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 20 results. Next

A139596 A033587(n) followed by even hexagonal number A014635(n+1).

Original entry on oeis.org

0, 6, 14, 28, 44, 66, 90, 120, 152, 190, 230, 276, 324, 378, 434, 496, 560, 630, 702, 780, 860, 946, 1034, 1128, 1224, 1326, 1430, 1540, 1652, 1770, 1890, 2016, 2144, 2278, 2414, 2556, 2700, 2850, 3002, 3160, 3320, 3486, 3654, 3828
Offset: 0

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Author

Omar E. Pol, May 03 2008

Keywords

Comments

Sequence found by reading the line from 0, in the direction 0, 6,... and the same line from 0, in the direction 0, 14,..., in the square spiral whose vertices are the triangular numbers A000217.

Examples

			Array begins:
0, 6
14, 28
44, 66
90, 120
		

Crossrefs

Programs

  • Mathematica
    LinearRecurrence[{2,0,-2,1},{0,6,14,28},50] (* Harvey P. Dale, Jan 20 2024 *)

Formula

Array read by rows: row n gives 8*n^2 + 6*n, 8*(n+1)^2 - 2(n+1).
O.g.f.: -2*x*(x+3)/((x-1)^3*(1+x)). - R. J. Mathar, May 06 2008
a(n) = 2*A156859(n). - R. J. Mathar, Feb 28 2018

A033588 Duplicate of A014635.

Original entry on oeis.org

0, 6, 28, 66, 120, 190, 276, 378, 496, 630, 780, 946, 1128, 1326, 1540, 1770, 2016, 2278
Offset: 0

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Author

Keywords

A033991 a(n) = n*(4*n-1).

Original entry on oeis.org

0, 3, 14, 33, 60, 95, 138, 189, 248, 315, 390, 473, 564, 663, 770, 885, 1008, 1139, 1278, 1425, 1580, 1743, 1914, 2093, 2280, 2475, 2678, 2889, 3108, 3335, 3570, 3813, 4064, 4323, 4590, 4865, 5148, 5439, 5738, 6045, 6360, 6683, 7014, 7353, 7700, 8055, 8418
Offset: 0

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Author

Keywords

Comments

Write 0,1,2,... in a clockwise spiral; sequence gives numbers on negative x axis. (See illustration in Example.)
This sequence is the number of expressions x generated for a given modulus n in finite arithmetic. For example, n=1 (modulus 1) generates 3 expressions: 0+0=0(mod 1), 0-0=0(mod 1), 0*0=0(mod 1). By subtracting n from 4n^2, we eliminate the counting of those expressions that would include division by zero, which would be, of course, undefined. - David Quentin Dauthier, Nov 04 2007
From Emeric Deutsch, Sep 21 2010: (Start)
a(n) is also the Wiener index of the windmill graph D(3,n).
The windmill graph D(m,n) is the graph obtained by taking n copies of the complete graph K_m with a vertex in common (i.e., a bouquet of n pieces of K_m graphs). The Wiener index of a connected graph is the sum of the distances between all unordered pairs of vertices in the graph.
Example: a(2)=14; indeed if the triangles are OAB and OCD, then, denoting distance by d, we have d(O,A)=d(O,B)=d(A,B)=d(O,C)=d(O,D)=d(C,D)=1 and d(A,C)=d(A,D)=d(B,C)=d(B,D)=2. The Wiener index of D(m,n) is (1/2)n(m-1)[(m-1)(2n-1)+1]. For the Wiener indices of D(4,n), D(5,n), and D(6,n) see A152743, A028994, and A180577, respectively. (End)
Even hexagonal numbers divided by 2. - Omar E. Pol, Aug 18 2011
For n > 0, a(n) equals the number of length 3*n binary words having exactly two 0's with the n first bits having at most one 0. For example a(2) = 14. Words are 010111, 011011, 011101, 011110, 100111, 101011, 101101, 101110, 110011, 110101, 110110, 111001, 111010, 111100. - Franck Maminirina Ramaharo, Mar 09 2018
For n >= 1, the continued fraction expansion of sqrt(a(n)) is [2n-1; {1, 2, 1, 4n-2}]. For n=1, this collapses to [1; {1, 2}]. - Magus K. Chu, Sep 06 2022

Examples

			Clockwise spiral (with sequence terms parenthesized) begins
   16--17--18--19
    |
   15   4---5---6
    |   |       |
  (14) (3) (0)  7
    |   |   |   |
   13   2---1   8
    |           |
   12--11--10---9
		

References

  • S. M. Ellerstein, The square spiral, J. Recreational Mathematics 29 (#3, 1998) 188; 30 (#4, 1999-2000), 246-250.
  • R. L. Graham, D. E. Knuth and O. Patashnik, Concrete Mathematics. Addison-Wesley, Reading, MA, 2nd ed., 1994, p. 99.

Crossrefs

Sequences on the four axes of the square spiral: Starting at 0: A001107, A033991, A007742, A033954; starting at 1: A054552, A054556, A054567, A033951.
Sequences on the four diagonals of the square spiral: Starting at 0: A002939 = 2*A000384, A016742 = 4*A000290, A002943 = 2*A014105, A033996 = 8*A000217; starting at 1: A054554, A053755, A054569, A016754.
Sequences obtained by reading alternate terms on the X and Y axes and the two main diagonals of the square spiral: Starting at 0: A035608, A156859, A002378 = 2*A000217, A137932 = 4*A002620; starting at 1: A317186, A267682, A002061, A080335.

Programs

Formula

a(n) = A007742(-n) = A074378(2n-1) = A014848(2n).
G.f.: x*(3+5*x)/(1-x)^3. - Michael Somos, Mar 03 2003
a(n) = A014635(n)/2. - Zerinvary Lajos, Jan 16 2007
From Zerinvary Lajos, Jun 12 2007: (Start)
a(n) = A000326(n) + A005476(n).
a(n) = A049452(n) - A001105(n). (End)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. - Harvey P. Dale, Oct 10 2011
a(n) = A118729(8n+2). - Philippe Deléham, Mar 26 2013
From Ilya Gutkovskiy, Dec 04 2016: (Start)
E.g.f.: x*(3 + 4*x)*exp(x).
Sum_{n>=1} 1/a(n) = 3*log(2) - Pi/2 = 0.50864521488... (End)
a(n) = Sum_{i=n..3n-1} i. - Wesley Ivan Hurt, Dec 04 2016
From Franck Maminirina Ramaharo, Mar 09 2018: (Start)
a(n) = binomial(2*n, 2) + 2*n^2.
a(n) = A054556(n+1) - 1. (End)
Sum_{n>=1} (-1)^(n+1)/a(n) = (Pi + log(3-2*sqrt(2)))/sqrt(2) - log(2). - Amiram Eldar, Mar 20 2022

Extensions

Two remarks combined into one by Emeric Deutsch, Oct 03 2010

A051870 18-gonal (or octadecagonal) numbers: a(n) = n*(8*n-7).

Original entry on oeis.org

0, 1, 18, 51, 100, 165, 246, 343, 456, 585, 730, 891, 1068, 1261, 1470, 1695, 1936, 2193, 2466, 2755, 3060, 3381, 3718, 4071, 4440, 4825, 5226, 5643, 6076, 6525, 6990, 7471, 7968, 8481, 9010, 9555, 10116, 10693, 11286, 11895, 12520
Offset: 0

Views

Author

N. J. A. Sloane, Dec 15 1999

Keywords

Comments

Also, sequence found by reading the segment (0, 1) together with the line from 1, in the direction 1, 18, ..., in the square spiral whose vertices are the triangular numbers A000217. - Omar E. Pol, Apr 26 2008
This sequence does not contain any triangular numbers other than 0 and 1. See A188892. - T. D. Noe, Apr 13 2011
Also sequence found by reading the line from 0, in the direction 0, 18, ... and the parallel line from 1, in the direction 1, 51, ..., in the square spiral whose vertices are the generalized 18-gonal numbers. - Omar E. Pol, Jul 18 2012
Partial sums of 16n + 1 (with offset 0), compare A005570. - Jeremy Gardiner, Aug 04 2012
All x values for Diophantine equation 32*x + 49 = y^2 are given by this sequence and A139278. - Bruno Berselli, Nov 11 2014
This is also a star enneagonal number: a(n) = A001106(n) + 9*A000217(n-1). - Luciano Ancora, Mar 30 2015

References

  • Albert H. Beiler, Recreations in the Theory of Numbers, Dover, N.Y., 1964, p. 189.
  • Elena Deza and Michel Marie Deza, Figurate numbers, World Scientific Publishing, 2012, page 6.

Crossrefs

Programs

Formula

G.f.: x*(1+15*x)/(1-x)^3. - Bruno Berselli, Feb 04 2011
a(n) = 16*n + a(n-1) - 15, with n > 0, a(0) = 0. - Vincenzo Librandi, Aug 06 2010
a(16*a(n)+121*n+1) = a(16*a(n)+121*n) + a(16*n+1). - Vladimir Shevelev, Jan 24 2014
E.g.f.: (8*x^2 + x)*exp(x). - G. C. Greubel, Jul 18 2017
Sum_{n>=1} 1/a(n) = ((1+sqrt(2))*Pi + 2*sqrt(2)*arccoth(sqrt(2)) + 8*log(2))/14. - Amiram Eldar, Oct 20 2020
Product_{n>=2} (1 - 1/a(n)) = 8/9. - Amiram Eldar, Jan 22 2021

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)

A139271 a(n) = 2*n*(4*n-3).

Original entry on oeis.org

0, 2, 20, 54, 104, 170, 252, 350, 464, 594, 740, 902, 1080, 1274, 1484, 1710, 1952, 2210, 2484, 2774, 3080, 3402, 3740, 4094, 4464, 4850, 5252, 5670, 6104, 6554, 7020, 7502, 8000, 8514, 9044, 9590, 10152, 10730, 11324, 11934, 12560, 13202, 13860, 14534, 15224
Offset: 0

Views

Author

Omar E. Pol, Apr 26 2008

Keywords

Comments

Sequence found by reading the line from 0, in the direction 0, 2, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the members of A033585 in the same spiral.
Twice decagonal numbers (or twice 10-gonal numbers). - Omar E. Pol, May 15 2008
a(n) is the number of walks in a cubic lattice of n dimensions that reach the point of origin for the first time after 4 steps. - Shel Kaphan, Mar 20 2023

Crossrefs

Cf. A001107.
Cf. numbers of the form n*(n*k-k+4)/2 listed in A226488 (this sequence is the case k=16). - Bruno Berselli, Jun 10 2013
Row n=2 of A361397.

Programs

Formula

a(n) = 8*n^2 - 6*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) = A001107(n)*2. - Omar E. Pol, May 15 2008
a(n) = 16*n + a(n-1) - 14 (with a(0)=0). - Vincenzo Librandi, Aug 03 2010
From G. C. Greubel, Jul 18 2017: (Start)
G.f.: (2*x)*(7*x+1)/(1-x)^3.
E.g.f.: (8*x^2 + 2*x)*exp(x). (End)
Sum_{n>=1} 1/a(n) = Pi/12 + log(2)/2. - Amiram Eldar, Mar 28 2023

Extensions

Corrected by Harvey P. Dale, Sep 26 2016

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

A033587 a(n) = 2*n*(4*n + 3).

Original entry on oeis.org

0, 14, 44, 90, 152, 230, 324, 434, 560, 702, 860, 1034, 1224, 1430, 1652, 1890, 2144, 2414, 2700, 3002, 3320, 3654, 4004, 4370, 4752, 5150, 5564, 5994, 6440, 6902, 7380, 7874, 8384, 8910, 9452, 10010, 10584, 11174, 11780, 12402, 13040, 13694, 14364, 15050
Offset: 0

Views

Author

Keywords

Comments

The inverse binomial transform is [0, 14, 16, 0, 0, 0, ...]. - R. J. Mathar, May 06 2008
Sequence found by reading the line from 0, in the direction 0, 14, ..., in the square spiral whose vertices are the triangular numbers A000217. Opposite numbers to the even hexagonal numbers A014635 in the same spiral. - Omar E. Pol, Sep 03 2011

Programs

Formula

a(n) = 2*A033954(n).
O.g.f.: 2*x*(7+x)/(1-x)^3. - R. J. Mathar, May 06 2008
a(n) = 16*n + a(n-1) - 2 with a(0)=0. - Vincenzo Librandi, Aug 05 2010
E.g.f.: (8*x^2 + 14*x)*exp(x). - G. C. Greubel, Jul 18 2017
From Vaclav Kotesovec, Aug 18 2018: (Start)
Sum_{n>=1} 1/a(n) = 2/9 + Pi/12 - log(2)/2.
Sum_{n>=1} (-1)^n/a(n) = 2/9 - Pi/(6*sqrt(2)) - log(2)/6 + log(1+sqrt(2))/(3*sqrt(2)). (End)
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