Search: a071053 -id:a071053
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1, 27, 27, 125, 27, 729, 125, 1331, 27, 729, 729, 3375, 125, 3375, 1331, 9261, 27, 729, 729, 3375, 729, 19683, 3375, 35937, 125, 3375, 3375, 15625, 1331, 35937, 9261, 79507, 27, 729, 729, 3375, 729, 19683, 3375, 35937, 729, 19683, 19683, 91125, 3375, 91125, 35937, 250047, 125, 3375, 3375, 15625
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OFFSET
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0,2
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COMMENTS
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Number of ON cells at n-th generation of 3-D CA defined by generalization of Rule 150, starting with a single ON cell at generation 0.
Number of odd coefficients in ((1/x+1+x)*(1/y+1+y)*(1/z+1+z))^n.
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LINKS
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N. J. A. Sloane, On the No. of ON Cells in Cellular Automata, Video of talk in Doron Zeilberger's Experimental Math Seminar at Rutgers University, Feb. 05 2015: Part 1, Part 2
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MATHEMATICA
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a71053[n_] := Total[CoefficientList[(x^2 + x + 1)^n, x, Modulus -> 2]];
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A246660
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Run Length Transform of factorials.
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+10
16
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1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 2, 2, 6, 24, 1, 1, 1, 2, 1, 1, 2, 6, 2, 2, 2, 4, 6, 6, 24, 120, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 2, 2, 6, 24, 2, 2, 2, 4, 2, 2, 4, 12, 6, 6, 6, 12, 24, 24, 120, 720, 1, 1, 1, 2, 1, 1, 2, 6, 1, 1, 1, 2, 2, 2, 6, 24, 1, 1, 1
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OFFSET
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0,4
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COMMENTS
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For the definition of the Run Length Transform see A246595.
Only Jordan-Polya numbers (A001013) are terms of this sequence.
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LINKS
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FORMULA
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a(2^n-1) = n!.
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MATHEMATICA
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Table[Times @@ (Length[#]!&) /@ Select[Split[IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 83}] (* Jean-François Alcover, Jul 11 2017 *)
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PROG
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(Sage)
def RLT(f, size):
L = lambda n: [a for a in Integer(n).binary().split('0') if a != '']
return [mul([f(len(d)) for d in L(n)]) for n in range(size)]
A246660_list = lambda len: RLT(factorial, len)
(PARI)
A246660(n) = { my(i=0, p=1); while(n>0, if(n%2, i++; p = p * i, i = 0); n = n\2); p; };
for(n=0, 8192, write("b246660.txt", n, " ", A246660(n)));
(Scheme)
;; A stand-alone loop version, like the Pari-program above:
(define (A246660 n) (let loop ((n n) (i 0) (p 1)) (cond ((zero? n) p) ((odd? n) (loop (/ (- n 1) 2) (+ i 1) (* p (+ 1 i)))) (else (loop (/ n 2) 0 p)))))
;; One based on given recurrence, utilizing memoizing definec-macro from my IntSeq-library:
;; Yet another implementation, using fold:
(define (A246660 n) (fold-left (lambda (a r) (* a (A000142 r))) 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))
(Python)
from operator import mul
from functools import reduce
from re import split
from math import factorial
return reduce(mul, (factorial(len(d)) for d in split('0+', bin(n)[2:]) if d)) if n > 0 else 1 # Chai Wah Wu, Sep 09 2014
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CROSSREFS
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Cf. A003714 (gives the positions of ones).
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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A038184
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State of one-dimensional cellular automaton 'sigma' (Rule 150): 000,001,010,011,100,101,110,111 -> 0,1,1,0,1,0,0,1 at generation n, converted to a decimal number.
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+10
15
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1, 7, 21, 107, 273, 1911, 5189, 28123, 65793, 460551, 1381653, 7039851, 17829905, 124809335, 340873541, 1840690907, 4295032833, 30065229831, 90195689493, 459568513131, 1172543963409, 8207807743863, 22286925370437
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OFFSET
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0,2
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COMMENTS
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Generation n (starting from the generation 0: 1) interpreted as a binary number, but written in base 10.
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LINKS
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Eric Weisstein's World of Mathematics, Rule 150
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EXAMPLE
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Bit patterns with "0" replaced by "." for visibilty [Georg Fischer, Dec 16 2021]:
0: 1
1: 111
2: 1.1.1
3: 11.1.11
4: 1...1...1
5: 111.111.111
6: 1.1...1...1.1
7: 11.11.111.11.11
8: 1.......1.......1
9: 111.....111.....111
10: 1.1.1...1.1.1...1.1.1
11: 11.1.11.11.1.11.11.1.11
12: 1...1.......1.......1...1
13: 111.111.....111.....111.111
14: 1.1...1.1...1.1.1...1.1...1.1
15: 11.11.11.11.11.1.11.11.11.11.11
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MAPLE
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bit_n := (x, n) -> `mod`(floor(x/(2^n)), 2);
sigmagen := proc(n) option remember: if (0 = n) then (1)
else sum('((bit_n(sigmagen(n-1), i)+bit_n(sigmagen(n-1), i-1)+bit_n(sigmagen(n-1), i-2)) mod 2)*(2^i)', 'i'=0..(2*n)) fi: end:
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MATHEMATICA
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f[n_] := Sum[2^k*Coefficient[ #, x, k], {k, 0, 2n}] & @ Expand[(1 + x + x^2)^n, Modulus -> 2] (* Jacob A. Siehler, Aug 25 2006 *)
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PROG
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(PARI)
a(n) = subst(lift(Pol(Mod([1, 1, 1], 2), 'x)^n), 'x, 2);
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CROSSREFS
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This sequence, A071036 and A118110 are equivalent descriptions of the Rule 150 automaton.
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A246035
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Number of odd terms in f^n, where f = (1/x+1+x)*(1/y+1+y).
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+10
13
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1, 9, 9, 25, 9, 81, 25, 121, 9, 81, 81, 225, 25, 225, 121, 441, 9, 81, 81, 225, 81, 729, 225, 1089, 25, 225, 225, 625, 121, 1089, 441, 1849, 9, 81, 81, 225, 81, 729, 225, 1089, 81, 729, 729, 2025, 225, 2025, 1089, 3969, 25, 225, 225, 625, 225, 2025, 625, 3025, 121, 1089, 1089, 3025, 441, 3969, 1849, 7225, 9, 81, 81, 225, 81, 729, 225
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OFFSET
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0,2
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COMMENTS
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This is the number of ON cells in a certain 2-D CA in which the neighborhood of a cell is defined by f, and in which a cell is ON iff there was an odd number of ON cells in the neighborhood at the previous generation.
This is the odd-rule cellular automaton defined by OddRule 777 (see Ekhad-Sloane-Zeilberger "Odd-Rule Cellular Automata on the Square Grid" link).
The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).
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LINKS
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N. J. A. Sloane, On the No. of ON Cells in Cellular Automata, Video of talk in Doron Zeilberger's Experimental Math Seminar at Rutgers University, Feb. 05 2015: Part 1, Part 2
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FORMULA
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EXAMPLE
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Here is the neighborhood:
[X, X, X]
[X, X, X]
[X, X, X]
which contains a(1) = 9 ON cells.
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From Omar E. Pol, Mar 17 2015: (Start)
Apart from the initial 1, the sequence can be written also as an irregular tetrahedron T(s,r,k) = A139818(r+2) * a(k), s>=1, 1<=r<=s, 0<=k<=(A011782(s-r)-1) as shown below:
..
9;
...
9;
25;
..........
9, 81;
25;
121;
....................
9, 81, 81, 225;
25, 225;
121;
441;
........................................
9, 81, 81, 225, 81, 729, 225, 1089;
25, 225, 225, 625;
121, 1089;
441;
1849;
...
Note that every row r is equal to A139818(r+2) times the beginning of the sequence itself, thus in 3D every column contains the same number: T(s,r,k) = T(s+1,r,k).
(End)
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MAPLE
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C:=f->subs({x=1, y=1}, f);
# Find number of ON cells in CA for generations 0 thru M defined by rule
# that cell is ON iff number of ON cells in nbd at time n-1 was odd
# where nbd is defined by a polynomial or Laurent series f(x, y).
OddCA:=proc(f, M) global C; local n, a, i, f2, p;
f2:=simplify(expand(f)) mod 2;
a:=[]; p:=1;
for n from 0 to M do a:=[op(a), C(p)]; p:=expand(p*f2) mod 2; od:
lprint([seq(a[i], i=1..nops(a))]);
end;
f:=(1/x+1+x)*(1/y+1+y);
OddCA(f, 70);
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MATHEMATICA
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b[0] = 1; b[n_] := b[n] = Expand[b[n - 1]*(x^2 + x + 1)];
a[n_] := Count[CoefficientList[b[n], x], _?OddQ]^2;
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A245562
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Table read by rows: row n gives list of lengths of runs of 1's in binary expansion of n, starting with high-order bits.
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+10
12
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0, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 1, 2, 2, 2, 1, 3, 4, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 3, 2, 2, 1, 2, 1, 2, 2, 3, 3, 1, 4, 5, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 1, 3, 1, 4, 2, 2, 1
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OFFSET
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0,4
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COMMENTS
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A formula for A071053(n) depends on this table.
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LINKS
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EXAMPLE
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Here are the run lengths for the numbers 0 through 21:
0, []
1, [1]
2, [1]
3, [2]
4, [1]
5, [1, 1]
6, [2]
7, [3]
8, [1]
9, [1, 1]
10, [1, 1]
11, [1, 2]
12, [2]
13, [2, 1]
14, [3]
15, [4]
16, [1]
17, [1, 1]
18, [1, 1]
19, [1, 2]
20, [1, 1]
21, [1, 1, 1]
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MAPLE
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for n from 0 to 128 do
lis:=[]; t1:=convert(n, base, 2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
elif out1 = 0 and t1[i] = 1 then c:=c+1;
elif out1 = 1 and t1[i] = 0 then c:=c;
elif out1 = 0 and t1[i] = 0 then lis:=[c, op(lis)]; out1:=1; c:=0;
fi;
if i = L1 and c>0 then lis:=[c, op(lis)]; fi;
od:
lprint(n, lis);
od:
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PROG
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(Python)
from re import split
for n in range(1, 100):
....A245562_list.extend(len(d) for d in split('0+', bin(n)[2:]) if d != '')
(PARI) row(n)=my(v=List(), k); while(n, n>>=valuation(n, 2); listput(v, k=valuation(n+1, 2)); n>>=k); Vecrev(v) \\ Charles R Greathouse IV, Oct 21 2016
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CROSSREFS
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Row sums = A000120 (the binary weight).
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KEYWORD
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nonn,base,tabf,easy
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AUTHOR
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STATUS
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approved
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A245563
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Table read by rows: row n gives list of lengths of runs of 1's in binary expansion of n, starting with low-order bits.
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+10
11
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0, 1, 1, 2, 1, 1, 1, 2, 3, 1, 1, 1, 1, 1, 2, 1, 2, 1, 2, 3, 4, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 2, 1, 2, 1, 2, 2, 2, 3, 1, 3, 4, 5, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 3, 1, 4, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 1, 2, 2, 2, 3, 2, 3, 1, 3, 1, 3, 2, 3, 4, 1, 4, 5, 6, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 1, 1
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OFFSET
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0,4
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COMMENTS
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A formula for A071053(n) depends on this table.
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LINKS
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EXAMPLE
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Here are the run lengths for the numbers 0 through 21:
0, []
1, [1]
2, [1]
3, [2]
4, [1]
5, [1, 1]
6, [2]
7, [3]
8, [1]
9, [1, 1]
10, [1, 1]
11, [2, 1]
12, [2]
13, [1, 2]
14, [3]
15, [4]
16, [1]
17, [1, 1]
18, [1, 1]
19, [2, 1]
20, [1, 1]
21, [1, 1, 1]
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MAPLE
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for n from 0 to 128 do
lis:=[]; t1:=convert(n, base, 2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
elif out1 = 0 and t1[i] = 1 then c:=c+1;
elif out1 = 1 and t1[i] = 0 then c:=c;
elif out1 = 0 and t1[i] = 0 then lis:=[op(lis), c]; out1:=1; c:=0;
fi;
if i = L1 and c>0 then lis:=[op(lis), c]; fi;
od:
lprint(n, lis);
od:
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MATHEMATICA
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Join@@Table[Length/@Split[Join@@Position[Reverse[IntegerDigits[n, 2]], 1], #2==#1+1&], {n, 0, 100}] (* Gus Wiseman, Nov 03 2019 *)
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PROG
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(Haskell)
import Data.List (group)
a245563 n k = a245563_tabf !! n !! k
a245563_row n = a245563_tabf !! n
a245563_tabf = [0] : map
(map length . (filter ((== 1) . head)) . group) (tail a030308_tabf)
(Python)
from re import split
for n in range(1, 100):
....A245563_list.extend(len(d) for d in split('0+', bin(n)[:1:-1]) if d != '')
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CROSSREFS
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Row sums = A000120 (the binary weight).
The version for prime indices (instead of binary indices) is A124010.
Numbers with distinct run-lengths are A328592.
Numbers with equal run-lengths are A164707.
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KEYWORD
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nonn,base,tabf
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AUTHOR
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STATUS
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approved
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A246588
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Run Length Transform of S(n) = wt(n) = 0,1,1,2,1,2,2,3,1,... (cf. A000120).
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+10
11
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1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 2, 1, 2, 1, 1, 1, 1, 1
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OFFSET
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0,8
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COMMENTS
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The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).
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LINKS
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MAPLE
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A000120 := proc(n) local w, m, i; w := 0; m :=n; while m > 0 do i := m mod 2; w := w+i; m := (m-i)/2; od; w; end: wt := A000120;
ans:=[];
for n from 0 to 100 do lis:=[]; t1:=convert(n, base, 2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
elif out1 = 0 and t1[i] = 1 then c:=c+1;
elif out1 = 1 and t1[i] = 0 then c:=c;
elif out1 = 0 and t1[i] = 0 then lis:=[c, op(lis)]; out1:=1; c:=0;
fi;
if i = L1 and c>0 then lis:=[c, op(lis)]; fi;
od:
a:=mul(wt(i), i in lis);
ans:=[op(ans), a];
od:
ans;
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MATHEMATICA
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f[n_] := DigitCount[n, 2, 1]; Table[Times @@ (f[Length[#]]&) /@ Select[ Split[ IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 100}] (* Jean-François Alcover, Jul 11 2017 *)
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PROG
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(Haskell)
import Data.List (group)
a246588 = product . map (a000120 . length) .
filter ((== 1) . head) . group . a030308_row
(Python)
from operator import mul
from functools import reduce
from re import split
return reduce(mul, (bin(len(d)).count('1') for d in split('0+', bin(n)[2:]) if d)) if n > 0 else 1 # Chai Wah Wu, Sep 07 2014
A246588_list = lambda len: RLT(lambda n: sum(Integer(n).digits(2)), len)
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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A246595
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Run Length Transform of squares.
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+10
11
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1, 1, 1, 4, 1, 1, 4, 9, 1, 1, 1, 4, 4, 4, 9, 16, 1, 1, 1, 4, 1, 1, 4, 9, 4, 4, 4, 16, 9, 9, 16, 25, 1, 1, 1, 4, 1, 1, 4, 9, 1, 1, 1, 4, 4, 4, 9, 16, 4, 4, 4, 16, 4, 4, 16, 36, 9, 9, 9, 36, 16, 16, 25, 36, 1, 1, 1, 4, 1, 1, 4, 9, 1, 1, 1, 4, 4, 4, 9, 16, 1, 1, 1, 4, 1, 1
(list;
graph;
refs;
listen;
history;
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OFFSET
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0,4
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COMMENTS
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The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g., 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).
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LINKS
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FORMULA
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EXAMPLE
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Written as an irregular triangle in which row lengths is A011782:
1;
1;
1,4;
1,1,4,9;
1,1,1,4,4,4,9,16;
1,1,1,4,1,1,4,9,4,4,4,16,9,9,16,25;
1,1,1,4,1,1,4,9,1,1,1,4,4,4,9,16,4,4,4,16,4,4,16,36,9,9,9,36,16,16,25,36;
...
Right border gives A253909: 1 together with the positive squares.
(End)
Also, the sequence can be written as an irregular tetrahedron T(s,r,k) as shown below:
1;
..
1;
..
1;
4;
.......
1, 1;
4;
9;
...............
1, 1, 1, 4;
4, 4;
9;
16;
.............................
1, 1, 1, 4, 1, 1, 4, 9;
4, 4, 4, 16;
9, 9;
16;
25;
......................................................
1, 1, 1, 4, 1, 1, 4, 9, 1, 1, 1, 4, 4, 4, 9, 16;
4, 4, 4, 16, 4, 4, 16, 36;
9, 9, 9, 36;
16, 16;
25;
36;
...
Apart from the initial 1, we have that T(s,r,k) = T(s+1,r,k).
(End)
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MAPLE
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ans:=[];
for n from 0 to 100 do lis:=[]; t1:=convert(n, base, 2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
elif out1 = 0 and t1[i] = 1 then c:=c+1;
elif out1 = 1 and t1[i] = 0 then c:=c;
elif out1 = 0 and t1[i] = 0 then lis:=[c, op(lis)]; out1:=1; c:=0;
fi;
if i = L1 and c>0 then lis:=[c, op(lis)]; fi;
od:
a:=mul(i^2, i in lis);
ans:=[op(ans), a];
od:
ans;
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MATHEMATICA
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Table[Times @@ (Length[#]^2&) /@ Select[Split[IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 85}] (* Jean-François Alcover, Jul 11 2017 *)
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PROG
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(Python)
from operator import mul
from functools import reduce
from re import split
return reduce(mul, (len(d)**2 for d in split('0+', bin(n)[2:]) if d != '')) if n > 0 else 1 # Chai Wah Wu, Sep 07 2014
A246595_list = lambda len: RLT(lambda n: n^2, len)
(Scheme) ; using MIT/GNU Scheme
(define (A246595 n) (fold-left (lambda (a r) (* a r r)) 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))
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CROSSREFS
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Cf. A003714 (gives the positions of ones).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 1, 2, 2, 2, 5, 14, 1, 1, 1, 2, 1, 1, 2, 5, 2, 2, 2, 4, 5, 5, 14, 42, 1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 1, 2, 2, 2, 5, 14, 2, 2, 2, 4, 2, 2, 4, 10, 5, 5, 5, 10, 14, 14, 42, 132, 1, 1, 1, 2, 1, 1, 2, 5, 1, 1, 1, 2, 2, 2, 5, 14, 1, 1, 1, 2, 1, 1, 2, 5
(list;
graph;
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internal format)
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OFFSET
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0,4
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COMMENTS
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The Run Length Transform of a sequence {S(n), n>=0} is defined to be the sequence {T(n), n>=0} given by T(n) = Product_i S(i), where i runs through the lengths of runs of 1's in the binary expansion of n. E.g. 19 is 10011 in binary, which has two runs of 1's, of lengths 1 and 2. So T(19) = S(1)*S(2). T(0)=1 (the empty product).
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LINKS
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FORMULA
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EXAMPLE
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Written as an irregular triangle in which row lengths are the terms of A011782:
1;
1;
1,2;
1,1,2,5;
1,1,1,2,2,2,5,14;
1,1,1,2,1,1,2,5,2,2,2,4,5,5,14,42;
1,1,1,2,1,1,2,5,1,1,1,2,2,2,5,14,2,2,2,4,2,2,4,10,5,5,5,10,14,14,42,132;
...
Right border gives the Catalan numbers. This is simply a restatement of the theorem that this sequence is the Run Length Transform of A000108.
(End)
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MAPLE
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Cat:=n->binomial(2*n, n)/(n+1);
ans:=[];
for n from 0 to 100 do lis:=[]; t1:=convert(n, base, 2); L1:=nops(t1); out1:=1; c:=0;
for i from 1 to L1 do
if out1 = 1 and t1[i] = 1 then out1:=0; c:=c+1;
elif out1 = 0 and t1[i] = 1 then c:=c+1;
elif out1 = 1 and t1[i] = 0 then c:=c;
elif out1 = 0 and t1[i] = 0 then lis:=[c, op(lis)]; out1:=1; c:=0;
fi;
if i = L1 and c>0 then lis:=[c, op(lis)]; fi;
od:
a:=mul(Cat(i), i in lis);
ans:=[op(ans), a];
od:
ans;
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MATHEMATICA
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f = CatalanNumber; Table[Times @@ (f[Length[#]]&) /@ Select[ Split[ IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 87}] (* Jean-François Alcover, Jul 11 2017 *)
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PROG
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(Python)
from operator import mul
from functools import reduce
from gmpy2 import divexact
from re import split
s, c = bin(n)[2:], [1, 1]
for m in range(1, len(s)):
c.append(divexact(c[-1]*(4*m+2), (m+2)))
return reduce(mul, (c[len(d)] for d in split('0+', s))) if n > 0 else 1
A246596_list = lambda len: RLT(lambda n: binomial(2*n, n)/(n+1), len)
(Scheme) ; using MIT/GNU Scheme
(define (A246596 n) (fold-left (lambda (a r) (* a (A000108 r))) 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))
(define A000108 (EIGEN-CONVOLUTION 1 *))
;; Note: EIGEN-CONVOLUTION can be found from my IntSeq-library and other functions are as in A227349. - Antti Karttunen, Sep 08 2014
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CROSSREFS
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Cf. A003714 (gives the positions of ones).
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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1, 1, 1, 3, 1, 1, 3, 7, 1, 1, 1, 3, 3, 3, 7, 15, 1, 1, 1, 3, 1, 1, 3, 7, 3, 3, 3, 9, 7, 7, 15, 31, 1, 1, 1, 3, 1, 1, 3, 7, 1, 1, 1, 3, 3, 3, 7, 15, 3, 3, 3, 9, 3, 3, 9, 21, 7, 7, 7, 21, 15, 15, 31, 63, 1, 1, 1, 3, 1, 1, 3, 7, 1, 1, 1, 3, 3, 3, 7, 15, 1, 1, 1, 3, 1, 1, 3, 7, 3, 3, 3, 9, 7, 7, 15, 31, 3, 3, 3, 9, 3, 3, 9, 21, 3, 3, 3, 9, 9, 9, 21, 45, 7, 7, 7, 21, 7, 7, 21, 49, 15, 15, 15, 45, 31, 31, 63, 127, 1
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OFFSET
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0,4
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COMMENTS
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a(n) can be also computed by replacing all consecutive runs of zeros in the binary expansion of n with * (multiplication sign), and then performing that multiplication, still in binary, after which the result is converted into decimal. See the example below.
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LINKS
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FORMULA
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EXAMPLE
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115 is '1110011' in binary. The run lengths of 1-runs are 2 and 3, thus a(115) = A000225(2) * A000225(3) = ((2^2)-1) * ((2^3)-1) = 3*7 = 21.
The same result can be also obtained more directly, by realizing that '111' and '11' are the binary representations of 7 and 3, and 7*3 = 21.
Written as an irregular triangle in which row lengths are the terms of A011782:
1;
1;
1,3;
1,1,3,7;
1,1,1,3,3,3,7,15;
1,1,1,3,1,1,3,7,3,3,3,9,7,7,15,31;
1,1,1,3,1,1,3,7,1,1,1,3,3,3,7,15,3,3,3,9,3,3,9,21,7,7,7,21,15,15,31,63;
...
Right border gives 1 together with the positive terms of A000225.
(End)
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MATHEMATICA
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f[n_] := 2^n - 1; Table[Times @@ (f[Length[#]]&) /@ Select[ Split[ IntegerDigits[n, 2]], #[[1]] == 1&], {n, 0, 100}] (* Jean-François Alcover, Jul 11 2017 *)
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PROG
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(MIT/GNU Scheme)
(define (A246674 n) (fold-left (lambda (a r) (* a (A000225 r))) 1 (bisect (reverse (binexp->runcount1list n)) (- 1 (modulo n 2)))))
(Python)
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CROSSREFS
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Cf. A003714 (gives the positions of ones).
A001316 is obtained when the same transformation is applied to A000079, the powers of two.
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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