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a(n) = 10^n*(10^n+1)/2.
6

%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&amp;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