OFFSET
0,2
COMMENTS
Partial sums of A117216. Binomial transform of 1,40,232,384,192,0,0,.. (0 continued). Convolution of the finite sequence 1,36,118,36,1 with A000332, dropping zeros.
Hypotenuse of Pythagorean triangles with smallest side a square: A016754(n)^2 + (a(n)-1)^2 = a(n)^2. - Martin Renner, Nov 12 2011
a(n) is also the first integer in a sum of (2*n + 1)^4 consecutive integers that equal (2*n + 1)^8. See A016756 and A016760. - Patrick J. McNab, Dec 26 2016
REFERENCES
Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 106, table 54.
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).
FORMULA
a(n) = 5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5).
G.f.: (1+36*x+118*x^2+36*x^3+x^4)/ (1-x)^5.
MAPLE
MATHEMATICA
CoefficientList[Series[(1 + 36*x + 118*x^2 + 36*x^3 + x^4)/(1-x)^5, {x, 0, 40}], x] (* Vincenzo Librandi, Dec 19 2012 *)
Table[((2 n + 1)^4 + 1)/2, {n, 0, 29}] (* Michael De Vlieger, Dec 26 2016 *)
LinearRecurrence[{5, -10, 10, -5, 1}, {1, 41, 313, 1201, 3281}, 40] (* Harvey P. Dale, Jan 01 2022 *)
PROG
(Magma) I:=[1, 41, 313, 1201, 3281]; [n le 5 select I[n] else 5*Self(n-1) - 10*Self(n-2) + 10*Self(n-3) - 5*Self(n-4) + Self(n-5): n in [1..40]]; // Vincenzo Librandi, Dec 19 2012
(PARI) a(n)=((2*n+1)^4+1)/2 \\ Charles R Greathouse IV, Oct 16 2015
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
R. J. Mathar, Feb 13 2010
STATUS
approved