A045944 -id:A045944 - cp's OEIS frontend

A179986 Second 9-gonal (or nonagonal) numbers: a(n) = n(7n+5)/2.

0, 6, 19, 39, 66, 100, 141, 189, 244, 306, 375, 451, 534, 624, 721, 825, 936, 1054, 1179, 1311, 1450, 1596, 1749, 1909, 2076, 2250, 2431, 2619, 2814, 3016, 3225, 3441, 3664, 3894, 4131, 4375, 4626, 4884, 5149, 5421, 5700, 5986, 6279, 6579, 6886

Offset: 0

Bruno Berselli, Jan 13 2011

This sequence is a bisection of A118277 (even part).
Sequence found by reading the line from 0, in the direction 0, 19... and the line from 6, in the direction 6, 39,..., in the square spiral whose vertices are the generalized 9-gonal numbers A118277. - Omar E. Pol, Jul 24 2012
The early part of this sequence is a strikingly close approximation to the early part of A100752. - Peter Munn, Nov 14 2019

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

Cf. second k-gonal numbers: A005449 (k=5), A014105 (k=6), A147875 (k=7), A045944 (k=8), this sequence (k=9), A033954 (k=10), A062728 (k=11), A135705 (k=12).

Cf. A001106, A022264, A022265, A024966, A055998, A100752, A174738, A186029, A218471.

[n*(7*n+5)/2: n in [0..50]]; // Bruno Berselli, Sep 23 2016

I:=[0, 6, 19]; [n le 3 select I[n] else 3*Self(n-1) -3*Self(n-2) +Self(n-3): n in [1..60]]; // Vincenzo Librandi, Oct 15 2012

f[n_] := n (7 n + 5)/2; f[Range[0, 60]] (* Vladimir Joseph Stephan Orlovsky, Feb 05 2011*)
LinearRecurrence[{3, -3, 1}, {0, 6, 19}, 60] (* or *) Array[(#(7# + 5))/2&, 60, 0] (* Harvey P. Dale, Aug 19 2011 *)
CoefficientList[Series[x (6 + x)/(1 - x)^3, {x, 0, 60}], x] (* Vincenzo Librandi, Oct 15 2012 *)

a(n)=n*(7*n+5)/2 \\ Charles R Greathouse IV, Sep 24 2015

G.f.: x*(6 + x)/(1 - x)^3.
a(n) = Sum_{i=0..(n-1)} A017053(i) for n>0.
a(-n) = A001106(n).
Sum_{i=0..n} (a(n)+i)^2 = ( Sum_{i=(n+1)..2*n} (a(n)+i)^2 ) + 21*A000217(n)^2 for n>0.
a(n) = a(n-1)+7*n-1 for n>0, with a(0)=0. - Vincenzo Librandi, Feb 05 2011
a(0)=0, a(1)=6, a(2)=19; for n>2, a(n) = 3*a(n-1)-3*a(n-2)+a(n-3). - Harvey P. Dale, Aug 19 2011
a(n) = A174738(7n+5). - Philippe Deléham, Mar 26 2013
a(n) = A001477(n) + 2*A000290(n) + 3*A000217(n). - J. M. Bergot, Apr 25 2014
a(n) = A055998(4*n) - A055998(3*n). - Bruno Berselli, Sep 23 2016
E.g.f.: (x/2)*(12 + 7*x)*exp(x). - G. C. Greubel, Aug 19 2017

A139273 a(n) = n(8n - 3).

Original entry on oeis.org

0, 5, 26, 63, 116, 185, 270, 371, 488, 621, 770, 935, 1116, 1313, 1526, 1755, 2000, 2261, 2538, 2831, 3140, 3465, 3806, 4163, 4536, 4925, 5330, 5751, 6188, 6641, 7110, 7595, 8096, 8613, 9146, 9695, 10260, 10841, 11438, 12051, 12680

Offset: 0

Omar E. Pol, Apr 26 2008

Sequence found by reading the line from 0, in the direction 0, 5, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139277 in the same spiral.
Also, sequence of numbers of the form d*A000217(n-1) + 5*n with generating functions x*(5+(d-5)*x)/(1-x)^3; the inverse binomial transform is 0,5,d,0,0,.. (0 continued). See Crossrefs. - Bruno Berselli, Feb 11 2011
Even decagonal numbers divided by 2. - Omar E. Pol, Aug 19 2011

G. C. Greubel, Table of n, a(n) for n = 0..5000
Omar E. Pol, Determinacion geometrica de los numeros primos y perfectos.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A000217, A014634, A014635, A033585, A033586, A033587, A035008, A051870, A069129, A085250, A072279, A139272, A139274, A139275, A139276, A139278, A139279, A139280, A139281, A139282.

Cf. numbers of the form n*(d*n+10-d)/2: A008587, A056000, A028347, A140090, A014106, A028895, A045944, A186029, A007742, A022267, A033429, A022268, A049452, A186030, A135703, A152734.

[ n*(8*n-3) : n in [0..40] ];  // Bruno Berselli, Feb 11 2011

Table[n (8 n - 3), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 26}, 40] (* Harvey P. Dale, Feb 02 2012 *)

a(n)=n*(8*n-3) \\ Charles R Greathouse IV, Sep 24 2015

a(n) = 8*n^2 - 3*n.
Sequences of the form a(n) = 8*n^2 + c*n have generating functions x{c+8+(8-c)x} / (1-x)^3 and recurrence a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). The inverse binomial transform is 0, c+8, 16, 0, 0, ... (0 continued). This applies to A139271-A139278, positive or negative c. - R. J. Mathar, May 12 2008
a(n) = 16*n + a(n-1) - 11 for n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From Bruno Berselli, Feb 11 2011: (Start)
G.f.: x*(5 + 11*x)/(1 - x)^3.
a(n) = 4*A000217(n) + A051866(n). (End)
a(n) = A028994(n)/2. - Omar E. Pol, Aug 19 2011
a(0)=0, a(1)=5, a(2)=26; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Feb 02 2012
E.g.f.: (8*x^2 + 5*x)*exp(x). - G. C. Greubel, Jul 18 2017
Sum_{n>=1} 1/a(n) = 4*log(2)/3 - (sqrt(2)-1)*Pi/6 - sqrt(2)*arccoth(sqrt(2))/3. - Amiram Eldar, Jul 03 2020

A061047 Numerator of 1/49 - 1/n^2.

Original entry on oeis.org

0, 15, 32, 51, 72, 95, 120, 3, 176, 207, 240, 275, 312, 351, 8, 435, 480, 527, 576, 627, 680, 15, 792, 851, 912, 975, 1040, 1107, 24, 1247, 1320, 1395, 1472, 1551, 1632, 5, 1800, 1887, 1976, 2067, 2160, 2255, 48, 2451, 2552, 2655, 2760, 2867

Offset: 7

N. J. A. Sloane, May 26 2001

a(n) = (n+7)^2-49 = n*(n+14) = A098848(n), except a(7p). The corresponding series of atomic transitions is named Hansen-Strong. It comes after Lyman (1906-1914), Balmer (1885), Paschen (1908), Brackett (1922), Pfund (1924) and Humphreys series (1952 not 1953, justified later). - Paul Curtz, Oct 07 2008

Vincenzo Librandi, Table of n, a(n) for n = 7..1000
Peter Hansen and John Strong, Seventh Series of Atomic Hydrogen, Applied Optics, vol 12, 2, (Feb 1973), pp. 429-430. [From _Paul Curtz_, Oct 07 2008]

Cf. A061035-A061050.

Cf. A061039, A061043, A045944, A098848.

[Numerator(1/49-1/n^2): n in [7..60]]; // Vincenzo Librandi, Sep 07 2016

Table[Numerator[1/49-1/n^2],{n,7,70}] (* Harvey P. Dale, Apr 26 2016 *)

a(n) = numerator(1/49 - 1/n^2); \\ Michel Marcus, Aug 15 2013

Edited by M. F. Hasler, Nov 17 2014

A062728 Second 11-gonal (or hendecagonal) numbers: a(n) = n(9n+7)/2.

Original entry on oeis.org

0, 8, 25, 51, 86, 130, 183, 245, 316, 396, 485, 583, 690, 806, 931, 1065, 1208, 1360, 1521, 1691, 1870, 2058, 2255, 2461, 2676, 2900, 3133, 3375, 3626, 3886, 4155, 4433, 4720, 5016, 5321, 5635, 5958, 6290, 6631, 6981, 7340, 7708, 8085, 8471, 8866, 9270

Offset: 0

Floor van Lamoen, Jul 21 2001

Old name: Write 0,1,2,3,4,... in a triangular spiral, then a(n) is the sequence found by reading the line from 0 in the direction 0,8,...

Sequence found by reading the line from 0, in the direction 0, 25, ... and the line from 8, in the direction 8, 51, ..., in the square spiral whose vertices are the generalized 11-gonal numbers A195160. - Omar E. Pol, Jul 24 2012

			The spiral begins:
          15
          / \
        16  14
        /     \
      17   3  13
      /   / \   \
    18   4   2  12
    /   /     \   \
  19   5   0---1  11
  /   /             \
20   6---7---8---9--10

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

Cf. A051682.

Second n-gonal numbers: A005449, A014105, A147875, A045944, A179986, A033954, this sequence, A135705.

List([0..50], n-> n*(9*n+7)/2); # G. C. Greubel, May 24 2019

[n*(9*n+7)/2: n in [0..50]]; // G. C. Greubel, May 24 2019

Table[n*(9*n+7)/2, {n,0,50}] (* G. C. Greubel, May 24 2019 *)
LinearRecurrence[{3,-3,1},{0,8,25},50] (* Harvey P. Dale, Sep 06 2019 *)

a(n)=n*(9*n+7)/2 \\ Charles R Greathouse IV, Jun 17 2017

[n*(9*n+7)/2 for n in (0..50)] # G. C. Greubel, May 24 2019

a(n) = n*(9*n+7)/2.
a(n) = 9*n + a(n-1) - 1 (with a(0)=0). - Vincenzo Librandi, Aug 07 2010
From Bruno Berselli, Jan 13 2011: (Start)
G.f.: x*(8 + x)/(1 - x)^3.
a(n) = Sum_{i=0..n-1} A017257(i) for n > 0. (End)
a(n) = A218470(9n+7). - Philippe Deléham, Mar 27 2013
E.g.f.: x*(16 + 9*x)*exp(x)/2. - G. C. Greubel, May 24 2019

New name from Bruno Berselli (with the original formula), Jan 13 2011

A033567 a(n) = (2n-1)(4*n-1).

Original entry on oeis.org

1, 3, 21, 55, 105, 171, 253, 351, 465, 595, 741, 903, 1081, 1275, 1485, 1711, 1953, 2211, 2485, 2775, 3081, 3403, 3741, 4095, 4465, 4851, 5253, 5671, 6105, 6555, 7021, 7503, 8001, 8515, 9045, 9591, 10153, 10731, 11325, 11935, 12561, 13203, 13861, 14535, 15225

Offset: 0

N. J. A. Sloane

a(n+1) = A005563(1), A061037(3), A061039(5), A061041(7), A061043(9), A061045(11), A061047(13), A061049(15). Lyman, Balmer, Paschen, Brackett, Pfund, Humphreys, Hansen-Strong, ... spectra of hydrogen. - Paul Curtz, Oct 08 2008

Sequence found by reading the segment [1, 3] together with the line from 3, in the direction 3, 21, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Sep 03 2011

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

Cf. A045944, A014634.

[(2*n-1)*(4*n-1): n in [0..50]]; // G. C. Greubel, Sep 19 2018

Table[(2*n - 1)*(4*n - 1), {n, 0, 50}] (* G. C. Greubel, Jul 06 2017 *)
LinearRecurrence[{3,-3,1},{1,3,21},50] (* Harvey P. Dale, Aug 25 2019 *)

vector(60, n, n--; (2*n-1)*(4*n-1)) \\ Michel Marcus, Apr 12 2015

a(n) = a(n-1) + 16*n - 14 (with a(0)=1). - Vincenzo Librandi, Nov 17 2010
From G. C. Greubel, Jul 06 2017: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-2).
E.g.f.: (1 + 2*x + 8*x^2)*exp(x).
G.f.: (1 + 15*x^2)/(1 - x)^3. (End)
From Amiram Eldar, Jan 03 2022: (Start)
Sum_{n>=0} 1/a(n) = 1 + Pi/4 - log(2)/2.
Sum_{n>=0} (-1)^n/a(n) = 1 + (sqrt(2)-1)*Pi/4 + log(sqrt(2)-1)/sqrt(2). (End)

More terms from Michel Marcus, Apr 12 2015

A144437 Period 3: repeat [3, 3, 1].

Original entry on oeis.org

3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3, 1, 3, 3

Offset: 1

Paul Curtz, Oct 05 2008

The sequence is generated from numerators in the energy differences of the hydrogen spectrum: A005563(1), A061037(4), A061039(6), A061041(8), A061043(10), A061045(12), A061047(14), A061049(16), ...
Conjecture: a(n) is the separatix. See A045944.
Also the decimal expansion of the constant 3310/999. - R. J. Mathar, May 21 2009
Continued fraction expansion of A171417.
Greatest common divisor of (n+1)^2-1 and (n+1)^2+2. - Bruno Berselli, Mar 08 2017

Index entries for linear recurrences with constant coefficients, signature (0,0,1).

Cf. A000217, A045944, A109007, A164306, A167192, A169609, A256095.

Numerators in the energy differences of the hydrogen spectrum: A005563(1), A061037(4), A061039(6), A061041(8), A061043(10), A061045(12), A061047(14), A061049(16), ...

&cat [[3, 3, 1]^^30]; // Wesley Ivan Hurt, Jul 02 2016

seq(op([3, 3, 1]), n=1..50); # Wesley Ivan Hurt, Jul 02 2016

A144437[n_]:=Denominator[n/3]; Array[A144437,100] (* Enrique Pérez Herrero, Oct 05 2011 *)
CoefficientList[Series[(3 + 3 x + x^2)/(1 - x^3), {x, 0, 120}], x] (* Michael De Vlieger, Jul 02 2016 *)
Table[Mod[2*n^2 + 1, 3,1], {n,1,50}] (* G. C. Greubel, Aug 24 2017 *)

a(n)=if(n%3,3,1) \\ Charles R Greathouse IV, Sep 28 2015

a(n) = (7-4*cos(2*Pi*n/3))/3. - Jaume Oliver Lafont, Nov 23 2008
G.f.: x*(3 + 3*x + x^2)/((1 - x)*(1 + x + x^2)). - R. J. Mathar, May 21 2009
a(n) = 3/gcd(n,3). - Reinhard Zumkeller, Oct 30 2009
a(n) = denominator(n^k/3), where k>0 is an integer. - Enrique Pérez Herrero, Oct 05 2011
a(n) = gcd(T(n+1), T(2)) = A256095(n+1, 2), with the triangular numbers T = A000217, for n >= 1. - Wolfdieter Lang, Mar 17 2015
a(n) = a(n-3) for n>3; a(n) = A169609(n) for n>0. - Wesley Ivan Hurt, Jul 02 2016
E.g.f.: (1/3)*(7*exp(x) - 4*exp(-x/2)*cos(sqrt(3)*x/2) - 3). - G. C. Greubel, Aug 24 2017
From Nicolas Bělohoubek, Nov 11 2021: (Start)
a(n) = 9/(a(n-2)*a(n-1)).
a(n) = 7 - a(n-2) - a(n-1). See also A052901 or A069705. (End)

Edited by R. J. Mathar, May 21 2009

A174709 Partial sums of floor(n/6).

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 1, 2, 3, 4, 5, 6, 8, 10, 12, 14, 16, 18, 21, 24, 27, 30, 33, 36, 40, 44, 48, 52, 56, 60, 65, 70, 75, 80, 85, 90, 96, 102, 108, 114, 120, 126, 133, 140, 147, 154, 161, 168, 176, 184, 192

Offset: 0

Mircea Merca, Nov 30 2010

Partial sums of A152467.

			a(7) = floor(0/6) + floor(1/6) + floor(2/6) + floor(3/6) + floor(4/6) + floor(5/6) + floor(6/6) + floor(7/6) = 0 + 0 + 0 + 0 + 0 + 0 + 1 + 1 = 2.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Mircea Merca, Inequalities and Identities Involving Sums of Integer Functions, J. Integer Sequences, Vol. 14 (2011), Article 11.9.1.
Index entries for linear recurrences with constant coefficients, signature (2,-1,0,0,0,1,-2,1).

Cf. A008724, A152467.

[Round(n*(n-4)/12): n in [0..60]]; // Vincenzo Librandi, Jun 22 2011

Maple
```
a(n):=round(n*(n-4)/12)
```

Table[Sum[Floor[k/6], {k,0,n}], {n,0,25}] (* G. C. Greubel, Dec 13 2016 *)

a(n)=(n-2)^2\12 \\ Charles R Greathouse IV, Oct 07 2015

a(n) = round(n*(n-4)/12) = round((2*n^2 - 8*n - 1)/24).
a(n) = floor((n-2)^2/12).
a(n) = ceiling((n+1)*(n-5)/12).
a(n) = a(n-6) + n - 5, n > 5.
From R. J. Mathar, Nov 30 2010: (Start)
a(n) = 2*a(n-1) - a(n-2) + a(n-6) - 2*a(n-7) + a(n-8).
G.f.: -x^6 / ( (1+x)*(x^2-x+1)*(1+x+x^2)*(x-1)^3 ).
a(n) = -n/3 + 5/72 + n^2/12 + (-1)^n/24 + A057079(n+5)/6 + A061347(n)/18. (End)
a(6n) = A000567(n), a(6n+1) = 2*A000326(n), a(6n+2) = A033428(n), a(6n+3) = A049451(n), a(6n+4) = A045944(n), a(6n+5) = A028896(n). - Philippe Deléham, Mar 26 2013
a(n) = A008724(n-2). - R. J. Mathar, Jul 10 2015
Sum_{n>=6} 1/a(n) = Pi^2/18 - Pi/(2*sqrt(3)) + 49/12. - Amiram Eldar, Aug 13 2022

A212194 Triangle T(n,k), n>=1, 0<=k<=n^2, read by rows: row n gives the coefficients of the chromatic polynomial of the staggered hexagonal square grid graph SH_(n,n), highest powers first.

Original entry on oeis.org

1, 0, 1, -5, 8, -4, 0, 1, -16, 112, -448, 1120, -1791, 1786, -1012, 248, 0, 1, -33, 510, -4898, 32703, -160859, 602408, -1749715, 3975561, -7068408, 9755858, -10265148, 7968348, -4304712, 1445104, -226720, 0, 1, -56, 1508, -25992, 321994, -3051871, 23000726, -141421592, 722137763, -3101089710

Offset: 1

Alois P. Heinz, May 03 2012

T differs from A212162 first at (n,k) = (5,10): T(5,10) = -3101089710, A212162(5,10) = -3101089711.

The staggered hexagonal square grid graph SH_(n,n) has n^2 = A000290(n) vertices and (n-1)*(3*n-1) = A045944(n-1) edges. The chromatic polynomial of SH_(n,n) has n^2+1 = A002522(n) coefficients.

			3 example graphs:                        o--o--o
.                                        | /|\ |
.                                        |/ | \|
.                            o--o        o--o--o
.                            | /|        | /|\ |
.                            |/ |        |/ | \|
.               o            o--o        o--o--o
Graph:       SH_(1,1)      SH_(2,2)      SH_(3,3)
Vertices:       1             4             9
Edges:          0             5            16
The staggered hexagonal square grid graph SH_(2,2) has chromatic polynomial q^4 -5*q^3 +8*q^2 -4*q => row 2 = [1, -5, 8, -4, 0].
Triangle T(n,k) begins:
1,    0;
1,   -5,     8,      -4,        0;
1,  -16,   112,    -448,     1120,      -1791, ...
1,  -33,   510,   -4898,    32703,    -160859, ...
1,  -56,  1508,  -25992,   321994,   -3051871, ... , -3101089710, ...
1,  -85,  3520,  -94620,  1855860,  -28306676, ...
1, -120,  7068, -272344,  7720110, -171656543, ...
1, -161, 12782, -667058, 25738055, -783003395, ...

Alois P. Heinz, Rows n = 1..8, flattened
Wikipedia, Chromatic Polynomial

Columns 1-2 give: A000012, (-1)*A045944(n-1).
Row sums (for n>1) and last elements of rows give: A000004, row lengths give: A002522.
Cf. A000290, A212162, A212195.

A135705 a(n) = 10binomial(n,2) + 9n.

Original entry on oeis.org

0, 9, 28, 57, 96, 145, 204, 273, 352, 441, 540, 649, 768, 897, 1036, 1185, 1344, 1513, 1692, 1881, 2080, 2289, 2508, 2737, 2976, 3225, 3484, 3753, 4032, 4321, 4620, 4929, 5248, 5577, 5916, 6265, 6624, 6993, 7372, 7761, 8160, 8569, 8988, 9417, 9856, 10305, 10764

Offset: 0

N. J. A. Sloane, Mar 04 2008

Also, second 12-gonal (or dodecagonal) numbers. Identity for the numbers b(n)=n*(h*n+h-2)/2 (see Crossrefs): Sum_{i=0..n} (b(n)+i)^2 = (Sum_{i=n+1..2*n} (b(n)+i)^2) + h*(h-4)*A000217(n)^2 for n>0. - Bruno Berselli, Jan 15 2011
Sequence found by reading the line from 0, in the direction 0, 28, ..., and the line from 9, in the direction 9, 57, ..., in the square spiral whose vertices are the generalized 12-gonal numbers A195162. - Omar E. Pol, Jul 24 2012
Bisection of A195162. - Omar E. Pol, Aug 04 2012

Ivan Panchenko, Table of n, a(n) for n = 0..1000
L. Hogben, Choice and Chance by Cardpack and Chessboard, Vol. 1, Max Parrish and Co, London, 1950, p. 36.
Leo Tavares, Illustration: Diamond Clipped Stars
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Second n-gonal numbers: A005449, A014105, A147875, A045944, A179986, A033954, A062728, this sequence.
Cf. A003154, A000290.
Cf. A195162.

List([0..50], n-> n*(5*n+4)); # G. C. Greubel, Jul 04 2019

[n*(5*n+4): n in [0..50]]; // G. C. Greubel, Jul 04 2019

LinearRecurrence[{3,-3,1}, {0,9,28}, 50] (* or *) Table[5*n^2 + 4*n, {n,0,50}] (* G. C. Greubel, Oct 29 2016 *)
Table[10 Binomial[n,2]+9n,{n,0,60}] (* Harvey P. Dale, Jun 14 2023 *)

a(n) = 10*binomial(n,2) + 9*n \\ Charles R Greathouse IV, Jun 11 2015

[n*(5*n+4) for n in (0..50)] # G. C. Greubel, Jul 04 2019

From R. J. Mathar, Mar 06 2008: (Start)
O.g.f.: x*(9+x)/(1-x)^3.
a(n) = n*(5*n+4). (End)
a(n) = a(n-1) + 10*n - 1 (with a(0)=0). - Vincenzo Librandi, Nov 24 2009
a(n) = Sum_{i=0..n-1} A017377(i) for n>0. - Bruno Berselli, Jan 15 2011
a(n) = A131242(10n+8). - Philippe Deléham, Mar 27 2013
Sum_{n>=1} 1/a(n) = 5/16 + sqrt(1 + 2/sqrt(5))*Pi/8 - 5*log(5)/16 - sqrt(5)*log((1 + sqrt(5))/2)/8 = 0.2155517745488486003038... . - Vaclav Kotesovec, Apr 27 2016
From G. C. Greubel, Oct 29 2016: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
E.g.f.: x*(9 + 5*x)*exp(x). (End)
a(n) = A003154(n+1) - A000290(n+1). - Leo Tavares, Mar 29 2022

A165355 a(n) = 3n + 1 if n is even, or a(n) = (3n + 1)/2 if n is odd.

Original entry on oeis.org

1, 2, 7, 5, 13, 8, 19, 11, 25, 14, 31, 17, 37, 20, 43, 23, 49, 26, 55, 29, 61, 32, 67, 35, 73, 38, 79, 41, 85, 44, 91, 47, 97, 50, 103, 53, 109, 56, 115, 59, 121, 62, 127, 65, 133, 68, 139, 71, 145, 74, 151, 77, 157, 80, 163, 83, 169, 86, 175, 89, 181, 92, 187, 95, 193, 98

Offset: 0

Paul Curtz, Sep 16 2009

Second trisection of A026741.
A111329(n+1) = A000041(a(n)). - Reinhard Zumkeller, Nov 19 2009
We observe that this sequence is a particular case of sequence for which there exists q: a(n+3) = (a(n+2)*a(n+1)+q)/a(n) for every n >= n0. Here q=-9 and n0=0. - Richard Choulet, Mar 01 2010
The entries are also encountered via the bilinear transform approximation to the natural log (unit circle). Specifically, evaluating 2(z-1)/(z+1) at z = 2, 3, 4, ..., A165355 entries stem from the pair (sums) seen 2 ahead of each new successive prime. For clarity, the evaluation output is 2, 3, 1, 1, 6, 5, 4, 3, 10, 7, 3, 2, 14, 9, 8, 5, 18, 11, ..., where (1+1), (4+3), (3+2), (8+5), ... generate the A165355 entries (after the first). As an aside, the same mechanism links A165355 to A140777. - Bill McEachen, Jan 08 2015
As a follow-up to the previous comment, it appears that the numerators and denominators of 2(z-1)/(z+1) are respectively given by A145979 and A060819, but with different offsets. - Michel Marcus, Jan 14 2015
Odd parts of the terms give A067745. E.g.: 1, 2/2, 7, 5, 13, 8/8 .... - Joe Slater, Nov 30 2016

G. C. Greubel, Table of n, a(n) for n = 0..1000
D. H. Lehmer, Continued fractions containing arithmetic progressions, Scripta Mathematica, 29 (1973): 17-24. [Annotated copy of offprint]
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A165351, A165367, A067745, A007814.

f[n_] := If[ OddQ@ n, (3n +1)/2, (3n +1)]; Array[f, 66, 0] (* Robert G. Wilson v, Jan 26 2015 *)
f[n_] := (3 (-1)^(2n) + (-1)^(1 + n)) (-2 + 3n)/4; Array[f, 66] (* or *)
CoefficientList[ Series[(x^3 + 5x^2 + 2x + 1)/(x^2 - 1)^2, {x, 0, 65}], x] (* or *)
LinearRecurrence[{0, 2, 0, -1}, {1, 2, 7, 5}, 66] (* Robert G. Wilson v, Apr 13 2017 *)

a(n)=n+=2*n+1; if(n%2,n,n/2) \\ Charles R Greathouse IV, Jan 13 2015

a(n) = A026741(3*n+1).
a(n)*A026741(n) = A005449(n).
a(n)*A022998(n+1) = A000567(n+1).
a(n) = A026741(n+1) + A022998(n).
a(2n) = A016921(n). a(2n+1) = A016789(n).
a(2n+1)*A026741(2n) = A045944(n).
G.f.: (1+2*x+5*x^2+x^3)/((x-1)^2 * (1+x)^2). - R. J. Mathar, Sep 26 2009
a(n) = (3+9*n)/4 + (-1)^n*(1+3*n)/4. - R. J. Mathar, Sep 26 2009
a(n) = 2*(3n+1)/(4-((2n+2) mod 4)). - Bill McEachen, Jan 09 2015
If a(2n-1) = x then a(2n) = 2x+3. - Robert G. Wilson v, Jan 26 2015
Let the reduced Collatz procedure be defined as Cr(n) = (3*n+1)/2. For odd n, a(n) = Cr(n). For even n, a(n) = Cr(4*n+1)/2. - Joe Slater, Nov 29 2016
a(n) = A067745(n+1) * 2^A007814((3n+1)/2). - Joe Slater, Nov 30 2016
a(n) = 2*a(n-2) - a(n-4). - G. C. Greubel, Apr 13 2017

All comments changed to formulas by R. J. Mathar, Sep 26 2009
New name from Charles R Greathouse IV, Jan 13 2015
Name corrected by Joe Slater, Nov 29 2016

cp's OEIS Frontend

A179986 Second 9-gonal (or nonagonal) numbers: a(n) = n*(7*n+5)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Magma

Mathematica

PARI

Formula

A139273 a(n) = n*(8*n - 3).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

A061047 Numerator of 1/49 - 1/n^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Extensions

A062728 Second 11-gonal (or hendecagonal) numbers: a(n) = n*(9*n+7)/2.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

GAP

Magma

Mathematica

PARI

Sage

Formula

Extensions

A033567 a(n) = (2*n-1)*(4*n-1).

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

PARI

Formula

Extensions

A144437 Period 3: repeat [3, 3, 1].

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A179986 Second 9-gonal (or nonagonal) numbers: a(n) = n(7n+5)/2.

A139273 a(n) = n(8n - 3).

A062728 Second 11-gonal (or hendecagonal) numbers: a(n) = n(9n+7)/2.

A033567 a(n) = (2n-1)(4*n-1).

A135705 a(n) = 10binomial(n,2) + 9n.