A140513 -id:A140513 - 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

A059268 Concatenate subsequences [2^0, 2^1, ..., 2^n] for n = 0, 1, 2, ...

Original entry on oeis.org

1, 1, 2, 1, 2, 4, 1, 2, 4, 8, 1, 2, 4, 8, 16, 1, 2, 4, 8, 16, 32, 1, 2, 4, 8, 16, 32, 64, 1, 2, 4, 8, 16, 32, 64, 128, 1, 2, 4, 8, 16, 32, 64, 128, 256, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048

Offset: 0

N. J. A. Sloane, Jan 23 2001

Triangular array T(n,k) read by rows, where T(n,k) = i!*j! times coefficient of x^n*y^k in exp(x+2y).
T(n,k) is the number of subsets of {0,1,...,n} whose largest element is k. To see this, let A be any subset of the 2^k subsets of {0,1,...,k-1}. Then there are 2^k subsets of the form (A U {k}). See example below. - Dennis P. Walsh, Nov 27 2011
Sequence B is called a reluctant sequence of sequence A, if B is triangle array read by rows: row number k coincides with first k elements. A059268 is reluctant sequence of sequence A000079. - Boris Putievskiy, Dec 17 2012

			T(4,3)=8 since there are 8 subsets of {0,1,2,3,4} whose largest element is 3, namely, {3}, {0,3}, {1,3}, {2,3}, {0,1,3}, {0,2,3}, {1,2,3}, and {0,1,2,3}.
Triangle starts:
  1;
  1, 2;
  1, 2, 4;
  1, 2, 4, 8;
  1, 2, 4, 8, 16;
  1, 2, 4, 8, 16, 32;
  ...

Reinhard Zumkeller, Rows n = 0..150 of triangle, flattened
J. L. Arregui, Tangent and Bernoulli numbers related to Motzkin and Catalan numbers by means of numerical triangles, arXiv:math/0109108 [math.NT], 2001.
Boris Putievskiy, Transformations Integer Sequences And Pairing Functions arXiv:1212.2732 [math.CO], 2012.

Cf. A140531.
Cf. A000079.
Cf. A131816.
Row sums give A126646.

a059268 n k = a059268_tabl !! n !! k
a059268_row n = a059268_tabl !! n
a059268_tabl = iterate (scanl (+) 1) [1]
-- Reinhard Zumkeller, Apr 18 2013, Jul 05 2012

Maple
```
seq(seq(2^k,k=0..n),n=0..10);
```

Table[2^k, {n, 0, 11}, {k, 0, n}] // Flatten (* Jean-François Alcover, Jun 10 2013 *)

from math import isqrt
def A059268(n):
    a = (m:=isqrt(k:=n+1<<1))-(k<=m*(m+1))
    return 1<>1) # Chai Wah Wu, Feb 24 2025

E.g.f.: exp(x+2*y) (T coordinates).
a(n) = A018900(n+1) - A140513(n). - Reinhard Zumkeller, Jun 24 2009
T(n,k) = A173786(n-1,k-1) - A173787(n-1,k-1), 0Reinhard Zumkeller, Feb 28 2010
T(n,k) = 2^k. - Reinhard Zumkeller, Jan 29 2010
As a linear array, the sequence is a(n) = 2^(n-1-t*(t+1)/2), where t = floor((-1+sqrt(8*n-7))/2), n>=1. - Boris Putievskiy, Dec 17 2012
As a linear array, the sequence is a(n) = 2^(n-1-t*(t+1)/2), where t = floor(sqrt(2*n)-1/2), n>=1. - Zhining Yang, Jun 09 2017

Formula corrected by Reinhard Zumkeller, Feb 23 2010

A173786 Triangle read by rows: T(n,k) = 2^n + 2^k, 0 <= k <= n.

Original entry on oeis.org

2, 3, 4, 5, 6, 8, 9, 10, 12, 16, 17, 18, 20, 24, 32, 33, 34, 36, 40, 48, 64, 65, 66, 68, 72, 80, 96, 128, 129, 130, 132, 136, 144, 160, 192, 256, 257, 258, 260, 264, 272, 288, 320, 384, 512, 513, 514, 516, 520, 528, 544, 576, 640, 768, 1024, 1025, 1026, 1028, 1032, 1040, 1056, 1088, 1152, 1280, 1536, 2048

Offset: 0

Reinhard Zumkeller, Feb 28 2010

Essentially the same as A048645. - T. D. Noe, Mar 28 2011

			Triangle begins as:
     2;
     3,    4;
     5,    6,    8;
     9,   10,   12,   16;
    17,   18,   20,   24,   32;
    33,   34,   36,   40,   48,   64;
    65,   66,   68,   72,   80,   96,  128;
   129,  130,  132,  136,  144,  160,  192,  256;
   257,  258,  260,  264,  272,  288,  320,  384,  512;
   513,  514,  516,  520,  528,  544,  576,  640,  768, 1024;
  1025, 1026, 1028, 1032, 1040, 1056, 1088, 1152, 1280, 1536, 2048;

T. D. Noe, Rows n = 0..100 of triangle, flattened
Wawrzyniec Bieniawski, Piotr Masierak, Andrzej Tomski, and Szymon Łukaszyk, Assembly Theory - Formalizing Assembly Spaces and Discovering Patterns and Bounds, Preprints.org (2025).

Cf. A048645, A118413, A118416.

Cf. also A087112, A370121.

[2^n + 2^k: k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 07 2021

Flatten[Table[2^n + 2^m, {n,0,10}, {m, 0, n}]] (* T. D. Noe, Jun 18 2013 *)

A173786(n) = { my(c = (sqrtint(8*n + 1) - 1) \ 2); 1 << c + 1 << (n - binomial(c + 1, 2)); }; \\ Antti Karttunen, Feb 29 2024, after David A. Corneth's PARI-program in A048645

from math import isqrt, comb
def A173786(n):
    a = (m:=isqrt(k:=n+1<<1))-(k<=m*(m+1))
    return (1<Chai Wah Wu, Jun 20 2025

flatten([[2^n + 2^k for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jul 07 2021

1 <= A000120(T(n,k)) <= 2.
For n>0, 0<=kA048645(n+1,k+2) and T(n,n) = A048645(n+2,1).
Row sums give A006589(n).
Central terms give A161168(n).
T(2*n+1,n) = A007582(n+1).
T(2*n+1,n+1) = A028403(n+1).
T(n,k) = A140513(n,k) - A173787(n,k), 0<=k<=n.
T(n,k) = A059268(n+1,k+1) + A173787(n,k), 0T(n,k) * A173787(n,k) = A173787(2*n,2*k), 0<=k<=n.
T(n,0) = A000051(n).
T(n,1) = A052548(n) for n>0.
T(n,2) = A140504(n) for n>1.
T(n,3) = A175161(n-3) for n>2.
T(n,4) = A175162(n-4) for n>3.
T(n,5) = A175163(n-5) for n>4.
T(n,n-4) = A110287(n-4) for n>3.
T(n,n-3) = A005010(n-3) for n>2.
T(n,n-2) = A020714(n-2) for n>1.
T(n,n-1) = A007283(n-1) for n>0.
T(n,n) = 2*A000079(n).

Typo in first comment line fixed by Reinhard Zumkeller, Mar 07 2010

A173787 Triangle read by rows: T(n,k) = 2^n - 2^k, 0 <= k <= n.

Original entry on oeis.org

0, 1, 0, 3, 2, 0, 7, 6, 4, 0, 15, 14, 12, 8, 0, 31, 30, 28, 24, 16, 0, 63, 62, 60, 56, 48, 32, 0, 127, 126, 124, 120, 112, 96, 64, 0, 255, 254, 252, 248, 240, 224, 192, 128, 0, 511, 510, 508, 504, 496, 480, 448, 384, 256, 0, 1023, 1022, 1020, 1016, 1008, 992, 960, 896, 768, 512, 0

Offset: 0

Reinhard Zumkeller, Feb 28 2010

			Triangle begins as:
   0;
   1,  0;
   3,  2,  0;
   7,  6,  4,  0;
  15, 14, 12,  8,  0;
  31, 30, 28, 24, 16, 0;

G. C. Greubel, Rows n = 0..50 of the triangle, flattened

[2^n -2^k: k in [0..n], n in [0..15]]; // G. C. Greubel, Jul 13 2021

Table[2^n -2^k, {n,0,15}, {k,0,n}]//Flatten (* G. C. Greubel, Jul 13 2021 *)

flatten([[2^n -2^k for k in (0..n)] for n in (0..15)]) # G. C. Greubel, Jul 13 2021

A000120(T(n,k)) = A025581(n,k).
Row sums give A000337.
Central terms give A020522.
T(2*n+1, n) = A006516(n+1).
T(2*n+3, n+2) = A059153(n).
T(n, k) = A140513(n,k) - A173786(n,k), 0 <= k <= n.
T(n, k) = A173786(n,k) - A059268(n+1,k+1), 0 < k <= n.
T(2*n, 2*k) = T(n,k) * A173786(n,k), 0 <= k <= n.
T(n, 0) = A000225(n).
T(n, 1) = A000918(n) for n>0.
T(n, 2) = A028399(n) for n>1.
T(n, 3) = A159741(n-3) for n>3.
T(n, 4) = A175164(n-4) for n>4.
T(n, 5) = A175165(n-5) for n>5.
T(n, 6) = A175166(n-6) for n>6.
T(n, n-4) = A110286(n-4) for n>3.
T(n, n-3) = A005009(n-3) for n>2.
T(n, n-2) = A007283(n-2) for n>1.
T(n, n-1) = A000079(n-1) for n>0.
T(n, n) = A000004(n).

A104709 Triangle read by rows: T(n,k) = Sum_{j=0..n} 2^(n-j)binomial(j,k) for n >= 0 and 0 <= k <= n; also, Riordan array (1/((1-x)(1-2*x)), x/(1-x)).

Original entry on oeis.org

1, 3, 1, 7, 4, 1, 15, 11, 5, 1, 31, 26, 16, 6, 1, 63, 57, 42, 22, 7, 1, 127, 120, 99, 64, 29, 8, 1, 255, 247, 219, 163, 93, 37, 9, 1, 511, 502, 466, 382, 256, 130, 46, 10, 1, 1023, 1013, 968, 848, 638, 386, 176, 56, 11, 1, 2047, 2036, 1981, 1816, 1486, 1024, 562, 232, 67

Offset: 0

Gary W. Adamson, Mar 19 2005

This array (A104709) is the mirror of the fission, A054143, of the polynomial sequence ((x+1)^n: n >= 0) by the polynomial sequence (q(n,x): n >= 0) given by q(n,x) = x^n + x^(n-1) + ... + x + 1.  See A193842 for the definition of fission. - Clark Kimberling, Aug 07 2011
The elements of the matrix inverse appear to be T^(-1)(n,k) = (-1)^(n+k)*A110813(n,k) assuming the same offset in both triangles. - R. J. Mathar, Mar 15 2013
From Paul Curtz, Jun 12 2019: (Start)
Numerators of the triangle [Curtz, page 15, triangle (E)]:
     1/2;
     3/4,  1/4;
     7/8,  4/8,    1/8;
   15/16, 11/16,  5/16,  1/16;
   31/32, 26/31, 16/32,  6/32, 1/32;
   63/64, 57/64, 42/64, 22/64, 7/64, 1/64;
   ...
Denominators - Numerators: Triangle A054143.
  1;
  1, 3;
  1, 4, 7;
  1, 5, 11, 15;
  ...
(E) is a transform which accelerates the convergence of series.
For log(2) = 1 - 1/2 + 1/3 - 1/4 ... = 0.6931..., we have
  1*(1/2) = 1/2,
  1*(3/4) - (1/2)*(1/4) = 5/8,
  1*(7/8) - (1/2)*(4/8) + (1/3)*(1/8) = 2/3,
  1*(15/16) - (1/2)*(11/16) + (1/3)*(5/16) - (1/4)*1/16 = 131/192,
  ...
This is A068566/A068565. (End)

			Triangle T(n,k) (with rows n >= 0 and columns k = 0..n) begins:
   1;
   3,  1;
   7,  4,  1;
  15, 11,  5,  1;
  31, 26, 16,  6,  1;
  63, 57, 42, 22,  7,  1;
  ...

Paul Curtz, Accélération de la convergence de certaines séries alternées à l'aide des fonctions de sommation, Thèse de 3ème Cycle d'Analyse Numérique, Faculté des Sciences de l'Université de Paris, 4 mai 1965.
Clark Kimberling, Fusion, Fission, and Factors, Fib. Q., 52(3) (2014), 195-202.

Cf. A000124, A000295, A001787 (row sums), A007318, A054143, A055248, A068565, A068566, A104709, A140513.

A104709_row := proc(n) add(add(binomial(n,n-i)*x^(n-k-1),i=0..k),k=0..n-1);
coeffs(sort(%)) end; seq(print(A104709_row(n)),n=1..6); # Peter Luschny, Sep 29 2011

z = 10;
p[n_, x_] := (x + 1)^n;
q[0, x_] := 1; q[n_, x_] := x*q[n - 1, x] + 1;
p1[n_, k_] := Coefficient[p[n, x], x^k];
p1[n_, 0] := p[n, x] /. x -> 0;
d[n_, x_] := Sum[p1[n, k]*q[n - 1 - k, x], {k, 0, n - 1}]
h[n_] := CoefficientList[d[n, x], {x}]
TableForm[Table[Reverse[h[n]], {n, 0, z}]]
Flatten[Table[Reverse[h[n]], {n, -1, z}]] (* A054143 *)
TableForm[Table[h[n], {n, 0, z}]]
Flatten[Table[h[n], {n, -1, z}]] (* A104709 *)
(* Clark Kimberling, Aug 07 2011 *)

Begin with A055248 as a triangle, delete leftmost column.
The Riordan array factors as (1/(1-2*x), x)*(1/(1-x), x/(1-x)) - the sequence array for 2^n times Pascal's triangle. - Paul Barry, Aug 05 2005
T(n,k) = Sum_{j=0..n-k} C(n-j, k)*2^j. - Paul Barry, Jan 12 2006
T(n,k) = 3*T(n-1,k) + T(n-1,k-1) - 2*T(n-2,k) - 2*T(n-2,k-1), T(0,0) = 1, T(n,k) = 0 if k < 0 or if k > n. - Philippe Deléham, Nov 30 2013
Working with an offset of 0, we have exp(x) * (e.g.f. for row n) = (e.g.f. for diagonal n). For example, for n = 3 we have exp(x)*(15 + 11*x + 5*x^2/2! + x^3/3!) = 15 + 26*x + 42*x^2/2! + 64*x^3/3! + 93*x^4/4! + .... The same property holds more generally for Riordan arrays of the form (f(x), x/(1 - x)). - Peter Bala, Dec 21 2014
From Petros Hadjicostas, Jun 05 2020: (Start)
Bivariate o.g.f.: A(x,y) = Sum_{n,k >= 0} T(n,k)*x^n*y^k = 1/(1 - 3*x - x*y + 2*x^2 + 2*x^2*y) = 1/((1 - 2*x)*(1 - x*(y+1))).
The o.g.f. of the n-th row is (2^(n+1) - (1 + y)^(n+1))/(1 - y).
Let B(x,y) be the bivariate o.g.f. of triangular array A054143. Because A054143 is the mirror image of the current array, we have A(x,y) = B(x*y, 1/y) and B(x,y) = A(x*y, 1/y). This makes it easy to identify lower diagonals of the array.
For example, if we want to identify the second lower diagonal of the array (i.e., 7, 11, 16, 22, ...), we take the 2nd derivative of B(x,y) with respect to y, set y = 0, and divide by 2!. (Note that columns in A054143 start at k = 0.) We get the g.f. x^2*(7 - 10*x + 4*x^2)/(1 - x)^3.
It is then easy to derive that T(n,n-2) = A000124(n+1) = (n+1)*(n+2)/2 + 1 for n >= 2 (by ignoring the first three terms of A000124). Of course, in the current case, it is much easier to use the formula for T(n,k) to find T(n,n-2). (End)
T(n,0) = 2^(n+1) - 1 for n >= 0; T(n,k) = T(n-1,k) + T(n-1,k-1) for 1 <= k <= n. - Peter Bala, Jan 30 2023
T(n,1) = 2^(n+1) - n - 2 = A000295(n+1) for n >= 1. - Bernard Schott, Feb 22 2023

Name edited and offset changed by Petros Hadjicostas, Jun 04 2020

A140589 Triangle A(k,n) = (-2)^k+2^n read by rows.

Original entry on oeis.org

2, -1, 0, 5, 6, 8, -7, -6, -4, 0, 17, 18, 20, 24, 32, -31, -30, -28, -24, -16, 0, 65, 66, 68, 72, 80, 96, 128, -127, -126, -124, -120, -112, -96, -64, 0, 257, 258, 260, 264, 272, 288, 320, 384, 512, -511, -510, -508, -504, -496, -480, -448, -384, -256, 0, 1025, 1026, 1028, 1032

Offset: 0

Paul Curtz, Jul 06 2008

The flattened sequence a(A000217(k)+j)=A(k,j) obeys a(n+1)-2a(n)= -5, 2, 5, -4, -4, -23, 8, 8, 8, 17, -16, -16, -16, -16, -95, ..., which is a dispersion of 2, -4, -4, 8, 8, 8, ... (a signed version of A140513) with -5, 5, -23, 17, -95, 65,... The latter sequence is A(k,0)-2*A(k-1,k-1), an alternation of the negative of A140529 with each second element of A000051.

			Rows starting at k=0: (2), (-1,0); (5, 6, 8); (-7,-6,-4,0); (17,18,20,24,32);...

Dana G. Korssjoen, Biyao Li, Stefan Steinerberger, Raghavendra Tripathi, and Ruimin Zhang, Finding structure in sequences of real numbers via graph theory: a problem list, arXiv:2012.04625, Dec 08, 2020

A(k,n) = A000079(n)+A122803(k).

Edited by R. J. Mathar, Jul 08 2008

A140946 Triangle T(n,k) = (-2)^n*(-1)^k if kA001045(n+1).

Original entry on oeis.org

1, -2, -1, 4, -4, 3, -8, 8, -8, -5, 16, -16, 16, -16, 11, -32, 32, -32, 32, -32, -21, 64, -64, 64, -64, 64, -64, 43, -128, 128, -128, 128, -128, 128, -128, -85, 256, -256, 256, -256, 256, -256, 256, -256, 171, -512, 512, -512, 512, -512, 512, -512, 512, -512, -341, 1024, -1024, 1024, -1024, 1024

Offset: 0

Paul Curtz, Jul 24 2008

The sequence appears if the values b(n+1)-2*b(n) are computed from the (flattened) sequence b(.)=A140944.

Reading the triangle by rows, taking absolute values and removing duplicates we obtain A112387.

			1;
-2,-1;
4,-4,3;
-8,8,-8,-5;
16,-16,16,-16,11;
-32,32,-32,32,-32,-21;
64,-64,64,-64,64,-64,43;
-128,128,-128,128,-128,128,-128,-85;

Cf. A140513, A140589.

(* A = A140944 *) A[0, 0] = 0; A[1, 0] = A[0, 1] = 1; A[0, k_] := A[0, k] = A[0, k-1] + 2*A[0, k-2]; A[n_, n_] = 0; A[n_, k_] := A[n, k] = A[n-1, k+1] - A[n-1, k];  T[n_, n_] := T[n, n] = A[n+1, 0] - 2*A[n, n]; T[n_, k_] := T[n, k] = A[n, k+1] - 2*A[n, k]; Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Dec 17 2014 *)

T(n,k) = A140944(n,k+1)-2*A140944(n,k), k

T(n,n) = A140944(n+1,0) -2*A140944(n,n).

Edited by R. J. Mathar, Jul 06 2011

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A018900 Sums of two distinct powers of 2.

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A059268 Concatenate subsequences [2^0, 2^1, ..., 2^n] for n = 0, 1, 2, ...

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A173786 Triangle read by rows: T(n,k) = 2^n + 2^k, 0 <= k <= n.

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A173787 Triangle read by rows: T(n,k) = 2^n - 2^k, 0 <= k <= n.

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A104709 Triangle read by rows: T(n,k) = Sum_{j=0..n} 2^(n-j)*binomial(j,k) for n >= 0 and 0 <= k <= n; also, Riordan array (1/((1-x)*(1-2*x)), x/(1-x)).

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A104709 Triangle read by rows: T(n,k) = Sum_{j=0..n} 2^(n-j)binomial(j,k) for n >= 0 and 0 <= k <= n; also, Riordan array (1/((1-x)(1-2*x)), x/(1-x)).