A000034 -id:A000034 - cp's OEIS frontend

A083906 Table read by rows: T(n, k) is the number of length n binary words with exactly k inversions.

1, 2, 3, 1, 4, 2, 2, 5, 3, 4, 3, 1, 6, 4, 6, 6, 6, 2, 2, 7, 5, 8, 9, 11, 9, 7, 4, 3, 1, 8, 6, 10, 12, 16, 16, 18, 12, 12, 8, 6, 2, 2, 9, 7, 12, 15, 21, 23, 29, 27, 26, 23, 21, 15, 13, 7, 4, 3, 1, 10, 8, 14, 18, 26, 30, 40, 42, 48, 44, 46, 40, 40, 30, 26, 18, 14, 8, 6, 2, 2

Offset: 0

Alford Arnold, Jun 19 2003

There are A033638(n) values in the n-th row, compliant with the order of the polynomial.
In the example for n=6 detailed below, the orders of [6, k]_q are 1, 6, 9, 10, 9, 6, 1 for k = 0..6,
the maximum order 10 defining the row length.
Note that 1 6 9 10 9 6 1 and related distributions are antidiagonals of A077028.
A083480 is a variation illustrating a relationship with numeric partitions, A000041.
The rows are formed by the nonzero entries of the columns of A049597.
The coefficient of q^j in the Gaussian polynomial [n, m]q is the number of binary words on alphabet {0,1} of length n having m 1's and j inversions. Hence T(n, k) is the number of length n binary words with exactly k inversions. - _Geoffrey Critzer, May 14 2017
If n is even the n-th row converges to n+1, n-1, n-4, ..., 19, 13, 7, 4, 3, 1 which is A029552 reversed, and if n is odd the sequence is twice A098613. - Michael Somos, Jun 25 2017

			When viewed as an array with A033638(r) entries per row, the table begins:
. 1 ............... : 1
. 2 ............... : 2
. 3 1 ............. : 3 + q = (1) + (1+q) + (1)
. 4 2 2 ........... : 4 + 2q + 2q^2 = 1 + (1+q+q^2) + (1+q+q^2) + 1
. 5 3 4 3 1 ....... : 5 + 3q + 4q^2 + 3q^3 + q^4
. 6 4 6 6 6 2 2
. 7 5 8 9 11 9 7 4 3 1
. 8 6 10 12 16 16 18 12 12 8 6 2 2
. 9 7 12 15 21 23 29 27 26 23 21 15 13 7 4 3 1
...
The second but last row is from the sum over 7 q-polynomials coefficients:
. 1 ....... : 1 = [6,0]_q
. 1 1 1 1 1 1 ....... : 1+q+q^2+q^3+q^4+q^5 = [6,1]_q
. 1 1 2 2 3 2 2 1 1 ....... : 1+q+2q^2+2q^3+3q^4+2q^5+2q^6+q^7+q^8 = [6,2]_q
. 1 1 2 3 3 3 3 2 1 1 ....... : 1+q+2q^2+3q^3+3q^4+3q^5+3q^6+2q^7+q^8+q^9 = [6,3]_q
. 1 1 2 2 3 2 2 1 1 ....... : 1+q+2q^2+2q^3+3q^4+2q^5+2q^6+q^7+q^8 = [6,4]_q
. 1 1 1 1 1 1 ....... : 1+q+q^2+q^3+q^4+q^5 = [6,5]_q
. 1 ....... : 1 = [6,6]_q

George E. Andrews, 'Theory of Partitions', 1976, page 242.

Seiichi Manyama, Rows n = 0..48, flattened
Geoffrey Critzer, Combinatorics of Vector Spaces over Finite Fields, Master's thesis, Emporia State University, 2018.
Alexander Gruber, "The Egg:" Bizarre behavior of the roots of a family of polynomials Mathematics StackExchange Oct 04 2012
MathOverflow, Partition numbers and Gaussian binomial coefficient
Eric Weisstein, q-Binomial Coefficient, Mathworld.
Wikipedia, q-binomial
Index entries for sequences related to binary linear codes

Cf. A000025, A000034, A000041, A016116, A029552, A033638, A060546, A063746, A077028, A083479, A083480, A098613, A260460.

R:=PowerSeriesRing(Rationals(), 100);
qBinom:= func< n,k,x | n eq 0 or k eq 0 select 1 else (&*[(1-x^(n-j))/(1-x^(j+1)): j in [0..k-1]]) >;
A083906:= func< n,k | Coefficient(R!((&+[qBinom(n,k,x): k in [0..n]]) ), k) >;
[A083906(n,k): k in [0..Floor(n^2/4)], n in [0..12]]; // G. C. Greubel, Feb 13 2024

QBinomial := proc(n,m,q) local i ; factor( mul((1-q^(n-i))/(1-q^(i+1)),i=0..m-1) ) ; expand(%) ; end:
A083906 := proc(n,k) add( QBinomial(n,m,q),m=0..n ) ; coeftayl(%,q=0,k) ; end:
for n from 0 to 10 do for k from 0 to A033638(n)-1 do printf("%d,",A083906(n,k)) ; od: od: # R. J. Mathar, May 28 2009
T := proc(n, k) if n < 0 or k < 0 or k > floor(n^2/4) then return 0 fi;
if n < 2 then return n + 1 fi; 2*T(n-1, k) - T(n-2, k) + T(n-2, k - n + 1) end:
seq(print(seq(T(n, k), k = 0..floor((n/2)^2))), n = 0..8);  # Peter Luschny, Feb 16 2024

Table[CoefficientList[Total[Table[FunctionExpand[QBinomial[n, k, q]], {k, 0, n}]],q], {n, 0, 10}] // Grid (* Geoffrey Critzer, May 14 2017 *)

{T(n, k) = polcoeff(sum(m=0, n, prod(k=0, m-1, (x^n - x^k) / (x^m - x^k))), k)}; /* Michael Somos, Jun 25 2017 */

def T(n,k): # T = A083906
    if k<0 or k> (n^2//4): return 0
    elif n<2 : return n+1
    else: return 2*T(n-1, k) - T(n-2, k) + T(n-2, k-n+1)
flatten([[T(n,k) for k in range(int(n^2//4)+1)] for n in range(13)]) # G. C. Greubel, Feb 13 2024

T(n, k) is the coefficient [q^k] of the Sum_{m=0..n} [n, m]_q over q-Binomial coefficients.
Row sums: Sum_{k=0..floor(n^2/4)} T(n,k) = 2^n.
For n >= k, T(n+1,k) = T(n, k) + A000041(k). - Geoffrey Critzer, Feb 12 2021
Sum_{k=0..floor(n^2/4)} (-1)^k*T(n, k) = A060546(n). - G. C. Greubel, Feb 13 2024
From Mikhail Kurkov, Feb 14 2024: (Start)
T(n, k) = 2*T(n-1, k) - T(n-2, k) + T(n-2, k - n + 1) for n >= 2 and 0 <= k <= floor(n^2/4).
Sum_{i=0..n} T(n-i, i) = A000041(n+1). Note that upper limit of the summation can be reduced to A083479(n) = (n+2) - ceiling(sqrt(4*n)).
Both results were proved (see MathOverflow link for details). (End)
From G. C. Greubel, Feb 17 2024: (Start)
T(n, floor(n^2/4)) = A000034(n).
Sum_{k=0..floor(n^2/4)} (-1)^k*T(n, k) = A016116(n+1).
Sum_{k=0..(n + 2) - ceiling(sqrt(4*n))} (-1)^k*T(n - k, k) = (-1)^n*A000025(n+1) = -A260460(n+1). (End)

Edited by R. J. Mathar, May 28 2009

New name using a comment from Geoffrey Critzer by Peter Luschny, Feb 17 2024

A218754 Number of ways to write n=p+q(3+(-1)^n)/2 with q<=n/2 and p, q, p^2+3pq+q^2 all prime.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 3, 2, 1, 1, 1, 2, 2, 1, 3, 1, 3, 1, 1, 2, 1, 0, 3, 3, 2, 3, 3, 0, 3, 0, 3, 2, 1, 1, 4, 1, 2, 2, 1, 2, 0, 2, 2, 2, 3, 0, 4, 1, 1, 2, 0, 1, 2, 3, 5, 0, 2, 1, 3, 4, 1, 1, 2, 2, 6, 2, 2, 4, 1, 2, 3, 2, 3, 3, 3, 2, 4, 1, 2, 5, 0, 3, 4, 2, 3, 4, 3, 1, 4, 3

Offset: 1

Zhi-Wei Sun, Nov 04 2012

nonn

Conjecture: a(n)>0 for all n>=1188.
Conjecture verified for n up to 10^9. - Mauro Fiorentini, Sep 23 2023
This conjecture is stronger than both Goldbach's conjecture and Lemoine's conjecture.
Zhi-Wei Sun also made the following conjecture: Given any positive odd integer d, there is a prime p(d) such that for any prime p>p(d) there is a prime q
Conjecture verified for d up to 100 and p up to 10^7. - Mauro Fiorentini, Sep 23 2023

			For n=72 we have a(72)=1 since the only primes p and q with p+q=72, q<=36 and p^2+3pq+q^2 prime are p=67 and q=5.

Zhi-Wei Sun, Table of n, a(n) for n = 1..20000
Zhi-Wei Sun, Conjectures involving primes and quadratic forms, arXiv preprint arXiv:1211.1588 [math.NT], 2012-2017.

Cf. A002372, A046927, A218585, A218654, A218656.

Cf. A000034 = 1,2,1,2,... = (3-(-1)^n)/2. (Note: Offset shifted w.r.t. use in the definition of this sequence.) - M. F. Hasler, Nov 05 2012

a[n_]:=a[n]=Sum[If[PrimeQ[q]==True&&PrimeQ[n-q(3-(-1)^n)/2]&&PrimeQ[q^2+3q(n-q(3-(-1)^n)/2)+(n-q(3-(-1)^n)/2)^2]==True,1,0],{q,1,n/2}]
Do[Print[n," ",a[n]],{n,1,20000}]

A068073 Period 4 sequence [ 1, 2, 3, 2, ...].

Original entry on oeis.org

1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1, 2, 3, 2, 1

Offset: 0

Robert G. Wilson v, Mar 01 2002

Continued fraction expansion of (2+sqrt(14))/4. - Klaus Brockhaus, May 01 2010

The sequence is like a sawtooth wave of period 4. - Michael Somos, Feb 13 2011

			G.f. = 1 + 2*x + 3*x^2 + 2*x^3 + x^4 + 2*x^5 + 3*x^6 + 2*x^7 + x^8 + 2*x^9 + ...

Index entries for linear recurrences with constant coefficients, signature (1,-1,1).

Cf. A000034, A000120, A001045, A028356, A164356.

Cf. A177033 (decimal expansion of (2+sqrt(14))/4). - Klaus Brockhaus, May 01 2010

CoefficientList[ Series[(1 + 2x + 3x^2 + 2x^3)/(1 - x^4), {x, 0, 85}], x]
a[ n_] := {2, 3, 2, 1}[[Mod[n, 4, 1]]]; (* Michael Somos, Apr 17 2015 *)
PadRight[{},120,{1,2,3,2}] (* Harvey P. Dale, Jun 13 2020 *)

{a(n) = [1, 2, 3, 2] [n%4 + 1]}; /* Michael Somos, Feb 13 2011 */

{a(n) = n%4 + 1 - 2 * (n%4 == 3)}; /* Michael Somos, Feb 13 2011 */

{a(n) = 2 + kronecker( -4, n-1)}; /* Michael Somos, Feb 13 2011 */

G.f.: (1 + 2*x + 3*x^2 + 2*x^3) / (1 - x^4).
Conjecture: a(n) = Sum_{k=0..n} e^(i*Pi*(A000120(A001045(n)) - A001045(A000120(n)))), i=sqrt(-1). - Paul Barry, Jan 14 2005
From Paul Barry, Jan 14 2005: (Start)
G.f.: (1 + x + 2x^2)/(1 - x + x^2 - x^3);
a(n) = 2 - cos(Pi*n/2). (End)
Moebius transform is length 4 sequence [2, 1, 0, -2]. - Michael Somos, Feb 13 2011
a(n) = 2 - A056594(n). - Bruno Berselli, Mar 10 2011
a(n) = a(-n) = a(n+4) for all n in Z. - Michael Somos, Apr 17 2015
2 * a(n) = A164356(n) unless n=0. - Michael Somos, Apr 17 2015
G.f.: 1 / (1 - 2*x / (1 + x / (2 - 5*x / (1 + 16*x / (5 - x))))). - Michael Somos, Jan 20 2017
G.f.: 2 / (1 - x) - 1 / (1 + x^2). - Michael Somos, Jan 07 2019
a(n) = abs(((n+2) mod 4)-2) + 1. - Daniel Jiménez, Jan 14 2023

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

A201908 Irregular triangle of 2^k mod (2n-1).

Original entry on oeis.org

0, 1, 2, 1, 2, 4, 3, 1, 2, 4, 1, 2, 4, 8, 7, 5, 1, 2, 4, 8, 5, 10, 9, 7, 3, 6, 1, 2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, 1, 2, 4, 8, 1, 2, 4, 8, 16, 15, 13, 9, 1, 2, 4, 8, 16, 13, 7, 14, 9, 18, 17, 15, 11, 3, 6, 12, 5, 10, 1, 2, 4, 8, 16, 11, 1, 2, 4, 8, 16, 9

Offset: 1

T. D. Noe, Dec 07 2011

The length of the rows is given by A002326. For n > 1, the first term of row n is 1 and the last term is n. Many sequences are in this one: starting at A036117 (mod 11) and A070335 (mod 23).

Row n, for n >= 2, divided elementwise by (2*n-1) gives the cycles of iterations of the doubling function D(x) = 2*x or 2*x-1 if 0 <= x < 1/2 or , 1/2 <= x < 1, respectively, with seed 1/(2*n-1). See the Devaney reference, pp. 25-26. D^[k](x) = frac(2^k/(2*n-1)), for k = 0, 1, ..., A002326(n-1) - 1. E.g., n = 3: 1/5, 2/5, 4/5, 3/5. - Gary W. Adamson and Wolfdieter Lang, Jul 29 2020.

			The irregular triangle T(n, k) begins:
n\k  0 1 2 3  4  5  6  7 8  9 10 11 12 13 14 15 16 17 ...
---------------------------------------------------------
1:   0
2:   1 2
3:   1 2 4 3
4:   1 2 4
5:   1 2 4 8  7  5
6:   1 2 4 8  5 10  9  7 3  6
7:   1 2 4 8  3  6 12 11 9  5 10  7
8:   1 2 4 8
9:   1 2 4 8 16 15 13  9
10:  1 2 4 8 16 13  7 14 9 18 17 15 11  3  6 12  5 10
... reformatted by _Wolfdieter Lang_, Jul 29 2020.

Robert L. Devaney, A First Course in Chaotic Dynamical Systems, Addison-Wesley., 1992. pp. 24-25

T. D. Noe, Rows n = 1..100, flattened

Cf. A002326, A201909 (3^k), A201910 (5^k), A201911 (7^k).

Cf. A000034 (3), A070402 (5), A069705 (7), A036117 (11), A036118 (13), A062116 (17), A036120 (19), A070347 (21), A070335 (23), A070336 (25), A070337 (27), A036122 (29), A070338 (33), A070339 (35), A036124 (37), A070340 (39), A070348 (41), A070349 (43), A070350 (45), A070351 (47), A036128 (53), A036129 (59), A036130 (61), A036131 (67), A036135 (83), A036138 (101), A036140 (107), A201920 (125), A036144 (131), A036146 (139), A036147 (149), A036150 (163), A036152 (173), A036153 (179), A036154 (181), A036157 (197), A036159 (211), A036161 (227).

R:=List([0..72],n->OrderMod(2,2*n+1));;
Flat(Concatenation([0],List([2..11],n->List([0..R[n]-1],k->PowerMod(2,k,2*n-1))))); # Muniru A Asiru, Feb 02 2019

nn = 30; p = 2; t = p^Range[0, nn]; Flatten[Table[If[IntegerQ[Log[p, n]], {0}, tm = Mod[t, n]; len = Position[tm, 1, 1, 2][[-1,1]]; Take[tm, len-1]], {n, 1, nn, 2}]]

T(n, k) = 2^k mod (2*n-1), n >= 1, k = 0, 1, ..., A002326(n-1) - 1.

T(n, k) = (2*n-1)*frac(2^k/(2*n-1)), n >= 1, k = 0, 1, ..., A002326(n-1) - 1, with the fractional part frac(x) = x - floor(x). - Wolfdieter Lang, Jul 29 2020

A007068 a(n) = a(n-1) + (3+(-1)^n)*a(n-2)/2.

Original entry on oeis.org

1, 3, 4, 10, 14, 34, 48, 116, 164, 396, 560, 1352, 1912, 4616, 6528, 15760, 22288, 53808, 76096, 183712, 259808, 627232, 887040, 2141504, 3028544, 7311552, 10340096, 24963200, 35303296, 85229696, 120532992, 290992384, 411525376, 993510144, 1405035520, 3392055808

Offset: 1

N. J. A. Sloane, Mira Bernstein

First row of spectral array W(sqrt 2).

Row sums of the square of the matrix with general term binomial(floor(n/2),n-k). - Paul Barry, Feb 14 2005

N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000
Aviezri S. Fraenkel and Clark Kimberling, Generalized Wythoff arrays, shuffles and interspersions, Discr. Math. 126 (1-3) (1994) 137-149.
Sean A. Irvine, Walks on Graphs.
Index entries for linear recurrences with constant coefficients, signature (0,4,0,-2).

Cf. A062112, A001882, A007070, A007052, A000034.

a007068 n = a007068_list !! (n-1)
a007068_list = 1 : 3 : zipWith (+)
   (tail a007068_list) (zipWith (*) a000034_list a007068_list)
-- Reinhard Zumkeller, Jan 21 2012

RecurrenceTable[{a[1]==1,a[2]==3,a[n]==a[n-1]+(3+(-1)^n) a[n-2]/2},a[n],{n,40}] (* Harvey P. Dale, Nov 12 2012 *)

a(2n+1) = a(2n)+a(2n-1); a(2n) = a(2n-1)+2*a(2n-2); same recurrence (mod parity) as A001882. - Len Smiley, Feb 05 2001
a(n) = Sum_{k=0..n} Sum_{j=0..n} C(floor(n/2), n-j)*C(floor(j/2), j-k). - Paul Barry, Feb 14 2005
a(n) = 4*a(n-2)-2*a(n-4). G.f.: -x*(1+x)*(2*x^2-2*x-1)/(1-4*x^2+2*x^4). a(2n+1)=A007070(n). a(2n)=A007052(n). [R. J. Mathar, Aug 17 2009]
a(n) = a(n-1) + a(n-2) * A000034(n-1). [Reinhard Zumkeller, Jan 21 2012]

Better description and more terms from Olivier Gérard, Jun 05 2001

A010694 Period 2: repeat (2,4).

Original entry on oeis.org

2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2, 4, 2

Offset: 0

N. J. A. Sloane

Simple continued fraction expansion of 1 + sqrt(3/2) = A176051. - R. J. Mathar, Mar 08 2012
Number of linear characters of dihedral group of order 2(n+1). - Eric M. Schmidt, Feb 12 2013
a(n) is the n-th increment between two consecutive elements of the wheel in the wheel factorization with the basis {2, 3}. See A038179. - Wojciech Raszka, May 10 2019
In base 3, make a sequence such that after the initial term 2, each term is the sum of the squares of the digits of the previous term. That's this sequence (see A000216 for the base 10 version). - Alonso del Arte, Mar 19 2020

INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 1073
Index entries for linear recurrences with constant coefficients, signature (0,1).

Cf. A000034, A038179, A176051.

ContinuedFraction[1 + Sqrt[6]/2, 100] (* Alonso del Arte, Mar 25 2020 *)

a(n)=2*(1+n%2) \\ Jaume Oliver Lafont, Mar 20 2009

(0 to 99).map( % 2 * 2 + 2) // _Alonso del Arte, Mar 25 2020

From R. J. Mathar, Aug 28 2008: (Start)
a(n) = 2 * A000034(n).
G.f.: 2(1 + 2x)/((1 - x)(1 + x)). (End)
a(n) = a(n-2) for n >= 2. - Jaume Oliver Lafont, Mar 20 2009
a(n) = 2^(n+1) mod 6. - Roderick MacPhee, Mar 31 2011

A070893 Let r, s, t be three permutations of the set {1,2,3,..,n}; a(n) = value of Sum_{i=1..n} r(i)s(i)t(i), with r={1,2,3,..,n}; s={n,n-1,..,1} and t={n,n-2,n-4,...,1,...,n-3,n-1}.

Original entry on oeis.org

1, 6, 19, 46, 94, 172, 290, 460, 695, 1010, 1421, 1946, 2604, 3416, 4404, 5592, 7005, 8670, 10615, 12870, 15466, 18436, 21814, 25636, 29939, 34762, 40145, 46130, 52760, 60080, 68136, 76976, 86649, 97206, 108699, 121182, 134710, 149340

Offset: 1

Wouter Meeussen, May 22 2002

See A070735 for the minimal values for these products. This sequence is an upper bound. The third permutation 't'= ceiling(abs(range(n-1/2,-n,-2))) is such that it associates its smallest factor with the largest factor of the product 'r'*'s'.

We observe that is the transform of A002717 by the following transform T: T(u_0,u_1,u_2,u_3,...) = (u_0,u_0+u_1, u_0+u_1+u_2, u_0+u_1+u_2+u_3+u_4,...). In other words, v_p = Sum_{k=0..p} u_k and the g.f. phi_v of v is given by phi_v = phi_u/(1-z). - Richard Choulet, Jan 28 2010

			{1,2,3,4,5,6,7}*{7,6,5,4,3,2,1}*{7,5,3,1,2,4,6} gives {49,60,45,16,30,48,42}, with sum 290, so a(7)=290.

Vincenzo Librandi, Table of n, a(n) for n = 1..10000
Index entries for two-way infinite sequences
Index entries for linear recurrences with constant coefficients, signature (4,-5,0,5,-4,1).

Cf. A070735, A082289. a(n)=A082290(2n-2).
Cf. A000034, A032766, A006578, A002717. - Richard Choulet, Jan 28 2010
Cf. A002717 (first differences). - Bruno Berselli, Aug 26 2011
Column k=3 of A166278. - Alois P. Heinz, Nov 02 2012

[(1/96)*(2*n*(n+2)*(3*n^2+10*n+4)+3*(-1)^n-3): n in [1..40]]; // Vincenzo Librandi, Aug 26 2011

Table[Plus@@(Range[n]*Range[n, 1, -1]*Ceiling[Abs[Range[n-1/2, -n, -2]]]), {n, 49}];
(* or *)
CoefficientList[Series[ -(1+2x)/(-1+x)^5/(1+x), {x, 0, 48}], x]//Flatten

a(n)=sum(i=1,n,i*(n+1-i)*ceil(abs(n+3/2-2*i)))

a(n)=polcoeff(if(n<0,x^4*(2+x)/((1+x)*(1-x)^5),x*(1+2*x)/((1+x)*(1-x)^5))+x*O(x^abs(n)),abs(n))

G.f.: x*(1+2*x)/((1+x)*(1-x)^5). - Michael Somos, Apr 07 2003
a(n) = 3*a(n-1) - 2*a(n-2) - 2*a(n-3) + 3*a(n-4) - a(n-5) + 3. If sequence is also defined for n <= 0 by this equation, then a(n)=0 for -3 <= n <= 0 and a(n)=A082289(-n) for n <= -4. - Michael Somos, Apr 07 2003
a(n) = (1/96)*(2*n*(n+2)*(3*n^2+10*n+4)+3*(-1)^n-3). a(n) - a(n-2) = A002411(n). - Bruno Berselli, Aug 26 2011

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A070893 Let r, s, t be three permutations of the set {1,2,3,..,n}; a(n) = value of Sum_{i=1..n} r(i)s(i)t(i), with r={1,2,3,..,n}; s={n,n-1,..,1} and t={n,n-2,n-4,...,1,...,n-3,n-1}.