A033567 -id:A033567 - cp's OEIS frontend

A330987 Alternatively add and half-multiply pairs of the nonnegative integers.

1, 3, 9, 21, 17, 55, 25, 105, 33, 171, 41, 253, 49, 351, 57, 465, 65, 595, 73, 741, 81, 903, 89, 1081, 97, 1275, 105, 1485, 113, 1711, 121, 1953, 129, 2211, 137, 2485, 145, 2775, 153, 3081, 161, 3403, 169, 3741, 177, 4095, 185, 4465, 193, 4851, 201, 5253, 209

Offset: 1

George E. Antoniou, Jan 05 2020

In groups of two, add and half-multiply the integers: 0+1, (2*3)/2, 4+5, (6*7)/2, ....
From Bernard Schott, Jan 06 2020: (Start)
The bisection of this sequence gives:
For n odd = 2*k+1, k >= 0: a(2*k+1) = 8*k+1 = A017077(k),
For n even = 2*k, k >= 1: a(2*k) = T(4*k-2) = A000217(4*k-2) = (2*k-1)*(4*k-1) = A033567(k) where T(j) is the j-th triangular number. (End)

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

Cf. A330983.

Interspersion of A017077 and A033567 (excluding first term). - Michel Marcus, Jan 06 2020

a[n_]:=If[OddQ[n],4n-3,(n-1)(2n-1)]; Array[a,53] (* Stefano Spezia, Jan 05 2020 *)

Vec(x*(1 + 3*x + 6*x^2 + 12*x^3 - 7*x^4 + x^5) / ((1 - x)^3*(1 + x)^3) + O(x^50)) \\ Colin Barker, Jan 06 2020

From Colin Barker, Jan 05 2020: (Start)
G.f.: x*(1 + 3*x + 6*x^2 + 12*x^3 - 7*x^4 + x^5) / ((1 - x)^3*(1 + x)^3).
a(n) = 3*a(n-2) - 3*a(n-4) + a(n-6) for n>6.
a(n) = -1 + 2*(-1)^n - (1/2)*(-1+7*(-1)^n)*n + (1+(-1)^n)*n^2.
(End)
E.g.f.: (1 + 4*x + 2*x^2)*cosh(x) - (3 + x)*sinh(x) - 1. - Stefano Spezia, Jan 05 2020 after Colin Barker

A256833 a(n) = (4n+3)(4*n+2).

Original entry on oeis.org

6, 42, 110, 210, 342, 506, 702, 930, 1190, 1482, 1806, 2162, 2550, 2970, 3422, 3906, 4422, 4970, 5550, 6162, 6806, 7482, 8190, 8930, 9702, 10506, 11342, 12210, 13110, 14042, 15006, 16002, 17030, 18090, 19182, 20306, 21462, 22650, 23870, 25122, 26406

Offset: 0

Bruce Zimov, Apr 10 2015

Since 0 = Sin(Pi) = Sum_{n>=0}(-1)^n*Pi^(2n+1)/(2n+1)!, we can move the negative terms to the other side of the equation to get: Sum_{n>=0} Pi^(4n+1)/(4n+1)! = Sum_{n>=0}Pi^(4n+3)/(4n+3)!.

Now, if we let f(n) = Pi^(4n+1)/(4n+1)!, then the previous equation can be written as Sum_{n>=0}f(n) = Sum_{n>=0}(Pi^2/((4*n+3)*(4*n+2)))*f(n); a(n) is the n-th denominator on the right hand side.

Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A002378, A004767, A016825, A033567.

[16*n^2 + 20*n + 6: n in [0..40]]; // Vincenzo Librandi, Apr 12 2015

CoefficientList[Series[(6 + 24 x + 2 x^2) / (1 - x)^3, {x, 0, 40}], x] (* Vincenzo Librandi, Apr 12 2015 *)

vector(50,n,(4*n-1)*(4*n-2)) \\ Derek Orr, Apr 13 2015

a(n) = 16*n^2 + 20*n + 6.
a(n) = 2*A033567(n+1).
G.f.: (6+24*x+2*x^2)/(1-x)^3. - Vincenzo Librandi, Apr 12 2015
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n>2. - Vincenzo Librandi, Apr 12 2015
a(n) = A016825(n)*A004767(n). - Tom Edgar, Apr 12 2015
a(n) = A002378(4*n+2) = 2*A000217(4*n+2). - Ivan N. Ianakiev, Apr 17 2015
E.g.f.: 2*exp(x)*(3+18*x+8*x^2). - Wesley Ivan Hurt, Apr 29 2020
From Amiram Eldar, Jan 03 2022: (Start)
Sum_{n>=0} 1/a(n) = Pi/8 - log(2)/4.
Sum_{n>=0} (-1)^n/a(n) = sqrt(2)*log(sqrt(2)+1)/4 - (sqrt(2)-1)*Pi/8. (End)

More terms from Vincenzo Librandi, Apr 12 2015

A257942 a(n) = (n+1)*(n+2)/A014695(n+1), where A014695 is repeat (1, 2, 2, 1).

Original entry on oeis.org

1, 3, 12, 20, 15, 21, 56, 72, 45, 55, 132, 156, 91, 105, 240, 272, 153, 171, 380, 420, 231, 253, 552, 600, 325, 351, 756, 812, 435, 465, 992, 1056, 561, 595, 1260, 1332, 703, 741, 1560, 1640, 861, 903, 1892, 1980, 1035, 1081, 2256, 2352, 1225, 1275, 2652

Offset: 0

Paul Curtz, Jul 14 2015

Consider, for n >= 0, a sequence s(n). A useful transform is wi(n) = s(0), s(2), s(3), ..., i.e., s(n) without s(1).
For s(n) = 1/(n+1), wi(n)= 1, 1/3, 1/4, 1/5, ..., whose inverse binomial transform is f(n) = 1, -2/3, 7/12, -11/20, 8/15, -11/21, 29/56, -37/72, 23/45, -28/55, 67/132, -79/156, 46/91, -53/105, 121/240, -137/272, ...
The denominator of f(n) is a(n), for n >= 0.
If the numerator of f(n) is b(n), then it can be seen that b(n+1) = -(-1)^n* A226089(n).
Alternating a(n) - b(n) with a(n) + b(n) yields 0, 1, 5, 9, ... = A160050(n+1).
a(4n+1) is linked to the Rydberg-Ritz spectra of hydrogen.
h(n) = 0, 0, 1, 1, 4, 4, 3, 3, 8, 8, 5, 5, ... = duplicated A022998(n).
A022998(n) is linked to the Balmer series (see A246943(n)).
With an initial 0 and offset=0, a(-n) = a(n). Then (a(n+10) - a(n-10))/10 = 1, 6, 10, 7, 9, 22, 26, ... = g(n). a(n) mod 9 is of period 20.

Colin Barker, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (3,-6,10,-12,12,-10,6,-3,1).

Cf. A002378(n+1), A014695(n+1)=A130658(n+2), A014634, A033567(n+1), A104188(n+1), 4*A007742(n+1), A160050 (in A226089), A022998, A109613, A000217(n+1), A246943.

A257942:=n->(n+1)*(n+2)/(3/2+(-1)^((2*n+7+(-1)^n)/4)/2): seq(A257942(n), n=0..100); # Wesley Ivan Hurt, Jul 18 2015

CoefficientList[Series[-(x^6 + 9 x^4 - 8 x^3 + 9 x^2 + 1)/((x - 1)^3 (x^2 + 1)^3), {x, 0, 50}], x] (* Michael De Vlieger, Jul 14 2015 *)
(* Using inverse binomial transform *) s[0]=1; s[n_] := 1/(n+2); f[n_] := Sum[(-1)^(n-k)*Binomial[n, k]*s[k], {k, 0, n}]; Table[f[n]//Denominator, {n, 0, 50}] (* Jean-François Alcover, Jul 14 2015 *)
LinearRecurrence[{3, -6, 10, -12, 12, -10, 6, -3, 1}, {1, 3, 12, 20, 15, 21, 56, 72, 45}, 55] (* Vincenzo Librandi, Jul 15 2015 *)

Vec(-(x^6+9*x^4-8*x^3+9*x^2+1)/((x-1)^3*(x^2+1)^3) + O(x^100)) \\ Colin Barker, Jul 14 2015

a(n)=(n+1)*(n+2)/if(n%4<2,2,1) \\ Charles R Greathouse IV, Jul 14 2015

a(4n)    = (2*n+1)*(4*n+1).
a(4n+1)  = (2*n+1)*(4*n+3).
a(4n+2)  = (4*n+3)*(4*n+4).
a(4n+3)  = (4*n+4)*(4*n+5).
a(n) = A064038(n+2) * (period 4: repeat 1, 1, 4, 4).
From Colin Barker, Jul 14 2015: (Start)
a(n) = (-1/8+i/8)*(((-3-i*3)+i*(-i)^n+i^n)*(2+3*n+n^2)) where i=sqrt(-1).
G.f.: -(x^6+9*x^4-8*x^3+9*x^2+1) / ((x-1)^3*(x^2+1)^3). (End)
a(n) = h(n+2) * A109613(n+1).
a(n) = (n+1)*(n+2) * period 4:repeat (1, 1, 2, 2) /2.
From Wesley Ivan Hurt, Jul 18 2015: (Start)
a(n) = (n+1)*(n+2)/(3/2+(-1)^((2*n+7+(-1)^n)/4)/2).
a(n) = 3*a(n-1)-6*a(n-2)+10*a(n-3)-12*a(n-4)+12*a(n-5)-10*a(n-6)+6*a(n-7)-3*a(n-8)+a(n-9), n>9. (End)
Sum_{n>=0} 1/a(n) = Pi/4 + 1. - Amiram Eldar, Aug 14 2022

A287057 a(n) = 2*n^2 + n - (n+1) mod 2.

Original entry on oeis.org

3, 9, 21, 35, 55, 77, 105, 135, 171, 209, 253, 299, 351, 405, 465, 527, 595, 665, 741, 819, 903, 989, 1081, 1175, 1275, 1377, 1485, 1595, 1711, 1829, 1953, 2079, 2211, 2345, 2485, 2627, 2775, 2925, 3081, 3239, 3403, 3569, 3741, 3915, 4095

Offset: 1

Dimitris Valianatos, Jun 24 2017

nonn

Let r(n) = (a(n)-1)/a(n) if n mod 2 = 1, (a(n)+1)/a(n) otherwise; then Product_{n>=1} r(n) = (2/3) * (10/9) * (20/21) * (36/35) * (54/55) * (78/77) * (104/105) * (136/135) * ... = agm(1,sqrt(2))^2/2 = 0.7177700110461299978211932237.

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

Cf. A033566, A033567.

[2*n^2+n-(n+1) mod 2: n in [1..60]]; // Vincenzo Librandi, Aug 12 2017

seq(2*n^2 + n - ((n+1) mod 2), n = 1 .. 30); # Robert Israel, Aug 11 2017

a[n_] := 2 n^2 + n - Mod[n + 1, 2]; Array[a, 50] (* Robert G. Wilson v, Aug 10 2017 *)

{for(n=1,100,print1(2*n^2+n-(n+1)%2", "))}

G.f.: x*(3+3*x+3*x^2-x^3)/((1+x)*(1-x)^3). - Robert Israel, Aug 11 2017

cp's OEIS Frontend

A330987 Alternatively add and half-multiply pairs of the nonnegative integers.

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A256833 a(n) = (4*n+3)*(4*n+2).

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A257942 a(n) = (n+1)*(n+2)/A014695(n+1), where A014695 is repeat (1, 2, 2, 1).

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A287057 a(n) = 2*n^2 + n - (n+1) mod 2.

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Maple

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A256833 a(n) = (4n+3)(4*n+2).