A247112 - OEIS

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A247112

Floor of sums of the cubes of the non-integer square roots of n, as partitioned by the integer roots: floor(Sum_{j=n^2+1..(n+1)^2-1} j^(3/2)).

0, 8, 67, 267, 746, 1690, 3333, 5957, 9892, 15516, 23255, 33583, 47022, 64142, 85561, 111945, 144008, 182512, 228267, 282131, 345010, 417858, 501677, 597517, 706476, 829700, 968383, 1123767, 1297142, 1489846, 1703265, 1938833, 2198032, 2482392, 2793491

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OFFSET

0,2

COMMENTS

The fractional portion of each sum converges to 1/40 as n --> infinity.

The corresponding sum for square roots of n is given by A014105 or n*(2n+1) with a fractional portion converging to 1/6.

See A248575 for the corresponding sums for the cube roots.

See A248621 for the corresponding sums of squares of the cube roots.

See A248698 for the corresponding sum of the fourth roots.

Conjecture: the corresponding sums for all fractional (rational) powers of n (e.g., 5/2, 7/2, 9/2, ..., 1/3, 2/3, 4/3, ..., 1/4, 3/4, 5/4, ..., 1/5, 2/5, 3/5, ...) will have polynomial integer formulas or recursive integer formulas for their floor, ceiling and/or rounded values, with convergence to a rational fractional portion, with possibly multiple fractional values in a repeating pattern as they converge. This was clear for some additional examples, less so for higher-order examples.

LINKS

Table of n, a(n) for n=0..34.

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

FORMULA

a(n) = floor(Sum_{j=n^2+1..(n+1)^2-1} j^(3/2)).

a(n) = n *(4*n^3 + 6*n^2 + 5*n + 1)/2.

G.f.: x*(8 + 27*x + 12*x^2 + x^3)/(1 - x)^5. - Bruno Berselli, Dec 03 2014

E.g.f.: exp(x)*x*(16 + 51*x + 30*x^2 + 4*x^3)/2. - Stefano Spezia, Jul 13 2024

MATHEMATICA

Table[1/2 (n + 5 n^2 + 6 n^3 + 4 n^4), {n, 0, 50}]

PROG

(Magma) [n eq 0 select 0 else Floor(&+[j^(3/2): j in [n^2+1..(n+1)^2-1]]): n in [0..50]]; // Bruno Berselli, Dec 03 2014