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A006995
Binary palindromes: numbers whose binary expansion is palindromic.
(Formerly M2403)
232
0, 1, 3, 5, 7, 9, 15, 17, 21, 27, 31, 33, 45, 51, 63, 65, 73, 85, 93, 99, 107, 119, 127, 129, 153, 165, 189, 195, 219, 231, 255, 257, 273, 297, 313, 325, 341, 365, 381, 387, 403, 427, 443, 455, 471, 495, 511, 513, 561, 585, 633, 645, 693, 717, 765, 771, 819, 843
OFFSET
1,3
COMMENTS
If b > 1 is a binary palindrome then both (2^(m+1) + 1)*b and 2^(m+1) + 2^m - b are also, where m = floor(log_2(b)). - Hieronymus Fischer, Feb 18 2012
Floor and ceiling: If d > 0 is any natural number, then A206913(d) is the greatest binary palindrome <= d and A206914(d) is the least binary palindrome >= d. - Hieronymus Fischer, Feb 18 2012
The greatest binary palindrome <= the n-th non-binary-palindrome is that binary palindrome with number A154809(n)-n+1. The corresponding formula identity is: A206913(A154809(n)) = A006995(A154809(n)-n+1). - Hieronymus Fischer, Mar 18 2012
From Hieronymus Fischer, Jan 23 2013: (Start)
The number of binary digits of a(n) is A070939(a(n)) = 1 + floor(log_2(n)) + floor(log_2(n/3)), for n > 1.
Also: A070939(a(n)) = A070939(n) + A070939(floor(n/3)) - 1, for n <> 2. (End)
Rajasekaran, Shallit, & Smith show that this is an additive basis of order 4. - Charles R Greathouse IV, Nov 06 2018
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
James Haoyu Bai, Joseph Meleshko, Samin Riasat, and Jeffrey Shallit, Quotients of Palindromic and Antipalindromic Numbers, arXiv:2202.13694 [math.NT], 2022.
William D. Banks and Igor E. Shparlinski, Average Value of the Euler Function on Binary Palindromes, Bulletin Polish Acad. Sci. Math., Vol. 54 (2006), pp. 95-101, alternative link.
Manfred Madritsch and Stephan Wagner, A central limit theorem for integer partitions, Monatshefte für Mathematik, Vol. 161, No. 1 (2010), pp. 85-114, alternative link. doi:10.1007/s00605-009-0126-y.
Kritkhajohn Onphaeng, Tammatada Khemaratchatakumthorn, Phakhinkon Napp Phunphayap, and Prapanpong Pongsriiam, Exact Formulas for the Number of Palindromes in Certain Arithmetic Progressions, Journal of Integer Sequences, Vol. 27 (2024), Article 24.4.8. See p. 2.
Aayush Rajasekaran, Using Automata Theory to Solve Problems in Additive Number Theory, MS thesis, University of Waterloo, 2018.
Aayush Rajasekaran, Jeffrey Shallit and Tim Smith, Sums of Palindromes: an Approach via Nested-Word Automata, preprint arXiv:1706.10206 [cs.FL], Jun 30 2017.
Aayush Rajasekaran, Jeffrey Shallit and Tim Smith, Additive Number Theory via Automata Theory, Theory of Computing Systems, Vol. 64 (2020), pp. 542-567.
FORMULA
A178225(a(n)) = 1; union of A048700 and A048701. - Reinhard Zumkeller, Oct 21 2011
From Hieronymus Fischer, Dec 31 2008, Jan 10 2012, Feb 18 2012: (Start)
Written as a decimal, a(10^n) has 2*n digits. For n > 1, the decimal expansion of a(10^n) starts with 22..., 23... or 24...:
a(1000) = 249903,
a(10^4) = 24183069,
a(10^5) = 2258634081,
a(10^6) = 249410097687,
a(10^7) = 24350854001805,
a(10^8) = 2229543293296319,
a(10^9) = 248640535848971067,
a(10^10)= 24502928886295666773.
Inequality: (2/9)*n^2 < a(n) < (1/4)*(n+1)^2, if n > 1.
lim sup_{n -> oo} a(n)/n^2 = 1/4, lim inf_{n -> oo} a(n)/n^2 = 2/9.
For n >= 2, a(2^n-1) = 2^(2n-2) - 1; a(2^n) = 2^(2n-2) + 1;
a(2^n+1) = 2^(2n-2) + 2^(n-1) + 1; a(2^n + 2^(n-1)) = 2^(2n-1) + 1.
Recursion for n > 2: a(n) = 2^(2k-q) + 1 + 2^p*a(m), where k = floor(log_2(n-1)), and p, q and m are determined as follows:
Case 1: If n = 2^(k+1), then p = 0, q = 0, m = 1;
Case 2: If 2^k < n < 2^k+2^(k-1), then p = k-floor(log_2(i))-1 with i = n-2^k, q = 2, m = 2^floor(log_2(i)) + i;
Case 3: If n = 2^k + 2^(k-1), then p = 0, q = 1, m = 1;
Case 4: If 2^k + 2^(k-1) < n < 2^(k+1), then p = k-floor(log_2(j))-1 with j = n-2^k-2^(k-1), q = 1, m = 2*2^floor(log_2(j))+j.
Non-recursive formula:
Let n >= 3, m = floor(log_2(n)), p = floor((3*2^(m-1)-1)/n), then
a(n) = 2^(2*m-1-p) + 1 + p*(1-(-1)^n)*2^(m-1-p) + sum_{k=1 .. m-1-p} (floor((n-(3-p)*2^(m-1))/2^(m-1-k)) mod 2)*(2^k+2^(2*m-1-p-k)). [Typo at the last exponent of the third sum term eliminated by the author, Sep 05 2018]
a(n) = 2^(2*m-2) + 1 + 2*floor((n-2^m)/2^(m-1)) + 2^(m-1)*floor((1/2)*min(n+1-2^m,2^(m-1)+1)) + 3*2^(m-1)*floor((1/2)*max(n+1-3*2^(m-1),0)) + 3*sum_{j=2 .. m-1} floor((n+2^(j-1)-2^m)/2^j)*2^(m-j). [Seems correct for n > 3. - The Editors]
Inversion formula: The index of any binary palindrome b = a(n) > 0 is n = palindromicIndex(b) = ((5-(-1)^m)/2 + Sum_{k=1..[m/2]} ([b/2^k] mod 2)/2^k)*2^[m/2], where [.] = floor(.) and m = [log_2(b)].
(End)
G.f.: g(x) = x^2 + 3x^3 + sum_{j=1..oo}( 3*2^j*(1-x^floor((j+1)/2))/(1-x)*x^((1/2)-floor((j+1)/2)) + f_j(x) - f_j(1/x))*x^(2*2^floor(j/2)+3*2^floor((j-1)/2)-(1/2)), where the f_j(x) are defined as follows:
f_1(x) = x^(1/2), and for j > 1,
f_j(x) = x^(1/2)*sum_{i=0..2^floor((j-1)/2)-1}((3+(1/2)*sum_{k=1..floor((j-1)/2)}(1-(-1)^floor(2i/2^k))*b(j,k))*x^i), where b(j,k) = 2^(floor((j-1)/2)-k)*((3+(-1)^j)*2^(2*k+1)+4) for k > 1, and b(j,1) = (2+(-1)^j)*2^(floor((j-1)/2)+1). - Hieronymus Fischer, Apr 04 2012
A044051(n) = (a(n)+1)/2 for n > 0. - Reinhard Zumkeller, Apr 20 2015
A145799(a(n)) = a(n). - Reinhard Zumkeller, Sep 24 2015
Sum_{n>=2} 1/a(n) = A244162. - Amiram Eldar, Oct 17 2020
EXAMPLE
a(3) = 3, since 3 = 11_2 is the 3rd symmetric binary number;
a(6) = 9, since 9 = 1001_2 is the 6th symmetric binary number.
MAPLE
dmax:= 15; # to get all terms with at most dmax binary digits
revdigs:= proc(n)
local L, Ln, i;
L:= convert(n, base, 2);
Ln:= nops(L);
add(L[i]*2^(Ln-i), i=1..Ln);
end proc;
A:= {0, 1}:
for d from 2 to dmax do
if d::even then
A:= A union {seq(2^(d/2)*x + revdigs(x), x=2^(d/2-1)..2^(d/2)-1)}
else
m:= (d-1)/2;
B:={seq(2^(m+1)*x + revdigs(x), x=2^(m-1)..2^m-1)};
A:= A union B union map(`+`, B, 2^m)
fi
od:
A; # Robert Israel, Aug 17 2014
MATHEMATICA
palQ[n_Integer, base_Integer] := Module[{idn=IntegerDigits[n, base]}, idn==Reverse[idn]]; Select[Range[1000], palQ[ #, 2]&]
Select[ Range[0, 1000], # == IntegerReverse[#, 2] &] (* Robert G. Wilson v, Feb 24 2018 *)
Select[Range[0, 1000], PalindromeQ[IntegerDigits[#, 2]]&] (* Jean-François Alcover, Mar 01 2018 *)
PROG
(PARI) for(n=0, 999, n-subst(Polrev(binary(n)), x, 2)||print1(n, ", ")) \\ Thomas Buchholz, Aug 16 2014
(PARI) for(n=0, 10^3, my(d=digits(n, 2)); if(d==Vecrev(d), print1(n, ", "))); \\ Joerg Arndt, Aug 17 2014
(PARI) is_A006995(n)=Vecrev(n=binary(n))==n \\ M. F. Hasler, Feb 23 2018
(PARI) A006995(n, m=logint(n, 2), c=1<<(m-1), a, d)={if(n>=3*c, a=n-3*c; d=2*c^2, a=n-2*c; n%2*c+d=c^2)+sum(k=1, m-2^(n<3*c), if(bittest(a, m-1-k), 1<<k+d>>k))+(n>2)} \\ Based on Fischer's smalltalk program. - M. F. Hasler, Feb 23 2018
(Magma) [n: n in [0..850] | Intseq(n, 2) eq Reverse(Intseq(n, 2))]; // Bruno Berselli, Aug 29 2011
(Haskell)
a006995 n = a006995_list !! (n-1)
a006995_list = 0 : filter ((== 1) . a178225) a005408_list
-- Reinhard Zumkeller, Oct 21 2011
(Smalltalk)
"Answer the n-th binary palindrome
(nonrecursive implementation)"
| m n a b c d k2 |
n := self.
n = 1 ifTrue: [^0].
n = 2 ifTrue: [^1].
m := n integerFloorLog: 2.
c := 2 raisedToInteger: m - 1.
n >= (3 * c)
ifTrue:
[a := n - (3 * c).
d := 2 * c * c.
b := d + 1.
k2 := 1.
1 to: m - 1
do:
[:k |
k2 := 2 * k2.
b := b + (a * k2 // c \\ 2 * (k2 + (d // k2)))]]
ifFalse:
[a := n - (2 * c).
d := c * c.
b := d + 1 + (n \\ 2 * c).
k2 := 1.
1 to: m - 2
do:
[:k |
k2 := 2 * k2.
b := b + (a * k2 // c \\ 2 * (k2 + (d // k2)))]].
^b // by Hieronymus Fischer, Feb 15 2013
(Sage)
def palgenbase2(): # generator of palindromes in base 2
yield 0
x, n, n2 = 1, 1, 2
while True:
for y in range(n, n2):
s = format(y, 'b')
yield int(s+s[-2::-1], 2)
for y in range(n, n2):
s = format(y, 'b')
yield int(s+s[::-1], 2)
x += 1
n *= 2
n2 *= 2 # Chai Wah Wu, Jan 07 2015
(Sage)
[n for n in (0..843) if Word(n.digits(2)).is_palindrome()] # Peter Luschny, Sep 13 2018
(Python)
from itertools import count, islice, product
def bin_pals(): # generator of binary palindromes in base 10
yield from [0, 1]
digits, midrange = 2, [[""], ["0", "1"]]
for digits in count(2):
for p in product("01", repeat=digits//2-1):
left = "1"+"".join(p)
for middle in midrange[digits%2]:
yield int(left + middle + left[::-1], 2)
print(list(islice(bin_pals(), 58))) # Michael S. Branicky, Jan 09 2023
(Python)
def A006995(n):
if n == 1: return 0
a = 1<<(l:=n.bit_length()-2)
m = a|(n&a-1)
return (m<<l+1)+int(bin(m)[-1:1:-1]or'0', 2) if a&n else (m<<l)+int(bin(m)[-2:1:-1]or'0', 2) # Chai Wah Wu, Jun 10 2024
CROSSREFS
See A057148 for the binary representations.
Cf. A178225, A005408, A164126, A154809 (complement).
Even numbers that are not the sum of two terms: A241491, A261678, A262556.
Cf. A145799.
Primes: A016041 and A117697.
Cf. A000051 (a subsequence).
Sequence in context: A359402 A329358 A180204 * A163410 A329419 A235264
KEYWORD
nonn,base,easy,nice,hear
EXTENSIONS
Edited and extended by Hieronymus Fischer, Feb 21 2012
Edited by M. F. Hasler, Feb 23 2018
STATUS
approved