A110591 -id:A110591 - cp's OEIS frontend

A110594 a(1) = 4, a(2) = 12, for n>1: a(n) = 3*4^(n-1).

4, 12, 48, 192, 768, 3072, 12288, 49152, 196608, 786432, 3145728, 12582912, 50331648, 201326592, 805306368, 3221225472, 12884901888, 51539607552, 206158430208, 824633720832, 3298534883328, 13194139533312, 52776558133248

Offset: 1

Jonathan Vos Post, Jul 29 2005

Vincenzo Librandi, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (4).

Cf. A000302, A007090, A081604, A110591, A110593.

Concatenation([4],List([2..25],n->3*4^(n-1))); # Muniru A Asiru, Oct 21 2018

[4] cat [3*4^(n-1): n in [2..30]]; // Vincenzo Librandi, May 29 2014

seq(coeff(series(4*x*(1-x)/(1-4*x),x,n+1), x, n), n = 1 .. 25); # Muniru A Asiru, Oct 21 2018

CoefficientList[Series[4 (1 - x)/(1 - 4 x), {x, 0, 40}], x] (* Vincenzo Librandi, May 29 2014 *)

x='x+O('x^50); Vec(4*x*(1 - x)/(1 - 4*x)) \\ G. C. Greubel, Sep 01 2017

a(n) = A002001(n), n>1. - R. J. Mathar, Aug 18 2008

G.f.: 4*x*(1 - x)/(1 - 4*x). - Vincenzo Librandi, May 29 2014

Definition corrected by R. J. Mathar, Aug 18 2008

A043000 Number of digits in all base-b representations of n, for 2 <= b <= n.

Original entry on oeis.org

2, 4, 7, 9, 11, 13, 16, 19, 21, 23, 25, 27, 29, 31, 35, 37, 39, 41, 43, 45, 47, 49, 51, 54, 56, 59, 61, 63, 65, 67, 70, 72, 74, 76, 79, 81, 83, 85, 87, 89, 91, 93, 95, 97, 99, 101, 103, 106, 108, 110, 112, 114, 116, 118, 120, 122, 124, 126

Offset: 2

Clark Kimberling

From A.H.M. Smeets, Dec 14 2019: (Start)
a(n)-a(n-1) >= 2 due to the fact that n = 10_n, so there is an increment of at least 2. If n can be written as a perfect power m^s, an additional +1 comes to it for the representation of n in each base m.
For instance, for n = 729 we have 729 = 3^6 = 9^3 = 27^2, so there is an additional increment of 3. For n = 1296 we have 1296 = 6^4 = 36^2, so there is an additional increment of 2. For n = 4096 we have 4096 = 2^12 = 4^6 = 8^4 = 16^3= 64^2, so there is an additional increment of 5. (End)

			5 = 101_2 = 12_3 = 11_4 = 10_5. Thus a(5) = 3+2+2+2 = 9.

A.H.M. Smeets, Table of n, a(n) for n = 2..20000
Vaclav Kotesovec, Plot of a(n)/(2*n) for n = 2..1000000

Cf. A001597, A043306, A068953, A191322, A253642, A255165.

[&+[Floor(Log(i,i*n)):k in [2..n]]:n in [1..70]]; // Marius A. Burtea, Nov 13 2019

A043000 := proc(n) add( nops(convert(n,base,b)),b=2..n) ; end proc: # R. J. Mathar, Jun 04 2011

Table[Total[IntegerLength[n,Range[2,n]]],{n,2,60}] (* Harvey P. Dale, Apr 23 2019 *)

a(n)=sum(b=2,n,#digits(n,b)) \\ Jeppe Stig Nielsen, Dec 14 2019

a(n)= n-1 +sum(b=2,n,logint(n,b)) \\ Jeppe Stig Nielsen, Dec 14 2019

a(n) = {2*n-2+sum(i=2, logint(n, 2), sqrtnint(n, i)-1)} \\ David A. Corneth, Dec 31 2019

first(n) = my(res = vector(n)); res[1] = 2; for(i = 2, n, inc = numdiv(gcd(factor(i+1)[,2]))+1; res[i] = res[i-1]+inc); res \\ David A. Corneth, Dec 31 2019

def count(n,b):
    c = 0
    while n > 0:
        n, c = n//b, c+1
    return c
n = 0
while n < 50:
    n = n+1
    a, b = 0, 1
    while b < n:
        b = b+1
        a = a + count(n,b)
    print(n,a) # A.H.M. Smeets, Dec 14 2019

a(n) = Sum_{i=2..n} floor(log_i(i*n)); a(n) ~ 2*n. - Vladimir Shevelev, Jun 03 2011 [corrected by Vaclav Kotesovec, Apr 05 2021]
a(n) = A070939(n) + A081604(n) + A110591(n) + ... + 1. - R. J. Mathar, Jun 04 2011
From Ridouane Oudra, Nov 13 2019: (Start)
a(n) = Sum_{i=1..n-1} floor(n^(1/i));
a(n) = n - 1 + Sum_{i=1..floor(log_2(n))} floor(n^(1/i) - 1);
a(n) = n - 1 + A255165(n). (End)
If n is in A001597 then a(A001597(m)) - a(A001597(m)-1) = 2 + A253642(m), otherwise a(n) - a(n-1) = 2. - A.H.M. Smeets, Dec 14 2019

A049803 a(n) = n mod 3 + n mod 9 + ... + n mod 3^k, where 3^k <= n < 3^(k+1).

Original entry on oeis.org

0, 0, 0, 1, 2, 0, 1, 2, 0, 2, 4, 3, 5, 7, 6, 8, 10, 0, 2, 4, 3, 5, 7, 6, 8, 10, 0, 3, 6, 6, 9, 12, 12, 15, 18, 9, 12, 15, 15, 18, 21, 21, 24, 27, 18, 21, 24, 24, 27, 30, 30, 33, 36, 0, 3, 6, 6, 9, 12, 12, 15, 18, 9, 12, 15, 15, 18, 21, 21, 24, 27, 18

Offset: 1

Clark Kimberling

nonn

From Petros Hadjicostas, Dec 11 2019: (Start)
Conjecture: For b >= 2, consider the function s(n,b) = Sum_{1 <= b^j <= n} (n mod b^j) from p. 8 in Dearden et al. (2011). Then s(b*n + r, b) = b*s(n,b) + r*N(n,b) for 0 <= r <= b-1, where N(n,b) = floor(log_b(n)) + 1 is the number of digits in the base-b representation of n. As initial conditions, we have s(n,b) = 0 for 1 <= n <= b. (This is a generalization of a result by Robert Israel in A049802.)
Here b = 3 and a(n) = s(n,3).
We have N(n,2) = A070939(n), N(n,3) = A081604(n), N(n,4) = A110591(n), and N(n,5) = A110592(n).
If A_b(x) = Sum_{n >= 1} s(n,b)*x^n is the g.f. of the sequence (s(n,b): n >= 1) and the above conjecture is correct, then it can be proved that A_b(x) = b * A_b(x^b) * (1-x^b)/(1-x) + x * ((b-1)*x^b - b*x^(b-1) + 1)/((1-x)^2 * (1-x^b)) * Sum_{k >= 1} x^(b^k). (End)

Metin Sariyar, Table of n, a(n) for n = 1..32000
B. Dearden, J. Iiams, and J. Metzger, A Function Related to the Rumor Sequence Conjecture, J. Int. Seq. 14 (2011), #11.2.3, Example 7.

Cf. A049802, A049804, A070939, A081604, A110591, A110592, A330358.

a:= n-> add(irem(n, 3^j), j=1..ilog[3](n)):
seq(a(n), n=1..105);  # Alois P. Heinz, Dec 13 2019

Table[n * Floor@Log[3, n] - Sum[Floor[n*3^-k]*3^k, {k, Log[3, n]}], {n, 100}] (* after Federico Provvedi in A049802*) (* Metin Sariyar, Dec 12 2019 *)

a(n) = sum(k=1, logint(n, 3), n % 3^k); \\ Michel Marcus, Dec 12 2019

From Petros Hadjicostas, Dec 11 2019: (Start)
Conjecture: a(3*n+r) = 3*a(n) + r*A081604(n) = 3*a(n) + r*(floor(log_3(n)) + 1) for n >= 1 and r = 0, 1, 2.
If the conjecture above is true, the g.f. A(x) satisfies A(x) = 3*(1+x+x^2)*A(x^3) + x*(2*x+1)/(1-x^3) * Sum_{k >= 1} x^(3^k). (End)

A049804 a(n) = n mod 4 + n mod 16 + ... + n mod 4^k, where 4^k <= n < 4^(k+1).

Original entry on oeis.org

0, 0, 0, 0, 1, 2, 3, 0, 1, 2, 3, 0, 1, 2, 3, 0, 2, 4, 6, 4, 6, 8, 10, 8, 10, 12, 14, 12, 14, 16, 18, 0, 2, 4, 6, 4, 6, 8, 10, 8, 10, 12, 14, 12, 14, 16, 18, 0, 2, 4, 6, 4, 6, 8, 10, 8, 10, 12, 14, 12, 14, 16, 18, 0, 3, 6, 9, 8, 11, 14, 17, 16, 19, 22

Offset: 1

Clark Kimberling

nonn

From Petros Hadjicostas, Dec 11 2019: (Start)
Conjecture: For b >= 2, consider the function s(n,b) = Sum_{1 <= b^j <= n} (n mod b^j) from p. 8 in Dearden et al. (2011). Then s(b*n + r, b) = b*s(n,b) + r*N(n,b) for 0 <= r <= b-1, where N(n,b) = floor(log_b(n)) + 1 is the number of digits in the base-b representation of n. As initial conditions, we have s(n,b) = 0 for 1 <= n <= b. (This is a generalization of a result by Robert Israel in A049802.)
Here b = 4 and a(n) = s(n,4).
We have N(n,2) = A070939(n), N(n,3) = A081604(n), N(n,4) = A110591(n), and N(n,5) = A110592(n).
If A_b(x) = Sum_{n >= 1} s(n,b)*x^n is the g.f. of the sequence (s(n,b): n >= 1) and the above conjecture is correct, then it can be proved that A_b(x) = b * A_b(x^b) * (1-x^b)/(1-x) + x * ((b-1)*x^b - b*x^(b-1) + 1)/((1-x)^2 * (1-x^b)) * Sum_{k >= 1} x^(b^k). (End)

Metin Sariyar, Table of n, a(n) for n = 1..32000
B. Dearden, J. Iiams, and J. Metzger, A Function Related to the Rumor Sequence Conjecture, J. Int. Seq. 14 (2011), #11.2.3, Example 7.

Cf. A049802, A049803, A070939, A081604, A110591, A110592, A330358.

a:= n-> add(irem(n, 4^j), j=1..ilog[4](n)):
seq(a(n), n=1..105);  # Petros Hadjicostas, Dec 13 2019 (after Alois P. Heinz's program for A330358)

Table[n * Floor@Log[4, n] - Sum[Floor[n*4^-k]*4^k, {k, Log[4, n]}], {n, 100}] (* Metin Sariyar, Dec 12 2019 *)
a[n_] := Sum[Mod[n, 4^j], {j, 1, Length[IntegerDigits[n, 4]] - 1}];
Array[a, 105] (* Jean-François Alcover, Dec 31 2021 *)

a(n) = sum(k=1, logint(n, 4), n % 4^k); \\ Michel Marcus, Dec 12 2019

From Petros Hadjicostas, Dec 11 2019: (Start)
Conjecture: a(4*n+r) = 4*a(n) + r*A110591(n) = 4*a(n) + r*(floor(log_4(n)) + 1) for n >= 1 and r = 0, 1, 2, 3.
If the conjecture above is true, the g.f. A(x) satisfies A(x) = 4*(1 + x + x^2 + x^3)*A(x^4) + x*(1 + 2*x + 3*x^2)/(1 - x^4) * Sum_{k >= 1} x^(4^k). (End)

A330358 a(n) = n mod 5 + n mod 25 + ... + n mod 5^k, where 5^k <= n < 5^(k+1).

Original entry on oeis.org

0, 0, 0, 0, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 1, 2, 3, 4, 0, 2, 4, 6, 8, 5, 7, 9, 11, 13, 10, 12, 14, 16, 18, 15, 17, 19, 21, 23, 20, 22, 24, 26, 28, 0, 2, 4, 6, 8, 5, 7, 9, 11, 13, 10, 12, 14, 16, 18, 15, 17, 19, 21, 23, 20, 22, 24, 26, 28, 0, 2, 4, 6, 8, 5, 7, 9, 11, 13, 10

Offset: 1

Petros Hadjicostas, Dec 12 2019

Conjecture: For b >= 2, consider the function s(n,b) = Sum_{1 <= b^j <= n} (n mod b^j) from p. 8 in Dearden et al. (2011). Then s(b*n + r, b) = b*s(n,b) + r*N(n,b) for 0 <= r <= b-1, where N(n,b) = floor(log_b(n)) + 1 is the number of digits in the base-b representation of n. As initial conditions, we have s(n,b) = 0 for 1 <= n <= b. (This is a generalization of a result by Robert Israel in A049802.)
Here b = 5 and a(n) = s(n,5).
We have N(n,2) = A070939(n), N(n,3) = A081604(n), N(n,4) = A110591(n), and N(n,5) = A110592(n).
If A_b(x) = Sum_{n >= 1} s(n,b)*x^n is the g.f. of the sequence (s(n,b): n >= 1) and the above conjecture is correct, then it can be proved that A_b(x) = b * A_b(x^b) * (1-x^b)/(1-x) + x * ((b-1)*x^b - b*x^(b-1) + 1)/((1-x)^2 * (1-x^b)) * Sum_{k >= 1} x^(b^k).

Robert Israel, Table of n, a(n) for n = 1..10000

Cf. A049802, A049803, A049804, A070939, A081604, A110591, A110592.

a:= n-> add(irem(n, 5^j), j=1..ilog[5](n)):
seq(a(n), n=1..105);  # Alois P. Heinz, Dec 13 2019

a[n_] := Sum[Mod[n, 5^j], {j, 1, Length[IntegerDigits[n, 5]] - 1}];
Array[a, 105] (* Jean-François Alcover, Dec 31 2021 *)

a(n) = sum(k=1, logint(n, 5), n % 5^k);
for(n=1, 100, print1(a(n), ", ")); \\ (after Michel Marcus's program in A049804)

Conjecture: a(5*n+r) = 5*a(n) + r*A110592(n) = 5*a(n) + r*(floor(log_5(n)) + 1) for n >= 1 and r = 0, 1, 2, 3, 4.

If the conjecture above is true, the g.f. A(x) satisfies A(x) = 5*(1 + x + x^2 + x^3 + x^4)*A(x^5) + x*(1 + 2*x + 3*x^2 + 4*x^3)/(1 - x^5) * Sum_{k >= 1} x^(5^k).

A048703 Numbers which in base 4 are palindromes and have an even number of digits.

Original entry on oeis.org

0, 5, 10, 15, 65, 85, 105, 125, 130, 150, 170, 190, 195, 215, 235, 255, 1025, 1105, 1185, 1265, 1285, 1365, 1445, 1525, 1545, 1625, 1705, 1785, 1805, 1885, 1965, 2045, 2050, 2130, 2210, 2290, 2310, 2390

Offset: 0

Antti Karttunen, Mar 07 1999

In quaternary base (base 4) the terms look like 0, 11, 22, 33, 1001, 1111, 1221, 1331, 2002, 2112, 2222, 2332, 3003, 3113, 3223, 3333, 100001, 101101, 102201, ..., which is a subsequence of A118595.

Zero is included as a(0) because we can consider it as having zero digits after leading zeros have been excluded.

			Each a(n) is obtained by concatenating the original base-4 expansion of n (which comes to the left hand, i.e., the most significant side) with its mirror-image (which comes to the right hand, i.e., the least significant side). For example, for a(4) we have 4 = '10' in base 4, which concatenated with its reversal '01' yields '1001', which when converted back to decimal yields 1*64 + 0*16 + 0*4 + 1*1 = 65, thus a(4)=65.

Amiram Eldar, Table of n, a(n) for n = 0..10000 (terms 0..4095 from Antti Karttunen)

Subsequence of A014192 (all numbers which are palindromes in base 4, including also those of odd number of digits).

Cf. also A048704 (this sequence divided by 5), A048701 (binary palindromes of even length), A055948, A110591, A118595, A030103, A007090, A000302.

A048703(n) := (n) -> (2^(floor_log_2_coarse(n)+1))*n + sum('(bit_i(n, i+((-1)^i))*(2^(floor_log_2_coarse(n)-i)))', 'i'=0..floor_log_2_coarse(n));
bit_i := (x,i) -> `mod`(floor(x/(2^i)),2);
# Following is like floor_log_2 but even results are incremented by one:
floor_log_2_coarse := proc(n) local nn,i: nn := n; for i from -1 to n do if(0 = nn) then RETURN(i+(1-(i mod 2))); fi: nn := floor(nn/2); od: end:

q[n_] := EvenQ[IntegerLength[n, 4]] && PalindromeQ[IntegerDigits[n, 4]]; Select[Range[0, 2400, 5], q] (* Amiram Eldar, May 27 2024 *)

def A048703(n):
    s = bin(n-1)[2:]
    if len(s) % 2: s = '0'+s
    t = [s[i:i+2] for i in range(0,len(s),2)]
    return int(''.join(t+t[::-1]),2) # Chai Wah Wu, Feb 26 2021

a(0) = 0, and for n >= 1, a(n) = A030103(n) + (n * A000302(A110591(n))). - Antti Karttunen, Oct 30 2013

a(n) = 5*A048704(n). [This is just a consequence of the definition of A048704.] - Antti Karttunen, Jul 25 2013

A260112 Minimal number of steps to get from 0 to n by (a) adding 1 or (b) multiplying by 4.

Original entry on oeis.org

0, 1, 2, 3, 2, 3, 4, 5, 3, 4, 5, 6, 4, 5, 6, 7, 3, 4, 5, 6, 4, 5, 6, 7, 5, 6, 7, 8, 6, 7, 8, 9, 4, 5, 6, 7, 5, 6, 7, 8, 6, 7, 8, 9, 7, 8, 9, 10, 5, 6, 7, 8, 6, 7, 8, 9, 7, 8, 9, 10, 8, 9, 10, 11, 4, 5, 6, 7, 5, 6, 7, 8, 6, 7, 8, 9, 7, 8, 9, 10, 5, 6, 7, 8, 6

Offset: 0

Peter Kagey, Jul 16 2015

a(n) = (Weight of quaternary expansion of n) + (length of quaternary expansion of n) - 1.

			For a(308) = 9, the nine steps are: 308 => 77 => 76 => 19 => 18 => 17 => 16 => 4 => 1 => 0.

Peter Kagey, Table of n, a(n) for n = 0..10000

Analogous sequences with a different multiplier k: A056792 (k=2), A061282 (k=3).

Cf. A053737, A110591, A007090: base 4 sequences.

c i = if i `mod` 4 == 0 then i `div` 4 else i - 1
b 0 foldCount = foldCount
b sheetCount foldCount = b (c sheetCount) (foldCount + 1)
a260112 n = b n 0 -- Peter Kagey, Sep 02 2015

a:= n-> (l-> nops(l)+add(i, i=l)-1)(convert(n, base, 4)):
seq(a(n), n=0..105);  # Alois P. Heinz, Jul 16 2015

a(n)=sumdigits(n,4)+#digits(n,4)-1 \\ Charles R Greathouse IV, Jul 16 2015

def a(n); n.to_s(4).length + n.to_s(4).split('').map(&:to_i).reduce(:+) - 1 end

a(n) = A053737(n) + A110591(n) - 1. - Michel Marcus, Jul 17 2015

A037863 a(n) = (number of digits <=1) - (number of digits >1) in base 4 representation of n.

Original entry on oeis.org

1, -1, -1, 2, 2, 0, 0, 0, 0, -2, -2, 0, 0, -2, -2, 3, 3, 1, 1, 3, 3, 1, 1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, 1, 1, -1, -1, -1, -1, -3, -3, -1, -1, -3, -3, 1, 1, -1, -1, 1, 1, -1, -1, -1, -1, -3, -3, -1, -1, -3, -3, 4, 4, 2, 2, 4, 4, 2, 2, 2, 2, 0, 0

Offset: 1

Clark Kimberling

Cf. A110591, A139352.

A037863 := proc(n)
    a := 0 ;
    dgs := convert(n,base,4);
    for i from 1 to nops(dgs) do
        if op(i,dgs)<=1 then
            a := a+1 ;
        else
            a := a-1 ;
        end if;
    end do:
    a ;
end proc: # R. J. Mathar, Oct 16 2015

a(n) = my(d=digits(n, 4)); #select(x->(x<=1), d) - #select(x->(x>1), d); \\ Michel Marcus, Jan 27 2025

a(n) = A110591(n)-2*A139352(n). - R. J. Mathar, Jan 27 2025

A020912 Number of terms in base 4 representation of n-th Fibonacci number.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5, 6, 6, 7, 7, 7, 8, 8, 8, 9, 9, 9, 10, 10, 10, 11, 11, 11, 12, 12, 12, 13, 13, 13, 14, 14, 14, 15, 15, 16, 16, 16, 17, 17, 17, 18, 18, 18, 19, 19, 19, 20, 20, 20, 21, 21, 21, 22, 22, 22, 23, 23, 24, 24, 24, 25, 25, 25, 26, 26, 26, 27, 27

Offset: 1

Clark Kimberling

A020912 := proc(n) nops(convert(combinat[fibonacci](n),base,4)) ; end: seq(A020912(n),n=1..120) ; # R. J. Mathar, Feb 18 2008

Length/@IntegerDigits[Fibonacci[Range[80]],4] (* Harvey P. Dale, Jul 27 2011 *)

a(n) = A110591(A000045(n)). - R. J. Mathar, Feb 18 2008

More terms from R. J. Mathar, Feb 18 2008

A110595 a(1)=5. For n > 1, a(n) = 4*5^(n-1) = A005054(n).

Original entry on oeis.org

5, 20, 100, 500, 2500, 12500, 62500, 312500, 1562500, 7812500, 39062500, 195312500, 976562500, 4882812500, 24414062500, 122070312500, 610351562500, 3051757812500, 15258789062500, 76293945312500, 381469726562500

Offset: 1

Jonathan Vos Post, Jul 29 2005

a(n) is the number of n-digit integers that contain only even digits (A014263). - Bernard Schott, Nov 11 2022

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5).

Cf. A000351, A007091, A014263, A081604, A110591, A110592, A110593, A110594.

Join[{5},NestList[5#&,20,20]] (* Harvey P. Dale, Jun 19 2013 *)
Rest[CoefficientList[Series[5 x (1 - x)/(1 - 5 x), {x,0,50}], x]] (* G. C. Greubel, Sep 01 2017 *)

my(x='x+O('x^50)); Vec(5*x*(1-x)/(1-5*x)) \\ G. C. Greubel, Sep 01 2017

O.g.f.: 5*x*(1-x)/(1-5*x). - Better definition from R. J. Mathar, May 13 2008

Sum_{n>=1} 1/a(n) = 21/80. - Bernard Schott, Nov 11 2022

Better definition from R. J. Mathar, May 13 2008

Incorrect comment removed by Michel Marcus, Nov 11 2022

cp's OEIS Frontend

A110594 a(1) = 4, a(2) = 12, for n>1: a(n) = 3*4^(n-1).

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A043000 Number of digits in all base-b representations of n, for 2 <= b <= n.

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A049803 a(n) = n mod 3 + n mod 9 + ... + n mod 3^k, where 3^k <= n < 3^(k+1).

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A049804 a(n) = n mod 4 + n mod 16 + ... + n mod 4^k, where 4^k <= n < 4^(k+1).

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A330358 a(n) = n mod 5 + n mod 25 + ... + n mod 5^k, where 5^k <= n < 5^(k+1).

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A048703 Numbers which in base 4 are palindromes and have an even number of digits.

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