A087323 -id:A087323 - cp's OEIS frontend

A018900 Sums of two distinct powers of 2.

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

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

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

			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)

T. D. Noe and Hieronymus Fischer, Table of n, a(n) for n = 1..10000 [terms 1..1000 from T. D. Noe]
Michael Beeler, R. William Gosper, and Richard Schroeppel, HAKMEM, MIT Artificial Intelligence Laboratory report AIM-239, February 1972. Item 175 page 81 by Gosper for iterating. Also HTML transcription.
Tilman Piesk, Square array in reverse binary

Cf. A000120, A001969, A048639, A048645, A057168.
Cf. A000217, A179951, 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

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

a018900 n = a018900_list !! (n-1)
a018900_list = elemIndices 2 a073267_list  -- Reinhard Zumkeller, Mar 07 2012

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

Select[ Range[ 1056 ], (Count[ IntegerDigits[ #, 2 ], 1 ]==2)& ]
Union[Total/@Subsets[2^Range[0,10],{2}]] (* Harvey P. Dale, Mar 04 2012 *)

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

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

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

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

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

A018900_list = [2**a+2**b for a in range(1,10) for b in range(a)] # Chai Wah Wu, Jan 24 2021

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

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]
------------

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]
------------

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]
------------

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]

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

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

Clark Kimberling, Aug 02 2011

nonn

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

A075362....A075362........A193665

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}.

			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.

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

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}]]

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

Ralf Stephan, Sep 28 2004

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

G. C. Greubel, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (5,-8,4).

First differences of A066524.

Cf. A135852, A128064, A000225.

[(n+1)*2^(n-1) -1: n in [1..30]]; // G. C. Greubel, Dec 31 2017

Table[(n + 1)*2^(n - 1) - 1, {n,1,30}] (* G. C. Greubel, Dec 31 2017 *)
LinearRecurrence[{5,-8,4},{1,5,15},30] (* Harvey P. Dale, Dec 28 2022 *)

a(n)=(n+1)*2^(n-1)-1 \\ Charles R Greathouse IV, Oct 07 2015

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) = 2n + 1. For 1 < k <= n, T(n, k) = 2T(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

Amarnath Murthy, Sep 03 2003

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

			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;
  ...

Paolo Xausa, Table of n, a(n) for n = 1..11325 (rows 1..150 of the triangle, flattened)

Cf. A087323, A099035, A190730.

A087322row[n_]:=NestList[2#+1&,2n+1,n-1];Array[A087322row,10] (* Paolo Xausa, Oct 17 2023 *)

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)

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

Gerasimov Sergey, Jan 30 2014

nonn

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

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

Cf. A002054, A003261, A196273, A230769.

More terms and corrections of terms and comments by Ralf Stephan, Feb 03 2014

Showing 1-5 of 5 results.

cp's OEIS Frontend

A018900 Sums of two distinct powers of 2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

C

Haskell

Maple

Mathematica

PARI

PARI

PARI

PARI

Python

Python

Python

Smalltalk

Smalltalk

Smalltalk

Smalltalk

Formula

Extensions

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.)

Views

Author

Keywords

Comments

Examples

Crossrefs

Programs

Mathematica

Formula

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

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

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.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

Mathematica

Formula

Extensions

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

Views

Author

Keywords

Comments

Examples

Crossrefs

Extensions

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