A195019 -id:A195019 - cp's OEIS frontend

A195020 Vertex number of a square spiral in which the length of the first two edges are the legs of the primitive Pythagorean triple [3, 4, 5]. The edges of the spiral have length A195019.

0, 3, 7, 13, 21, 30, 42, 54, 70, 85, 105, 123, 147, 168, 196, 220, 252, 279, 315, 345, 385, 418, 462, 498, 546, 585, 637, 679, 735, 780, 840, 888, 952, 1003, 1071, 1125, 1197, 1254, 1330, 1390, 1470, 1533, 1617, 1683, 1771, 1840, 1932, 2004, 2100

Offset: 0

Omar E. Pol, Sep 07 2011 - Sep 12 2011

Zero together with the partial sums of A195019.
The spiral contains infinitely many Pythagorean triples in which the hypotenuses on the main diagonal are the positives A008587. The vertices on the main diagonal are the numbers A024966 = (3+4)*A000217 = 7*A000217, where both 3 and 4 are the first two edges in the spiral. The distance "a" between nearest edges that are perpendicular to the initial edge of the spiral is 3, while the distance "b" between nearest edges that are parallel to the initial edge is 4, so the distance "c" between nearest vertices on the same axis is 5 because from the Pythagorean theorem we can write c = (a^2+b^2)^(1/2) = sqrt(3^2+4^2) = sqrt(9+16) = sqrt(25) = 5.
Let an array have m(0,n)=m(n,0)=n*(n-1)/2 and m(n,n)=n*(n+1)/2. The first n+1 terms in row(n) are the numbers in the closed interval m(0,n) to m(n,n). The terms in column(n) are the same from m(n,0) to m(n,n). The first few antidiagonals are 0; 0,0; 1,1,1; 3,2,2,3; 6,4,3,4,6; 10,7,5,5,7,10. a(n) is the difference between the sum of the terms in the n+1 X n+1 matrices and those in the n X n matrices. - J. M. Bergot, Jul 05 2013 [The first five rows are: 0,0,1,3,6; 0,1,2,4,7; 1,2,3,5,8; 3,4,5,6,9; 6,7,8,9,10]

Vincenzo Librandi, Table of n, a(n) for n = 0..10000
Ron Knott, Pythagorean triangles and triples
Eric Weisstein's World of Mathematics, Pythagorean Triple
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A024966, A008585, A008586, A001106, A022264, A024966, A033572, A144555, A158482, A158485, A195018, A195032, A195034, A195036.

[(2*n*(7*n+13)+(2*n-5)*(-1)^n+5)/16: n in [0..50]]; // Vincenzo Librandi, Oct 14 2011

With[{r = Range[50]}, Join[{0}, Accumulate[Riffle[3*r, 4*r]]]] (* or *)
LinearRecurrence[{1, 2, -2, -1, 1}, {0, 3, 7, 13, 21}, 100] (* Paolo Xausa, Feb 09 2024 *)

From Bruno Berselli, Oct 13 2011: (Start)
G.f.: x*(3+4*x)/((1+x)^2*(1-x)^3).
a(n) = (1/2)*A004526(n+2)*A047335(n+1) = (2*n*(7*n+13) + (2*n-5)*(-1)^n+5)/16.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5).
a(n) - a(n-2) = A047355(n+1). (End)

A001106 9-gonal (or enneagonal or nonagonal) numbers: a(n) = n(7n-5)/2.

Original entry on oeis.org

0, 1, 9, 24, 46, 75, 111, 154, 204, 261, 325, 396, 474, 559, 651, 750, 856, 969, 1089, 1216, 1350, 1491, 1639, 1794, 1956, 2125, 2301, 2484, 2674, 2871, 3075, 3286, 3504, 3729, 3961, 4200, 4446, 4699, 4959, 5226, 5500, 5781, 6069, 6364

Offset: 0

N. J. A. Sloane

Sequence found by reading the line from 0, in the direction 0, 9, ... and the parallel line from 1, in the direction 1, 24, ..., in the square spiral whose vertices are the generalized 9-gonal (enneagonal) numbers A118277. Also sequence found by reading the same lines in the square spiral whose edges have length A195019 and whose vertices are the numbers A195020. - Omar E. Pol, Sep 10 2011
Number of ordered pairs of integers (x,y) with abs(x) < n, abs(y) < n and x+y <= n. - Reinhard Zumkeller, Jan 23 2012
Partial sums give A007584. - Omar E. Pol, Jan 15 2013

Albert H. Beiler, Recreations in the Theory of Numbers, Dover, NY, 1964, p. 189.
E. Deza and M. M. Deza, Figurate numbers, World Scientific Publishing (2012), page 6.
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

T. D. Noe and William A. Tedeschi, Table of n, a(n) for n = 0..10000 (1000 terms were computed by T. D. Noe)
S. Barbero, U. Cerruti and N. Murru, Transforming Recurrent Sequences by Using the Binomial and Invert Operators, J. Int. Seq. 13 (2010) # 10.7.7, section 4.4.
C. K. Cook and M. R. Bacon, Some polygonal number summation formulas, Fib. Q., 52 (2014), 336-343.
INRIA Algorithms Project, Encyclopedia of Combinatorial Structures 343
Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992
Eric Weisstein's World of Mathematics, Nonagonal Number.
Index to sequences related to polygonal numbers
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A093564 ((7, 1) Pascal, column m=2). Partial sums of A016993.

Cf. A131875, A057655, A069099, A244646.

a001106 n = length [(x,y) | x <- [-n+1..n-1], y <- [-n+1..n-1], x + y <= n]
-- Reinhard Zumkeller, Jan 23 2012

a001106 n = n*(7*n-5) `div` 2 -- James Spahlinger, Oct 18 2012

Table[n(7n - 5)/2, {n, 0, 50}] (* or *) LinearRecurrence[{3, -3, 1}, {0, 1, 9}, 50] (* Harvey P. Dale, Nov 06 2011 *)
(* For Mathematica 10.4+ *) Table[PolygonalNumber[RegularPolygon[9], n], {n, 0, 43}] (* Arkadiusz Wesolowski, Aug 27 2016 *)
PolygonalNumber[9,Range[0,50]] (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Nov 19 2019 *)

a(n)=n*(7*n-5)/2 \\ Charles R Greathouse IV, Jun 10 2011

def aList(): # Intended to compute the initial segment of the sequence, not isolated terms.
     x, y = 1, 1
     yield 0
     while True:
         yield x
         x, y = x + y + 7, y + 7
A001106 = aList()
print([next(A001106) for i in range(49)]) # Peter Luschny, Aug 04 2019

a(n) = (7*n - 5)*n/2.
G.f.: x*(1+6*x)/(1-x)^3. - Simon Plouffe in his 1992 dissertation.
a(n) = n + 7*A000217(n-1). - Floor van Lamoen, Oct 14 2005
Starting (1, 9, 24, 46, 75, ...) gives the binomial transform of (1, 8, 7, 0, 0, 0, ...). - Gary W. Adamson, Jul 22 2007
Row sums of triangle A131875 starting (1, 9, 24, 46, 75, 111, ...). A001106 = binomial transform of (1, 8, 7, 0, 0, 0, ...). - Gary W. Adamson, Jul 22 2007
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3), a(0) = 0, a(1) = 1, a(2) = 9. - Jaume Oliver Lafont, Dec 02 2008
a(n) = 2*a(n-1) - a(n-2) + 7. - Mohamed Bouhamida, May 05 2010
a(n) = a(n-1) + 7*n - 6 (with a(0) = 0). - Vincenzo Librandi, Nov 12 2010
a(n) = A174738(7n). - Philippe Deléham, Mar 26 2013
a(7*a(n) + 22*n + 1) =  a(7*a(n) + 22*n) + a(7*n+1). - Vladimir Shevelev, Jan 24 2014
E.g.f.: x*(2 + 7*x)*exp(x)/2. - Ilya Gutkovskiy, Jul 28 2016
a(n+2) + A000217(n) = (2*n+3)^2. - Ezhilarasu Velayutham, Mar 18 2020
Product_{n>=2} (1 - 1/a(n)) = 7/9. - Amiram Eldar, Jan 21 2021
Sum_{n>=1} 1/a(n) = A244646. - Amiram Eldar, Nov 12 2021
a(n) = A000217(3*n-2) - (n-1)^2. - Charlie Marion, Feb 27 2022
a(n) = 3*A000217(n) + 2*A005563(n-2). In general, if P(k,n) = the n-th k-gonal number, then P(m*k,n) = m*P(k,n) + (m-1)*A005563(n-2). - Charlie Marion, Feb 21 2023

A022264 a(n) = n(7n - 1)/2.

Original entry on oeis.org

0, 3, 13, 30, 54, 85, 123, 168, 220, 279, 345, 418, 498, 585, 679, 780, 888, 1003, 1125, 1254, 1390, 1533, 1683, 1840, 2004, 2175, 2353, 2538, 2730, 2929, 3135, 3348, 3568, 3795, 4029, 4270, 4518, 4773, 5035, 5304, 5580, 5863, 6153, 6450, 6754, 7065, 7383

Offset: 0

N. J. A. Sloane

Sequence found by reading the line from 0, in the direction 0, 13, ..., and the parallel line from 3, in the direction 3, 30, ..., in the square spiral whose edges have length A195019 and whose vertices are the numbers A195020. - Omar E. Pol, Sep 09 2011

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

Cf. A000217, A022265.
Cf. sequences listed in A254963.
Cf. similar sequences listed in A022288.
Cf. A016742, A049450, A174738, A195019, A195020.
Cf. A005449, A000384.

[n*(7*n-1)/2 : n in [0..50]]; // Wesley Ivan Hurt, Dec 04 2016

[seq(binomial(7*n,2)/7, n=0..37)]; # Zerinvary Lajos, Jan 02 2007

Table[n (7*n - 1)/2, {n, 0, 40}] (* Zerinvary Lajos, Jul 10 2009 *)

a(n)=n*(7*n-1)/2 \\ Charles R Greathouse IV, Mar 08 2013

a(n) = C(7*n,2)/7, n >= 0. - Zerinvary Lajos, Jan 02 2007
a(n) = A049450(n) + A000217(n). - Reinhard Zumkeller, Oct 09 2008
a(n) = 7*n + a(n-1) - 4 for n > 0, a(0)=0. - Vincenzo Librandi, Aug 04 2010
a(n) = (2*n)^2 - n*(n+1)/2 = A016742(n) - A000217(n). - Philippe Deléham, Mar 08 2013
a(n) = A174738(7*n+2). - Philippe Deléham, Mar 26 2013
G.f.: x*(3 + 4*x)/(1 - x)^3. - R. J. Mathar, Aug 04 2016
a(n) = A000217(4*n-1) - A000217(3*n-1). - Bruno Berselli, Oct 17 2016
a(n) = (1/5) * Sum_{i=n..(6*n-1)} i. - Wesley Ivan Hurt, Dec 04 2016
E.g.f.: (1/2)*x*(7*x + 6)*exp(x). - G. C. Greubel, Aug 19 2017
a(n) = A005449(n) + A000384(n). See Crysta-gons illustration. - Leo Tavares, Nov 21 2021

A144555 a(n) = 14*n^2.

Original entry on oeis.org

0, 14, 56, 126, 224, 350, 504, 686, 896, 1134, 1400, 1694, 2016, 2366, 2744, 3150, 3584, 4046, 4536, 5054, 5600, 6174, 6776, 7406, 8064, 8750, 9464, 10206, 10976, 11774, 12600, 13454, 14336, 15246, 16184, 17150, 18144, 19166, 20216, 21294, 22400, 23534, 24696

Offset: 0

N. J. A. Sloane, Jan 01 2009

Sequence found by reading the line from 0, in the direction 0, 14, ..., in the square spiral whose vertices are the generalized enneagonal numbers A118277. Also sequence found by reading the same line and direction in the square spiral whose edges have length A195019 and whose vertices are the numbers A195020. - Omar E. Pol, Sep 10 2011

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

See also A033428, A033429, A033581, A033582, A033583, A033584, ... and A249327 for the whole table.
Cf. A000290, A001105, A064761, A118277, A152742, A195019, A195020.
Cf. A008596, A195145.

Table[14*n^2, {n, 0, 45}] (* Amiram Eldar, Feb 03 2021 *)

A144555(n)=14*n^2 \\ M. F. Hasler, Oct 31 2014

a(n) = 14*A000290(n) = 7*A001105(n) = 2*A033582(n). - Omar E. Pol, Jan 01 2009
a(n) = a(n-1) + 14*(2*n-1), with a(0) = 0. - Vincenzo Librandi, Nov 25 2010
From Amiram Eldar, Feb 03 2021: (Start)
Sum_{n>=1} 1/a(n) = Pi^2/84.
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi^2/168.
Product_{n>=1} (1 + 1/a(n)) = sqrt(14)*sinh(Pi/sqrt(14))/Pi.
Product_{n>=1} (1 - 1/a(n)) = sqrt(14)*sin(Pi/sqrt(14))/Pi. (End)
From Elmo R. Oliveira, Nov 30 2024: (Start)
G.f.: 14*x*(1 + x)/(1-x)^3.
E.g.f.: 14*x*(1 + x)*exp(x).
a(n) = n*A008596(n) = A195145(2*n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

A024966 7 times triangular numbers: 7n(n+1)/2.

Original entry on oeis.org

0, 7, 21, 42, 70, 105, 147, 196, 252, 315, 385, 462, 546, 637, 735, 840, 952, 1071, 1197, 1330, 1470, 1617, 1771, 1932, 2100, 2275, 2457, 2646, 2842, 3045, 3255, 3472, 3696, 3927, 4165, 4410, 4662, 4921, 5187, 5460, 5740, 6027, 6321, 6622

Offset: 0

Joe Keane (jgk(AT)jgk.org), Dec 11 1999

Sequence found by reading the line from 0, in the direction 0, 7, ... and the same line from 0, in the direction 1, 21, ..., in the square spiral whose edges have length A195019 and whose vertices are the numbers A195020. This is the main diagonal in the spiral. - Omar E. Pol, Sep 09 2011
Also sequence found by reading the same line mentioned above in the square spiral whose vertices are the generalized enneagonal numbers A118277. Axis perpendicular to A195145 in the same spiral. - Omar E. Pol, Sep 18 2011
Sequence provides all integers m such that 56*m + 49 is a square. - Bruno Berselli, Oct 07 2015
Sum of the numbers from 3*n to 4*n. - Wesley Ivan Hurt, Dec 22 2015

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

Cf. A000217, A001106, A002378, A022264, A022265, A028895, A028896, A033996.
Cf. A045943, A046092, A069099, A118277, A174738, A179986, A186029, A193053.
Cf. A195019, A195020, A195145, A218471.

[ (7*n^2 + 7*n)/2 : n in [0..50] ]; // Wesley Ivan Hurt, Jun 09 2014

[seq(7*binomial(n,2), n=1..44)]; # Zerinvary Lajos, Nov 24 2006

7 Table[n (n + 1)/2, {n, 0, 43}] (* or *)
Table[Sum[i, {i, 3 n, 4 n}], {n, 0, 43}] (* or *)
Table[SeriesCoefficient[7 x/(1 - x)^3, {x, 0, n}], {n, 0, 43}] (* Michael De Vlieger, Dec 22 2015 *)
7*Accumulate[Range[0,50]] (* or *) LinearRecurrence[{3,-3,1},{0,7,21},50] (* Harvey P. Dale, Jul 20 2025 *)

x='x+O('x^100); concat(0, Vec(7*x/(1-x)^3)) \\ Altug Alkan, Dec 23 2015

a(n) = (7/2)*n*(n+1).
G.f.: 7*x/(1-x)^3.
a(n) = (7*n^2 + 7*n)/2 = 7*A000217(n). - Omar E. Pol, Dec 12 2008
a(n) = a(n-1) + 7*n with n > 0, a(0)=0. - Vincenzo Librandi, Nov 19 2010
a(n) = A069099(n+1) - 1. - Omar E. Pol, Oct 03 2011
a(n) = a(-n-1), a(n+2) = A193053(n+2) + 2*A193053(n+1) + A193053(n). - Bruno Berselli, Oct 21 2011
From Philippe Deléham, Mar 26 2013: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 0, a(1) = 7, a(2) = 21.
a(n) = A174738(7*n+6).
a(n) = A179986(n) + n = A186029(n) + 2*n = A022265(n) + 3*n = A022264(n) + 4*n = A218471(n) + 5*n = A001106(n) + 6*n. (End)
a(n) = Sum_{i=3*n..4*n} i. - Wesley Ivan Hurt, Dec 22 2015
E.g.f.: (7/2)*x*(x+2)*exp(x). - G. C. Greubel, Aug 19 2017
From Amiram Eldar, Feb 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 2/7.
Sum_{n>=1} (-1)^(n+1)/a(n) = (2/7)*(2*log(2) - 1). (End)
From Amiram Eldar, Feb 21 2023: (Start)
Product_{n>=1} (1 - 1/a(n)) = -(7/(2*Pi))*cos(sqrt(15/7)*Pi/2).
Product_{n>=1} (1 + 1/a(n)) = (7/(2*Pi))*cosh(Pi/(2*sqrt(7))). (End)

A195013 Multiples of 2 and of 3 interleaved: a(2n-1) = 2n, a(2n) = 3n.

Original entry on oeis.org

2, 3, 4, 6, 6, 9, 8, 12, 10, 15, 12, 18, 14, 21, 16, 24, 18, 27, 20, 30, 22, 33, 24, 36, 26, 39, 28, 42, 30, 45, 32, 48, 34, 51, 36, 54, 38, 57, 40, 60, 42, 63, 44, 66, 46, 69, 48, 72, 50, 75, 52, 78, 54, 81, 56, 84, 58, 87, 60, 90, 62, 93, 64, 96, 66, 99, 68, 102

Offset: 1

Omar E. Pol, Sep 09 2011

First differences of A195014.

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000
D. H. Bailey, J. M. Borwein, and J. S. Kimberley, Discovery of large Poisson polynomials using a new arbitrary precision software package, Slides, 2015.
D. H. Bailey, J. M. Borwein, and J. S. Kimberley, Computer discovery and analysis of large Poisson polynomials, 2016.
Index entries for linear recurrences with constant coefficients, signature (0,2,0,-1).

Cf. A005843, A008585, A195014, A195019.

Cf. A111712 (partial sums of this sequence prepended with 1).

import Data.List (transpose)
a195013 n = a195013_list !! (n-1)
a195013_list = concat $ transpose [[2, 4 ..], [3, 6 ..]]
-- Reinhard Zumkeller, Apr 06 2015

&cat[[2*n,3*n]: n in [1..34]]; // Bruno Berselli, Sep 25 2011

With[{r = Range[50]}, Riffle[2*r, 3*r]] (* or *)
LinearRecurrence[{0, 2, 0, -1}, {2, 3, 4, 6}, 100] (* Paolo Xausa, Feb 09 2024 *)

a(n)=(5*n+(n-2)*(-1)^n+2)/4 \\ Charles R Greathouse IV, Sep 24 2015

Pair(2*n, 3*n).
From Bruno Berselli, Sep 26 2011: (Start)
G.f.: x*(2+3*x)/(1-x^2)^2.
a(n) = (5*n+(n-2)*(-1)^n+2)/4.
a(n) = 2*a(n-2) - a(n-4) = a(n-2) + A010693(n-1).
a(n)+a(-n) = A010673(n).
a(n)-a(-n) = A106832(n). (End)

A195320 7 times hexagonal numbers: a(n) = 7n(2*n-1).

Original entry on oeis.org

0, 7, 42, 105, 196, 315, 462, 637, 840, 1071, 1330, 1617, 1932, 2275, 2646, 3045, 3472, 3927, 4410, 4921, 5460, 6027, 6622, 7245, 7896, 8575, 9282, 10017, 10780, 11571, 12390, 13237, 14112, 15015, 15946, 16905, 17892, 18907, 19950, 21021, 22120, 23247, 24402, 25585

Offset: 0

Omar E. Pol, Sep 18 2011

Sequence found by reading the line from 0, in the direction 0, 7, ..., in the square spiral whose vertices are the generalized enneagonal numbers A118277.

Also sequence found by reading the same line (mentioned above) 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, which is related to the primitive Pythagorean triple [3, 4, 5]. - Omar E. Pol, Oct 13 2011

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

Bisection of A024966.

Cf. A000384, A118277, A152746, A152750, A185019, A193053, A195019, A195020, A198017, A316466.

[7*n*(2*n-1): n in [0..50]]; // Vincenzo Librandi, Sep 28 2011

Table[7*n*(2*n - 1), {n, 0, 50}] (* Paolo Xausa, Jan 29 2025 *)
7*PolygonalNumber[6, Range[0, 50]] (* Paolo Xausa, Jan 29 2025 *)

a(n)=7*n*(2*n-1) \\ Charles R Greathouse IV, Sep 28 2015

a(n) = 14*n^2 - 7*n = 7*A000384(n).
G.f.: -7*x*(1+3*x)/(x-1)^3. - R. J. Mathar, Sep 27 2011
From Elmo R. Oliveira, Dec 27 2024: (Start)
E.g.f.: 7*exp(x)*x*(2*x + 1).
a(n) = A316466(n) - n = A024966(2*n+1).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (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

cp's OEIS Frontend

A195020 Vertex number of a square spiral in which the length of the first two edges are the legs of the primitive Pythagorean triple [3, 4, 5]. The edges of the spiral have length A195019.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Mathematica

Formula

A001106 9-gonal (or enneagonal or nonagonal) numbers: a(n) = n*(7*n-5)/2.

Views

Author

Keywords

Comments

References

Links

Crossrefs

Programs

Haskell

Haskell

Mathematica

PARI

Python

Formula

A022264 a(n) = n*(7*n - 1)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A144555 a(n) = 14*n^2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Mathematica

PARI

Formula

A024966 7 times triangular numbers: 7*n*(n+1)/2.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Magma

Maple

Mathematica

PARI

Formula

A195013 Multiples of 2 and of 3 interleaved: a(2n-1) = 2n, a(2n) = 3n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Haskell

Magma

Mathematica

PARI

Formula

A195320 7 times hexagonal numbers: a(n) = 7*n*(2*n-1).

Views

Author

A001106 9-gonal (or enneagonal or nonagonal) numbers: a(n) = n(7n-5)/2.

A022264 a(n) = n(7n - 1)/2.

A024966 7 times triangular numbers: 7n(n+1)/2.

A195320 7 times hexagonal numbers: a(n) = 7n(2*n-1).

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

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

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