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 A158482 a(n) = 14*n^2 + 1. 5
 15, 57, 127, 225, 351, 505, 687, 897, 1135, 1401, 1695, 2017, 2367, 2745, 3151, 3585, 4047, 4537, 5055, 5601, 6175, 6777, 7407, 8065, 8751, 9465, 10207, 10977, 11775, 12601, 13455, 14337, 15247, 16185, 17151, 18145, 19167, 20217, 21295, 22401 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS The identity (14*n^2+1)^2-(49*n^2+7)*(2*n)^2=1 can be written as a(n)^2-A158481(n)*A005843(n)^2=1. Sequence found by reading the line from 15, in the direction 15, 57,..., in the square spiral whose vertices are the generalized enneagonal numbers numbers A118277. Also sequence found by reading the same line in the square spiral whose edges have length A195019 and whose vertices are the numbers A195020. - Omar E. Pol, Sep 13 2011 LINKS Vincenzo Librandi, Table of n, a(n) for n = 1..10000 Index entries for linear recurrences with constant coefficients, signature (3,-3,1). FORMULA a(n) = 3*a(n-1) -3*a(n-2) +a(n-3). G.f: x*(15+12*x+x^2)/(1-x)^3. From Amiram Eldar, Feb 05 2021: (Start) Sum_{n>=0} 1/a(n) = (1 - (Pi/sqrt(14))*coth(Pi/sqrt(14)))/2. Sum_{n>=0} (-1)^n/a(n) = (1 + (Pi/sqrt(14))*csch(Pi/sqrt(14)))/2. Product_{n>=0} (1 + 1/a(n)) = sqrt(2)*csch(Pi/sqrt(14))*sinh(Pi/sqrt(7)). Product_{n>=1} (1 - 1/a(n)) = (Pi/sqrt(14))*csch(Pi/sqrt(14)). (End) MATHEMATICA LinearRecurrence[{3, -3, 1}, {15, 57, 127}, 50] PROG (Magma) I:=[15, 57, 127]; [n le 3 select I[n] else 3*Self(n-1)-3*Self(n-2)+1*Self(n-3): n in [1..40]]; (PARI) a(n) = 14*n^2+1; CROSSREFS Cf. A005843, A158481. Sequence in context: A020222 A256867 A336251 * A184223 A084815 A240152 Adjacent sequences: A158479 A158480 A158481 * A158483 A158484 A158485 KEYWORD nonn,easy AUTHOR Vincenzo Librandi, Mar 20 2009 STATUS approved

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Last modified December 3 09:50 EST 2022. Contains 358517 sequences. (Running on oeis4.)