

A001317


Sierpiński's triangle (Pascal's triangle mod 2) converted to decimal.
(Formerly M2495 N0988)


67



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

0,2


COMMENTS

The members are all palindromic in binary, i.e., a subset of A006995.  Ralf 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 oddsided constructible polygons. See also A047999.  M. Dauchez (mdzzdm(AT)yahoo.fr), Sep 19 2005
Decimal number generated by the binary bits of the nth 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(n1) XOR (2a(n1)), 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).  Gary W. Adamson, Oct 16 2009
Equals row sums of triangle A166555.  Gary W. Adamson, Oct 17 2009
For n>=1, all terms are in A001969.  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).  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.  Vladimir Shevelev, Nov 29 2010
Trisection of this sequence gives rows of A008287 mod 2 converted to decimal. See also A177897, A177960.  Vladimir Shevelev, Jan 02 2011
Converting the rows of the powers of the knomial (k=2^e where e>=1) termwise to binary and reading the concatenation as binary number gives every (k1)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 kth row, treated as binary fraction, seems to be equal to 2^k / a(k).  Katarzyna Matylla, Mar 12 2011
From Daniel Forgues, Jun 1618 2011: (Start)
Since there are 5 known Fermat primes, there are 32 products of distinct Fermat primes (thus there are 31 constructible oddsided polygons, since a polygon has at least 3 sides.) a(0)=1 (empty product) and a(1) to a(31) are those 31 nonproducts 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..n1) F_i^(alpha_i), alpha_i in {0, 1},
this implies
a(2^n+k) = prod_(i=0..n1) F_i^(alpha_i) * F_n, alpha_i in {0, 1}.
(Cf. OEIS Wiki links below.) (End)


REFERENCES

J.P. Allouche & J. Shallit, Automatic sequences, Cambridge University Press, 2003, p. 113.
H. W. Gould, Exponential Binomial Coefficient Series. Tech. Rep. 4, Math. Dept., West Virginia Univ., Morgantown, WV, Sept. 1961.
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).
J. F. Sweeney, Clifford Clock and the Moolakaprithi Cube, http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.404.5350&rep=rep1&type=pdf, 2014. See p. 26.  N. J. A. Sloane, Mar 20 2014


LINKS

T. D. Noe and Tilman Piesk, Table of n, a(n) for n = 0..1023 (first 300 terms by T. D. Noe)
C. Cobeli and A. Zaharescu, Promenade around Pascal TriangleNumber Motives<:a>, Bull. Math. Soc. Sci. Math. Roumanie, Tome 56(104) No. 1, 2013, 7398.  From N. J. A. Sloane, Feb 16 2013
R. K. Guy, The Second Strong Law of Small Numbers, Math. Mag, 63 (1990), no. 1, 320.
D. Hewgill, A relationship between Pascal's triangle and Fermat numbers, Fib. Quart., 15 (1977), 183184.
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, arXiv:1011.6083 [math.NT], 20102012.
OEIS Wiki, Sierpinski's triangle
OEIS Wiki, Constructible oddsided polygons
Index entries for sequences related to cellular automata


FORMULA

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 jth 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 and Ralf 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).  Vladimir Shevelev, Oct 25 2010
a(2^n)=A000215(n); a(2^n1)=a(2^n)2; for n>=1,m>=0,
a(2^(n1)1)*a(2^n*m+2^(n1))= 3*a(2^(n1))*a(2^n*m+2^(n1)2).  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 ThueMorse 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 2nd one by 11/z.  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


EXAMPLE

Given a(5)=51, a(6)=85 since a(5) XOR 2*a(5) = 51 XOR 102 = 85.
From Daniel Forgues, Jun 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)


MAPLE

A001317 := proc(n) local k; add((binomial(n, k) mod 2)*2^k, k=0..n); end;


MATHEMATICA

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 *)


PROG

(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 */


CROSSREFS

Cf. A000215 (Fermat numbers).
Oddnumbered terms give A038183 (1D Cellular Automata Rule 90, "sigma minus").
Not the same as A053576 nor as A045544.
Cf. A047999, A054432, A177882, A177897, A177960, A227434, A230116.
Cf. A249184.
Sequence in context: A003527 A004729 A045544 * A053576 A197818 A077406
Adjacent sequences: A001314 A001315 A001316 * A001318 A001319 A001320


KEYWORD

nonn,base,easy,nice,hear,changed


AUTHOR

N. J. A. Sloane


STATUS

approved



