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%I #91 Jan 15 2023 18:33:05
%S 0,5,39,150,410,915,1785,3164,5220,8145,12155,17490,24414,33215,44205,
%T 57720,74120,93789,117135,144590,176610,213675,256289,304980,360300,
%U 422825,493155,571914,659750,757335,865365,984560,1115664
%N a(n) = floor((n + 1/2)^4).
%C 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.
%C -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
%C Numbers k such that 16*k+1 is a fourth power. - _Bruno Berselli_, May 29 2018
%C 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
%H Clark Kimberling, <a href="/A219086/b219086.txt">Table of n, a(n) for n = 0..10000</a>
%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (5,-10,10,-5,1).
%F G.f.: (5*x^3 + 14*x^2 + 5*x)/(1 - x)^5.
%F a(n) = 5*a(n-1) - 10*a(n-2) + 10*a(n-3) - 5*a(n-4) + a(n-5).
%F a(n) = (2*n^4 + 4*n^3 + 3*n^2 + n)/2. - _J. M. Bergot_, Apr 05 2014
%F 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
%F a(n) = lcm((2*n + 1)^2 - 1, (2*n + 1)^2 + 1)/8 for n>=1. - _Lechoslaw Ratajczak_, Mar 26 2017
%F a(n) = A000217(n) * A001844(n). - _Bruce J. Nicholson_, May 14 2017
%F E.g.f.: (1/2)*exp(x)*x*(10 + 29*x + 16*x^2 + 2*x^3). - _Stefano Spezia_, Dec 27 2019
%F a(n) = ((2*n+1)^4 - 1)/16. - _Jianing Song_, Jan 03 2023
%F Sum_{n>=1} 1/a(n) = 6 - 2*Pi*tanh(Pi/2). - _Amiram Eldar_, Jan 08 2023
%e 0^(1/4) = 0.000...; 1^(1/4) = 1.000...
%e 5^(1/4) = 1.495...; 6^(1/4) = 1.565...
%e 39^(1/4) = 2.499...; 40^(1/4) = 2.514...
%p A219086:=n->floor((n + (1/2))^4); seq(A219086(n), n=0..50); # _Wesley Ivan Hurt_, Apr 05 2014
%t Table[Floor[(n + 1/2)^4], {n, 0, 100}]
%t LinearRecurrence[{5,-10,10,-5,1},{0,5,39,150,410},40] (* _Harvey P. Dale_, Jan 15 2023 *)
%o (PARI) a(n)=floor((n + 1/2)^4) \\ _Charles R Greathouse IV_, Apr 15 2014
%Y Cf. A123865, A219085, A234459, A000217, A001844.
%Y Cf. A317297, A006003, A005917, A000583, A011863, A019298.
%K nonn,easy
%O 0,2
%A _Clark Kimberling_, Jan 01 2013
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