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%I #122 Sep 13 2022 02:16:52
%S 7,23,47,79,119,167,223,287,359,439,527,623,727,839,959,1087,1223,
%T 1367,1519,1679,1847,2023,2207,2399,2599,2807,3023,3247,3479,3719,
%U 3967,4223,4487,4759,5039,5327,5623,5927,6239,6559,6887,7223,7567,7919,8279,8647
%N a(n) = 4*n^2 + 4*n - 1.
%C The sum of the squares of two consecutive terms multiplied (or divided) by 2 is always a perfect square. In general, numbers represented by the quadratic form a(n) = (2*i*n + j)^2 - 2*i^2 for any i and j have 2(a(n)^2 + a(n+1)^2)) and (a(n)^2 + a(n+1)^2)/2 as perfect squares: in this case, i=j=1.
%C The terms of this sequence may be seen to be 2 less than the odd squares. As such they run parallel to those in the square spiral as well as the Ulam square spiral. - Stuart M. Ellerstein (ellerstein(AT)aol.com), Oct 01 2002
%C Primes in the sequence are in A028871. - _Russ Cox_, Aug 26 2019
%C The continued fraction expansion of sqrt(4*a(n)) is [4n+1; {1, n-1, 2, 2n, 2, n-1, 1, 8n+2}]. For n=1, this collapses to [5; {3, 2, 3, 10}]. - _Magus K. Chu_, Sep 12 2022
%H G. C. Greubel, <a href="/A073577/b073577.txt">Table of n, a(n) for n = 1..5000</a>
%H Soren Laing Aletheia-Zomlefer, Lenny Fukshansky, and Stephan Ramon Garcia, <a href="https://doi.org/10.1016/j.exmath.2019.04.005">The Bateman-Horn Conjecture: Heuristics, History, and Applications</a>, Expositiones Mathematicae, Vol. 38, No. 4 (2020), pp. 430-479; <a href="https://arxiv.org/abs/1807.08899">arXiv preprint</a>, arXiv:1807.08899 [math.NT], 2018-2019. See 6.6.7, p. 36 (p. 35 in the preprint).
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).
%F a(n) = FrobeniusNumber(2*n+1, 2*n+3). - _Darrell Minor_, Jul 29 2008
%F a(n) = 8*n + a(n-1) (with a(1)=7). - _Vincenzo Librandi_, Aug 08 2010
%F G.f.: x*(7+2*x-x^2)/(1-x)^3. - _Robert Israel_, Jan 13 2015
%F E.g.f.: 1 - (1-8*x-4*x^2)*exp(x). - _Robert Israel_, Jan 13 2015
%F a(n+1) = a(n) + A008590(n+1), a(1) = 7. - _Altug Alkan_, Sep 28 2015
%F a(n) = (2*n+1)+(2*n-1) + (2*n+1)*(2*n-1). - _J. M. Bergot_, Apr 17 2016
%F a(n) = (2*n+1)^2 - 2. - _Zhandos Mambetaliyev_, Jun 13 2017
%F From _Stefano Spezia_, Nov 04 2018: (Start)
%F L.g.f.: 4*x*(2+x)/(1+x)^2-log(1+x).
%F L.h.g.f.: -4*(-2+x)*x/(-1+x)^2+log(1-x).
%F (End)
%F Sum_{n>=1} 1/a(n) = 1 + sqrt(2)*Pi*tan(Pi/sqrt(2))/8. - _Amiram Eldar_, Jan 03 2021
%e a(2) = 8*2 + 7 = 23;
%e a(3) = 8*3 + 23 = 47;
%e a(4) = 8*4 + 47 = 79. - _Vincenzo Librandi_, Aug 08 2010
%p seq(4*n^2+4*n-1,n=1..100); # _Robert Israel_, Jan 13 2015
%t Table[4*n^2+4*n-1,{n,60}] (* _Vladimir Joseph Stephan Orlovsky_, Nov 18 2009 *)
%t LinearRecurrence[{3,-3,1},{7,23,47},50] (* _Harvey P. Dale_, Dec 04 2018 *)
%o (Maxima) A073577(n):=4*n^2+4*n-1$
%o makelist(A073577(n),n,1,30); /* _Martin Ettl_, Nov 01 2012 */
%o (PARI) vector(50, n, 4*n^2 + 4*n - 1) \\ _Michel Marcus_, Jan 14 2015
%o (Magma) [4*n^2 + 4*n - 1: n in [1..50]]; // _Wesley Ivan Hurt_, Apr 18 2016
%o (Python) for n in range(1,50): print(4*n**2+4*n-1, end=', ') # _Stefano Spezia_, Nov 01 2018
%o (GAP) List([1..50],n->4*n^2+4*n-1); # _Muniru A Asiru_, Nov 01 2018
%Y Cf. A008590, A028871, A214345.
%K nonn,easy
%O 1,1
%A M. N. Deshpande (dpratap(AT)nagpur.dot.net.in), Aug 27 2002
%E Edited and extended by _Henry Bottomley_, Oct 10 2002
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