A175110 a(n) = ((2*n+1)^4+1)/2.

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

Offset: 0

Author

R. J. Mathar, Feb 13 2010

Keywords

• easy
• nonn

Comments

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).

Crossrefs

Cf. A000332, A016756, A016760. Partial sums of A117216.

Programs

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

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 *)

PARI

a(n)=((2*n+1)^4+1)/2 \\ Charles R Greathouse IV, Oct 16 2015

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)-a(n-1) = A117216(n).

a(n) = 8*A001844(n) * A000217(n) + 1 = 8*A219086(n) + 1. - Bruce J. Nicholson, Apr 13 2017