cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A001316 Gould's sequence: a(n) = Sum_{k=0..n} (binomial(n,k) mod 2); number of odd entries in row n of Pascal's triangle (A007318); a(n) = 2^A000120(n).

Original entry on oeis.org

1, 2, 2, 4, 2, 4, 4, 8, 2, 4, 4, 8, 4, 8, 8, 16, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 4, 8, 8, 16, 8, 16, 16, 32, 8, 16, 16, 32, 16, 32, 32, 64, 2, 4, 4, 8, 4, 8, 8, 16, 4, 8, 8, 16, 8, 16, 16, 32, 4, 8, 8, 16, 8, 16, 16, 32
Offset: 0

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Author

N. J. A. Sloane

Keywords

Comments

Also called Dress's sequence.
This sequence might be better called Glaisher's sequence, since James Glaisher showed that odd binomial coefficients are counted by 2^A000120(n) in 1899. - Eric Rowland, Mar 17 2017 [However, the name "Gould's sequence" is deeply entrenched in the literature. - N. J. A. Sloane, Mar 17 2017] [Named after the American mathematician Henry Wadsworth Gould (b. 1928). - Amiram Eldar, Jun 19 2021]
All terms are powers of 2. The first occurrence of 2^k is at n = 2^k - 1; e.g., the first occurrence of 16 is at n = 15. - Robert G. Wilson v, Dec 06 2000
a(n) is the highest power of 2 dividing binomial(2n,n) = A000984(n). - Benoit Cloitre, Jan 23 2002
Also number of 1's in n-th row of triangle in A070886. - Hans Havermann, May 26 2002. Equivalently, number of live cells in generation n of a one-dimensional cellular automaton, Rule 90, starting with a single live cell. - Ben Branman, Feb 28 2009. Ditto for Rule 18. - N. J. A. Sloane, Aug 09 2014. This is also the odd-rule cellular automaton defined by OddRule 003 (see Ekhad-Sloane-Zeilberger "Odd-Rule Cellular Automata on the Square Grid" link). - N. J. A. Sloane, Feb 25 2015
Also number of numbers k, 0<=k<=n, such that (k OR n) = n (bitwise logical OR): a(n) = #{k : T(n,k)=n, 0<=k<=n}, where T is defined as in A080098. - Reinhard Zumkeller, Jan 28 2003
To construct the sequence, start with 1 and use the rule: If k >= 0 and a(0),a(1),...,a(2^k-1) are the first 2^k terms, then the next 2^k terms are 2*a(0),2*a(1),...,2*a(2^k-1). - Benoit Cloitre, Jan 30 2003
Also, numerator((2^k)/k!). - Mohammed Bouayoun (mohammed.bouayoun(AT)sanef.com), Mar 03 2004
The odd entries in Pascal's triangle form the Sierpiński Gasket (a fractal). - Amarnath Murthy, Nov 20 2004
Row sums of Sierpiński's Gasket A047999. - Johannes W. Meijer, Jun 05 2011
Fixed point of the morphism "1" -> "1,2", "2" -> "2,4", "4" -> "4,8", ..., "2^k" -> "2^k,2^(k+1)", ... starting with a(0) = 1; 1 -> 12 -> 1224 -> = 12242448 -> 122424482448488(16) -> ... . - Philippe Deléham, Jun 18 2005
a(n) = number of 1's of stage n of the one-dimensional cellular automaton with Rule 90. - Andras Erszegi (erszegi.andras(AT)chello.hu), Apr 01 2006
a(33)..a(63) = A117973(1)..A117973(31). - Stephen Crowley, Mar 21 2007
Or the number of solutions of the equation: A000120(x) + A000120(n-x) = A000120(n). - Vladimir Shevelev, Jul 19 2009
For positive n, a(n) equals the denominator of the permanent of the n X n matrix consisting entirely of (1/2)'s. - John M. Campbell, May 26 2011
Companions to A001316 are A048896, A105321, A117973, A151930 and A191488. They all have the same structure. We observe that for all these sequences a((2*n+1)*2^p-1) = C(p)*A001316(n), p >= 0. If C(p) = 2^p then a(n) = A001316(n), if C(p) = 1 then a(n) = A048896(n), if C(p) = 2^p+2 then a(n) = A105321(n+1), if C(p) = 2^(p+1) then a(n) = A117973(n), if C(p) = 2^p-2 then a(n) = (-1)*A151930(n) and if C(p) = 2^(p+1)+2 then a(n) = A191488(n). Furthermore for all a(2^p - 1) = C(p). - Johannes W. Meijer, Jun 05 2011
a(n) = number of zeros in n-th row of A219463 = number of ones in n-th row of A047999. - Reinhard Zumkeller, Nov 30 2012
This is the Run Length Transform of S(n) = {1,2,4,8,16,...} (cf. A000079). 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). - N. J. A. Sloane, Sep 05 2014
A105321(n+1) = a(n+1) + a(n). - Reinhard Zumkeller, Nov 14 2014
a(n) = A261363(n,n) = number of distinct terms in row n of A261363 = number of odd terms in row n+1 of A261363. - Reinhard Zumkeller, Aug 16 2015
From Gary W. Adamson, Aug 26 2016: (Start)
A production matrix for the sequence is lim_{k->infinity} M^k, the left-shifted vector of M:
1, 0, 0, 0, 0, ...
2, 0, 0, 0, 0, ...
0, 1, 0, 0, 0, ...
0, 2, 0, 0, 0, ...
0, 0, 1, 0, 0, ...
0, 0, 2, 0, 0, ...
0, 0, 0, 1, 0, ...
...
The result is equivalent to the g.f. of Apr 06 2003: Product_{k>=0} (1 + 2*z^(2^k)). (End)
Number of binary palindromes of length n for which the first floor(n/2) symbols are themselves a palindrome (Ji and Wilf 2008). - Jeffrey Shallit, Jun 15 2017

Examples

			Has a natural structure as a triangle:
  1,
  2,
  2,4,
  2,4,4,8,
  2,4,4,8,4,8,8,16,
  2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,
  2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,4,8,8,16,8,16,16,32,8,16,16,32,16,32,32,64,
  ...
The rows converge to A117973.
From _Omar E. Pol_, Jun 07 2009: (Start)
Also, triangle begins:
   1;
   2,2;
   4,2,4,4;
   8,2,4,4,8,4,8,8;
  16,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16;
  32,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,4,8,8,16,8,16,16,32,8,16,16,32,16,32,32;
  64,2,4,4,8,4,8,8,16,4,8,8,16,8,16,16,32,4,8,8,16,8,16,16,32,8,16,16,32,16,32,...
(End)
G.f. = 1 + 2*x + 2*x^2 + 4*x^3 + 2*x^4 + 4*x^5 + 4*x^6 + 8*x^7 + 2*x^8 + ... - _Michael Somos_, Aug 26 2015

References

  • Arthur T. Benjamin and Jennifer J. Quinn, Proofs that really count: the art of combinatorial proof, M.A.A., 2003, p. 75ff.
  • Steven R. Finch, Mathematical Constants, Cambridge, 2003, pp. 145-151.
  • James W. L. Glaisher, On the residue of a binomial-theorem coefficient with respect to a prime modulus, Quarterly Journal of Pure and Applied Mathematics, Vol. 30 (1899), pp. 150-156.
  • H. W. Gould, Exponential Binomial Coefficient Series. Tech. Rep. 4, Math. Dept., West Virginia Univ., Morgantown, WV, Sep 1961.
  • Olivier Martin, Andrew M. Odlyzko, and Stephen Wolfram, Algebraic properties of cellular automata, Comm. Math. Physics, Vol. 93 (1984), pp. 219-258. Reprinted in Theory and Applications of Cellular Automata, S Wolfram, Ed., World Scientific, 1986, pp. 51-90 and in Cellular Automata and Complexity: Collected Papers of Stephen Wolfram, Addison-Wesley, 1994, pp. 71-113
  • Manfred R. Schroeder, Fractals, Chaos, Power Laws, W. H. Freeman, NY, 1991, page 383.
  • 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).
  • Andrew Wuensche, Exploring Discrete Dynamics, Luniver Press, 2011. See Fig. 2.3.

Links

Crossrefs

Equals left border of triangle A166548. - Gary W. Adamson, Oct 16 2009
For generating functions Product_{k>=0} (1+a*x^(b^k)) for the following values of (a,b) see: (1,2) A000012 and A000027, (1,3) A039966 and A005836, (1,4) A151666 and A000695, (1,5) A151667 and A033042, (2,2) A001316, (2,3) A151668, (2,4) A151669, (2,5) A151670, (3,2) A048883, (3,3) A117940, (3,4) A151665, (3,5) A151671, (4,2) A102376, (4,3) A151672, (4,4) A151673, (4,5) A151674.
For partial sums see A006046. For first differences see A151930.
This is the numerator of 2^n/n!, while A049606 gives the denominator.
Cf. A051638, A048967, A007318, A094959, A048896, A117973, A008977, A139541, A048883, A102376, A038573, A159913, A000079, A166548, A006047, A089898, A105321, A061142.
Cf. A047999, A261363, A261366.
If we subtract 1 from the terms we get a pair of essentially identical sequences, A038573 and A159913.
A163000 and A163577 count binomial coefficients with 2-adic valuation 1 and 2. A275012 gives a measure of complexity of these sequences. - Eric Rowland, Mar 15 2017
Cf. A286575 (run-length transform), A368655 (binomial transform), also A037445.

Programs

  • Haskell
    import Data.List (transpose)
a001316 = sum . a047999_row  -- Reinhard Zumkeller, Nov 24 2012
a001316_list = 1 : zs where
   zs = 2 : (concat $ transpose [zs, map (* 2) zs])
-- Reinhard Zumkeller, Aug 27 2014, Sep 16 2011
(Sage, Python)
from functools import cache
@cache
def A001316(n):
    if n <= 1: return n+1
    return A001316(n//2) << n%2
print([A001316(n) for n in range(88)])  # Peter Luschny, Nov 19 2012
  • Maple
    A001316 := proc(n) local k; add(binomial(n,k) mod 2, k=0..n); end;
S:=[1]; S:=[op(S),op(2*s)]; # repeat ad infinitum!
a := n -> 2^add(i,i=convert(n,base,2)); # Peter Luschny, Mar 11 2009
  • Mathematica
    Table[ Sum[ Mod[ Binomial[n, k], 2], {k, 0, n} ], {n, 0, 100} ]
Nest[ Join[#, 2#] &, {1}, 7] (* Robert G. Wilson v, Jan 24 2006 and modified Jul 27 2014 *)
Map[Function[Apply[Plus,Flatten[ #1]]], CellularAutomaton[90,{{1},0},100]] (* Produces counts of ON cells. N. J. A. Sloane, Aug 10 2009 *)
ArrayPlot[CellularAutomaton[90, {{1}, 0}, 20]] (* Illustration of first 20 generations. - N. J. A. Sloane, Aug 14 2014 *)
Table[2^(RealDigits[n - 1, 2][[1]] // Total), {n, 1, 100}] (* Gabriel C. Benamy, Dec 08 2009 *)
CoefficientList[Series[Exp[2*x], {x, 0, 100}], x] // Numerator (* Jean-François Alcover, Oct 25 2013 *)
Count[#,?OddQ]&/@Table[Binomial[n,k],{n,0,90},{k,0,n}] (* _Harvey P. Dale, Sep 22 2015 *)
2^DigitSum[Range[0, 100], 2] (* Paolo Xausa, Jul 31 2025 *)
  • PARI
    {a(n) = if( n<0, 0, numerator(2^n / n!))};
  • PARI
    A001316(n)=1<M. F. Hasler, May 03 2009
  • PARI
    a(n)=2^hammingweight(n) \\ Charles R Greathouse IV, Jan 04 2013
  • Python
    def A001316(n):
    return 2**bin(n)[2:].count("1") # Indranil Ghosh, Feb 06 2017
  • Python
    def A001316(n): return 1<Karl-Heinz Hofmann, Aug 01 2025
  • Python
    import numpy # (version >= 2.0.0)
n_up_to = 2**22
A000079 = 1 << numpy.arange(n_up_to.bit_length())
A001316 = A000079[numpy.bitwise_count(numpy.arange(n_up_to))]
print(A001316[0:100]) # Karl-Heinz Hofmann, Aug 01 2025
  • Scheme
    (define (A001316 n) (let loop ((n n) (z 1)) (cond ((zero? n) z) ((even? n) (loop (/ n 2) z)) (else (loop (/ (- n 1) 2) (* z 2)))))) ;; Antti Karttunen, May 29 2017

Formula

a(n) = 2^A000120(n).
a(0) = 1; for n > 0, write n = 2^i + j where 0 <= j < 2^i; then a(n) = 2*a(j).
a(n) = 2*a(n-1)/A006519(n) = A000079(n)*A049606(n)/A000142(n).
a(n) = A038573(n) + 1.
G.f.: Product_{k>=0} (1+2*z^(2^k)). - Ralf Stephan, Apr 06 2003
a(n) = Sum_{i=0..2*n} (binomial(2*n, i) mod 2)*(-1)^i. - Benoit Cloitre, Nov 16 2003
a(n) mod 3 = A001285(n). - Benoit Cloitre, May 09 2004
a(n) = 2^n - 2*Sum_{k=0..n} floor(binomial(n, k)/2). - Paul Barry, Dec 24 2004
a(n) = Product_{k=0..log_2(n)} 2^b(n, k), b(n, k) = coefficient of 2^k in binary expansion of n. - Paul D. Hanna
Sum_{k=0..n-1} a(k) = A006046(n).
a(n) = n/2 + 1/2 + (1/2)*Sum_{k=0..n} (-(-1)^binomial(n,k)). - Stephen Crowley, Mar 21 2007
G.f. for a(n)/A156769(n): (1/2)*z^(1/2)*sinh(2*z^(1/2)). - Johannes W. Meijer, Feb 20 2009
Equals infinite convolution product of [1,2,0,0,0,0,0,0,0] aerated (A000079 - 1) times, i.e., [1,2,0,0,0,0,0,0,0] * [1,0,2,0,0,0,0,0,0] * [1,0,0,0,2,0,0,0,0]. - Mats Granvik, Gary W. Adamson, Oct 02 2009
a(n) = f(n, 1) with f(x, y) = if x = 0 then y otherwise f(floor(x/2), y*(1 + x mod 2)). - Reinhard Zumkeller, Nov 21 2009
a(n) = 2^(number of 1's in binary form of (n-1)). - Gabriel C. Benamy, Dec 08 2009
a((2*n+1)*2^p-1) = (2^p)*a(n), p >= 0. - Johannes W. Meijer, Jun 05 2011
a(n) = A000120(A001317(n)). - Reinhard Zumkeller, Nov 24 2012
a(n) = A226078(n,1). - Reinhard Zumkeller, May 25 2013
a(n) = lcm(n!, 2^n) / n!. - Daniel Suteu, Apr 28 2017
a(n) = A061142(A005940(1+n)). - Antti Karttunen, May 29 2017
a(0) = 1, a(2*n) = a(n), a(2*n+1) = 2*a(n). - Daniele Parisse, Feb 15 2024
a(n*m) <= a(n)^A000120(m). - Joe Amos, Mar 27 2025

Extensions

Additional comments from Henry Bottomley, Mar 12 2001
Further comments from N. J. A. Sloane, May 30 2009

A001317 Sierpiński's triangle (Pascal's triangle mod 2) converted to decimal.

Original entry on oeis.org

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
Offset: 0

Views

Author

N. J. A. Sloane

Keywords

Comments

The members are all palindromic in binary, i.e., a subset of A006995. - Ralf Stephan, 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 [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 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
Limit_{n->oo} 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 the 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 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 a binary fraction, seems to be equal to 2^k / a(k). - Katarzyna Matylla, Mar 12 2011
From Daniel Forgues, Jun 16-18 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) = Product_{i=0..n-1} F_i^(alpha_i), alpha_i in {0, 1},
this implies
a(2^n+k) = Product_{i=0..n-1} 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 n-th 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

Examples

			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)

References

  • Jean-Paul Allouche and Jeffrey Shallit, Automatic sequences, Cambridge University Press, 2003, p. 113.
  • Henry Wadsworth 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).
  • James J. Tattersall, Elementary Number Theory in Nine Chapters, Cambridge University Press, 1999, pages 136-137.

Links

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.
Cf. A045544.

Programs

  • Haskell
    a001317 = foldr (\u v-> 2*v + u) 0 . map toInteger . a047999_row
-- Reinhard Zumkeller, Nov 24 2012
(Scheme, with memoization-macro 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
  • Maple
    A001317 := proc(n) local k; add((binomial(n,k) mod 2)*2^k, k=0..n); end;
  • Mathematica
    a[n_] := Nest[ BitXor[#, BitShiftLeft[#, 1]] &, 1, n]; Array[a, 42, 0] (* Joel Madigan (dochoncho(AT)gmail.com), Dec 03 2007 *)
NestList[BitXor[#,2#]&,1,50] (* Harvey P. Dale, Aug 02 2021 *)
  • 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
  • Python
    from sympy import binomial
def a(n): return sum([(binomial(n, i)%2)*2**i for i in range(n + 1)]) # Indranil Ghosh, Apr 11 2017
  • Python
    def A001317(n): return int(''.join(str(int(not(~n&k))) for k in range(n+1)),2) # Chai Wah Wu, Feb 04 2022

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 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 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^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). - 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 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)^{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 1-1/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(n-1)) = A048724(a(n-1)).
a(n) = A048723(3,n).
a(n) = A193231(A000079(n)).
For all n >= 0: A268389(a(n)) = n.
(End)
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