A062128 -id:A062128 - cp's OEIS frontend

A066057 'Reverse and Add' carried out in base 2 (cf. A062128); number of steps needed to reach a palindrome, or -1 if no palindrome is ever reached.

0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 1, 2, 1, 2, 1, 0, 1, 0, 1, 4, 5, 0, -1, 2, 1, 4, -1, 0, -1, 2, 1, 0, 1, 0, 1, -1, 1, -1, 1, 2, 1, -1, 1, 2, 3, 0, -1, -1, 1, -1, 3, 0, 1, 2, 3, 2, 1, 2, 3, 2, -1, -1, 1, 0, 1, 0, 1, -1, 1, 2, 1, 4, 3, 0, 11, -1, 5, -1, -1, 2, 1, 2, 1, 4, -1, 0, -1, 2, 5, -1, -1, 2, 3, 0, -1, -1, 1, -1, 3, 0, 1, 4, 1, 10, 11, -1, -1, 0, -1, 2, -1, 4

Offset: 0

Klaus Brockhaus, Dec 04 2001

The analog of A033665 in base 2.

			10011 (19 in base 10) -> 10011 + 11001 = 101100 -> 101100 + 1101 = 111001 -> 111001 + 100111 = 1100000 -> 1100000 + 11 = 1100011 (palindrome) requires 4 steps, so a(19) = 4.

Index entries for sequences related to Reverse and Add!
Klaus Brockhaus, On the'Reverse and Add!' algorithm in base 2

Cf. A033665, A058042, A062128, A062130, A033865, A061561, A066058, A006995, A057148.

function b2reverse(a: integer): integer; var n,i,rev: integer; begin n := bit_length(a); for i := 0 to n-1 do if bit_test(a,i) = 1 then rev := bit_set(rev,n-1-i); end; end; return rev; end; function a066057(mx,stop: integer); var c,k,m,rev: integer; begin for k := 0 to mx do c := 0; m := k; rev := b2reverse(m); while m <> rev and c < stop do inc(c); m := m + rev; rev := b2reverse(m); end; if c < stop then write(c); else write(-1); end; write(" "); end; end; a066057(120,300);

limit = 10^4; (* Assumes that there is no palindrome if none is found before "limit" iterations *)
Table[np = n; i = 0;
 While[np != IntegerReverse[np, 2] && i < limit,
  np = np + IntegerReverse[np, 2]; i++];
If[i >= limit, -1, i], {n, 0, 111}] (* Robert Price, Oct 14 2019 *)

A066059 Integers such that the 'Reverse and Add!' algorithm in base 2 (cf. A062128) does not lead to a palindrome.

Original entry on oeis.org

22, 26, 28, 35, 37, 41, 46, 47, 49, 60, 61, 67, 75, 77, 78, 84, 86, 89, 90, 94, 95, 97, 105, 106, 108, 110, 116, 120, 122, 124, 125, 131, 135, 139, 141, 147, 149, 152, 155, 157, 158, 163, 164, 166, 169, 172, 174, 177, 180, 182, 185, 186, 190, 191, 193, 197, 199

Offset: 1

Klaus Brockhaus, Dec 04 2001

The analog of A023108 in base 2.

It seems that for all these numbers it can be proven that they never reach a palindrome. For this it is sufficient to prove this for all seeds as given in A075252. As observed, for all numbers in A075252, lim_{n -> inf} t(n+1)/t(n) is 1 or 2 (1 for even n, 2 for odd n or reverse); i.e., lim_{n -> inf} t(n+2)/t(n) = 2, t(n) being the n-th term of the trajectory. - A.H.M. Smeets, Feb 10 2019

A.H.M. Smeets, Table of n, a(n) for n = 1..20000
Klaus Brockhaus, On the'Reverse and Add!' algorithm in base 2
Index entries for sequences related to Reverse and Add!

Cf. A062128, A023108, A062130, A033865, A058042, A061561, A066057.

: For function b2reverse see A066057; function a066059(mx,stop: integer); var k,c,m,rev: integer; begin for k := 1 to mx do c := 0; m := k; rev := b2reverse(m); while m <> rev and c < stop do inc(c); m := m + rev; rev := b2reverse(m); end; if c >= stop then write(k," "); end; end; end; a066059(210,300).

limit = 10^4; (* Assumes that there is no palindrome if none is found before "limit" iterations *)
Select[Range[200],
Length@NestWhileList[# + IntegerReverse[#, 2] &, #, # !=
IntegerReverse[#, 2]  &, 1, limit] == limit + 1 &] (* Robert Price, Oct 14 2019 *)

A062130 A062128 written in base 10.

Original entry on oeis.org

0, 1, 3, 3, 5, 5, 9, 7, 9, 9, 15, 27, 15, 27, 21, 15, 17, 17, 27, 99, 99, 21, -1, 63, 27, 99, -1, 27, -1, 63, 45, 31, 33, 33, 51, -1, 45, -1, 63, 99, 45, -1, 63, 99, 99, 45, -1, -1, 51, -1, 255, 51, 63, 99, 255, 153, 63, 99, 255, 153, -1, -1, 93, 63, 65, 65, 99, -1, 85, 255, 119, 387, 255, 73, 13299, -1, 387, -1, -1, 219, 85

Offset: 0

Klaus Brockhaus, Jun 06 2001

			23 -> 23 + 29 = 52 -> 52 + 11 = 63, so a(23) = 63.

Index entries for sequences related to Reverse and Add!
Klaus Brockhaus, On the'Reverse and Add!' algorithm in base 2

Cf. A033865, A062128, A062131, A061561.

stop := 500; for k := 0 to 80 do c := 0; m := k; rev := bit_reverse(m); while m <> rev and c < stop do inc(c); m := m + rev; rev := bit_reverse(m); end; if c < stop then write(m); else write(-1); end; write(" "); end;.

limit = 10^4; (* Assumes that there is no palindrome if none is found before "limit" iterations *)
Table[np = n; i = 0;
 While[np != IntegerReverse[np, 2] && i < limit,
  np = np + IntegerReverse[np, 2]; i++];
If[i >= limit, -1, np], {n, 0, 80}] (* Robert Price, Oct 14 2019 *)

A066058 In base 2: smallest integer which requires n 'Reverse and Add' steps to reach a palindrome.

Original entry on oeis.org

0, 2, 11, 44, 19, 20, 275, 326, 259, 202, 103, 74, 1027, 1070, 1049, 1072, 1547, 1310, 1117, 794, 569, 398, 3083, 2154, 1177, 1064, 4697, 4264, 4443, 2678, 2169, 1422, 779, 3226, 1551, 1114, 1815, 1062, 4197, 3106, 8697, 7238, 16633, 12302, 6683

Offset: 0

Klaus Brockhaus, Dec 04 2001

The analog of A023109 in base 2.

			11 is the smallest integer which requires two steps to reach a base 2 palindrome (cf. A066057), so a(2) = 11; written in base 10: 11 -> 11 + 13 = 24 -> 24 + 3 = 27; written in base 2: 1011 -> 1011 + 1101 = 11000 -> 11000 + 11 = 11011.

Cf. A066057, A023109, A062128, A062130, A033865, A006995, A057148.

(* For function b2reverse see A066057. *) function a066058(mx: integer); var k,m,n,rev,steps: integer; begin for k := 0 to mx do n := 0; steps := 0; m := n; rev := b2reverse(m); while not(steps = k and m = rev) do inc(n); m := n; rev := b2reverse(m); steps := 0; while steps < k and m <> rev do m := m + rev; rev := b2reverse(m); inc(steps); end; end; write(n,","); end; end; a066058(45);

Table[ SelectFirst[Range[0, 20000], (np = #; i = 0;
    While[ np != IntegerReverse[np, 2] && i <= n,
     np = np + IntegerReverse[np, 2]; i++];
i == n ) &] , {n, 0, 44}] (* Robert Price, Oct 16 2019 *)

def A066058(n):
    if n > 0:
        k = 0
        while True:
            m = k
            for i in range(n):
                s1 = format(m,'b')
                s2 = s1[::-1]
                if s1 == s2:
                    break
                m += int(s2,2)
            else:
                s1 = format(m,'b')
                if s1 == s1[::-1]:
                    return k
            k += 1
    else:
        return 0 # Chai Wah Wu, Jan 06 2015

A062129 In base 2: start with n; add to itself with digits reversed; if palindrome, stop; otherwise repeat; a(n) gives palindrome at which it stops, or -1 if no palindrome is ever reached.

Original entry on oeis.org

0, 11, 11, 1001, 101, 1111, 1001, 10101, 1001, 11011, 1111, 11011, 1111, 11011, 10101, 101101, 10001, 110011, 11011, 1100011, 1100011, 111111, -1, 111111, 11011, 1100011, -1, 11111111, -1, 111111, 101101, 1011101, 100001, 1100011, 110011, -1, 101101, -1, 111111, 1100011, 101101, -1, 111111

Offset: 0

Klaus Brockhaus, Jun 06 2001

The analog of A061563 in base 2. Differs from A062128 only for those n, which are palindromes in base 2.

			23: 10111 -> 10111 + 11101 = 110100 -> 110100 + 1011 = 111111, so a(23) = 111111.

Cf. A061563, A062128, A062131, A058042.

stop := 500; for k := 0 to 60 do c := 0; m := k; test := true; while test and c < stop do inc(c); m := m + bit_reverse(m); test := m <> bit_reverse(m); end; if c < stop then bit_write(m); else write(-1); end; write(" "); end;

cp's OEIS Frontend

A066057 'Reverse and Add' carried out in base 2 (cf. A062128); number of steps needed to reach a palindrome, or -1 if no palindrome is ever reached.

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A066059 Integers such that the 'Reverse and Add!' algorithm in base 2 (cf. A062128) does not lead to a palindrome.

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A062130 A062128 written in base 10.

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A066058 In base 2: smallest integer which requires n 'Reverse and Add' steps to reach a palindrome.

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A062129 In base 2: start with n; add to itself with digits reversed; if palindrome, stop; otherwise repeat; a(n) gives palindrome at which it stops, or -1 if no palindrome is ever reached.

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