|
%I #24 Dec 23 2020 07:29:24
%S 1000,10648,39304,97336,195112,343000,551368,830584,1191016,1643032,
%T 2197000,2863288,3652264,4574296,5639752,6859000,8242408,9800344,
%U 11543176,13481272,15625000,17984728,20570824,23393656,26463592,29791000,33386248,37259704
%N a(n) = (12n+10)^3.
%C 6n + 5 = (12n + 10) / 2 is never a square, as 5 is not a quadratic residue modulo 6. Using this, we can show that each term has an even square part and an even squarefree part, neither part being a power of 2. (Less than 2% of integers have this property - see A339245.) - _Peter Munn_, Dec 14 2020
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/QuadraticResidue.html">Quadratic Residue</a>.
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1).
%F a(n) = 4*a(n-1) - 6*a(n-2) + 4*a(n-3) - a(n-4) with a(0)=1000, a(1)=10648, a(2)=39304, a(3)=97336. [_Harvey P. Dale_, Sep 30 2011]
%F a(n) = A017641(n)^3 = A000578(A017641(n)). - _Michel Marcus_, Nov 25 2013
%p A017643:=(12*n+10)^3; seq(A017643(n), n=0..100); # _Wesley Ivan Hurt_, Nov 25 2013
%t (12Range[0,30]+10)^3 (* or *) LinearRecurrence[{4,-6,4,-1},{1000,10648,39304,97336},30] (* _Harvey P. Dale_, Sep 30 2011 *)
%Y A000578, A017641 are used in a formula defining this sequence.
%Y Subsequence of A339245.
%Y Cf. A017642, A017644.
%K nonn,easy
%O 0,1
%A _N. J. A. Sloane_.
|
