OFFSET
0,1
COMMENTS
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
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
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
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Milan Janjic, Two Enumerative Functions.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).
FORMULA
G.f.: (25 + 46*x + x^2) / (1-x)^3. - R. J. Mathar, Mar 10 2011
a(n) = 24 * A000326(n+1) + 1. - Jean-Bernard François, Oct 12 2014
a(n) = 6*A033579(n+1) + 1. - Miquel Cerda, Jul 28 2016
E.g.f.: exp(x)*(25 + 96*x + 36*x^2). - Stefano Spezia, Dec 07 2018
Sum_{n>=0} 1/a(n) = A086731. - Amiram Eldar, Nov 17 2020
MAPLE
[(6*n+5)^2$n=0..40]; # Muniru A Asiru, Dec 06 2018
MATHEMATICA
Array[(6 # + 5)^2 &, 38, 0] (* or *)
CoefficientList[Series[(-25 - 46 x - x^2)/(x - 1)^3, {x, 0, 37}], x] (* Michael De Vlieger, Dec 06 2018 *)
CoefficientList[Series[E^x (25 + 96 x + 36 x^2), {x, 0, 50}], x]*Table[n!, {n, 0, 50}] (* Stefano Spezia, Dec 07 2018 *)
PROG
(Magma) [(6*n+5)^2: n in [0..50]]; // Vincenzo Librandi, May 07 2011
(PARI) a(n)=(6*n+5)^2 \\ Charles R Greathouse IV, Jul 28 2016
(GAP) List([0..40], n->(6*n+5)^2); # Muniru A Asiru, Dec 06 2018
(Sage) s=((25+46*x+x^2)/(1-x)^3).series(x, 20); s.coefficients(x, sparse=False) # G. C. Greubel, Dec 07 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved