%I #40 Oct 11 2023 17:55:11
%S 1,55,5050,500500,50005000,5000050000,500000500000,50000005000000,
%T 5000000050000000,500000000500000000,50000000005000000000,
%U 5000000000050000000000,500000000000500000000000,50000000000005000000000000,5000000000000050000000000000
%N a(n) = 10^n*(10^n+1)/2.
%C Sum of first 10^n positive integers. - _Omar E. Pol_, May 03 2015
%D C. Pickover, Wonders of Numbers, Oxford University Press, NY, 2001, p. 328.
%H Brian Hayes, <a href="http://www.americanscientist.org/issues/pub/gausss-day-of-reckoning">Gauss's Day of Reckoning</a>, American Scientist
%H Bill Johnson, <a href="http://www.wbilljohnson.com/journal/math/gauss.htm">The great Gauss summation trick</a>
%H C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," <a href="http://www.zentralblatt-math.org/zmath/en/search/?q=an:0983.00008&format=complete">Zentralblatt review</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (110, -1000).
%F a(n) = A000533(n) * A093143(n). - _Omar E. Pol_, May 03 2015
%F From _Chai Wah Wu_, May 28 2016: (Start)
%F a(n) = 110*a(n-1) - 1000*a(n-2).
%F G.f.: (1 - 55*x)/((10*x - 1)*(100*x - 1)).
%F (End)
%F a(n) = sqrt(A038544(n)). - _Bernard Schott_, Jan 20 2022
%e From _Omar E. Pol_, May 03 2015: (Start)
%e For n = 0; a(0) = 1 = 1 * 1 = 1
%e For n = 1; a(1) = 1 + 2 + ...... + 9 + 10 = 11 * 5 = 55
%e For n = 2; a(2) = 1 + 2 + .... + 99 + 100 = 101 * 50 = 5050
%e For n = 3; a(3) = 1 + 2 + .. + 999 + 1000 = 1001 * 500 = 500500
%e ...
%e (End)
%t LinearRecurrence[{110,-1000},{1,55},20] (* _Harvey P. Dale_, Oct 11 2023 *)
%Y A subsequence of the triangular numbers A000217.
%Y Cf. A038544.
%K easy,nonn
%O 0,2
%A _Marvin Ray Burns_
%E Corrected by _T. D. Noe_, Nov 07 2006