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A018900 Sum of two distinct powers of 2. 51
3, 5, 6, 9, 10, 12, 17, 18, 20, 24, 33, 34, 36, 40, 48, 65, 66, 68, 72, 80, 96, 129, 130, 132, 136, 144, 160, 192, 257, 258, 260, 264, 272, 288, 320, 384, 513, 514, 516, 520, 528, 544, 576, 640, 768, 1025, 1026, 1028, 1032, 1040, 1056, 1088, 1152, 1280, 1536, 2049, 2050, 2052, 2056, 2064, 2080, 2112, 2176, 2304, 2560, 3072 (list; table; graph; refs; listen; history; text; internal format)
OFFSET

1,1

COMMENTS

Appears to give all n such that 8 is the highest power of 2 dividing A005148 (n). General conjecture: numbers k such that 8^a is the highest power of 2 dividing A005148 (k) is the same sequence as numbers k such that k has exactly (a+1) 1's in its binary representation. - Benoit Cloitre, Jun 22 2002

Seen as a triangle read by rows, T(n,k) = 2^(k-1) + 2^n, 1 <= k <= n,

   the sum of n-th row equals A087323(n). - Reinhard Zumkeller, Jun 24 2009

A073267(a(n)) = 2. - Reinhard Zumkeller, Mar 07 2012

Numbers whose base-2 sum of digits is 2. - Tom Edgar, Aug 31 2013

All odd terms are A000051. - Robert G. Wilson v, Jan 03 2014

A239708 holds the subsequence of terms m such that m - 1 is prime. - Hieronymus Fischer, Apr 20 2014

LINKS

T. D. Noe and Hieronymus Fischer, Table of n, a(n) for n = 1..10000 [terms 1..1000 from T. D. Noe]

Beeler, M., Gosper, R. W. and Schroeppel, R., HAKMEM. MIT AI Memo 239, Feb 29 1972 (Item 175).

Tilman Piesk, Square array in reverse binary

FORMULA

a(n) = 2^trinv(n-1) + 2^((n-1)-((trinv(n-1)*(trinv(n-1)-1))/2)), i.e., 2^A002024(n)+2^A002262(n-1). - Antti Karttunen

a(n) = A059268(n-1) + A140513(n-1). A000120(a(n)) = 2. Complement of A161989. A151774(a(n)) = 1. - Reinhard Zumkeller, Jun 24 2009

Start with A000051. If n is in sequence, then so is 2n. - Ralf Stephan, Aug 16 2013

a(n) = A057168(a(n-1)) for n>1 and a(1) = 3. - Marc LeBrun, Jan 01 2014

From Hieronymus Fischer, Apr 20 2014 (Start):

Formulas for a general parameter b according to a(n) = b^i + b^j, i>j>=0; b = 2 for this sequence.

a(n) = b^i + b^j, where i = floor((sqrt(8n - 1) + 1)/2), j = n - 1 - i*(i - 1)/2 [for a Smalltalk implementation see Prog section, method distinctPowersOf: b (2 versions)].

a(A000217(n)) = (b + 1)*b^(n-1) = b^n + b^(n-1).

a(A000217(n)+1) = 1 + b^(n+1).

a(n + 1 + floor((sqrt(8n - 1) + 1)/2)) = b*a(n).

a(n + 1 + floor(log_b(a(n)))) = b*a(n).

a(n + 1) = b^2/(b+1) * a(n) + 1, if n is a triangular number (s. A000217).

a(n + 1) = b*a(n) + (1-b)* b^floor((sqrt(8n - 1) + 1)/2), if n is not a triangular number.

The next term can also be calculated without using the index n. Let m be a term and i = floor(log_b(m)), then:

a(n + 1) = b*m + (1-b)* b^i, if floor(log_b(m/(b+1))) + 1 < i,

a(n + 1) = b^2/(b+1) * m + 1, if floor(log_b(m/(b+1))) + 1 = i.

Partial sum:

sum_{k=1..n} a(k) = (((b-1)*(j+1)+i-1)*b^(i-j) + b)*b^j - i)/(b-1), where i = floor((sqrt(8n - 1) + 1)/2), j = n - 1 - i*(i - 1)/2.

Inverse:

For each sequence term m, the index n such that a(n) = m is determined by n := i*(i-1)/2 + j + 1, where i := floor(log_b(m)), j := floor(log_b(m - b^floor(log_b(m)))) [for a Smalltalk implementation see Prog section, method invertedDistinctPowersOf: b].

Inequalities:

a(n) <= (b+1)/b * b^floor(sqrt(2n)+1/2), equality holds for triangular numbers.

a(n) > b^floor(sqrt(2n)+1/2).

a(n) < b^sqrt(2n)*sqrt(b).

a(n) > b^sqrt(2n)/sqrt(b).

Asymptotic behavior:

lim sup a(n)/b^sqrt(2n) = sqrt(b).

lim inf a(n)/b^sqrt(2n) = 1/sqrt(b).

lim sup a(n)/b^(floor(sqrt(2n))) = b.

lim inf a(n)/b^(floor(sqrt(2n))) = 1.

lim sup a(n)/b^(floor(sqrt(2n)+1/2)) = (b+1)/b.

lim inf a(n)/b^(floor(sqrt(2n)+1/2)) = 1.

(End)

EXAMPLE

From Hieronymus Fischer, Apr 27 2014: (Start)

a(1) = 3, since 2 = 2^1 + 2^0.

a(5) = 10, since 10 = 2^3 + 2^1.

a(10^2) = 16640

a(10^3) = 35184372089344

a(10^4) = 2788273714550169769618891533295908724670464 = 2.788273714550...*10^42

a(10^5) = 3.6341936214780344527466190...*10^134

a(10^6) = 4.5332938264998904048012398...*10^425

a(10^7) = 1.6074616084721302346802429...*10^1346

a(10^8) = 1.4662184497310967196301632...*10^4257

a(10^9) = 2.3037539289782230932863807...*10^13462

a(10^10) = 9.1836811272250798973464436...*10^42571

(End)

MATHEMATICA

Select[ Range[ 1056 ], (Count[ IntegerDigits[ #, 2 ], 1 ]==2)& ]

Union[Total/@Subsets[2^Range[0, 10], {2}]] (* Harvey P. Dale, Mar 04 2012 *)

PROG

(PARI) for(m=1, 9, for(n=0, m-1, print1(2^m+2^n", "))) \\ Charles R Greathouse IV, Sep 09 2011

(PARI) is(n)=hammingweight(n)==2 \\ Charles R Greathouse IV, Mar 03 2014

(PARI) for(n=0, 10^5, if(hammingweight(n)==2, print1(n, ", "))); \\ Joerg Arndt, Mar 04 2014

(Haskell)

a018900 n = a018900_list !! (n-1)

a018900_list = elemIndices 2 a073267_list  -- Reinhard Zumkeller, Mar 07 2012

(C)

unsigned hakmem175(unsigned x) {

    unsigned s, o, r;

    s = x & -x; r = x + s;

    o = x ^ r;  o = (o >> 2) / s;

    return r | o;

}

unsigned A018900(int n) {

    if (n == 1) return 3;

    return hakmem175(A018900(n - 1));

} // Peter Luschny, Jan 01 2014

(Smalltalk)

distinctPowersOf: b

  "Version 1: Answers the n-th number of the form b^i + b^j, i>j>=0, where n is the receiver.

  b > 1 (b = 2, for this sequence).

  Usage: n distinctPowersOf: 2

  Answer: a(n)"

  | n i j |

  n := self.

  i := (8*n - 1) sqrtTruncated + 1 // 2.

  j := n - (i*(i - 1)/2) - 1.

  ^(b raisedToInteger: i) + (b raisedToInteger: j)

[by Hieronymus Fischer, Apr 20 2014]

------------

(Smalltalk)

distinctPowersOf: b

  "Version 2: Answers an array which holds the first n numbers of the form b^i + b^j, i>j>=0, where n is the receiver. b > 1 (b = 2, for this sequence).

  Usage: n distinctPowersOf: 2

  Answer: #(3 5 6 9 10 12 ...) [first n terms]"

  | k p q terms |

  terms := OrderedCollection new.

  k := 0.

  p := b.

  q := 1.

  [k < self] whileTrue:

         [[q < p and: [k < self]] whileTrue:

                   [k := k + 1.

                   terms add: p + q.

                   q := b * q].

         p := b * p.

         q := 1].

  ^terms as Array

[by Hieronymus Fischer, Apr 20 2014]

------------

(Smalltalk)

floorDistinctPowersOf: b

  "Answers an array which holds all the numbers b^i + b^j < n, i>j>=0, where n is the receiver.

  b > 1 (b = 2, for this sequence).

  Usage: n floorDistinctPowersOf: 2

  Answer: #(3 5 6 9 10 12 ...) [all terms < n]"

  | a n p q terms |

  terms := OrderedCollection new.

  n := self.

  p := b.

  q := 1.

  a := p + q.

  [a < n] whileTrue:

         [[q < p and: [a < n]] whileTrue:

                   [terms add: a.

                   q := b * q.

                   a := p + q].

         p := b * p.

         q := 1.

         a := p + q].

  ^terms as Array

[by Hieronymus Fischer, Apr 20 2014]

------------

(Smalltalk)

invertedDistinctPowersOf: b

  "Given a number m which is a distinct power of b, this method answers the index n such that there are uniquely defined i>j>=0 for which b^i + b^j = m, where m is the receiver;  b > 1 (b = 2, for this sequence).

  Usage: m invertedDistinctPowersOf: 2

  Answer: n such that a(n) = m, or, if no such n exists, min (k | a(k) >= m)"

  | n i j k m |

  m := self.

  i := m integerFloorLog: b.

  j := m - (b raisedToInteger: i) integerFloorLog: b.

  n := i * (i - 1) / 2 + 1 + j.

  ^n

[by Hieronymus Fischer, Apr 20 2014]

(Python)

print [n for n in xrange(1, 3001) if bin(n)[2:].count("1")==2] # Indranil Ghosh, Jun 03 2017

CROSSREFS

Cf. A001969, A048639, A048645, A057168.

Cf. A000217, A187813, A239708.

Cf. A000079, A014311, A014312, A014313, A023688, A023689, A023690, A023691 (Hamming weight = 1, 3, 4, ..., 9).

Sum of base-b digits equal b: A226636 (b = 3), A226969 (b = 4), A227062 (b = 5), A227080 (b = 6), A227092 (b = 7), A227095 (b = 8), A227238 (b = 9), A052224 (b = 10). [M. F. Hasler, Dec 23 2016]

Sequence in context: A080309 A281746 A287162 * A126590 A140584 A085705

Adjacent sequences:  A018897 A018898 A018899 * A018901 A018902 A018903

KEYWORD

nonn,easy,nice,tabl,look

AUTHOR

Jonn Dalton (jdalton(AT)vnet.ibm.com), Dec 11 1996

EXTENSIONS

Edited by M. F. Hasler, Dec 23 2016

STATUS

approved

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Last modified August 21 17:52 EDT 2017. Contains 290892 sequences.