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.

Showing 1-4 of 4 results.

A286574 Sum of the binary weights of the lengths of 1-runs in base-2 representation of n: a(n) = A000523(A286575(n)).

Original entry on oeis.org

0, 1, 1, 1, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 2, 1, 1, 2, 2, 2, 2, 3, 2, 3, 1, 2, 2, 2, 2, 3, 1, 2, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 2, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 1, 2, 2, 2, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 2, 2, 3, 3, 3, 3, 4, 3, 4, 2, 3, 3, 3, 3, 4, 2, 3, 1, 2, 2, 2, 2, 3, 2, 3, 2, 3, 3, 3, 2, 3, 3, 2, 2, 3, 3, 3, 3, 4, 3, 4, 1
Offset: 0

Views

Author

Antti Karttunen, May 28 2017

Keywords

Comments

a(0) = 0 (an empty sum).

Examples

			For n = 27, "11011" in binary, there are two 1-runs, both of length 2, thus a(27) = A000120(2) + A000120(2) = 1 + 1 = 2.
For n = 29, "11101" in binary, there are two 1-runs, of lengths 1 and 3, thus a(29) = A000120(1) + A000120(3) = 1 + 2 = 3.
For n = 61, "111101" in binary, there are two 1-runs, of lengths 1 and 4, thus a(61) = A000120(1) + A000120(4) = 1 + 1 = 2.
		

Crossrefs

Programs

  • Python
    from sympy import factorint, prime, log
    import math
    def wt(n): return bin(n).count("1")
    def a037445(n):
        f=factorint(n)
        return 2**sum([wt(f[i]) for i in f])
    def A(n): return n - 2**int(math.floor(log(n, 2)))
    def b(n): return n + 1 if n<2 else prime(1 + (len(bin(n)[2:]) - bin(n)[2:].count("1"))) * b(A(n))
    def a286575(n): return a037445(b(n))
    def a(n): return int(math.floor(log(a286575(n), 2))) # Indranil Ghosh, May 30 2017
    
  • Python
    # uses RLT function from A278159
    def A286574(n): return len(bin(RLT(n,lambda m: 2**(bin(m).count('1')))))-3 # Chai Wah Wu, Feb 04 2022
  • Scheme
    (define (A286574 n) (A000523 (A286575 n)))
    

Formula

a(n) = A000523(A286575(n)). [Log_2 of run-length transform of A001316.]
a(n) = A064547(A005940(1+n)).

A156552 Unary-encoded compressed factorization of natural numbers.

Original entry on oeis.org

0, 1, 2, 3, 4, 5, 8, 7, 6, 9, 16, 11, 32, 17, 10, 15, 64, 13, 128, 19, 18, 33, 256, 23, 12, 65, 14, 35, 512, 21, 1024, 31, 34, 129, 20, 27, 2048, 257, 66, 39, 4096, 37, 8192, 67, 22, 513, 16384, 47, 24, 25, 130, 131, 32768, 29, 36, 71, 258, 1025, 65536, 43, 131072, 2049, 38, 63, 68, 69, 262144
Offset: 1

Views

Author

Leonid Broukhis, Feb 09 2009

Keywords

Comments

The primes become the powers of 2 (2 -> 1, 3 -> 2, 5 -> 4, 7 -> 8); the composite numbers are formed by taking the values for the factors in the increasing order, multiplying them by the consecutive powers of 2, and summing. See the Example section.
From Antti Karttunen, Jun 27 2014: (Start)
The odd bisection (containing even terms) halved gives A244153.
The even bisection (containing odd terms), when one is subtracted from each and halved, gives this sequence back.
(End)
Question: Are there any other solutions that would satisfy the recurrence r(1) = 0; and for n > 1, r(n) = Sum_{d|n, d>1} 2^A033265(r(d)), apart from simple variants 2^k * A156552(n)? See also A297112, A297113. - Antti Karttunen, Dec 30 2017

Examples

			For 84 = 2*2*3*7 -> 1*1 + 1*2 + 2*4 + 8*8 =  75.
For 105 = 3*5*7 -> 2*1 + 4*2 + 8*4 = 42.
For 137 = p_33 -> 2^32 = 4294967296.
For 420 = 2*2*3*5*7 -> 1*1 + 1*2 + 2*4 + 4*8 + 8*16 = 171.
For 147 = 3*7*7 = p_2 * p_4 * p_4 -> 2*1 + 8*2 + 8*4 = 50.
		

Crossrefs

One less than A005941.
Inverse permutation: A005940 with starting offset 0 instead of 1.
Cf. also A297106, A297112 (Möbius transform), A297113, A153013, A290308, A300827, A323243, A323244, A323247, A324201, A324812 (n for which a(n) is a square), A324813, A324822, A324823, A324398, A324713, A324815, A324819, A324865, A324866, A324867.

Programs

  • Mathematica
    Table[Floor@ Total@ Flatten@ MapIndexed[#1 2^(#2 - 1) &, Flatten[ Table[2^(PrimePi@ #1 - 1), {#2}] & @@@ FactorInteger@ n]], {n, 67}] (* Michael De Vlieger, Sep 08 2016 *)
  • PARI
    a(n) = {my(f = factor(n), p2 = 1, res = 0); for(i = 1, #f~, p = 1 << (primepi(f[i, 1]) - 1); res += (p * p2 * (2^(f[i, 2]) - 1)); p2 <<= f[i, 2]); res}; \\ David A. Corneth, Mar 08 2019
    
  • PARI
    A064989(n) = {my(f); f = factor(n); if((n>1 && f[1,1]==2), f[1,2] = 0); for (i=1, #f~, f[i,1] = precprime(f[i,1]-1)); factorback(f)};
    A156552(n) = if(1==n, 0, if(!(n%2), 1+(2*A156552(n/2)), 2*A156552(A064989(n)))); \\ (based on the given recurrence) - Antti Karttunen, Mar 08 2019
    
  • Perl
    # Program corrected per instructions from Leonid Broukhis. - Antti Karttunen, Jun 26 2014
    # However, it gives correct answers only up to n=136, before corruption by a wrap-around effect.
    # Note that the correct answer for n=137 is A156552(137) = 4294967296.
    $max = $ARGV[0];
    $pow = 0;
    foreach $i (2..$max) {
    @a = split(/ /, `factor $i`);
    shift @a;
    $shift = 0;
    $cur = 0;
    while ($n = int shift @a) {
    $prime{$n} = 1 << $pow++ if !defined($prime{$n});
    $cur |= $prime{$n} << $shift++;
    }
    print "$cur, ";
    }
    print "\n";
    (Scheme, with memoization-macro definec from Antti Karttunen's IntSeq-library, two different implementations)
    (definec (A156552 n) (cond ((= n 1) 0) (else (+ (A000079 (+ -2 (A001222 n) (A061395 n))) (A156552 (A052126 n))))))
    (definec (A156552 n) (cond ((= 1 n) (- n 1)) ((even? n) (+ 1 (* 2 (A156552 (/ n 2))))) (else (* 2 (A156552 (A064989 n))))))
    ;; Antti Karttunen, Jun 26 2014
    
  • Python
    from sympy import primepi, factorint
    def A156552(n): return sum((1<Chai Wah Wu, Mar 10 2023

Formula

From Antti Karttunen, Jun 26 2014: (Start)
a(1) = 0, a(n) = A000079(A001222(n)+A061395(n)-2) + a(A052126(n)).
a(1) = 0, a(2n) = 1+2*a(n), a(2n+1) = 2*a(A064989(2n+1)). [Compare to the entanglement recurrence A243071].
For n >= 0, a(2n+1) = 2*A244153(n+1). [Follows from the latter clause of the above formula.]
a(n) = A005941(n) - 1.
As a composition of related permutations:
a(n) = A003188(A243354(n)).
a(n) = A054429(A243071(n)).
For all n >= 1, A005940(1+a(n)) = n and for all n >= 0, a(A005940(n+1)) = n. [The offset-0 version of A005940 works as an inverse for this permutation.]
This permutations also maps between the partition-lists A112798 and A125106:
A056239(n) = A161511(a(n)). [The sums of parts of each partition (the total sizes).]
A003963(n) = A243499(a(n)). [And also the products of those parts.]
(End)
From Antti Karttunen, Oct 09 2016: (Start)
A161511(a(n)) = A056239(n).
A029837(1+a(n)) = A252464(n). [Binary width of terms.]
A080791(a(n)) = A252735(n). [Number of nonleading 0-bits.]
A000120(a(n)) = A001222(n). [Binary weight.]
For all n >= 2, A001511(a(n)) = A055396(n).
For all n >= 2, A000120(a(n))-1 = A252736(n). [Binary weight minus one.]
A252750(a(n)) = A252748(n).
a(A250246(n)) = A252754(n).
a(A005117(n)) = A277010(n). [Maps squarefree numbers to a permutation of A003714, fibbinary numbers.]
A085357(a(n)) = A008966(n). [Ditto for their characteristic functions.]
For all n >= 0:
a(A276076(n)) = A277012(n).
a(A276086(n)) = A277022(n).
a(A260443(n)) = A277020(n).
(End)
From Antti Karttunen, Dec 30 2017: (Start)
For n > 1, a(n) = Sum_{d|n, d>1} 2^A033265(a(d)). [See comments.]
More linking formulas:
A106737(a(n)) = A000005(n).
A290077(a(n)) = A000010(n).
A069010(a(n)) = A001221(n).
A136277(a(n)) = A181591(n).
A132971(a(n)) = A008683(n).
A106400(a(n)) = A008836(n).
A268411(a(n)) = A092248(n).
A037011(a(n)) = A010052(n) [conjectured, depends on the exact definition of A037011].
A278161(a(n)) = A046951(n).
A001316(a(n)) = A061142(n).
A277561(a(n)) = A034444(n).
A286575(a(n)) = A037445(n).
A246029(a(n)) = A181819(n).
A278159(a(n)) = A124859(n).
A246660(a(n)) = A112624(n).
A246596(a(n)) = A069739(n).
A295896(a(n)) = A053866(n).
A295875(a(n)) = A295297(n).
A284569(a(n)) = A072411(n).
A286574(a(n)) = A064547(n).
A048735(a(n)) = A292380(n).
A292272(a(n)) = A292382(n).
A244154(a(n)) = A048673(n), a(A064216(n)) = A244153(n).
A279344(a(n)) = A279339(n), a(A279338(n)) = A279343(n).
a(A277324(n)) = A277189(n).
A037800(a(n)) = A297155(n).
For n > 1, A033265(a(n)) = 1+A297113(n).
(End)
From Antti Karttunen, Mar 08 2019: (Start)
a(n) = A048675(n) + A323905(n).
a(A324201(n)) = A000396(n), provided there are no odd perfect numbers.
The following sequences are derived from or related to the base-2 expansion of a(n):
A000265(a(n)) = A322993(n).
A002487(a(n)) = A323902(n).
A005187(a(n)) = A323247(n).
A324288(a(n)) = A324116(n).
A323505(a(n)) = A323508(n).
A079559(a(n)) = A323512(n).
A085405(a(n)) = A323239(n).
The following sequences are obtained by applying to a(n) a function that depends on the prime factorization of its argument, which goes "against the grain" because a(n) is the binary code of the factorization of n, which in these cases is then factored again:
A000203(a(n)) = A323243(n).
A033879(a(n)) = A323244(n) = 2*a(n) - A323243(n),
A294898(a(n)) = A323248(n).
A000005(a(n)) = A324105(n).
A000010(a(n)) = A324104(n).
A083254(a(n)) = A324103(n).
A001227(a(n)) = A324117(n).
A000593(a(n)) = A324118(n).
A001221(a(n)) = A324119(n).
A009194(a(n)) = A324396(n).
A318458(a(n)) = A324398(n).
A192895(a(n)) = A324100(n).
A106315(a(n)) = A324051(n).
A010052(a(n)) = A324822(n).
A053866(a(n)) = A324823(n).
A001065(a(n)) = A324865(n) = A323243(n) - a(n),
A318456(a(n)) = A324866(n) = A324865(n) OR a(n),
A318457(a(n)) = A324867(n) = A324865(n) XOR a(n),
A318458(a(n)) = A324398(n) = A324865(n) AND a(n),
A318466(a(n)) = A324819(n) = A323243(n) OR 2*a(n),
A318467(a(n)) = A324713(n) = A323243(n) XOR 2*a(n),
A318468(a(n)) = A324815(n) = A323243(n) AND 2*a(n).
(End)

Extensions

More terms from Antti Karttunen, Jun 28 2014

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

Views

Author

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.

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.
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

A037445 Number of infinitary divisors (or i-divisors) of n.

Original entry on oeis.org

1, 2, 2, 2, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 4, 2, 2, 4, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 2, 8, 2, 4, 4, 4, 4, 4, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 2, 4, 2, 4, 4, 4, 2, 8, 4, 8, 4, 4, 2, 8, 2, 4, 4, 4, 4, 8, 2, 4, 4, 8, 2, 8, 2, 4, 4, 4, 4, 8, 2, 4, 2, 4, 2, 8, 4, 4, 4, 8, 2, 8, 4, 4, 4, 4, 4, 8, 2, 4, 4, 4, 2, 8, 2, 8, 8
Offset: 1

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Author

Keywords

Comments

A divisor of n is called infinitary if it is a product of divisors of the form p^{y_a 2^a}, where p^y is a prime power dividing n and sum_a y_a 2^a is the binary representation of y.
The smallest number m with exactly 2^n infinitary divisors is A037992(n); for these values m, a(m) increases also to a new record. - Bernard Schott, Mar 09 2023

Examples

			For n = 8, n = 2^3 = 2^"11" (writing 3 in binary) so the infinitary divisors are 2^"00" = 1, 2^"01" = 2, 2^"10" = 4 and 2^"11" = 8, so a(8) = 4.
For n = 90, n = 2*5*9 where 2,5,9 are in A050376, so a(90) = 2^3 = 8.
		

Crossrefs

Programs

  • Haskell
    a037445 = product . map (a000079 . a000120) . a124010_row
    -- Reinhard Zumkeller, Mar 19 2013
    
  • Maple
    A037445 := proc(n)
        local a,p;
        a := 1 ;
        for p in ifactors(n)[2] do
            a := a*2^wt(p[2]) ;
        end do:
        a ;
    end proc: # R. J. Mathar, May 16 2016
  • Mathematica
    Table[Length@((Times @@ (First[it]^(#1 /. z -> List)) & ) /@
    Flatten[Outer[z, Sequence @@ bitty /@
    Last[it = Transpose[FactorInteger[k]]], 1]]), {k, 2, 240}]
    bitty[k_] := Union[Flatten[Outer[Plus, Sequence @@ ({0, #1} & ) /@ Union[2^Range[0, Floor[Log[2, k]]]*Reverse[IntegerDigits[k, 2]]]]]]
    y[n_] := Select[Range[0, n], BitOr[n, # ] == n & ] divisors[Infinity][1] := {1}
    divisors[Infinity][n_] := Sort[Flatten[Outer[Times, Sequence @@ (FactorInteger[n] /. {p_, m_Integer} :> p^y[m])]]] Length /@ divisors[Infinity] /@ Range[105] (* Paul Abbott (paul(AT)physics.uwa.edu.au), Apr 29 2005 *)
    a[1] = 1; a[n_] := Times @@ Flatten[ 2^DigitCount[#, 2, 1]&  /@ FactorInteger[n][[All, 2]] ]; Table[a[n], {n, 1, 105}] (* Jean-François Alcover, Aug 19 2013, after Reinhard Zumkeller *)
  • PARI
    A037445(n) = factorback(apply(a -> 2^hammingweight(a), factorint(n)[,2])) \\ Andrew Lelechenko, May 10 2014
    
  • Python
    from sympy import factorint
    def wt(n): return bin(n).count("1")
    def a(n):
        f=factorint(n)
        return 2**sum([wt(f[i]) for i in f]) # Indranil Ghosh, May 30 2017
  • Scheme
    (define (A037445 n) (if (= 1 n) n (* (A001316 (A067029 n)) (A037445 (A028234 n))))) ;; Antti Karttunen, May 28 2017
    

Formula

Multiplicative with a(p^e) = 2^A000120(e). - David W. Wilson, Sep 01 2001
Let n = q_1*...*q_k, where q_1,...,q_k are different terms of A050376. Then a(n) = 2^k (the number of subsets of a set with k elements is 2^k). - Vladimir Shevelev, Feb 19 2011.
a(n) = Product_{k=1..A001221(n)} A000079(A000120(A124010(n,k))). - Reinhard Zumkeller, Mar 19 2013
From Antti Karttunen, May 28 2017: (Start)
a(n) = A286575(A156552(n)). [Because multiplicative with a(p^e) = A001316(e).]
a(n) = 2^A064547(n). (End)
a(A037992(n)) = 2^n. - Bernard Schott, Mar 10 2023

Extensions

Corrected and extended by Naohiro Nomoto, Jun 21 2001
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