

A174420


Let A(n) = floor((3/2)^n), B(n)=3^n2^n*A(n); then a(n)=2^nA(n)B(n)2.


2



2, 2, 1, 0, 8, 4, 26, 98, 68, 245, 284, 941, 908, 2921, 866, 3038, 9773, 95842, 26864, 82811, 776048, 235984, 715436, 2157533, 14878043, 27882168, 16575521, 116892244, 82326503, 515542801, 1009949246, 882651721, 500902958, 1503356036, 4511038850, 13534572662, 40605902153, 53101505973, 21870478820, 65618808017
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OFFSET

0,1


COMMENTS

Related to Waring's problem.
a(n) is conjectured to be >= 0 for all n >= 3. If this were proved it would settle Waring's problem (see A002804). It is known that a(n) >= 0 for 3 <= n <= 471600000.
If we rewrite the formula as (2^n1)*ceiling((3/2)^n)  3^n  1, we see more clearly a comparison between 3^n and the product of an undervaluation of 2^n and an overvaluation of (3/2)^n. If the undervaluation is proportionately smaller than the ceiling overvaluation, a(n) is nonnegative. 2^n grows faster than (3/2)^n, so for a negative value to occur the target difference between (3/2)^n and ceiling((3/2)^n) gets smaller as n gets larger, and the sum of these target differences (for n > 0) is finite.  Peter Munn, Dec 08 2022


REFERENCES

G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers. 3rd ed., Oxford Univ. Press, 1954, p. 337.
R. C. Vaughan and T. D. Wooley, Waring's problem: a survey, pp. 285324 of Surveys in Number Theory (Urbana, May 21, 2000), ed. M. A. Bennett et al., Peters, 2003.


LINKS



MATHEMATICA

a[n_] := Floor[(3/2)^n]; b[n_] := 3^n  2^n*a[n]; Table[2^n  a[n]  b[n]  2, {n, 0, 39}] (* JeanFrançois Alcover, Apr 05 2011 *)


PROG

(Python)


CROSSREFS



KEYWORD

sign


AUTHOR



STATUS

approved



