

A001317


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


90



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. This observation was also made in 1982 by N. L. White (see letter).  N. J. A. Sloane, Jun 15 2015
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
lim_{n>inf} log(a(n))/n = log(2).  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) = Product_{i=0..n1} F_i^(alpha_i), alpha_i in {0, 1},
this implies
a(2^n+k) = Product_{i=0..n1} F_i^(alpha_i) * F_n, alpha_i in {0, 1}.
(Cf. OEIS Wiki links below.) (End)
The bits in the binary expansion of a(n) give the coefficients of the nth power of polynomial (X+1) in ring GF(2)[X]. E.g. 3 ("11" in binary) stands for (X+1)^1, 5 ("101" in binary) stands for (X+1)^2 = (X^2 + 1), and so on.  Antti Karttunen, Feb 10 2016


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).


LINKS

T. D. Noe and Tilman Piesk, Table of n, a(n) for n = 0..1023 (first 300 terms from T. D. Noe)
Gary W. Adamson, Letter to N. J. A. Sloane, Apr 29 1994, and MS "Algorithm for the Chinese Remainder Theorem"
Gary W. Adamson and N. J. A. Sloane, Correspondence, May 1994, including Adamson's MSS "Algorithm for Generating nth Row of Pascal's Triangle, mod 2, from n", and "The Tower of Hanoi Wheel".
C. Cobeli and A. Zaharescu, Promenade around Pascal TriangleNumber Motives, Bull. Math. Soc. Sci. Math. Roumanie, Tome 56(104) No. 1, 2013, 7398.
A. J. Macfarlane, Generating functions for integer sequences defined by the evolution of cellular automata..., Fig. 4.
R. K. Guy, The Second Strong Law of Small Numbers, Math. Mag, 63 (1990), no. 1, 320. [Annotated scanned copy]
R. K. Guy and N. J. A. Sloane, Correspondence, 1988.
D. Hewgill, A relationship between Pascal's triangle and Fermat numbers, Fib. Quart., 15 (1977), 183184.
A. J. Macfarlane, Generating functions for integer sequences defined by the evolution of cellular automata...
Dr. Math, Regular polygon formulas
OEIS Wiki, Constructible oddsided polygons and Sierpinski's triangle.
Eric Weisstein's World of Mathematics, Rule 60 and Rule 102.
V. Shevelev, On Stephan's conjectures concerning Pascal triangle modulo 2, arXiv:1011.6083 [math.NT], 20102012.
J. F. Sweeney, Clifford Clock and the Moolakaprithi Cube, 2014. See p. 26.  N. J. A. Sloane, Mar 20 2014
Neil L. White, Letter to N. J. A. Sloane, Apr 15 1982
R. G. Wilson, V, Letter to N. J. A. Sloane, Aug. 1993
Index entries for sequences related to cellular automata
Index entries for sequences operating on polynomials in ring GF(2)[X]


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) = Product_{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) = Product_{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)^{binomial(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
From Antti Karttunen, Feb 10 2016: (Start)
a(0) = 1, and for n > 1, a(n) = A048720(3, a(n1)) = A048724(a(n1)).
a(n) = A048723(3,n).
a(n) = A193231(A000079(n)).
For all n >= 0: A268389(a(n)) = n.
(End)


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)
(PARI) a=1; for(n=0, 66, print1(a, ", "); a=bitxor(a, a<<1) ); \\ Joerg Arndt, Mar 27 2013
(PARI) A001317(n, a=1)={for(k=1, n, a=bitxor(a, a<<1)); a} \\ M. F. Hasler, Jun 06 2016
(PARI) a(n) = subst(lift(Mod(1+'x, 2)^n), 'x, 2); \\ Gheorghe Coserea, Nov 09 2017
(Haskell)
a001317 = foldr (\u v> 2*v + u) 0 . map toInteger . a047999_row
 Reinhard Zumkeller, Nov 24 2012
(Scheme, with memoizationmacro definec, two variants)
(definec (A001317 n) (if (zero? n) 1 (A048724 (A001317 ( n 1)))))
(definec (A001317 n) (if (zero? n) 1 (A048720bi 3 (A001317 ( n 1))))) ;; Where A048720bi implements the dyadic function given in A048720.
;; Antti Karttunen, Feb 10 2016
(MAGMA) [&+[(Binomial(n, i) mod 2)*2^i: i in [0..n]]: n in [0..41]]; // Vincenzo Librandi, Feb 12 2016
(Python)
from sympy import binomial
def a(n): return sum([(binomial(n, i)%2)*2**i for i in xrange(n + 1)]) # Indranil Ghosh, Apr 11 2017


CROSSREFS

Cf. A000079, A000215 (Fermat numbers), A047999, A048720, A054432, A173019, A177882, A177897, A177960, A193231, A230116, A247282, A249184, A268391.
Cf. A038183 (odd bisection, 1D Cellular Automata Rule 90).
Iterates of A048724 (starting from 1).
Row 3 of A048723.
Positions of records in A268389.
Positions of ones in A268669 and A268384 (characteristic function).
Not the same as A045544 nor as A053576.
Sequence in context: A003527 A004729 A045544 * A053576 A197818 A077406
Adjacent sequences: A001314 A001315 A001316 * A001318 A001319 A001320


KEYWORD

nonn,base,easy,nice,hear


AUTHOR

N. J. A. Sloane


STATUS

approved



