A061917 -id:A061917 - cp's OEIS frontend

A338827 For any number with decimal representation (d(1), d(2), ..., d(k)), the decimal representation of a(n) is (abs(d(1)-d(k)), abs(d(2)-d(k-1)), ..., abs(d(k)-d(1))).

0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 11, 0, 11, 22, 33, 44, 55, 66, 77, 88, 22, 11, 0, 11, 22, 33, 44, 55, 66, 77, 33, 22, 11, 0, 11, 22, 33, 44, 55, 66, 44, 33, 22, 11, 0, 11, 22, 33, 44, 55, 55, 44, 33, 22, 11, 0, 11, 22, 33, 44, 66, 55, 44, 33, 22, 11, 0, 11, 22

Offset: 0

Rémy Sigrist, Nov 11 2020

Leading zeros are ignored.

All terms belong to A061917.

			For n = 1021:
- abs(1-1) = 0,
- abs(0-2) = 2,
- abs(2-0) = 2,
- abs(1-1) = 0,
- so a(1021) = 220.

Rémy Sigrist, Table of n, a(n) for n = 0..10000

Cf. A002113, A004086, A056965, A061917, A175919 (binary analog), A330240, A338828 (ternary analog).

a:= n-> (l-> (h-> add(h[j]*10^(j-1), j=1..nops(h)))([seq(
    abs(l[i]-l[-i]), i=1..nops(l))]))(convert(n, base, 10)):
seq(a(n), n=0..70);  # Alois P. Heinz, Nov 12 2020

a(n, base=10) = my (d=digits(n, base)); fromdigits(abs(d-Vecrev(d)), base)

a(n) = 0 iff n is a palindrome (A002113).

a(n) = A330240(n, A004086(n)).

A069024 Numbers that are palindromic in base 2 as well as in base 10 (initial zeros may be prepended).

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 20, 30, 33, 40, 60, 66, 80, 90, 99, 252, 272, 292, 313, 330, 585, 626, 660, 717, 990, 2112, 2720, 2772, 2920, 4224, 5850, 6336, 7447, 7470, 8448, 8580, 9009, 15351, 21120, 22122, 25752, 32223, 39993, 40904, 42240, 44244, 48384

Offset: 1

Amarnath Murthy, Apr 02 2002

			66 in base 2 is 1000010, which is palindromic if rewritten as 01000010.

Robert Israel, Table of n, a(n) for n = 1..243

Cf. A007632.

Intersection of A061917 and A057890.

nextpal:= proc(p,d,V,b)
  local i,i2,pp,m,m2;
  pp:=p;
  V[1]:= V[1]+1;
  m2:= floor(d/2);
  i2:= ceil(d/2);
  if d::odd then pp:= pp + b^m2 else pp:= pp + b^m2 + b^(m2-1) fi;
  for i from 1 while V[i] = b do
    V[i]:= 0:
    if i = i2 then
      if d::even then
        ArrayTools:-Extend(V,[1],inplace);
        return b^d+1, d+1, V
      else
        V[i2]:= 1;
        return b^d+1, d+1, V;
      fi;
    fi;
    V[i+1]:= V[i+1]+1;
    if (d::odd and i=1) then pp:= pp + b^(i2-i-1) else
      pp:= pp + b^(i2-i-1) - b^(i2-i+1) fi;
  od;
  return pp, d, V
end proc:
count:= 1:
S:= 0:
p2[0]:=1: V2[0]:= <1>: d2[0]:= 1:m2:= 0:
p10[0]:= 1: V10[0]:= <1>: d10[0]:= 1: m10:= 0:
while count < 100 do
  i2:= min[index]([seq(p2[i],i=0..m2)])-1; p2o:= p2[i2];
  i10:= min[index]([seq(p10[i],i=0..m10)])-1; p10o:= p10[i10];
  if p2o = p10o then
    S:= S, p2o; count:= count+1;
  fi;
  if p2o <= p10o then x, d2[i2], V2[i2]:= nextpal(p2o/2^i2, d2[i2], V2[i2],2); p2[i2]:= 2^i2 *x;
    if i2 = m2 then m2:= m2+1; p2[m2]:= 2^m2; V2[m2]:= <1>; d2[m2]:= 1;
 fi;
  else
    x, d10[i10], V10[i10]:= nextpal(p10o/10^i10, d10[i10], V10[i10],10);
    p10[i10]:= 10^i10 * x;
    if i10 = m10 then m10:= m10+1; p10[m10]:= 10^m10; V10[m10]:= <1>; d10[m10]:= 1
fi fi od:
S; # Robert Israel, Apr 01 2024

pal[n_, b_] := (z=IntegerDigits[n, b]) == Reverse[z]; extpal[n_, b_] := If[Mod[n, b]>0, pal[n, b], extpal[n/b, b]]; Select[Range[50000], extpal[ #, 10]&&extpal[ #, 2]&]

Edited by Dean Hickerson, Apr 06 2002

0 inserted by Sean A. Irvine, Mar 29 2024

A225416 Number of iterations of the map n -> f(n) needed to reach 0 and starting with n, where f(n) is given by the following definition: f(n) = u(n) mod v(n) where u(n) = max (n, reverse(n)) and v(n) = min(n, reverse(n)).

Original entry on oeis.org

0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 3, 2, 3, 2, 2, 3, 1, 2, 1, 2, 3, 2, 2, 3, 3, 2, 1, 2, 2, 1, 2, 3, 4, 5, 2, 3, 1, 3, 3, 2, 1, 2, 3, 4, 5, 3, 1, 2, 2, 3, 2, 1, 2, 3, 4, 5, 1, 3, 2, 4, 3, 2, 1, 2, 3, 4, 1, 2, 3, 5, 4, 3, 2, 1, 2, 3, 1, 2, 3, 2, 5, 4, 3

Offset: 0

Michel Lagneau, May 07 2013

The fixed points are in A061917 (either a palindrome or becomes a palindrome if trailing 0's are omitted). The number of iterations needed to reach a fixed point equals a(n) - 1 with n > 0, and 0 for n = 0.
The smallest k such that a(k) = n iterations are {0, 1, 12, 14, 36, 37, 103, 118, 238, 257, 1282, 2165, 2459, 11908, 100673, 113233, 144104, 300768, 1329025, ...}, and it seems that 3*log_10(k)/n ~ 1 where n tends into infinity.
The Maple program below gives two sequences: the number of iterations of this sequence and the fixed points in increasing order (sequence A061917).

			The trajectory of 37 is 37 -> 36 -> 27 -> 18 -> 9 -> 0, so a(37) = 5. The fixed point is 9 = A061917(10).
73 mod 37 = 36, 63 mod 36 = 27, 72 mod 27 = 18, 81 mod 18 = 9 and 9 mod 9 = 0.

Michel Lagneau, Table of n, a(n) for n = 0..10000

Cf. A061917.

lst1:={}:for n from 1 to 494 do:nn:=n:ii:=0:r:=1:lst:={n}:for it from 1 to 20 while(r<>0) do: V:=convert(nn, base, 10): n1:=nops(V):s:=0:for a from n1 by -1  to 1 do:s:=s+V[a]*10^(n1-a): od:m1:=min(nn,s):m2:=max(nn,s):r:=irem(m2,m1): lst:=lst union {r}:nn:=r: od: printf(`%d, `,it-1): lst1:=lst1 union { lst[2]}: od:print(lst1):

A233574 a(n) is the smallest term of either A233010 or A233572 such that |n-a(n)|

Original entry on oeis.org

0, 1, 2, 2, 4, 3, 4, 6, 8, 6, 6, 8, 8, 7, 12, 8, 10, 9, 10, 13, 12, 13, 16, 20, 16, 18, 26, 18, 18, 20, 18, 18, 20, 20, 18, 26, 20, 24, 26, 21, 24, 21, 26, 43, 32, 24, 28, 26, 28, 43, 30, 27, 28, 27, 28, 54, 30, 39, 40, 32, 32, 43, 32, 39, 40, 39, 40, 43, 36

Offset: 0

Lei Zhou, Dec 13 2013

It is conjectured that a(n) exists for all n >= 0.

			a(19)=13 since 19=13+6=A233010(9)+A233572(3) and 13>6. There is no number in A233010 or A233572 smaller than 13 that satisfies the same condition.

Lei Zhou, Table of n, a(n) for n = 0..10000

Cf. A002113, A061917, A006995, A057890, A134027, A233010, A233571, A233572, A233573

BTDigits[m_Integer, g_] :=
(*This is to determine digits of a number in balanced ternary notation.*)
Module[{n = m, d, sign, t = g},
  If[n != 0, If[n > 0, sign = 1, sign = -1; n = -n];
   d = Ceiling[Log[3, n]]; If[3^d - n <= ((3^d - 1)/2), d++];
   While[Length[t] < d, PrependTo[t, 0]];
   t[[Length[t] + 1 - d]] = sign;
   t = BTDigits[sign*(n - 3^(d - 1)), t]]; t];
BTpaleQ[n_Integer] :=
(*This is to query if a number is an element of sequence A233010.*)
Module[{t, trim = n/3^IntegerExponent[n, 3]},
  t = BTDigits[trim, {0}]; t == Reverse[t]];
BTrteQ[n_Integer] :=
(*This is to query if a number is an element of sequence A233572.*)
Module[{t, trim = n/3^IntegerExponent[n, 3]},
  t = BTDigits[trim, {0}]; DeleteDuplicates[t + Reverse[t]] == {0}];
sa = Select[Range[0, 30000], BTpaleQ[#] &];
(*This is to generate a limited list of A233010.*)
sb = Select[Range[0, 30000], BTrteQ[#] &];
(*This is to generate a limited list of A233572.*)
range = 68; Table[i1 = 0; i2 = 0;
While[If[sa[[i1 + 1]] < sb[[i2 + 1]], i1++; nh = sa[[i1]]; isa = 1,
   i2++; nh = sb[[i2]]; isa = 0]; (2*nh) < n];
While[If[isa == 0, chk = MemberQ[sa, Abs[n - nh]],
   chk = MemberQ[sb, Abs[n - nh]]]; ! chk,
  If[sa[[i1 + 1]] < sb[[i2 + 1]], i1++; nh = sa[[i1]]; isa = 1, i2++;
   nh = sb[[i2]]; isa = 0]];
If[isa == 0, m = sb[[i2]], m = sa[[i1]]]; m, {n, 0, range}]

A233580 In balanced ternary notation, zerofree non-repdigit numbers that are either palindromes or sign reversed palindromes.

Original entry on oeis.org

2, 7, 16, 20, 32, 43, 61, 103, 124, 146, 182, 196, 292, 302, 338, 367, 421, 547, 601, 859, 913, 1039, 1096, 1172, 1280, 1312, 1600, 1640, 1748, 1816, 2560, 2624, 2732, 2776, 3064, 3092, 3200, 3283, 3445, 3823, 3985, 4759, 4921, 5299, 5461, 7663, 7825, 8203

Offset: 1

Lei Zhou, Dec 14 2013

Zerofree numbers in balanced ternary notation can be used as reversible sign operators. This sequence collects such operators that are either in palindrome form or sign reversed palindrome form (which is defined as (n)_bt+Reverse((n)_bt)=0).

			2 = (1T)_bt in balanced ternary notation, where we use T to represent -1.
1T + T1 = 0, matches the definition of sign reversed palindrome form. So 2 is in the sequence.
Other examples:
7 = (1T1_bt) - palindrome; in the sequence.
13 = (111)_bt - palindrome but repdigit; not in the sequence.
16 = (1TT1)_bt - palindrome; in the sequence.
...
52 = (1T0T1)_bt - palindrome but not zerofree; not in the sequence.

Lei Zhou, Table of n, a(n) for n = 1..10000

Cf. A002113, A061917, A006995, A057890, A134027, A233010, A233571, A233572.

BTDigits[m_Integer, g_] :=
(* This is to determine digits of a number in balanced ternary notation. *)
Module[{n = m, d, sign, t = g},  If[n != 0, If[n > 0, sign = 1,
  sign = -1; n = -n];   d = Ceiling[Log[3, n]]; If[3^d - n <= ((3^d - 1)/2), d++];   While[Length[t] < d, PrependTo[t, 0]]; t[[Length[t] + 1 - d]] = sign;   t = BTDigits[sign*(n - 3^(d - 1)), t]]; t];
BTnum[g_]:=Module[{bo=Reverse[g],data=0,i},Do[data=data+3^(i-1)*bo[[i]],{i,1,Length[bo]}];data];
ct=0;n=0;dg=0;spool={};res={};While[ct<50,n++; nbits = BTDigits[n, {0}];cdg=Length[nbits];If[cdg>dg,If[Length[spool]>0,Do[bits=spool[[j]];If[!MemberQ[bits,0],rb=Reverse[bits]; sign=rb[[1]];bo=Join[bits,-sign*rb];If[MemberQ[bo,-1],data=BTnum[bo];ct++;AppendTo[res,data]];bo=Join[bits,sign*rb];If[MemberQ[bo,-1],data=BTnum[bo];ct++;AppendTo[res,data]]],{j,1,Length[spool]}];Do[bits=spool[[j]];If[!MemberQ[bits,0],rb=Reverse[bits];bo=Join[bits,{-1},rb];If[MemberQ[bo,-1],data=BTnum[bo];ct++;AppendTo[res,data]];bo=Join[bits,{1},rb];If[MemberQ[bo,-1],data=BTnum[bo];ct++;AppendTo[res,data]]],{j,1,Length[spool]}];spool={};dg=cdg]];AppendTo[spool,nbits]];Print[res]

A325152 Numbers whose squares can be expressed as the product of a number and its reversal.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 20, 22, 30, 33, 40, 44, 50, 55, 60, 66, 70, 77, 80, 88, 90, 99, 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, 363, 373, 383, 393, 400, 403, 404, 414, 424, 434

Offset: 1

Bernard Schott, Apr 11 2019

The corresponding squares are in A325148 and the numbers k such that k * rev(k) is a square are in A306273.
The squares of the first 47 terms of this sequence (from 0 to 242) can be expressed as the product of a number and its reversal in only one way; then a(48) = 252 and 252^2 = 252 * 252 = 144 * 441.
The first 65 terms of this sequence (from 0 to 400) are exactly the first 65 terms of A061917; then a(66) = 403, non-palindrome, is the first term of the sequence A325151.

			One way: 20^2 = 400 = 200 * 2.
Two ways: 2772^2 = 7683984 = 2772 * 2772 = 1584 * 4851.
Three ways: 2520^2 = 14400 * 441 = 25200 * 252 = 44100 * 144.
403 is a member since 403^2 = 162409 = 169*961 (note that 403 is not a member of A281625).

Cf. A325148, A325149, A083408, A325150, A307019.
Cf. also A061917, A325151.
Similar to but different from A281625.

a(n) = sqrt(A325148(n)).

cp's OEIS Frontend

A338827 For any number with decimal representation (d(1), d(2), ..., d(k)), the decimal representation of a(n) is (abs(d(1)-d(k)), abs(d(2)-d(k-1)), ..., abs(d(k)-d(1))).

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A069024 Numbers that are palindromic in base 2 as well as in base 10 (initial zeros may be prepended).

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A225416 Number of iterations of the map n -> f(n) needed to reach 0 and starting with n, where f(n) is given by the following definition: f(n) = u(n) mod v(n) where u(n) = max (n, reverse(n)) and v(n) = min(n, reverse(n)).

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A233574 a(n) is the smallest term of either A233010 or A233572 such that |n-a(n)|

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A233580 In balanced ternary notation, zerofree non-repdigit numbers that are either palindromes or sign reversed palindromes.

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A325152 Numbers whose squares can be expressed as the product of a number and its reversal.

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