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A001317
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Sierpinski's triangle (Pascal's triangle mod 2) converted to decimal.
(Formerly M2495 N0988)
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57
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1, 3, 5, 15, 17, 51, 85, 255, 257, 771, 1285, 3855, 4369, 13107, 21845, 65535, 65537, 196611, 327685, 983055, 1114129, 3342387, 5570645, 16711935, 16843009, 50529027, 84215045, 252645135, 286331153, 858993459, 1431655765, 4294967295, 4294967297, 12884901891, 21474836485, 64424509455, 73014444049, 219043332147, 365072220245, 1095216660735, 1103806595329, 3311419785987
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OFFSET
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0,2
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COMMENTS
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The members are all palindromic in binary, i.e. a subset of A006995. - R. Stephan, Sep 28 2004
a(2n+1) = 3 * a(2n), as follows from a(n)=product_{k in K} (1+2^(2^k)), where K is the set of integers such that n=sum_{k in K} 2^k. - Emmanuel Ferrand, Sep 28 2004
J. H. Conway writes (in Math Forum): at least the first 31 numbers give odd-sided constructible polygons. See also A047999. - M. Dauchez (mdzzdm(AT)yahoo.fr), Sep 19 2005
Decimal number generated by the binary bits of the n-th generation of the Rule 60 elementary cellular automaton. Thus: 1; 0, 1, 1; 0, 0, 1, 0, 1; 0, 0, 0, 1, 1, 1, 1; 0, 0, 0, 0, 1, 0, 0, 0, 1; .. - Eric W. Weisstein, Apr 08, 2006
One can generate this sequence using simple bitwise operations: a(n) = a(n-1) XOR (2a(n-1)), a(0)=1 where XOR is bitwise XOR. - Joel Madigan (dochoncho(AT)gmail.com), Dec 03 2007. Corrected by Dmitry Kamenetsky, Jun 08 2010
limit(n->inf) log(a(n)-1)/log(3) = n*log(2)/log(3) - Bret Mulvey, May 17 2008
Equals row sums of triangle A166548; e.g., 17 = (2 + 4 + 6 + 4 + 1). [From Gary W. Adamson, Oct 16 2009]
Equals row sums of triangle A166555 [From Gary W. Adamson, Oct 17 2009]
For n>=1, all terms are in A001969. [From Vladimir Shevelev, Oct 25 2010]
Let n,m>=0 be such that no carries occur when adding them. Then a(n+m)=a(n)*a(m). [From Vladimir Shevelev, Nov 28 2010]
Let phi_a(n) be the number of a(k)<=a(n) and respectively prime to a(n) (i.e.totient function over {a(n)}). Then, for n>=1, phi_a(n)= 2^(v(n)), where v(n) is number of 0's in the binary representation of n. [From Vladimir Shevelev, Nov 29 2010]
Trisection of this sequence gives rows of A008287 mod 2 converted to decimal. See also A177897, A177960. [From Vladimir Shevelev, Jan 02 2011]
Converting the rows of the powers of the k-nomial (k=2^e where e>=1) term-wise to binary and reading the concatenation as binary number gives every (k-1)st term of this sequence. Similarly with powers p^k of any prime. It might be interesting to study how this fails for powers of composites. - Joerg Arndt, Jan 07 2011
This sequence appears in Pascal's triangle mod 2 in another way, too. If we write it as
1111111...
10101010...
11001100...
10001000...
we get (taking the period part in each row):
.(1) (base 2) = 1
.(10) = 2/3
.(1100) = 12/15 = 4/5
.(1000) = 8/15
The k-th row, treated as binary fraction, seems to be equal to 2^k / a(k). - Katarzyna Matylla, Mar 12 2011
From Daniel Forgues, June 16-17-18 2011: (Start)
Since there are 5 known Fermat primes, there are 32 products of distinct
Fermat primes (thus there are 31 constructible odd-sided polygons, since a polygon has at least 3 sides.) a(0)=1 (empty product) and a(1) to a(31) are those 31 non-products of distinct Fermat primes.
It can be proved by induction that all terms of this sequence are products of distinct Fermat numbers (A000215):
a(0)=1 (empty product) are products of distinct Fermat numbers in { };
a(2^n+k) = a(k) * (2^(2^n)+1) = a(k) * F_n, n >= 0, 0 <= k <= 2^n - 1.
Thus for n >= 1, 0 <= k <= 2^n - 1, and
a(k) = prod_(i=0..n-1) F_i^(alpha_i), alpha_i in {0, 1},
this implies
a(2^n+k) = prod_(i=0..n-1) F_i^(alpha_i) * F_n, alpha_i in {0, 1}.
(Cf. OEIS Wiki links below) (End)
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REFERENCES
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J.-P. Allouche & J. Shallit, Automatic sequences, Cambridge Univeristy Press, 2003, p 113
C. Cobeli and A. Zaharescu, Promenade around Pascal Triangle-Number Motives, Bull. Math. Soc. Sci. Math. Roumanie, Tome 56(104) No. 1, 2013, 73-98; http://rms.unibuc.ro/bulletin/pdf/56-1/PromenadePascalPart1.pdf. - From N. J. A. Sloane, Feb 16 2013
H. W. Gould, Exponential Binomial Coefficient Series. Tech. Rep. 4, Math. Dept., West Virginia Univ., Morgantown, WV, Sept. 1961.
R. K. Guy, The second strong law of small numbers. Math. Mag. 63 (1990), no. 1, 3-20.
D. Hewgill, A relationship between Pascal's triangle and Fermat numbers, Fib. Quart., 15 (1977), 183-184.
N. J. A. Sloane, A Handbook of Integer Sequences, Academic Press, 1973 (includes this sequence).
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
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LINKS
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Tilman Piesk, Table of n, a(n) for n = 0..1023 (first 300 terms by T. D. Noe).
Index entries for sequences related to cellular automata
Dr. Math, Regular polygon formulas
Eric Weisstein's World of Mathematics, Rule 60
Eric Weisstein's World of Mathematics, Rule 102
V. Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2
OEIS Wiki, Sierpinski's triangle
OEIS Wiki, Constructible odd-sided polygons
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FORMULA
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a(n+1) = a(n) XOR 2*a(n), where XOR is binary exclusive OR operator. - Paul D. Hanna, Apr 27 2003
a(n)=prod(e(j, n)=1, 2^(2^j)+1) where e(j, n) is the j-th least significant digit in the binary representation of n (Roberts : see Allouche & Shallit) - Benoit Cloitre, Jun 08 2004
a(2*n+1) = 3*a(2*n). Proof: Since a(n)=product_{k in K} (1+2^(2^k)), where K is the set of integers such that n=sum_{k in K} 2^k, clearly K(2*n+1) = K(2*n) union {0}, hence a(2*n+1)=(1+2^(2^0))*a(2*n)=3*a(2*n). - Emmanuel Ferrand, Sep 28 2004 - R. Stephan, Sep 28 2004
a(32*n) = 3 ^ (32 * n * log(2) / log(3)) + 1 - Bret Mulvey, May 17 2008
For n>=1, A000120(a(n))=2^A000120(n). [From Vladimir Shevelev, Oct 25 2010]
a(2^n)=A000215(n); a(2^n-1)=a(2^n)-2; for n>=1,m>=0,
a(2^(n-1)-1)*a(2^n*m+2^(n-1))= 3*a(2^(n-1))*a(2^n*m+2^(n-1)-2). [From Vladimir Shevelev, Nov 28 2010]
Sum{k>=0} 1/a(k) = Prod{n>=0} (1+1/F_n), where F_n=A000215(n);
Sum{k>=0} (-1)^(m(k))/a(k) = 1/2, where {m(n)} is Thue-Morse sequence (A010060).
If F_n is defined by F_n(z)=z^(2^n)+1 and a(n) by (1/2)*Sum{i>=0}(1-(-1)^{binom(n,i)})*z^i, then, for z>1, the latter two identities hold as well with the replacement 1/2 in the right hand side of the 2-nd one by 1-1/z.[From Vladimir Shevelev, Nov 29 2010]
G.f.: Product_{k>=0} ( 1 + z^(2^k) + (2*z)^(2^k) ). [Conjectured by Shamil Shakirov, proved by Vladimir Shevelev]
a(n) = A000225(n+1) - A219843(n). - Reinhard Zumkeller, Nov 30 2012
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EXAMPLE
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Given a(5)=51, a(6)=85 since a(5) XOR 2*a(5) = 51 XOR 102 = 85.
From Daniel Forgues, June 18 2011: (Start)
a(0) = 1 (empty product;)
a(1) = 3 = 1 * F_0 = a(2^0+0) = a(0) * F_0;
a(2) = 5 = 1 * F_1 = a(2^1+0) = a(0) * F_1;
a(3) = 15 = 3 * 5 = F_0 * F_1 = a(2^1+1) = a(1) * F_1;
a(4) = 17 = 1 * F_2 = a(2^2+0) = a(0) * F_2;
a(5) = 51 = 3 * 17 = F_0 * F_2 = a(2^2+1) = a(1) * F_2;
a(6) = 85 = 5 * 17 = F_1 * F_2 = a(2^2+2) = a(2) * F_2;
a(7) = 255 = 3 * 5 * 17 = F_0 * F_1 * F_2 = a(2^2+3) = a(3) * F_2;
... (End)
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MAPLE
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A001317 := proc(n) local k; add((binomial(n, k) mod 2)*2^k, k=0..n); end;
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MATHEMATICA
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f[n_] := Nest[ BitXor[#, BitShiftLeft[#, 1]] &, 1, n]; Array[f, 42, 0] - Joel Madigan (dochoncho(AT)gmail.com), Dec 03 2007
f[n_] := FromDigits[ Table[ Mod[ Binomial[n, k], 2], {k, 0, n}], 2]; Array[f, 42, 0] (* Robert G. Wilson v *)
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PROG
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(PARI) a(n)=sum(i=0, n, (binomial(n, i)%2)*2^i)
(Haskell)
a001317 = foldr (\u v-> 2*v + u) 0 . map toInteger . a047999_row
-- Reinhard Zumkeller, Nov 24 2012
(PARI) a=1; for(n=0, 66, print1(a, ", "); a=bitxor(a, a<<1) ); /* Joerg Arndt, Mar 27 2013 */
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CROSSREFS
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Cf. A000215 (Fermat numbers).
Odd-numbered terms give A038183 (1D Cellular Automata Rule 90, "sigma minus")
Not the same as A053576 nor as A045544.
Cf. A047999, A054432, A177882, A177897, A177960
Sequence in context: A003527 A004729 A045544 * A053576 A197818 A077406
Adjacent sequences: A001314 A001315 A001316 * A001318 A001319 A001320
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KEYWORD
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nonn,base,easy,nice
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AUTHOR
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N. J. A. Sloane.
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EXTENSIONS
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Corrected Joel Madigan's formula Dmitry Kamenetsky, Jun 08 2010
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STATUS
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approved
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