)), where A007088 = write in binary. - M. F. Hasler, Oct 06 2022

A101706 Numbers n such that reversal(n)=(7/3)*n.

Original entry on oeis.org

0, 3267, 32967, 329967, 3299967, 32673267, 32999967, 326703267, 329999967, 3267003267, 3296732967, 3299999967, 32670003267, 32967032967, 32999999967, 326700003267, 326732673267, 329670032967, 329967329967, 329999999967, 3267000003267, 3267329673267, 3296700032967, 3299670329967, 3299999999967

Offset: 1

Farideh Firoozbakht, Jan 01 2005

If m is in the sequence then all numbers of the form g(m,s,t) for nonnagative integers s and t are in the sequence (the function g has been defined in the sequence A101704), for example g(3267,1,1)= 326703267 is in the sequence. If n=0 or n>1 then 33*(10^n-1) is in the sequence.

There are Fibonacci(floor((n-2)/2)) terms with n digits, n>1 (this is essentially A103609). - Ray Chandler, Oct 12 2017

			g(3267,10,2) = 32670000000000326700000000003267 is in the sequence
because reversal(32670000000000326700000000003267) =
76230000000000762300000000007623 =
(7/3)*32670000000000326700000000003267, g(3267,0,4) =
32673267326732673267 is in the sequence because
reversal(32673267326732673267) = 76237623762376237623 =
(7/3)*32673267326732673267.

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

Cf. A101704, A101705, A001232, A008918.

Do[If[FromDigits[Reverse[IntegerDigits[n]]] == (7/3)*n, Print[n]], {n, 100000000}]

Terms a(8) onward from Max Alekseyev, Aug 18 2013

A222811 Number of n-digit numbers N such that the reversal of N divides N.

Original entry on oeis.org

9, 18, 111, 212, 1128, 2067, 11123, 20270, 110440, 200971, 1101475, 2003592, 11005388

Offset: 1

N. J. A. Sloane, Mar 10 2013

Suggested by A214927.

Even terms are roughly double the previous term. - David Consiglio, Jr., Mar 22 2013

			Some of the smallest solutions are:
[1, 2, 3, 4, 5, 6, 7, 8, 9] (so a(1) = 9),
[10, 11, 20, 22, 30, 33, 40, 44, 50, 55, 60, 66, 70, 77, 80, 88, 90, 99] (so a(2) = 18),
[100, 101, 110, 111, 121, 131, 141, 151, 161, 171, 181, 191, 200, 202, 212, 220, 222, 232, 242, 252, 262, 272, 282, 292, 300, 303, 313, 323, 330, 333, 343, 353, ...]

David Consiglio, Jr., C++ program
N. J. A. Sloane, 2178 And All That, Fib. Quart., 52 (2014), 99-120.
N. J. A. Sloane, 2178 And All That [Local copy]

Cf. A214927, A222809, A222810, A222812.

a(8) - a(11) from David Consiglio, Jr., Mar 22 2013

a(12)-a(13) from Giovanni Resta, Apr 01 2013

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

Original entry on oeis.org

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.

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

A193434 6*n/5 = (n written backwards), n > 0.

Original entry on oeis.org

45, 495, 4545, 4995, 45045, 49995, 450045, 454545, 495495, 499995, 4500045, 4549545, 4950495, 4999995, 45000045, 45045045, 45454545, 45499545, 49500495, 49545495, 49954995, 49999995, 450000045, 450495045, 454504545, 454999545, 495000495, 495495495, 499504995

Offset: 1

Arkadiusz Wesolowski, Aug 01 2011

			495 belongs to this sequence because 6*495/5 = 594.

Ray Chandler, Table of n, a(n) for n = 1..10000 (first 4000 terms from Arkadiusz Wesolowski)

Cf. A001232, A008918, A057148, A101705.

Rest@Select[FromDigits /@ Tuples[{0, 45}, 8], IntegerDigits[6*#/5] == Reverse@IntegerDigits[#] &] (* Arkadiusz Wesolowski, Aug 14 2012 *)

def A193434(n):
    a = 1<<(m:=n+1).bit_length()-2
    s = bin(a|(m&a-1))[2:]
    return 45*int(s+(s[::-1] if a&m else s[-2::-1])) # Chai Wah Wu, Jul 23 2024

a(n) = 45*A057148(n+1). - Ray Chandler, Oct 09 2017

cp's OEIS Frontend

A222817 Irregular triangle read by rows: row n gives list of nontrivial reverse multipliers for base n.

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A061851 Digital representation of n contains only 1's and 0's, is palindromic and contains no singleton 1's or 0's.

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PARI

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A101706 Numbers n such that reversal(n)=(7/3)*n.

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A222811 Number of n-digit numbers N such that the reversal of N divides N.

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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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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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Maple

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

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