A016970 - OEIS

%I #87 Sep 08 2022 08:44:41

%S 25,121,289,529,841,1225,1681,2209,2809,3481,4225,5041,5929,6889,7921,

%T 9025,10201,11449,12769,14161,15625,17161,18769,20449,22201,24025,

%U 25921,27889,29929,32041,34225,36481,38809,41209,43681,46225,48841,51529

%N a(n) = (6*n + 5)^2.

%C The product of 4 successive terms of an arithmetic progression + square of the common difference is a square: a(n) = the square arising as the sum of first four terms of an arithmetic progression + n^2 where 1 is the first term and n is the common difference. a(1) = 25 = 1*2*3*4+1 a(2) = 121 = 1*3*5*7 +2^2 a(3) = 289 = 1*4*7*10 + 3^2, etc. - _Amarnath Murthy_, Mar 25 2004

%C If Y is a fixed 2-subset of a (6n+1)-set X then a(n-1) is the number of 3-subsets of X intersecting Y. - _Milan Janjic_, Oct 21 2007

%C Sequence found by reading the line from 25 in the direction 25, 121,... in the square spiral whose vertices are the generalized 20-gonal numbers. - _Omar E. Pol_, Jul 28 2016

%H Vincenzo Librandi, <a href="/A016970/b016970.txt">Table of n, a(n) for n = 0..1000</a>

%H Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Two Enumerative Functions</a>.

%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3,1).

%F G.f.: (25 + 46*x + x^2) / (1-x)^3. - _R. J. Mathar_, Mar 10 2011

%F a(n) = 24 * A000326(n+1) + 1. - _Jean-Bernard François_, Oct 12 2014

%F a(n) = 6*A033579(n+1) + 1. - _Miquel Cerda_, Jul 28 2016

%F E.g.f.: exp(x)*(25 + 96*x + 36*x^2). - _Stefano Spezia_, Dec 07 2018

%F a(n) = A003215(3*n+2) + A002378(3*n+2). - _Klaus Purath_, Jun 09 2020

%F Sum_{n>=0} 1/a(n) = A086731. - _Amiram Eldar_, Nov 17 2020

%p [(6*n+5)^2$n=0..40]; # _Muniru A Asiru_, Dec 06 2018

%t Array[(6 # + 5)^2 &, 38, 0] (* or *)

%t CoefficientList[Series[(-25 - 46 x - x^2)/(x - 1)^3, {x, 0, 37}], x] (* _Michael De Vlieger_, Dec 06 2018 *)

%t CoefficientList[Series[E^x (25 + 96 x + 36 x^2), {x, 0, 50}], x]*Table[n!, {n, 0, 50}] (* _Stefano Spezia_, Dec 07 2018 *)

%o (Magma) [(6*n+5)^2: n in [0..50]]; // _Vincenzo Librandi_, May 07 2011

%o (PARI) a(n)=(6*n+5)^2 \\ _Charles R Greathouse IV_, Jul 28 2016

%o (GAP) List([0..40],n->(6*n+5)^2); # _Muniru A Asiru_, Dec 06 2018

%o (Sage) s=((25+46*x+x^2)/(1-x)^3).series(x, 20); s.coefficients(x, sparse=False) # _G. C. Greubel_, Dec 07 2018

%Y Cf. A016969 (6*n+5), A086731, A174371.

%K nonn,easy

%O 0,1

%A _N. J. A. Sloane_