A219086 a(n) = floor((n + 1/2)^4).

0, 5, 39, 150, 410, 915, 1785, 3164, 5220, 8145, 12155, 17490, 24414, 33215, 44205, 57720, 74120, 93789, 117135, 144590, 176610, 213675, 256289, 304980, 360300, 422825, 493155, 571914, 659750, 757335, 865365, 984560, 1115664

Offset: 0

Author

Clark Kimberling, Jan 01 2013

Keywords

• nonn
• easy

Comments

a(n) is the number k such that {k^p} < 1/2 < {(k+1)^p}, where p = 1/4 and { } = fractional part. Equivalently, the jump sequence of f(x) = x^(1/4), in the sense that these are the nonnegative integers k for which round(k^p) < round((k+1)^p). For details and a guide to related sequences, see A219085.

-4*a(n) gives the real part of (n+n*i)*((n+1)+n*i)*(n+(n+1)*i)*((n+1)+(n+1)*i). The imaginary part is always zero. - Jon Perry, Feb 05 2014

Numbers k such that 16*k+1 is a fourth power. - Bruno Berselli, May 29 2018

The row sums of "Floyd's Triangle", which is a triangular array of natural numbers beginning with the number 1, produce the sequence A006003. A006003 can be bisected to get the Rhombic Dodecahedron Sequence A005917, whose n-th partial sum is n^4, and A317297, whose n-th partial sum is a(n). Interleave n^4 or A000583 back with {a(n)} to get A011863, whose first differences are A019298. Finally, A011863(n)-A011863(n-2) = A006003(n-1). - Bruce J. Nicholson, Dec 22 2019

Examples

			0^(1/4) = 0.000...; 1^(1/4) = 1.000...
5^(1/4) = 1.495...; 6^(1/4) = 1.565...
39^(1/4) = 2.499...; 40^(1/4) = 2.514...
		

Links

• Clark Kimberling, Table of n, a(n) for n = 0..10000
• Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Crossrefs

Cf. A123865, A219085, A234459, A000217, A001844.

Cf. A317297, A006003, A005917, A000583, A011863, A019298.

Programs

Maple

A219086:=n->floor((n + (1/2))^4); seq(A219086(n), n=0..50); # Wesley Ivan Hurt, Apr 05 2014

Mathematica

Table[Floor[(n + 1/2)^4], {n, 0, 100}]
LinearRecurrence[{5,-10,10,-5,1},{0,5,39,150,410},40] (* Harvey P. Dale, Jan 15 2023 *)

PARI

a(n)=floor((n + 1/2)^4) \\ Charles R Greathouse IV, Apr 15 2014

Formula

G.f.: (5*x^3 + 14*x^2 + 5*x)/(1 - x)^5.

a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).

a(n) = (2*n^4 + 4*n^3 + 3*n^2 + n)/2. - J. M. Bergot, Apr 05 2014

a(n) = Sum_{i=0..n} i*(4*i^2 + 1) = n*(n + 1)*(2*n^2 + 2*n + 1)/2. - Bruno Berselli, Feb 09 2017

a(n) = lcm((2*n + 1)^2 - 1, (2*n + 1)^2 + 1)/8 for n>=1. - Lechoslaw Ratajczak, Mar 26 2017

a(n) = A000217(n) * A001844(n). - Bruce J. Nicholson, May 14 2017

E.g.f.: (1/2)*exp(x)*x*(10 + 29*x + 16*x^2 + 2*x^3). - Stefano Spezia, Dec 27 2019

a(n) = ((2*n+1)^4 - 1)/16. - Jianing Song, Jan 03 2023

Sum_{n>=1} 1/a(n) = 6 - 2*Pi*tanh(Pi/2). - Amiram Eldar, Jan 08 2023