A195320 -id:A195320 - cp's OEIS frontend

A193053 a(n) = (14n(n+3) + (2n-5)(-1)^n + 21)/16.

1, 5, 10, 17, 26, 36, 49, 62, 79, 95, 116, 135, 160, 182, 211, 236, 269, 297, 334, 365, 406, 440, 485, 522, 571, 611, 664, 707, 764, 810, 871, 920, 985, 1037, 1106, 1161, 1234, 1292, 1369, 1430, 1511, 1575, 1660, 1727, 1816, 1886, 1979, 2052, 2149, 2225, 2326

Offset: 0

Bruno Berselli, Oct 20 2011 - based on remarks and sequences by Omar E. Pol

For an origin of this sequence, see the numerical spiral illustrated in the Links section.

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A195020 (vertices of the numerical spiral in link).

Cf. A001106, A022264, A033572, A144555, A152760, A158482, A158485, A185019, A195021, A195023-A195030, A195320, A198017 [incomplete list].

[(14*n*(n+3)+(2*n-5)*(-1)^n+21)/16: n in [0..50]];

Table[(14*n*(n + 3) + (2*n - 5)*(-1)^n + 21)/16, {n, 0, 50}] (* Vincenzo Librandi, Mar 26 2013 *)
LinearRecurrence[{1,2,-2,-1,1},{1,5,10,17,26},60] (* Harvey P. Dale, Jun 19 2020 *)

for(n=0, 50, print1((14*n*(n+3)+(2*n-5)*(-1)^n+21)/16", "));

O.g.f.: (1 + 4*x + 3*x^2 - x^3)/((1 + x)^2*(1 - x)^3).
E.g.f.: (1/16)*((21 + 56*x + 14*x^2)*exp(x) - (5 + 2*x)*exp(-x)). - G. C. Greubel, Aug 19 2017
a(n) = A195020(n) + n + 1.
a(n)   - a(-n-1)  = A047336(n+1).
a(n+1) - a(-n)    = A113804(n+1).
a(n+2) - a(n)     = A047385(n+3).
a(n+4) - a(n)     = A113803(n+4).
a(2*n) + a(2*n-1) = A069127(n+1).
a(2*n) - a(2*n-1) = A016813(n).
a(2*n+1) - a(2*n) = A016777(n+1).
a(n+2) + 2*a(n+1) + a(n) = A024966(n+2).

A198017 a(n) = n(7n + 11)/2 + 1.

Original entry on oeis.org

1, 10, 26, 49, 79, 116, 160, 211, 269, 334, 406, 485, 571, 664, 764, 871, 985, 1106, 1234, 1369, 1511, 1660, 1816, 1979, 2149, 2326, 2510, 2701, 2899, 3104, 3316, 3535, 3761, 3994, 4234, 4481, 4735, 4996, 5264, 5539, 5821, 6110, 6406, 6709, 7019, 7336, 7660, 7991

Offset: 0

Bruno Berselli, Oct 21 2011 - based on remarks and sequences by Omar E. Pol

First bisection of A193053 (see also the numerical spiral illustrated in the Links section).

The inverse binomial transform yields 1, 9, 7, 0, 0 (0 continued).

Bruno Berselli, Table of n, a(n) for n = 0..1000
Bruno Berselli, Illustration of initial terms.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A195020 (vertices of the numerical spiral in link).
Cf. A017005 (first differences).
Cf. A069099, A001106.
Cf. A001106, A022264, A033572, A144555, A152760, A158482, A158485, A185019, A193053, A195021, A195023-A195030, A195320 [incomplete list].

Magma
```
[n*(7*n+11)/2+1: n in [0..47]];
```

Table[(n(7n+11))/2+1,{n,0,60}] (* or *) LinearRecurrence[{3,-3,1},{1,10,26},60] (* Harvey P. Dale, Mar 03 2013 *)

for(n=0, 47, print1(n*(7*n+11)/2+1", "));

G.f.: (1 + 7*x - x^2)/(1-x)^3.
a(n) = A195020(2*n) + 2*n + 1.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) = 2*a(n-1) - a(n-2) + 7.
From Elmo R. Oliveira, Dec 24 2024: (Start)
E.g.f.: exp(x)*(2 + 18*x + 7*x^2)/2.
a(n) = n + A001106(n+1). (End)

A195021 a(n) = n(14n - 11).

Original entry on oeis.org

0, 3, 34, 93, 180, 295, 438, 609, 808, 1035, 1290, 1573, 1884, 2223, 2590, 2985, 3408, 3859, 4338, 4845, 5380, 5943, 6534, 7153, 7800, 8475, 9178, 9909, 10668, 11455, 12270, 13113, 13984, 14883, 15810, 16765, 17748, 18759, 19798, 20865, 21960, 23083, 24234, 25413

Offset: 0

Omar E. Pol, Sep 07 2011

Sequence found by reading the first two vertices [0, 3] together with the line from 34, in the direction 34, 93, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020, which is related to the primitive Pythagorean triple [3, 4, 5]. For another version see A195030.

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A144555, A185019, A193053, A195019, A195020, A195023, A195024, A195025, A195030, A195320, A198017.

Cf. numbers of the form n*(n*k - k + 6)/2, this sequence is the case k=28: see Comments lines of A226492.

[14*n^2 - 11*n: n in [0..50]]; // Vincenzo Librandi, Oct 14 2011

Table[n*(14*n - 11), {n, 0, 50}] (* Paolo Xausa, Jan 15 2025 *)

a(n)=n*(14*n-11) \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 14*n^2 - 11*n.
From Colin Barker, Apr 09 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(3+25*x)/(1-x)^3. (End)
E.g.f.: exp(x)*x*(3 + 14*x). - Elmo R. Oliveira, Dec 30 2024

Edited by Bruno Berselli, Oct 18 2011

A195023 a(n) = 14n^2 - 4n.

Original entry on oeis.org

0, 10, 48, 114, 208, 330, 480, 658, 864, 1098, 1360, 1650, 1968, 2314, 2688, 3090, 3520, 3978, 4464, 4978, 5520, 6090, 6688, 7314, 7968, 8650, 9360, 10098, 10864, 11658, 12480, 13330, 14208, 15114, 16048, 17010, 18000, 19018, 20064, 21138, 22240, 23370, 24528, 25714

Offset: 0

Omar E. Pol, Oct 13 2011

Sequence found by reading the line from 0, in the direction 0, 10, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the one of the semi-axis of the square spiral, which is related to the primitive Pythagorean triple [3, 4, 5].

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A135703, A144555, A152760, A193053, A195019, A195020, A195024, A195320, A198017.

[14*n^2 - 4*n: n in [0..50]]; // Vincenzo Librandi, Oct 14 2011

Table[14n^2-4n,{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,10,48},50] (* Harvey P. Dale, Sep 05 2012 *)

a(n)=14*n^2-4*n \\ Charles R Greathouse IV, Apr 10 2012

a(n) = 2*A135703(n). - Bruno Berselli, Oct 13 2011
From Colin Barker, Apr 09 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: 2*x*(5+9*x)/(1-x)^3. (End)
E.g.f.: 2*exp(x)*x*(5 + 7*x). - Elmo R. Oliveira, Dec 30 2024

Corrected by Vincenzo Librandi, Oct 14 2011

A195024 a(n) = n(14n - 1).

Original entry on oeis.org

0, 13, 54, 123, 220, 345, 498, 679, 888, 1125, 1390, 1683, 2004, 2353, 2730, 3135, 3568, 4029, 4518, 5035, 5580, 6153, 6754, 7383, 8040, 8725, 9438, 10179, 10948, 11745, 12570, 13423, 14304, 15213, 16150, 17115, 18108, 19129, 20178, 21255, 22360, 23493, 24654, 25843

Offset: 0

Omar E. Pol, Oct 13 2011

Related to the primitive Pythagorean triple [3, 4, 5].
Sequence found by reading the line from 0, in the direction 0, 13, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the one of the semi-diagonals of the square spiral.
Also sequence found by reading the line from 0, in the direction 0, 13, ..., in the square spiral whose vertices are the generalized 9-gonal numbers A118277. - Omar E. Pol, Jul 28 2012

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A118277, A144555, A152760, A185019, A193053, A195019, A195020, A195320, A198017.

[14*n^2 - n: n in [0..50]]; // Vincenzo Librandi, Oct 14 2011

Table[n(14n-1),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,13,54},50] (* Harvey P. Dale, Jul 28 2012 *)

a(n)=n*(14*n-1) \\ Charles R Greathouse IV, Oct 07 2015

a(n) = 14*n^2 - n.
From Colin Barker, Apr 09 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3).
G.f.: x*(13+15*x)/(1-x)^3. (End)
E.g.f.: exp(x)*x*(13 + 14*x). - Elmo R. Oliveira, Jan 12 2025

A195025 a(n) = n(14n + 3).

Original entry on oeis.org

0, 17, 62, 135, 236, 365, 522, 707, 920, 1161, 1430, 1727, 2052, 2405, 2786, 3195, 3632, 4097, 4590, 5111, 5660, 6237, 6842, 7475, 8136, 8825, 9542, 10287, 11060, 11861, 12690, 13547, 14432, 15345, 16286, 17255, 18252, 19277, 20330, 21411, 22520, 23657, 24822, 26015

Offset: 0

Omar E. Pol, Oct 13 2011

Sequence found by reading the line from 0, in the direction 0, 17, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the one of the semi-axis of the square spiral, which is related to the primitive Pythagorean triple [3, 4, 5].

a(k) is a square for k = (3/56)*((449 + 120*sqrt(14))^n + (449 - 120*sqrt(14))^n - 2). - Bruno Berselli, Oct 18 2011

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A144555, A152760, A185019, A193053, A195019, A195020, A195023, A195024, A195320.

[14*n^2 +3*n: n in [0..50]]; // Vincenzo Librandi, Oct 14 2011

Table[n(14n+3),{n,0,40}] (* or *) LinearRecurrence[{3,-3,1},{0,17,62},50] (* Harvey P. Dale, Jul 17 2023 *)

a(n)=n*(14*n+3) \\ Charles R Greathouse IV, Oct 07 2015

a(n) =  14*n^2 + 3*n.
G.f.: x*(17+11*x)/(1-x)^3. - Bruno Berselli, Oct 18 2011
From Elmo R. Oliveira, Dec 30 2024: (Start)
E.g.f.: exp(x)*x*(17 + 14*x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)

Name suggested by Bruno Berselli, Oct 13 2011

A195026 a(n) = 7n(2*n + 1).

Original entry on oeis.org

0, 21, 70, 147, 252, 385, 546, 735, 952, 1197, 1470, 1771, 2100, 2457, 2842, 3255, 3696, 4165, 4662, 5187, 5740, 6321, 6930, 7567, 8232, 8925, 9646, 10395, 11172, 11977, 12810, 13671, 14560, 15477, 16422, 17395, 18396, 19425, 20482, 21567, 22680, 23821, 24990

Offset: 0

Omar E. Pol, Oct 13 2011

Sequence found by reading the line from 0, in the direction 0, 21, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. Semi-diagonal opposite to A195320 in the same square spiral, which is related to the primitive Pythagorean triple [3, 4, 5].

Sum of the numbers from 6*n to 8*n. - Wesley Ivan Hurt, Dec 23 2015

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A014105, A144555, A152760, A195019, A195020, A195021, A195023, A195024, A195025, A195320.

Cf. A185019, A193053, A198017.

[14*n^2 +7*n: n in [0..50]]; // Vincenzo Librandi, Oct 14 2011

A195026:=n->7*n*(2*n+1): seq(A195026(n), n=0..50); # Wesley Ivan Hurt, Dec 23 2015

Table[7*n*(2*n + 1), {n, 0, 50}] (* Wesley Ivan Hurt, Dec 23 2015 *)
LinearRecurrence[{3,-3,1},{0,21,70},50] (* Harvey P. Dale, Apr 26 2017 *)

a(n)=7*n*(2*n+1) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 14*n^2 + 7*n.
a(n) = 7*A014105(n). - Bruno Berselli, Oct 13 2011
From Colin Barker, Apr 09 2012: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
G.f.: 7*x*(3+x)/(1-x)^3. (End)
a(n) = Sum_{i=6*n..8*n} i. - Wesley Ivan Hurt, Dec 23 2015
E.g.f.: 7*exp(x)*x*(3 + 2*x). - Elmo R. Oliveira, Dec 29 2024

A195027 a(n) = 2n(7*n + 5).

Original entry on oeis.org

0, 24, 76, 156, 264, 400, 564, 756, 976, 1224, 1500, 1804, 2136, 2496, 2884, 3300, 3744, 4216, 4716, 5244, 5800, 6384, 6996, 7636, 8304, 9000, 9724, 10476, 11256, 12064, 12900, 13764, 14656, 15576, 16524, 17500, 18504, 19536, 20596, 21684, 22800, 23944, 25116, 26316

Offset: 0

Omar E. Pol, Oct 13 2011

Sequence found by reading the line from 0, in the direction 0, 24, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. Semi-axis opposite to A195023 in the same square spiral, which is related to the primitive Pythagorean triple [3, 4, 5].

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A144555, A152760, A179986, A195019, A195020, A195021, A195023, A195024, A195025, A195026, A195320.

Cf. A185019, A193053, A198017.

[14*n^2 +10*n: n in [0..50]]; // Vincenzo Librandi, Oct 14 2011

Table[2n(7n+5),{n,0,50}] (* or *) LinearRecurrence[{3,-3,1},{0,24,76},50] (* Harvey P. Dale, Jul 24 2012 *)

a(n)=2*n*(7*n+5) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 14*n^2 + 10*n.
a(n) = 4*A179986(n). - Bruno Berselli, Oct 13 2011
G.f.: 4*x*(6+x)/(1-x)^3. - Colin Barker, Jan 09 2012
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3); a(0)=0, a(1)=24, a(2)=76. - Harvey P. Dale, Jul 24 2012
E.g.f.: 2*exp(x)*x*(12 + 7*x). - Elmo R. Oliveira, Dec 30 2024

A195028 a(n) = n(14n + 13).

Original entry on oeis.org

0, 27, 82, 165, 276, 415, 582, 777, 1000, 1251, 1530, 1837, 2172, 2535, 2926, 3345, 3792, 4267, 4770, 5301, 5860, 6447, 7062, 7705, 8376, 9075, 9802, 10557, 11340, 12151, 12990, 13857, 14752, 15675, 16626, 17605, 18612, 19647, 20710, 21801, 22920, 24067, 25242

Offset: 0

Omar E. Pol, Oct 13 2011

Sequence found by reading the line from 0, in the direction 0, 27, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. Numbers opposite to the semi-diagonal A195024 in the same square spiral, which is related to the primitive Pythagorean triple [3, 4, 5].

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A144555, A152760, A185019, A193053, A195019, A195020, A195021, A195023, A195024, A195025, A195026, A195027, A195320, A198017.

[14*n^2 +13*n: n in [0..50]]; // Vincenzo Librandi, Oct 14 2011

a(n)=n*(14*n+13) \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 14*n^2 + 13*n.
G.f.: x*(27+x)/(1-x)^3. - Colin Barker, Jan 09 2012
From Elmo R. Oliveira, Dec 30 2024: (Start)
E.g.f.: exp(x)*x*(27 + 14*x).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n >= 3. (End)

Name suggested by Bruno Berselli, Oct 13 2011

A195029 a(n) = n(14n + 13) + 3.

Original entry on oeis.org

3, 30, 85, 168, 279, 418, 585, 780, 1003, 1254, 1533, 1840, 2175, 2538, 2929, 3348, 3795, 4270, 4773, 5304, 5863, 6450, 7065, 7708, 8379, 9078, 9805, 10560, 11343, 12154, 12993, 13860, 14755, 15678, 16629, 17608, 18615, 19650, 20713, 21804, 22923, 24070, 25245

Offset: 0

Omar E. Pol, Sep 07 2011

Sequence found by reading the line from 3, in the direction 3, 30, ..., in the Pythagorean spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the semi-diagonal parallel to A195024 and also parallel to A195028 in the same square spiral, which is related to the primitive Pythagorean triple [3, 4, 5].

56*a(n) + 1 is a perfect square. - Bruno Berselli, Feb 14 2017

Matthew House, Table of n, a(n) for n = 0..10000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Bisection of A022264.
Cf. A144555, A152760, A195019, A195020, A195021, A195023, A195024, A195025, A195026, A195027, A195028, A195320.
Cf. A005408, A017017, A185019, A193053, A198017.

Table[n (14 n + 13) + 3, {n, 0, 40}] (* Bruno Berselli, Feb 14 2017 *)
LinearRecurrence[{3,-3,1},{3,30,85},50] (* Harvey P. Dale, May 03 2018 *)

a(n)=n*(14*n+13)+3 \\ Charles R Greathouse IV, Jun 17 2017

a(n) = 14*n^2 + 13*n + 3 = A195028(n) + 3 = (2*n + 1)*(7*n + 3).
From Colin Barker, Apr 09 2012: (Start)
G.f.: (3 + 21*x + 4*x^2)/(1 - x)^3.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3). (End)
From Elmo R. Oliveira, Dec 29 2024: (Start)
E.g.f.: exp(x)*(3 + 27*x + 14*x^2).
a(n) = A005408(n)*A017017(n) = A022264(2*n+1). (End)

Edited by Bruno Berselli, Feb 14 2017

cp's OEIS Frontend

A193053 a(n) = (14*n*(n+3) + (2*n-5)*(-1)^n + 21)/16.

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A198017 a(n) = n*(7*n + 11)/2 + 1.

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A195021 a(n) = n*(14*n - 11).

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A195023 a(n) = 14*n^2 - 4*n.

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A195024 a(n) = n*(14*n - 1).

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A195025 a(n) = n*(14*n + 3).

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A195026 a(n) = 7*n*(2*n + 1).

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A193053 a(n) = (14n(n+3) + (2n-5)(-1)^n + 21)/16.

A198017 a(n) = n(7n + 11)/2 + 1.

A195021 a(n) = n(14n - 11).

A195023 a(n) = 14n^2 - 4n.

A195024 a(n) = n(14n - 1).

A195025 a(n) = n(14n + 3).

A195026 a(n) = 7n(2*n + 1).

A195027 a(n) = 2n(7*n + 5).

A195028 a(n) = n(14n + 13).

A195029 a(n) = n(14n + 13) + 3.