A168489 -id:A168489 - cp's OEIS frontend

A222182 Numbers m such that 2*m + 11 is a square.

-5, -1, 7, 19, 35, 55, 79, 107, 139, 175, 215, 259, 307, 359, 415, 475, 539, 607, 679, 755, 835, 919, 1007, 1099, 1195, 1295, 1399, 1507, 1619, 1735, 1855, 1979, 2107, 2239, 2375, 2515, 2659, 2807, 2959, 3115, 3275, 3439, 3607, 3779, 3955, 4135, 4319, 4507, 4699

Offset: 1

Bruno Berselli, Mar 01 2013

Except the first term, main diagonal of A155546. - Vincenzo Librandi, Mar 04 2013

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

Cf. numbers n such that 2*n + 2*k + 1 is a square: A046092 (k=0), A142463 (k=1), A090288 (k=2), A059993 (k=3), A139570 (k=4), this sequence (k=5), A181510 (k=6).
Cf. A005408 (square roots of 2*a(n)+11), A155546.
After a(2), subsequence of A168489.

[m: m in [-5..5000] | IsSquare(2*m+11)];

I:=[-5,-1,7]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+Self(n-3): n in [1..50]]; // Vincenzo Librandi, Mar 04 2013

Mathematica
```
Table[2 n^2 - 2 n - 5, {n, 50}]
```

makelist(coeff(taylor(-(5-14*x+5*x^2)/(1-x)^3, x, 0, n), x, n), n, 0, 50);

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

G.f.: -x*(5 - 14*x + 5*x^2)/(1-x)^3.
a(n) = a(-n+1) = 2*n^2 - 2*n - 5.
a(n) = A046092(n-1) - 5.
Sum_{n>=1} 1/a(n) = Pi*tan(sqrt(11)*Pi/2)/(2*sqrt(11)). - Amiram Eldar, Dec 23 2022
From Elmo R. Oliveira, Nov 17 2024: (Start)
E.g.f.: exp(x)*(2*x^2 - 5) + 5.
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 3. (End)

A259750 Numbers that are congruent to {14, 22} mod 24.

Original entry on oeis.org

14, 22, 38, 46, 62, 70, 86, 94, 110, 118, 134, 142, 158, 166, 182, 190, 206, 214, 230, 238, 254, 262, 278, 286, 302, 310, 326, 334, 350, 358, 374, 382, 398, 406, 422, 430, 446, 454, 470, 478, 494, 502, 518, 526, 542, 550, 566, 574, 590, 598, 614, 622, 638

Offset: 1

José María Grau Ribas, Jul 04 2015

Original name: Numbers n such that n/A259748(n) = 2.

Danny Rorabaugh, Table of n, a(n) for n = 1..10000
Index entries for linear recurrences with constant coefficients, signature (1,1,-1).

Cf. A000914, A168489, A259748.

A[n_] := A[n] = Sum[a b, {a, 1,  n}, {b, a + 1, n}] ; Select[Range[600], Mod[A[#], #]/# == 1/2 & ]

vector(100, n, 2*(6*n-(-1)^n)) \\ Altug Alkan, Oct 23 2015

Vec(2*x*(7+4*x+x^2)/((1-x)^2*(1+x)) + O(x^100)) \\ Colin Barker, Aug 26 2016

A259748(a(n))/a(n) = 1/2.
a(n) = 2*A168489(n) - Danny Rorabaugh, Oct 22 2015
From Colin Barker, Aug 26 2016: (Start)
a(n) = a(n-1)+a(n-2)-a(n-3) for n>3.
G.f.: 2*x*(7+4*x+x^2) / ((1-x)^2*(1+x)).
(End)
Sum_{n>=1} (-1)^(n+1)/a(n) = sqrt(3)*Pi/24 - log(2+sqrt(3))/(4*sqrt(3)). - Amiram Eldar, Dec 31 2021
E.g.f.: 2*(1 + 6*x*exp(x) - exp(-x)). - David Lovler, Sep 06 2022

Better name from Danny Rorabaugh, Oct 22 2015

A164578 Integers of the form (k+1)*(2k+1)/12.

Original entry on oeis.org

10, 23, 65, 94, 168, 213, 319, 380, 518, 595, 765, 858, 1060, 1169, 1403, 1528, 1794, 1935, 2233, 2390, 2720, 2893, 3255, 3444, 3838, 4043, 4469, 4690, 5148, 5385, 5875, 6128, 6650, 6919, 7473, 7758, 8344, 8645, 9263, 9580, 10230, 10563, 11245, 11594

Offset: 1

Vladimir Joseph Stephan Orlovsky, Aug 16 2009

This can also be defined as integer averages of the first k halved squares, 1^2/2, 2^2/2, 3^2/2,... , 3^k/2, because sum_{j=1..k} j^2/2 = k*(k+1)*(2k+1)/12. The generating k are in A168489.

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

Cf. A078617, A078618, A154293, A164576, A164577.

s=0;lst={};Do[a=(s+=(n^2)/2)/n;If[Mod[a,1]==0,AppendTo[lst,a]],{n,2*6!}];lst
Select[Table[((n+1)(2n+1))/12,{n,300}],IntegerQ] (* or *) LinearRecurrence[ {1,2,-2,-1,1},{10,23,65,94,168},60] (* Harvey P. Dale, Jun 14 2017 *)

Vec(x*(10+13*x+22*x^2+3*x^3)/((1-x)^3*(1+x)^2) + O(x^100)) \\ Colin Barker, Jan 26 2016

a(n) = +a(n-1) +2*a(n-2) -2*a(n-3) -a(n-4) +a(n-5). G.f. x*(-10-13*x-22*x^2-3*x^3) / ((1+x)^2*(x-1)^3). - R. J. Mathar, Jan 25 2011
From Colin Barker, Jan 26 2016: (Start)
a(n) = (24*n^2+6*n-(-1)^n*(8*n+1)+1)/4.
a(n) = (12*n^2-n)/2 for n even.
a(n) = (12*n^2+7*n+1)/2 for n odd.
(End)

A268063 Primes of the form (k^3 - k^2 - k - 1)/2 for some integer k > 0.

Original entry on oeis.org

7, 47, 599, 1567, 5807, 7487, 9463, 20807, 24623, 28879, 33599, 81647, 111599, 123007, 161839, 225263, 262399, 282407, 397807, 541007, 573247, 606743, 641519, 922807, 1115399, 1513727, 1577383, 1709999, 1779007, 1849847, 1997119, 2399039, 2573807, 2948399

Offset: 1

Emre APARI, Jan 25 2016

Also primes of the form 4*k^3 + 4*k^2 - 1.

			k=15: (15^3 - 15^2 - 15 - 1)/2 = 1567 (is prime).

G. C. Greubel, Table of n, a(n) for n = 1..3240

Cf. A004771, A042991, A168489.

[a: n in [0..200] | IsPrime(a) where a is (n^3-n^2-n-1) div 2 ]; // Vincenzo Librandi, Jan 26 2016

Select[Table[(n^3 - n^2 - n - 1) / 2, {n, 200}], PrimeQ] (* Vincenzo Librandi, Jan 26 2016 *)

lista(nn) = for(n=1, nn, if(ispseudoprime(p=4*n^3+4*n^2-1), print1(p, ", "))); \\ Altug Alkan, Mar 14 2016

[(k^3-k^2-k-1)/2 for k in [2*i+1 for i in [1..100]] if is_prime(Integer((k^3-k^2-k-1)/2))] # Tom Edgar, Jan 25 2016

More terms from Tom Edgar, Jan 25 2016

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A222182 Numbers m such that 2*m + 11 is a square.

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A259750 Numbers that are congruent to {14, 22} mod 24.

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A164578 Integers of the form (k+1)*(2k+1)/12.

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A268063 Primes of the form (k^3 - k^2 - k - 1)/2 for some integer k > 0.

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