cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

Showing 1-10 of 16 results. Next

A139592 A033585(n) followed by A139271(n+1).

Original entry on oeis.org

0, 2, 10, 20, 36, 54, 78, 104, 136, 170, 210, 252, 300, 350, 406, 464, 528, 594, 666, 740, 820, 902, 990, 1080, 1176, 1274, 1378, 1484, 1596, 1710, 1830, 1952, 2080, 2210, 2346, 2484, 2628, 2774, 2926, 3080, 3240, 3402, 3570, 3740
Offset: 0

Views

Author

Omar E. Pol, May 03 2008

Keywords

Comments

Sequence found by reading the line from 0, in the direction 0, 2,... and the same line from 0, in the direction 0, 10,..., in the square spiral whose vertices are the triangular numbers A000217.
a(n) = 2*A006578(n) - A002378(n)/2 = 2*A035608(n). [From Reinhard Zumkeller, Feb 07 2010]

Examples

			Array begins:
0, 2
10, 20
36, 54
78, 104
		

Crossrefs

Formula

Array read by rows: row n gives 8*n^2 + 2n, 8*(n+1)^2 - 6(n+1).
a(n) = 2*floor((n + 1/4)^2). [From Reinhard Zumkeller, Feb 07 2010]
G.f.: 2*x*(1+3*x)/((1-x)^3*(1+x)). [Colin Barker, Apr 26 2012]

A226488 a(n) = n*(13*n - 9)/2.

Original entry on oeis.org

0, 2, 17, 45, 86, 140, 207, 287, 380, 486, 605, 737, 882, 1040, 1211, 1395, 1592, 1802, 2025, 2261, 2510, 2772, 3047, 3335, 3636, 3950, 4277, 4617, 4970, 5336, 5715, 6107, 6512, 6930, 7361, 7805, 8262, 8732, 9215, 9711, 10220, 10742, 11277, 11825, 12386, 12960
Offset: 0

Views

Author

Bruno Berselli, Jun 09 2013

Keywords

Comments

Sum of n-th octagonal number and n-th 9-gonal (nonagonal) number.
Sum of reciprocals of a(n), for n>0: 0.629618994194109711163742089971688...

Crossrefs

Cf. A000567, A001106, A153080 (first differences).
Cf. numbers of the form n*(n*k-k+4)/2 listed in A005843 (k=0), A000096 (k=1), A002378 (k=2), A005449 (k=3), A001105 (k=4), A005476 (k=5), A049450 (k=6), A218471 (k=7), A002939 (k=8), A062708 (k=9), A135706 (k=10), A180223 (k=11), A139267 (n=12), this sequence (k=13), A139268 (k=14), A226489 (k=15), A139271 (k=16), A180232 (k=17), A152995 (k=18), A226490 (k=19), A152965 (k=20), A226491 (k=21), A152997 (k=22).

Programs

  • GAP
    List([0..50], n-> n*(13*n-9)/2); # G. C. Greubel, Aug 30 2019
  • Magma
    [n*(13*n-9)/2: n in [0..50]];
    
  • Magma
    I:=[0,2,17]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2) +Self(n-3): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013
    
  • Maple
    A226488:=n->n*(13*n - 9)/2; seq(A226488(n), n=0..50); # Wesley Ivan Hurt, Feb 25 2014
  • Mathematica
    Table[n(13n-9)/2, {n, 0, 50}]
    LinearRecurrence[{3, -3, 1}, {0, 2, 17}, 50] (* Harvey P. Dale, Jun 19 2013 *)
    CoefficientList[Series[x(2+11x)/(1-x)^3, {x, 0, 45}], x] (* Vincenzo Librandi, Aug 18 2013 *)
  • PARI
    a(n)=n*(13*n-9)/2 \\ Charles R Greathouse IV, Sep 24 2015
    
  • Sage
    [n*(13*n-9)/2 for n in (0..50)] # G. C. Greubel, Aug 30 2019
    

Formula

G.f.: x*(2+11*x)/(1-x)^3.
a(n) + a(-n) = A152742(n).
a(0)=0, a(1)=2, a(2)=17; for n>2, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, Jun 19 2013
E.g.f.: x*(4 + 13*x)*exp(x)/2. - G. C. Greubel, Aug 30 2019
a(n) = A000567(n) + A001106(n). - Michel Marcus, Aug 31 2019

A139275 a(n) = n*(8*n+1).

Original entry on oeis.org

0, 9, 34, 75, 132, 205, 294, 399, 520, 657, 810, 979, 1164, 1365, 1582, 1815, 2064, 2329, 2610, 2907, 3220, 3549, 3894, 4255, 4632, 5025, 5434, 5859, 6300, 6757, 7230, 7719, 8224, 8745, 9282, 9835, 10404, 10989, 11590, 12207, 12840
Offset: 0

Views

Author

Omar E. Pol, Apr 26 2008

Keywords

Comments

Sequence found by reading the line from 0, in the direction 0, 9,..., in the square spiral whose vertices are the triangular numbers A000217.

Crossrefs

Programs

  • Mathematica
    Table[n (8 n + 1), {n, 0, 40}] (* Bruno Berselli, Sep 21 2016 *)
    LinearRecurrence[{3,-3,1},{0,9,34},50] (* Harvey P. Dale, Apr 21 2020 *)
  • PARI
    a(n) = n*(8*n+1); \\ Altug Alkan, Sep 21 2016

Formula

a(n) = 8*n^2 + 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) - 7 with n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
a(n) = A000217(5*n) - A000217(3*n). - Bruno Berselli, Sep 21 2016
Sum_{n>=1} 1/a(n) = 8 - (1+sqrt(2))*Pi/2 - 4*log(2) - sqrt(2) * log(1+sqrt(2)) = 0.1887230016056779928... . - Vaclav Kotesovec, Sep 21 2016
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(7*x + 9)/(1-x)^3.
E.g.f.: (8*x^2 + 9*x)*exp(x). (End)

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

Views

Author

Omar E. Pol, Apr 26 2008

Keywords

Comments

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

Crossrefs

Programs

  • Magma
    [ n*(8*n-3) : n in [0..40] ];  // Bruno Berselli, Feb 11 2011
    
  • Mathematica
    Table[n (8 n - 3), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 5, 26}, 40] (* Harvey P. Dale, Feb 02 2012 *)
  • PARI
    a(n)=n*(8*n-3) \\ Charles R Greathouse IV, Sep 24 2015

Formula

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

A139278 a(n) = n*(8*n+7).

Original entry on oeis.org

0, 15, 46, 93, 156, 235, 330, 441, 568, 711, 870, 1045, 1236, 1443, 1666, 1905, 2160, 2431, 2718, 3021, 3340, 3675, 4026, 4393, 4776, 5175, 5590, 6021, 6468, 6931, 7410, 7905, 8416, 8943, 9486, 10045, 10620, 11211, 11818, 12441, 13080
Offset: 0

Views

Author

Omar E. Pol, Apr 26 2008

Keywords

Comments

Sequence found by reading the segment (0, 15) together with the line from 15, in the direction 15, 46, ..., in the square spiral whose vertices are the triangular numbers A000217.

Crossrefs

Programs

  • Mathematica
    Table[n (8 n + 7), {n, 0, 40}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 15, 46}, 50] (* Harvey P. Dale, Oct 07 2015 *)
  • PARI
    a(n)=n*(8*n+7) \\ Charles R Greathouse IV, Jun 17 2017

Formula

a(n) = 8*n^2 + 7*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)= 3a(n-1)-3a(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)-1 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(x+15)/(1-x)^3.
E.g.f.: (8*x^2 + 15*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 8/49 + (sqrt(2)+1)*Pi/14 - 4*log(2)/7 - sqrt(2)*log(sqrt(2)+1)/7. - Amiram Eldar, Mar 17 2022

A139272 a(n) = n*(8*n-5).

Original entry on oeis.org

0, 3, 22, 57, 108, 175, 258, 357, 472, 603, 750, 913, 1092, 1287, 1498, 1725, 1968, 2227, 2502, 2793, 3100, 3423, 3762, 4117, 4488, 4875, 5278, 5697, 6132, 6583, 7050, 7533, 8032, 8547, 9078, 9625, 10188, 10767, 11362, 11973, 12600
Offset: 0

Views

Author

Omar E. Pol, Apr 26 2008

Keywords

Comments

Sequence found by reading the line from 0, in the direction 0, 3, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139276 in the same spiral.

Crossrefs

Cf. numbers of the form n*(n*k-k+6)/2, this sequence is the case k=16: see Comments lines of A226492.

Programs

Formula

a(n) = 8*n^2 - 5*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) = 3a(n-1) - 3a(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) - 13 with n>0, a(0)=0. - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(13*x + 3)/(1-x)^3.
E.g.f.: (8*x^2 + 3*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = ((sqrt(2)-1)*Pi + 8*log(2) - 2*sqrt(2)*log(sqrt(2)+1))/10. - Amiram Eldar, Mar 17 2022

A139274 a(n) = n*(8*n-1).

Original entry on oeis.org

0, 7, 30, 69, 124, 195, 282, 385, 504, 639, 790, 957, 1140, 1339, 1554, 1785, 2032, 2295, 2574, 2869, 3180, 3507, 3850, 4209, 4584, 4975, 5382, 5805, 6244, 6699, 7170, 7657, 8160, 8679, 9214, 9765, 10332, 10915, 11514, 12129, 12760, 13407, 14070, 14749
Offset: 0

Views

Author

Omar E. Pol, Apr 26 2008

Keywords

Comments

Sequence found by reading the line from 0, in the direction 0, 7, ..., in the square spiral whose vertices are the triangular numbers A000217.
Polygonal number connection: 2*P_n + 5*S_n where P_n is the n-th pentagonal number and S_n is the n-th square. - William A. Tedeschi, Sep 12 2010

Examples

			a(1) = 16*1 + 0 - 9 = 7; a(2) = 16*2 + 7 - 9 = 30; a(3) = 16*3 + 30 - 9 = 69. - _Vincenzo Librandi_, Aug 03 2010
		

Crossrefs

Programs

Formula

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) = 3a(n-1) - 3a(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) - 9 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
a(n) = (1/3) * Sum_{i=n..(7*n-1)} i. - Wesley Ivan Hurt, Dec 04 2016
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(9*x+7)/(1-x)^3.
E.g.f.: (8*x^2 + 7*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 4*log(2) + sqrt(2)*log(sqrt(2)+1) - (sqrt(2)+1)*Pi/2. - Amiram Eldar, Mar 18 2022

A152767 3 times 10-gonal (or decagonal) numbers: a(n) = 3*n*(4*n-3).

Original entry on oeis.org

0, 3, 30, 81, 156, 255, 378, 525, 696, 891, 1110, 1353, 1620, 1911, 2226, 2565, 2928, 3315, 3726, 4161, 4620, 5103, 5610, 6141, 6696, 7275, 7878, 8505, 9156, 9831, 10530, 11253, 12000, 12771, 13566, 14385, 15228, 16095, 16986, 17901, 18840, 19803, 20790, 21801
Offset: 0

Views

Author

Omar E. Pol, Dec 15 2008

Keywords

Comments

3*A172078(n) = n*a(n) - Sum_{k=0..n-1} a(k). - Bruno Berselli, Dec 12 2010

Examples

			For n=8, a(8) = (1*3 + 5*7 + 9*11 +..+ 29*31) - (2*4 + 6*8 + 10*12 +..+ 26*28) = 696 (see Problem 1052 in References). - _Bruno Berselli_, Dec 12 2010
		

References

  • "Supplemento al Periodico di Matematica", Raffaello Giusti Editore (Livorno), Jan. 1910 p. 47 (Problem 1052).

Crossrefs

Cf. numbers of the form n*(n*k - k + 6)/2, this sequence is the case k=24: see Comments lines of A226492.

Programs

Formula

a(n) = 12*n^2 - 9*n = 3*A001107(n).
a(n) = a(n-1) + 24*n - 21, n > 0. - Vincenzo Librandi, Nov 26 2010
a(n) = Sum_{k=0..n-1} A001539(k) - Sum_{k=0..n-1} 4*A002939(k) if n > 0 (see References, Problem 1052). - Bruno Berselli, Dec 08 2010 - Jan 21 2011
G.f.: -3*x*(1+7*x)/(x-1)^3.
a(0)=0, a(1)=3, a(2)=30, a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). - Harvey P. Dale, May 26 2012
From Elmo R. Oliveira, Dec 15 2024: (Start)
E.g.f.: 3*exp(x)*x*(1 + 4*x).
a(n) = A153794(n) - n. (End)

A139276 a(n) = n*(8*n+3).

Original entry on oeis.org

0, 11, 38, 81, 140, 215, 306, 413, 536, 675, 830, 1001, 1188, 1391, 1610, 1845, 2096, 2363, 2646, 2945, 3260, 3591, 3938, 4301, 4680, 5075, 5486, 5913, 6356, 6815, 7290, 7781, 8288, 8811, 9350, 9905, 10476, 11063, 11666, 12285, 12920
Offset: 0

Views

Author

Omar E. Pol, Apr 26 2008

Keywords

Comments

Sequence found by reading the line from 0, in the direction 0, 11,..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A139272 in the same spiral.

Examples

			a(1)=16*1+0-5=11; a(2)=16*2+11-5=38; a(3)=16*3+38-5=81. - _Vincenzo Librandi_, Aug 03 2010
		

Crossrefs

Programs

Formula

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)= 3a(n-1)-3a(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)-5 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: x*(5*x + 11)/(1-x)^3.
E.g.f.: (8*x^2 + 11*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = 8/9 - (sqrt(2)-1)*Pi/6 - 4*log(2)/3 + sqrt(2)*log(sqrt(2)+1)/3. - Amiram Eldar, Mar 17 2022

A211377 T(n,k) = ((k + n)^2 - 4*k + 3 + (-1)^k - (k + n - 2)*(-1)^(k + n))/2; n, k > 0, read by antidiagonals.

Original entry on oeis.org

1, 3, 4, 2, 5, 6, 8, 9, 12, 13, 7, 10, 11, 14, 15, 17, 18, 21, 22, 25, 26, 16, 19, 20, 23, 24, 27, 28, 30, 31, 34, 35, 38, 39, 42, 43, 29, 32, 33, 36, 37, 40, 41, 44, 45, 47, 48, 51, 52, 55, 56, 59, 60, 63, 64, 46, 49, 50, 53, 54, 57, 58, 61, 62, 65, 66, 68
Offset: 1

Views

Author

Boris Putievskiy, Feb 07 2013

Keywords

Comments

Permutation of the natural numbers.
a(n) is a pairing function: a function that reversibly maps Z^{+} x Z^{+} onto Z^{+}, where Z^{+} is the set of integer positive numbers.
Enumeration table T(n,k). The order of the list:
T(1,1)=1;
T(1,3), T(1,2), T(2,1), T(2,2), T(3,1);
...
T(1,n), T(1,n-1), T(2,n-2), T(2,n-1), T(3,n-2), T(3,n-3)...T(n,1);
...
Descent by snake along two adjacent antidiagonal - step to the west, step to the southwest, step to the east, step to the southwest and so on. The length of each step is 1.
Table contains:
row 1 is alternation of elements A130883 and A033816,
row 2 accommodates elements A100037 in odd places;
column 1 is alternation of elements A000384 and A091823,
column 2 is alternation of elements A071355 and A014106,
column 3 accommodates elements A130861 in even places;
main diagonal accommodates elements A188135 in odd places,
diagonal 1, located above the main diagonal, is alternation of elements A033567 and A033566,
diagonal 2, located above the main diagonal, is alternation of elements A139271 and A033585.

Examples

			The start of the sequence as a table:
   1,  3,  2,   8,   7,  17,  16,  30,  29,  47,  46, ...
   4,  5,  9,  10,  18,  19,  31,  32,  48,  49,  69, ...
   6, 12, 11,  21,  20,  34,  33,  51,  50,  72,  71, ...
  13, 14, 22,  23,  35,  36,  52,  53,  73,  74,  98, ...
  15, 25, 24,  38,  37,  55,  54,  76,  75, 101, 100, ...
  26, 27, 39,  40,  56,  57,  77,  78, 102, 103, 131, ...
  28, 42, 41,  59,  58,  80,  79, 105, 104, 134, 133, ...
  43, 44, 60,  61,  81,  82, 106, 107, 135, 136, 168, ...
  45, 63, 62,  84,  83, 109, 108, 138, 137, 171, 170, ...
  64, 65, 85,  86, 110, 111, 139, 140, 172, 173, 209, ...
  66, 88, 87, 113, 112, 142, 141, 175, 174, 212, 211, ...
  ...
The start of the sequence as triangle array read by rows:
   1;
   3,  4;
   2,  5,  6;
   8,  9, 12, 13;
   7, 10, 11, 14, 15;
  17, 18, 21, 22, 25, 26;
  16, 19, 20, 23, 24, 27, 28;
  30, 31, 34, 35, 38, 39, 42, 43;
  29, 32, 33, 36, 37, 40, 41, 44, 45;
  47, 48, 51, 52, 55, 56, 59, 60, 63, 64;
  46, 49, 50, 53, 54, 57, 58, 61, 62, 65, 66;
  ...
The start of the sequence as an array read by rows, the length of row r is 4*r-3.
First 2*r-2 numbers are from row number 2*r-2 of the triangular array above.
Last  2*r-1 numbers are from row number 2*r-1 of the triangular array above.
   1;
   3,  4,  2,  5,  6;
   8,  9, 12, 13,  7, 10, 11, 14, 15;
  17, 18, 21, 22, 25, 26, 16, 19, 20, 23, 24, 27, 28;
  30, 31, 34, 35, 38, 39, 42, 43, 29, 32, 33, 36, 37, 40, 41, 44, 45;
  47, 48, 51, 52, 55, 56, 59, 60, 63, 64, 46, 49, 50, 53, 54, 57, 58, 61, 62, 65, 66;
  ...
Row number r contains permutation 4*r-3 numbers from 2*r*r-5*r+4 to 2*r*r-r:
2*r*r-5*r+5, 2*r*r-5*r+6,...2*r*r-r-4, 2*r*r-r-1, 2*r*r-r.
		

Crossrefs

Programs

  • Mathematica
    T[n_, k_] := ((k+n)^2 - 4k + 3 + (-1)^k - (k+n-2)(-1)^(k+n))/2;
    Table[T[n-k+1, k], {n, 1, 12}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Nov 29 2018 *)
  • Python
    t=int((math.sqrt(8*n-7) - 1)/ 2)
    i=n-t*(t+1)/2
    j=(t*t+3*t+4)/2-n
    result=((t+2)**2-4*j+3+(-1)**j-t*(-1)**(t+2))/2

Formula

As a table:
T(n,k) = ((k + n)^2 - 4*k + 3 + (-1)^k - (k + n - 2)*(-1)^(k + n))/2.
As a linear sequence:
a(n) = ((t + 2)^2 - 4*j + 3 + (-1)^j - t*(-1)^t)/2, where j = (t*t + 3*t + 4)/2 - n and t = int((sqrt(8*n - 7) - 1)/ 2).
Showing 1-10 of 16 results. Next