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A175110 a(n) = ((2*n+1)^4+1)/2. 7
1, 41, 313, 1201, 3281, 7321, 14281, 25313, 41761, 65161, 97241, 139921, 195313, 265721, 353641, 461761, 592961, 750313, 937081, 1156721, 1412881, 1709401, 2050313, 2439841, 2882401, 3382601, 3945241, 4575313, 5278001, 6058681 (list; graph; refs; listen; history; text; internal format)
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
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.
a(n) = 8*A001844(n) * A000217(n) + 1 = 8*A219086(n) + 1. - Bruce J. Nicholson, Apr 13 2017
MAPLE
A175110:=n->((2*n+1)^4+1)/2: seq(A175110(n), n=0..50); # Wesley Ivan Hurt, Apr 13 2017
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
Sequence in context: A201043 A364269 A002646 * A096170 A277201 A340465
KEYWORD
easy,nonn
AUTHOR
R. J. Mathar, Feb 13 2010
STATUS
approved

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Last modified May 14 17:50 EDT 2024. Contains 372533 sequences. (Running on oeis4.)