A219191 -id:A219191 - 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

A219190 Numbers of the form k(5k+1), where k = 0,-1,1,-2,2,-3,3,...

Original entry on oeis.org

0, 4, 6, 18, 22, 42, 48, 76, 84, 120, 130, 174, 186, 238, 252, 312, 328, 396, 414, 490, 510, 594, 616, 708, 732, 832, 858, 966, 994, 1110, 1140, 1264, 1296, 1428, 1462, 1602, 1638, 1786, 1824, 1980, 2020, 2184, 2226, 2398, 2442, 2622, 2668, 2856, 2904, 3100

Offset: 1

Bruno Berselli, Nov 14 2012

Equivalently, numbers m such that 20*m+1 is a square.
Also, integer values of h*(h+1)/5.
More generally, for the numbers of the form n*(k*n+1) with n in A001057, we have:
. generating function (offset 1): x^2*(k-1+2*x+(k-1)*x^2)/((1+x)^2*(1-x)^3);
. n-th term: b(n) = (2*k*n*(n-1)+(k-2)*(-1)^n*(2*n-1)+k-2)/8;
. first differences: (n-1)*((-1)^n*(k-2)+k)/2;
. b(2n+1)-b(2n) = 2*n (independent from k);
. (4*k)*b(n)+1 = (2*k*n+(k-2)*(-1)^n-k)^2/4.

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).

Subsequence of A011858.
Cf. A090771: square roots of 20*a(n)+1 (see the first comment).
Cf. numbers of the form n*(k*n+1) with n in A001057: k=0, A001057; k=1, A110660; k=2, A000217; k=3, A152749; k=4, A074378; k=5, this sequence; k=6, A036498; k=7, A219191; k=8, A154260.
Cf. similar sequences listed in A219257.

k:=5; f:=func; [0] cat [f(n*m): m in [-1,1], n in [1..25]];

I:=[0,4,6,18,22]; [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

Rest[Flatten[{# (5 # - 1), # (5 # + 1)} & /@ Range[0, 25]]]
CoefficientList[Series[2 x (2 + x + 2 x^2) / ((1 + x)^2 (1 - x)^3), {x, 0, 50}], x] (* Vincenzo Librandi, Aug 18 2013 *)
LinearRecurrence[{1,2,-2,-1,1},{0,4,6,18,22},50] (* Harvey P. Dale, Jan 21 2015 *)

G.f.: 2*x^2*(2 + x + 2*x^2)/((1 + x)^2*(1 - x)^3).
a(n) = a(-n+1) = (10*n*(n-1) + 3*(-1)^n*(2*n - 1) + 3)/8.
a(n) = 2*A057569(n) = A008851(n+1)*A047208(n)/5.
a(n) = a(n-1) + 2*a(n-2) - 2*a(n-3) - a(n-4) + a(n-5). - Harvey P. Dale, Jan 21 2015
Sum_{n>=2} 1/a(n) = 5 - sqrt(1+2/sqrt(5))*Pi. - Amiram Eldar, Mar 15 2022
a(n) = A132356(n-1)/2, n >= 1. - Bernard Schott, Mar 15 2022

A219390 Numbers k such that 14*k+1 is a square.

Original entry on oeis.org

0, 12, 16, 52, 60, 120, 132, 216, 232, 340, 360, 492, 516, 672, 700, 880, 912, 1116, 1152, 1380, 1420, 1672, 1716, 1992, 2040, 2340, 2392, 2716, 2772, 3120, 3180, 3552, 3616, 4012, 4080, 4500, 4572, 5016, 5092, 5560, 5640, 6132, 6216, 6732, 6820

Offset: 1

Bruno Berselli, Nov 19 2012

Equivalently, numbers of the form m*(14*m+2), where m = 0,-1,1,-2,2,-3,3,...

Also, integer values of 2*h*(h+1)/7.

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. similar sequences listed in A219257.

Cf. A113801 (square roots of 14*a(n)+1).

[n: n in [0..7000] | IsSquare(14*n+1)];

I:=[0,12,16,52,60]; [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

A219390:=proc(q)
local n;
for n from 1 to q do if type(sqrt(14*n+1), integer) then print(n);
fi; od; end:
A219390(1000); # Paolo P. Lava, Feb 19 2013

Select[Range[0, 7000], IntegerQ[Sqrt[14 # + 1]] &]
CoefficientList[Series[4 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},{0,12,16,52,60},50] (* Harvey P. Dale, Feb 05 2019 *)

G.f.: 4*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)+1)/4 +1.
a(n) = 2*A219191(n).
Sum_{n>=2} 1/a(n) = 7/2 - cot(Pi/7)*Pi/2. - Amiram Eldar, Mar 15 2022

A092277 a(n) = 7*n^2 + n.

Original entry on oeis.org

0, 8, 30, 66, 116, 180, 258, 350, 456, 576, 710, 858, 1020, 1196, 1386, 1590, 1808, 2040, 2286, 2546, 2820, 3108, 3410, 3726, 4056, 4400, 4758, 5130, 5516, 5916, 6330, 6758, 7200, 7656, 8126, 8610, 9108, 9620, 10146, 10686, 11240, 11808, 12390, 12986, 13596

Offset: 0

Evgeniy A. Chukhlomin (dkea(AT)yandex.ru), Feb 18 2004

First bisection of A219191. - Bruno Berselli, Nov 15 2012

			From _Bruno Berselli_, Oct 27 2017: (Start)
After 0:
8   =       -(1) + (2+3+4).
30  =     -(1+2) + (3+4+5+6+7+8).
66  =   -(1+2+3) + (4+5+6+7+8+9+10+11+12).
116 = -(1+2+3+4) + (5+6+7+8+9+10+11+12+13+14+15+16). (End)

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

Cf. A000290, A033582. - Omar E. Pol, Dec 22 2008

Cf. A022265, A219191.

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

a(n) = 7*A000290(n) + n = A033582(n) + n. - Omar E. Pol, Dec 22 2008
a(n) = a(n-1) + 14*n - 6 with n > 0, a(0)=0. - Vincenzo Librandi, Nov 17 2010
From Elmo R. Oliveira, Nov 29 2024: (Start)
G.f.: 2*x*(4 + 3*x)/(1-x)^3.
E.g.f.: exp(x)*x*(8 + 7*x).
a(n) = 2*A022265(n).
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2. (End)

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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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A219190 Numbers of the form k*(5*k+1), where k = 0,-1,1,-2,2,-3,3,...

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A219390 Numbers k such that 14*k+1 is a square.

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A092277 a(n) = 7*n^2 + n.

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

A219190 Numbers of the form k(5k+1), where k = 0,-1,1,-2,2,-3,3,...