A084640 -id:A084640 - cp's OEIS frontend

A171501 Inverse binomial transform of A084640.

0, 1, 3, -1, 7, -9, 23, -41, 87, -169, 343, -681, 1367, -2729, 5463, -10921, 21847, -43689, 87383, -174761, 349527, -699049, 1398103, -2796201, 5592407, -11184809, 22369623, -44739241, 89478487, -178956969, 357913943, -715827881

Offset: 0

Paul Curtz, Dec 10 2009

a(n) and differences are
0,     1,   3,  -1,    7,   -9,
1,     2,  -4,   8,  -16,   32,  =(-1)^(n+1) * A171449(n),
1,    -6,  12, -24,   48,  -96,
-7,   18, -36,  72, -144,  288,
25,  -54, 108, -216, 432, -864,
Vertical: 1) 0 followed with A168589(n).
          2) (-1 followed with A008776(n) ) signed. See A025192(n).
Main diagonal: see A167747(1+n). - Paul Curtz, Jun 16 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (-1,2).

I:=[0, 1, 3]; [n le 3 select I[n] else -Self(n-1) + 2*Self(n-2): n in [1..40]]; // Vincenzo Librandi, Oct 18 2012

CoefficientList[Series[x*(1 + 4*x)/((1 + 2*x)*(1 - x)), {x, 0, 30}], x]
LinearRecurrence[{-1,2},{0,1,3},40] (* Harvey P. Dale, Jan 14 2020 *)

a(n) = A140966(n), n>0.
G.f.: x*(1+4*x) / ( (1+2*x)*(1-x) ). - R. J. Mathar, Jun 14 2011
a(1+n)= (-1)^(1+n) * A001045(1+n) + 2. - Paul Curtz, Jun 16 2011

Mathematica program by Olivier Gérard, Jul 06 2011

A084639 Expansion of x(1+2x)/((1+x)(1-x)(1-2*x)).

Original entry on oeis.org

0, 1, 4, 9, 20, 41, 84, 169, 340, 681, 1364, 2729, 5460, 10921, 21844, 43689, 87380, 174761, 349524, 699049, 1398100, 2796201, 5592404, 11184809, 22369620, 44739241, 89478484, 178956969, 357913940, 715827881, 1431655764, 2863311529, 5726623060, 11453246121

Offset: 0

Paul Barry, Jun 06 2003

Original name was: Generalized Jacobsthal numbers.
This is the sequence A(0,1;1,2;3) of the family of sequences [a,b:c,d:k] considered by G. Detlefs, and treated as A(a,b;c,d;k) in the W. Lang link given below. - Wolfdieter Lang, Oct 18 2010
Entries correspond to value bound adjustment for an N-bit string having M bits set and a(n+1) bit transitions. Wolfram Alpha can easily generate an entry. a(5)=41 stems from input as 1111110_2 - 1010101_2. The subtraction pattern alternates (begins at 1), and bit count is ptr+2 both terms, with the lead term having only its LSB clear. - Bill McEachen, Jul 15 2011
Also a(n) = 2*A000975(n) if n even, a(n) = 2*A000975(n) - 1 if n odd. - Michel Lagneau, Jan 11 2012
In the above comment by Bill McEachen the binary pattern (in an obvious notation) is for even n 1^(n+1)0 - (10)^((n+2)/2) and for odd n 1^(n+1)0 - (10)^((n+1)/2)1. That is for even n a(n) = sum(2^k, k=1..(n+1)) - sum(2^(2*k-1), k=1..(n+2)/2)  = (2^(n+2) - 4)/3, and for odd n a(n) = sum(2^k , k=1..(n+1)) - sum(2^(2*k), k=0..(n+1)/2) = (2^(n+2) - 5)/3. This checks with the formula a(n) = (2^(n+3) + (-1)^n - 9)/6 given below. After a correspondence with Bill McEachen. - Wolfdieter Lang, Jan 24 2014
Michel Lagneau's comment above is equal to the fact that a(n) = A000975(n)-1, or in other words, this sequence gives the partial sums of Jacobsthal sequence, starting from its second 1, A001045(2). From this also follows that this sequence gives the positions of repunits in "Jacobsthal greedy base", A265747. - Antti Karttunen, Dec 17 2015
From Kensuke Matsuoka, Aug 11 2020: (Start)
This sequence is the sum of diagonally arranged powers of 2 repeated in an L shape. For example, a(1)=1, a(2) = 4, a(3)=9, a(4)= 20, a(5)=41, a(6)=84 are obtained from the figure below.
  32
  16  8
   8  4  2
   4  2  1  2
   2  1  2  4  8
   1  2  4  8 16 32
From this figure, a(n) = a(n-2) + 2^n is obtained. (End)
For n > 0, also the total distance that the disks travel from the leftmost peg to the middle peg in the Tower of Hanoi puzzle, in the unique solution with 2^n - 1 moves (see links). - Sela Fried, Dec 17 2023

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Sela Fried, Economically solving the Tower of Hanoi puzzle.
A. F. Horadam, Jacobsthal Representation Numbers, Fib Quart. 34, 40-54, 1996.
Wolfdieter Lang, Notes on certain inhomogeneous three term recurrences.
El-Mehdi Mehiri, Saad Mneimneh, and Hacène Belbachir, The Towers of Fibonacci, Lucas, Pell, and Jacobsthal, arXiv:2502.11045 [math.CO], 2025. See p. 12.
Index entries for linear recurrences with constant coefficients, signature (2,1,-2).

Cf. A000975, A000225, A001045.

Cf. A265747.

[2^(n+2)/3+(-1)^n/6-3/2: n in [0..35]]; // Vincenzo Librandi, Aug 08 2011

a:=proc(n) (2^(n+3) + (-1)^n - 9)/6 end proc: [seq(a(n), n=0..33)]; # Wolfdieter Lang, Jan 24 2014

a[0] = 0; a[1] = 1; a[n_] := a[n] = a[n - 1] + 2 a[n - 2] + 3; Array[a, 32, 0] (* Or *)
a[0] = 0; a[1] = 1; a[n_] := a[n] = 3 a[n - 1] - 2 a[n - 2] + (-1)^n; Array[a, 32, 0]
CoefficientList[Series[x*(1+2*x)/((1+x)*(1-x)*(1-2*x)),{x,0,40}],x] (* or *) LinearRecurrence[{2,1,-2},{0,1,4},40]  (* Vladimir Joseph Stephan Orlovsky, Jan 30 2012 *)

a(n)=2^(n+2)/3-if(n%2,5,4)/3 \\ Charles R Greathouse IV, Aug 08 2011

concat(0, Vec(x*(1+2*x)/((1+x)*(1-x)*(1-2*x)) + O(x^100))) \\ Altug Alkan, Dec 17 2015

def A084639(n): return (4<Chai Wah Wu, Apr 25 2025

G.f.: x*(1+2*x)/((1+x)*(1-x)*(1-2*x)).
E.g.f.: 4*exp(2*x)/3-3*exp(x)/2+exp(-x)/6.
a(n) = a(n-1)+2*a(n-2)+3, a(0)=0, a(1)=1.
a(n) = 2^(n+2)/3+(-1)^n/6-3/2.
a(n) = A001045(n+2) - A000034(n).
a(n) = 5*a(n-2)-4*a(n-4). Cf. A084640, A101622. - Paul Curtz, Apr 03 2008
a(n) = 2*a(n-1) + a(n-2) -2*a(n-3). - R. J. Mathar, Jun 28 2010
a(n) = a(n-1)+2*a(n-2)+3, n>1. - Gary Detlefs, Dec 19 2010
a(n) = 3*a(n-1)-2*a(n-2) +(-1)^n, n>1. - Gary Detlefs, Dec 19 2010
a(n) = a(n-2) + 2^n for n >= 2. - Kensuke Matsuoka, Aug 11 2020

Replaced duplicate of a formula by another recurrence - R. J. Mathar, Jun 28 2010

A171507 a(n) = (52^(n+1)-9-(-1)^n)/6-2n.

Original entry on oeis.org

0, 0, 1, 6, 17, 42, 93, 198, 409, 834, 1685, 3390, 6801, 13626, 27277, 54582, 109193, 218418, 436869, 873774, 1747585, 3495210, 6990461, 13980966, 27961977, 55924002, 111848053, 223696158, 447392369, 894784794, 1789569645, 3579139350, 7158278761, 14316557586

Offset: 0

Paul Curtz, Dec 10 2009

Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3, -1, -3, 2).

Cf. A042948, A084640.

[(5*2^(n+1)-9-(-1)^n)/6 -2*n: n in [0..40]]; // Vincenzo Librandi, Aug 05 2011

A171507:=n->(5*2^(n+1)-9-(-1)^n)/6 -2*n: seq(A171507(n), n=0..50); # Wesley Ivan Hurt, May 03 2017

a(n) = 3*a(n-1)-a(n-2)-3*a(n-3)+2*a(n-4). G.f.: x^2*(1+3*x)/((1+x)*(1-2*x)*(1-x)^2).
a(n) = A084640(n) - A042948(n).
a(n+1)-2*a(n) = A042948(n+1).
First differences: a(n+1)-a(n) = A084640(n).
Last digits: a(n) == a(n+10) (mod 10), n>=1.

Edited and extended by R. J. Mathar, Dec 15 2009

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A171501 Inverse binomial transform of A084640.

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A084639 Expansion of x(1+2x)/((1+x)(1-x)(1-2*x)).

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A171507 a(n) = (52^(n+1)-9-(-1)^n)/6-2n.

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cp's OEIS Frontend

A171501 Inverse binomial transform of A084640.

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Author

Keywords

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Links

Programs

Magma

Mathematica

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A084639 Expansion of x*(1+2*x)/((1+x)*(1-x)*(1-2*x)).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

PARI

Python

Formula

Extensions

A171507 a(n) = (5*2^(n+1)-9-(-1)^n)/6-2*n.

Views

Author

Keywords

Links

Crossrefs

Programs

Magma

Maple

Formula

Extensions

A084639 Expansion of x(1+2x)/((1+x)(1-x)(1-2*x)).

A171507 a(n) = (52^(n+1)-9-(-1)^n)/6-2n.