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 1-5 of 5 results.

A018900 Sums of two distinct powers of 2.

Original entry on oeis.org

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
Offset: 1

Views

Author

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

Keywords

Comments

Appears to give all k such that 8 is the highest power of 2 dividing A005148(k). - 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 the n-th row equals A087323(n). - Reinhard Zumkeller, Jun 24 2009
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

Examples

			From _Hieronymus Fischer_, Apr 27 2014: (Start)
a(1) = 3, since 3 = 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)
		

Crossrefs

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

Programs

  • 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
    
  • Haskell
    a018900 n = a018900_list !! (n-1)
    a018900_list = elemIndices 2 a073267_list  -- Reinhard Zumkeller, Mar 07 2012
    
  • Maple
    a:= n-> (i-> 2^i+2^(n-1-i*(i-1)/2))(floor((sqrt(8*n-1)+1)/2)):
    seq(a(n), n=1..100);  # Alois P. Heinz, Feb 01 2022
  • Mathematica
    Select[ Range[ 1056 ], (Count[ IntegerDigits[ #, 2 ], 1 ]==2)& ]
    Union[Total/@Subsets[2^Range[0,10],{2}]] (* Harvey P. Dale, Mar 04 2012 *)
  • 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
    
  • PARI
    a(n)= my(t=sqrtint(n*8)\/2); 2^t + 2^(n-1-t*(t-1)/2); \\ Ruud H.G. van Tol, Nov 30 2024
    
  • Python
    print([n for n in range(1, 3001) if bin(n)[2:].count("1")==2]) # Indranil Ghosh, Jun 03 2017
    
  • Python
    A018900_list = [2**a+2**b for a in range(1,10) for b in range(a)] # Chai Wah Wu, Jan 24 2021
    
  • Python
    from math import isqrt, comb
    def A018900(n): return (1<<(m:=isqrt(n<<3)+1>>1))+(1<<(n-1-comb(m,2))) # Chai Wah Wu, Oct 30 2024
  • 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]
    

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
A073267(a(n)) = 2. - Reinhard Zumkeller, Mar 07 2012
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(8*n - 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)
Sum_{n>=1} 1/a(n) = A179951. - Amiram Eldar, Oct 06 2020

Extensions

Edited by M. F. Hasler, Dec 23 2016

A193649 Q-residue of the (n+1)st Fibonacci polynomial, where Q is the triangular array (t(i,j)) given by t(i,j)=1. (See Comments.)

Original entry on oeis.org

1, 1, 3, 5, 15, 33, 91, 221, 583, 1465, 3795, 9653, 24831, 63441, 162763, 416525, 1067575, 2733673, 7003971, 17938661, 45954543, 117709185, 301527355, 772364093, 1978473511
Offset: 0

Views

Author

Clark Kimberling, Aug 02 2011

Keywords

Comments

Suppose that p=p(0)*x^n+p(1)*x^(n-1)+...+p(n-1)*x+p(n) is a polynomial of positive degree and that Q is a sequence of polynomials: q(k,x)=t(k,0)*x^k+t(k,1)*x^(k-1)+...+t(k,k-1)*x+t(k,k), for k=0,1,2,... The Q-downstep of p is the polynomial given by D(p)=p(0)*q(n-1,x)+p(1)*q(n-2,x)+...+p(n-1)*q(0,x)+p(n).
Since degree(D(p))
Example: let p(x)=2*x^3+3*x^2+4*x+5 and q(k,x)=(x+1)^k.
D(p)=2(x+1)^2+3(x+1)+4(1)+5=2x^2+7x+14
D(D(p))=2(x+1)+7(1)+14=2x+23
D(D(D(p)))=2(1)+23=25;
the Q-residue of p is 25.
We may regard the sequence Q of polynomials as the triangular array formed by coefficients:
t(0,0)
t(1,0)....t(1,1)
t(2,0)....t(2,1)....t(2,2)
t(3,0)....t(3,1)....t(3,2)....t(3,3)
and regard p as the vector (p(0),p(1),...,p(n)). If P is a sequence of polynomials [or triangular array having (row n)=(p(0),p(1),...,p(n))], then the Q-residues of the polynomials form a numerical sequence.
Following are examples in which Q is the triangle given by t(i,j)=1 for 0<=i<=j:
Q.....P...................Q-residue of P
1.....1...................A000079, 2^n
1....(x+1)^n..............A007051, (1+3^n)/2
1....(x+2)^n..............A034478, (1+5^n)/2
1....(x+3)^n..............A034494, (1+7^n)/2
1....(2x+1)^n.............A007582
1....(3x+1)^n.............A081186
1....(2x+3)^n.............A081342
1....(3x+2)^n.............A081336
1.....A040310.............A193649
1....(x+1)^n+(x-1)^n)/2...A122983
1....(x+2)(x+1)^(n-1).....A057198
1....(1,2,3,4,...,n)......A002064
1....(1,1,2,3,4,...,n)....A048495
1....(n,n+1,...,2n).......A087323
1....(n+1,n+2,...,2n+1)...A099035
1....p(n,k)=(2^(n-k))*3^k.A085350
1....p(n,k)=(3^(n-k))*2^k.A090040
1....A008288 (Delannoy)...A193653
1....A054142..............A101265
1....cyclotomic...........A193650
1....(x+1)(x+2)...(x+n)...A193651
1....A114525..............A193662
More examples:
Q...........P.............Q-residue of P
(x+1)^n...(x+1)^n.........A000110, Bell numbers
(x+1)^n...(x+2)^n.........A126390
(x+2)^n...(x+1)^n.........A028361
(x+2)^n...(x+2)^n.........A126443
(x+1)^n.....1.............A005001
(x+2)^n.....1.............A193660
A094727.....1.............A193657
(k+1).....(k+1)...........A001906 (even-ind. Fib. nos.)
(k+1).....(x+1)^n.........A112091
(x+1)^n...(k+1)...........A029761
(k+1)......A049310........A193663
(In these last four, (k+1) represents the triangle t(n,k)=k+1, 0<=k<=n.)
A051162...(x+1)^n.........A193658
A094727...(x+1)^n.........A193659
A049310...(x+1)^n.........A193664
Changing the notation slightly leads to the Mathematica program below and the following formulation for the Q-downstep of p: first, write t(n,k) as q(n,k). Define r(k)=Sum{q(k-1,i)*r(k-1-i) : i=0,1,...,k-1} Then row n of D(p) is given by v(n)=Sum{p(n,k)*r(n-k) : k=0,1,...,n}.

Examples

			First five rows of Q, coefficients of Fibonacci polynomials (A049310):
1
1...0
1...0...1
1...0...2...0
1...0...3...0...1
To obtain a(4)=15, downstep four times:
D(x^4+3*x^2+1)=(x^3+x^2+x+1)+3(x+1)+1: (1,1,4,5) [coefficients]
DD(x^4+3*x^2+1)=D(1,1,4,5)=(1,2,11)
DDD(x^4+3*x^2+1)=D(1,2,11)=(1,14)
DDDD(x^4+3*x^2+1)=D(1,14)=15.
		

Crossrefs

Cf. A192872 (polynomial reduction), A193091 (polynomial augmentation), A193722 (the upstep operation and fusion of polynomial sequences or triangular arrays).

Programs

  • Mathematica
    q[n_, k_] := 1;
    r[0] = 1; r[k_] := Sum[q[k - 1, i] r[k - 1 - i], {i, 0, k - 1}];
    f[n_, x_] := Fibonacci[n + 1, x];
    p[n_, k_] := Coefficient[f[n, x], x, k]; (* A049310 *)
    v[n_] := Sum[p[n, k] r[n - k], {k, 0, n}]
    Table[v[n], {n, 0, 24}]    (* A193649 *)
    TableForm[Table[q[i, k], {i, 0, 4}, {k, 0, i}]]
    Table[r[k], {k, 0, 8}]  (* 2^k *)
    TableForm[Table[p[n, k], {n, 0, 6}, {k, 0, n}]]

Formula

Conjecture: G.f.: -(1+x)*(2*x-1) / ( (x-1)*(4*x^2+x-1) ). - R. J. Mathar, Feb 19 2015

A099035 a(n) = (n+1)*2^(n-1) - 1.

Original entry on oeis.org

1, 5, 15, 39, 95, 223, 511, 1151, 2559, 5631, 12287, 26623, 57343, 122879, 262143, 557055, 1179647, 2490367, 5242879, 11010047, 23068671, 48234495, 100663295, 209715199, 436207615, 905969663, 1879048191, 3892314111, 8053063679
Offset: 1

Author

Ralf Stephan, Sep 28 2004

Keywords

Comments

Row sums of triangle A135852. - Gary W. Adamson, Dec 01 2007
Binomial transform of [1, 4, 6, 8, 10, 12, 14, 16, ...]. Equals A128064 * A000225, (A000225 starting 1, 3, 7, 15, ...). - Gary W. Adamson, Dec 28 2007

Crossrefs

First differences of A066524.

Programs

Formula

a(n) = A057711(n+1) - 1 = A058966(n+3)/2 = (A087323(n)-1)/2 = (A074494(n+1)-2)/3 = (A003261(n+1)-3)/4 = A036289(n+1)/4 - 1, n>0.
a(n) = A131056(n+1) - 2. - Juri-Stepan Gerasimov, Oct 02 2011
From Colin Barker, Mar 23 2012: (Start)
a(n) = 5*a(n-1) - 8*a(n-2) + 4*a(n-3).
G.f.: x*(1-2*x^2)/((1-x)*(1-2*x)^2). (End)
E.g.f.: ((2*x+1)*exp(2*x) - 2*exp(x) + 1)/2. - G. C. Greubel, Dec 31 2017

A087322 Triangle T read by rows: T(n, 1) = 2*n + 1. For 1 < k <= n, T(n, k) = 2*T(n,k-1) + 1.

Original entry on oeis.org

3, 5, 11, 7, 15, 31, 9, 19, 39, 79, 11, 23, 47, 95, 191, 13, 27, 55, 111, 223, 447, 15, 31, 63, 127, 255, 511, 1023, 17, 35, 71, 143, 287, 575, 1151, 2303, 19, 39, 79, 159, 319, 639, 1279, 2559, 5119, 21, 43, 87, 175, 351, 703, 1407, 2815, 5631, 11263, 23, 47, 95
Offset: 1

Author

Amarnath Murthy, Sep 03 2003

Keywords

Comments

With T(n,0) = n for n >= 0, this becomes J. M. Bergot's triangular array in the definition of A190730. - Petros Hadjicostas, Feb 15 2021

Examples

			Triangle T(n,k) (with rows n >= 1 and columns k = 1..n) begins:
   3;
   5, 11;
   7, 15, 31;
   9, 19, 39,  79;
  11, 23, 47,  95, 191;
  13, 27, 55, 111, 223, 447;
  15, 31, 63, 127, 255, 511, 1023;
  17, 35, 71, 143, 287, 575, 1151, 2303;
  19, 39, 79, 159, 319, 639, 1279, 2559, 5119;
  ...
		

Crossrefs

Programs

Formula

T(n, k) = (n + 1)*2^k - 1 for n >= 1 and 1 <= k <= n.
From Petros Hadjicostas, Feb 15 2021: (Start)
Sum_{k=1..n} T(n,k) = A190730(n).
T(n,2) = 4*n + 3 for n >= 2.
T(n,n) = A087323(n).
T(n,n-1) = A099035(n) = (n+1)*2^(n-1) - 1 for n >= 2.
Recurrence: T(n,k) = 3*T(n,k-1) - 2*T(n,k-2) for n >= 2 and 2 <= k <= n with initial conditions the values of T(n, 1) and T(n,2).
Bivariate o.g.f.: Sum_{n,k>=1} T(n,k)*x^n*y^k = (4*x^3*y^2 - 2*x^2*y - 4*x*y - x + 3)*x*y/((1 - 2*x*y)^2*(1 - x*y)*(1 - x)^2). (End)

Extensions

Edited and extended by David Wasserman, May 06 2005
Name edited by Petros Hadjicostas, Feb 15 2021

A236752 Primes of the form k*2^(k-1) - 1.

Original entry on oeis.org

3, 11, 31, 79, 191, 5119, 245759, 524287, 1114111, 3758096383, 1618481116086271, 653980173926178609468673073657929531391, 5359447279004780799548150067050349330431
Offset: 1

Author

Gerasimov Sergey, Jan 30 2014

Keywords

Comments

Primes in A087323.
Corresponding values of k: 2, 3, 4, 5, 6, 10, 15, 16, 17, 28, 46, 123, ...
The values of k-1 are listed in A230769. - Jeppe Stig Nielsen, Oct 16 2019

Examples

			79 is in this sequence because it is prime and for k = 5, k*2^(k-1) - 1 = 5*2^(5-1) - 1 = 79.
		

Crossrefs

Extensions

More terms and corrections of terms and comments by Ralf Stephan, Feb 03 2014
Showing 1-5 of 5 results.