A129388 -id:A129388 - cp's OEIS frontend

A102305 a(n) = n^2 + 2*n + 3.

6, 11, 18, 27, 38, 51, 66, 83, 102, 123, 146, 171, 198, 227, 258, 291, 326, 363, 402, 443, 486, 531, 578, 627, 678, 731, 786, 843, 902, 963, 1026, 1091, 1158, 1227, 1298, 1371, 1446, 1523, 1602, 1683, 1766, 1851, 1938, 2027, 2118, 2211, 2306, 2403

Offset: 1

Ralf Stephan, Jan 03 2005

Essentially a duplicate of A059100.

G. C. Greubel, Table of n, a(n) for n = 1..5000
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A010000, A027578, A059100, A129388.

[(n+1)^2+2: n in [1..60]]; // G. C. Greubel, Feb 03 2024

A102305:=n->n^2+2*n+3: seq(A102305(n), n=1..100); # Wesley Ivan Hurt, Jan 22 2017

Table[n^2+2n+3,{n,50}] (* or *) LinearRecurrence[{3,-3,1},{6,11,18},50] (* Harvey P. Dale, Aug 05 2015 *)

a(n)=n^2+2*n+3 \\ Charles R Greathouse IV, Oct 16 2015

[n^2+2*n+3 for n in range(1,61)] # G. C. Greubel, Feb 03 2024

a(n) = (1/5) * A027578(n-1).
a(n) = 2*n + a(n-1) + 1 (with a(1)=6). - Vincenzo Librandi, Nov 16 2010
a(n) = A059100(n+1). - Reinhard Zumkeller, Mar 21 2008
a(n) = A010000(n+1) for n >= 1. - Georg Fischer, Nov 02 2018
From Amiram Eldar, Sep 14 2022: (Start)
Sum_{n>=1} 1/a(n) = Pi * coth(sqrt(2)*Pi)/(2*sqrt(2)) - 7/12.
Sum_{n>=1} (-1)^(n+1)/a(n) = cosech(sqrt(2)*Pi)*Pi/(2*sqrt(2)) + 1/12. (End)
From G. C. Greubel, Feb 03 2024: (Start)
G.f.: (3 - 3*x + 2*x^2)/(1-x)^3.
E.g.f.: (3 + 3*x + x^2)*exp(x). (End)

A129389 Numbers k such that the mean of 5 consecutive squares starting with k^2 is prime.

Original entry on oeis.org

1, 7, 13, 19, 31, 37, 43, 55, 79, 97, 103, 109, 115, 121, 145, 169, 217, 223, 235, 241, 247, 253, 271, 295, 301, 307, 319, 343, 349, 361, 367, 373, 385, 415, 421, 427, 439, 445, 451, 475, 499, 511, 547, 553, 559, 571, 601, 607, 649, 673, 679, 697, 709, 751

Offset: 1

Zak Seidov, Apr 12 2007

nonn

Sum of 5 consecutive squares starting with k^2 is equal to 5*(6 + 4*k + k^2) and mean is (6 + 4*k + k^2) = (k+2)^2 + 2. Hence a(n) = A067201(n+2).

Also, numbers k such that A000217(k) + A000217(k+3) is prime. - Bruno Berselli, Apr 17 2013

			(1^2 + ... + 5^2)/5 = 11, which is prime;
(7^2 + ... + 11^2)/5 = 83, which is prime;
(13^2 + ... + 17^2)/5 = 227, which is prime.

Bruno Berselli, Table of n, a(n) for n = 1..1000

Cf. A056899, A067201, A129388.

Cf. A000217, A128815 (numbers n such that A000217(n)+A000217(n+2) is prime). [Bruno Berselli, Apr 17 2013]

[n: n in [1..800] | IsPrime(n^2+4*n+6)]; /* or, from the second comment: */ A000217:=func; [n: n in [1..800] | IsPrime(A000217(n)+A000217(n+3))]; // Bruno Berselli, Apr 17 2013

Mathematica

Select[Range[800], PrimeQ[#^2 + 4 # + 6] &] (* Bruno Berselli, Apr 17 2012 *)

SageMath

[n for n in (1..1000) if is_prime(n^2+4*n+6)] # G. C. Greubel, Feb 04 2024

cp's OEIS Frontend

A102305 a(n) = n^2 + 2*n + 3.

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