A219190 -id:A219190 - cp's OEIS frontend

A219257 Numbers k such that 11*k+1 is a square.

0, 9, 13, 40, 48, 93, 105, 168, 184, 265, 285, 384, 408, 525, 553, 688, 720, 873, 909, 1080, 1120, 1309, 1353, 1560, 1608, 1833, 1885, 2128, 2184, 2445, 2505, 2784, 2848, 3145, 3213, 3528, 3600, 3933, 4009, 4360, 4440, 4809, 4893, 5280, 5368, 5773, 5865

Offset: 1

Bruno Berselli, Nov 16 2012

Equivalently, numbers of the form m*(11*m+2), where m = 0,-1,1,-2,2,-3,3,...

Also, integer values of h*(h+2)/11.

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

Cf. numbers k such that h*k+1 is a square: A005563 (h=1), A046092 (h=2), A001082 (h=3), A002378 (h=4), A036666 (h=5), A062717 (h=6), A132354 (h=7), A000217 (h=8), A132355 (h=9), A132356 (h=10), A152749 (h=12), A219389 (h=13), A219390 (h=14), A204221 (h=15), A074378 (h=16), A219394 (h=17), A219395 (h=18), A219396 (h=19), A219190 (h=20), A219391 (h=21), A219392 (h=22), A219393 (h=23), A001318 (h=24), A219259 (h=25), A217441 (h=26), A219258 (h=27), A219191 (h=28).

Cf. A175885 (square roots of 11*a(n)+1).

[n: n in [0..7000] | IsSquare(11*n+1)];

I:=[0,9,13,40,48]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013

Select[Range[0, 7000], IntegerQ[Sqrt[11 # + 1]] &]
CoefficientList[Series[x (9 + 4 x + 9 x^2)/((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)

G.f.: x^2*(9+4*x+9*x^2)/((1+x)^2*(1-x)^3).
a(n) = a(-n+1) = (22*n*(n-1)+7*(-1)^n*(2*n-1)-1)/8 + 1 = (1/176)*(22*n+7*(-1)^n-15)*(22*n+7*(-1)^n-7).
Sum_{n>=2} 1/a(n) = 11/4 - cot(2*Pi/11)*Pi/2. - Amiram Eldar, Mar 15 2022

A132356 a(2k) = k(10k+2), a(2k+1) = 10k^2 + 18k + 8, with k >= 0.

Original entry on oeis.org

0, 8, 12, 36, 44, 84, 96, 152, 168, 240, 260, 348, 372, 476, 504, 624, 656, 792, 828, 980, 1020, 1188, 1232, 1416, 1464, 1664, 1716, 1932, 1988, 2220, 2280, 2528, 2592, 2856, 2924, 3204, 3276, 3572, 3648, 3960, 4040, 4368, 4452, 4796, 4884, 5244, 5336, 5712

Offset: 0

Mohamed Bouhamida, Nov 08 2007

X values of solutions to the equation 10*X^3 + X^2 = Y^2.
Polygonal number connection: 2*H_n + 6S_n, where H_n is the n-th hexagonal number and S_n is the n-th square number. This is the base formula that is expanded upon to achieve the full series. See contributing formula below. - William A. Tedeschi, Sep 12 2010
Equivalently, numbers of the form 2*h*(5*h+1), where h = 0, -1, 1, -2, 2, -3, 3, -4, 4, ... . - Bruno Berselli, Feb 02 2017

G. C. Greubel, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (1,2,-2,-1,1).

Cf. A054000, A056220, A087475, A028347, A046092, A132209.
Cf. numbers m such that k*m+1 is a square: A005563 (k=1), A046092 (k=2), A001082 (k=3), A002378 (k=4), A036666 (k=5), A062717 (k=6), A132354 (k=7), A000217 (k=8), A132355 (k=9), A219257 (k=11), A152749 (k=12), A219389 (k=13), A219390 (k=14), A204221 (k=15), A074378 (k=16), A219394 (k=17), A219395 (k=18), A219396 (k=19), A219190 (k=20), A219391 (k=21), A219392 (k=22), A219393 (k=23), A001318 (k=24), A219259 (k=25), A217441 (k=26), A219258 (k=27), A219191 (k=28).
Cf. A220082 (numbers k such that 10*k-1 is a square).

CoefficientList[Series[4*x*(2*x^2 + x + 2)/((1 - x)^3*(1 + x)^2), {x, 0, 50}], x] (* G. C. Greubel, Jun 12 2017 *)
LinearRecurrence[{1,2,-2,-1,1},{0,8,12,36,44},50] (* Harvey P. Dale, Dec 15 2023 *)

my(x='x+O('x^50)); concat([0], Vec(4*x*(2*x^2+x+2)/((1-x)^3*(1+x)^2))) \\ G. C. Greubel, Jun 12 2017

a(n) = n^2 + n + 6*((n+1)\2)^2 \\ Charles R Greathouse IV, Sep 11 2022

G.f.: 4*x*(2*x^2+x+2)/((1-x)^3*(1+x)^2). - R. J. Mathar, Apr 07 2008
a(n) = 10*x^2 - 2*x, where x = floor(n/2)*(-1)^n for n >= 1. - William A. Tedeschi, Sep 12 2010
a(n) = ((2*n+1-(-1)^n)*(10*(2*n+1)-2*(-1)^n))/16. - Luce ETIENNE, Sep 13 2014
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5) for n > 4. - Chai Wah Wu, May 24 2016
Sum_{n>=1} 1/a(n) = 5/2 - sqrt(1+2/sqrt(5))*Pi/2. - Amiram Eldar, Mar 15 2022
a(n) = n^2 + n + 6*ceiling(n/2)^2. - Ridouane Oudra, Aug 06 2022

More terms from Max Alekseyev, Nov 13 2009

A202803 a(n) = n(5n+1).

Original entry on oeis.org

0, 6, 22, 48, 84, 130, 186, 252, 328, 414, 510, 616, 732, 858, 994, 1140, 1296, 1462, 1638, 1824, 2020, 2226, 2442, 2668, 2904, 3150, 3406, 3672, 3948, 4234, 4530, 4836, 5152, 5478, 5814, 6160, 6516, 6882, 7258, 7644, 8040, 8446, 8862, 9288, 9724, 10170

Offset: 0

Jeremy Gardiner, Dec 24 2011

First bisection of A219190. - Bruno Berselli, Nov 15 2012

a(n)*Pi is the total length of 5 points circle center spiral after n rotations. The spiral length at each rotation (L(n)) is A017341. The spiral length ratio rounded down [floor(L(n)/L(1))] is A032793. See illustration in links. - Kival Ngaokrajang, Dec 27 2013

			G.f. = 6*x + 22*x^2 + 48*x^3 + 84*x^4 + 130*x^5 +186*x^6 + 252*x^7 + 328*x^8 + ...

Jeremy Gardiner, Table of n, a(n) for n = 0..3000
Kival Ngaokrajang, Illustration of 5 points circle center spiral.
Leo Tavares, Illustration: Twin Trapezoids.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A033429, A168668, A126264, A135705, A124080.
Cf. sequences listed in A254963.
Cf. A001622, A005475.

List([0..50], n-> n*(5*n+1)); # G. C. Greubel, Jul 04 2019

[n*(5*n+1):n in [0..50]]; // Vincenzo Librandi, Aug 11 2014

CoefficientList[Series[2x(3+2x)/(1-x)^3, {x, 0, 50}] ,x] (* Vincenzo Librandi, Aug 11 2014 *)
Table[5*n^2+n, {n,0,50}] (* G. C. Greubel, Jul 04 2019 *)

a(n)=n*(5*n+1) \\ Charles R Greathouse IV, Jun 17 2017

[n*(5*n+1) for n in (0..50)] # G. C. Greubel, Jul 04 2019

a(n) = 5*n^2 + n.
a(n) = A033429(n) + n. - Omar E. Pol, Dec 24 2011
G.f.: 2*x*(3+2*x)/(1-x)^3. - Philippe Deléham, Mar 27 2013
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) with a(0) = 0, a(1) = 6, a(2) = 22. - Philippe Deléham, Mar 27 2013
a(n) = A131242(10n+5). - Philippe Deléham, Mar 27 2013
a(n) = 2*A005475(n). - Philippe Deléham, Mar 27 2013
a(n) = A168668(n) - n. - Philippe Deléham, Mar 27 2013
a(n) = (n+1)^3 - (1 + n + n*(n-1) + n*(n-1)*(n-2)). - Michael Somos, Aug 10 2014
E.g.f.: x*(6+5*x)*exp(x). - G. C. Greubel, Aug 22 2017
Sum_{n>=1} 1/a(n) = 5*(1-log(5)/4) - sqrt(1+2/sqrt(5))*Pi/2 -sqrt(5)*log(phi)/2, where phi is the golden ratio (A001622). - Amiram Eldar, Jul 19 2022

A219191 Numbers of the form k(7k+1), where k = 0,-1,1,-2,2,-3,3,...

Original entry on oeis.org

0, 6, 8, 26, 30, 60, 66, 108, 116, 170, 180, 246, 258, 336, 350, 440, 456, 558, 576, 690, 710, 836, 858, 996, 1020, 1170, 1196, 1358, 1386, 1560, 1590, 1776, 1808, 2006, 2040, 2250, 2286, 2508, 2546, 2780, 2820, 3066, 3108, 3366, 3410, 3680, 3726, 4008

Offset: 1

Bruno Berselli, Nov 14 2012

Equivalently, numbers m such that 28*m+1 is a square.
Also, integer values of h*(h+1)/7.
Let F(r) = Product_{n >= 1} 1 - q^(14*n-r). The sequence terms are the exponents in the expansion of F(0)*F(6)*F(8) = 1 - q^6 - q^8 + q^26 + q^30 - q^60 - q^66 + + - - ... (by the triple product identity).- Peter Bala, Dec 25 2024

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

Cf. numbers of the form k*(i*k+1) with k in A001057: i=0, A001057; i=1, A110660; i=2, A000217; i=3, A152749; i=4, A074378; i=5, A219190; i=6, A036498; i=7, this sequence; i=8, A154260.
Cf. A113801 (square roots of 28*a(n)+1, see the comment).
Cf. similar sequences listed in A219257.
Subsequence of A011860.

k:=7; f:=func; [0] cat [f(n*m): m in [-1,1], n in [1..25]];

I:=[0,6,8,26,30]; [n le 5 select I[n] else Self(n-1)+2*Self(n-2)-2*Self(n-3)-Self(n-4)+Self(n-5): n in [1..50]]; // Vincenzo Librandi, Aug 18 2013

A := proc (q) local n; for n from 0 to q do if type(sqrt(28*n+1), integer) then print(n) fi; od; end: A(4100); # Peter Bala, Dec 25 2024

Rest[Flatten[{# (7 # - 1), # (7 # + 1)} & /@ Range[0, 25]]]
CoefficientList[Series[2 x (3 + x + 3 x^2) / ((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)
LinearRecurrence[{1,2,-2,-1,1},{0,6,8,26,30},50] (* Harvey P. Dale, Sep 14 2022 *)

G.f.: 2*x^2*(3+x+3*x^2)/((1+x)^2*(1-x)^3).
a(n) = a(-n+1) = (14*n*(n-1)+5*(-1)^n*(2*n-1)+5)/8.
a(n) = 2*A057570(n) = (1/7)*A047335(n)*A047274(n+1).
Sum_{n>=2} 1/a(n) = 7 - cot(Pi/7)*Pi. - Amiram Eldar, Mar 17 2022

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A219257 Numbers k such that 11*k+1 is a square.

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A132356 a(2*k) = k*(10*k+2), a(2*k+1) = 10*k^2 + 18*k + 8, with k >= 0.

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A202803 a(n) = n*(5*n+1).

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A219191 Numbers of the form k*(7*k+1), where k = 0,-1,1,-2,2,-3,3,...

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A132356 a(2k) = k(10k+2), a(2k+1) = 10k^2 + 18k + 8, with k >= 0.

A202803 a(n) = n(5n+1).

A219191 Numbers of the form k(7k+1), where k = 0,-1,1,-2,2,-3,3,...