A037156 a(n) = 10^n*(10^n+1)/2.

1, 55, 5050, 500500, 50005000, 5000050000, 500000500000, 50000005000000, 5000000050000000, 500000000500000000, 50000000005000000000, 5000000000050000000000, 500000000000500000000000, 50000000000005000000000000, 5000000000000050000000000000

Offset: 0

Author

Marvin Ray Burns

Keywords

• easy
• nonn

Comments

Sum of first 10^n positive integers. - Omar E. Pol, May 03 2015

Examples

			From _Omar E. Pol_, May 03 2015: (Start)
For n = 0; a(0) = 1                       =    1 * 1   = 1
For n = 1; a(1) = 1 + 2 + ...... + 9 + 10 =   11 * 5   = 55
For n = 2; a(2) = 1 + 2 + .... + 99 + 100 =  101 * 50  = 5050
For n = 3; a(3) = 1 + 2 + .. + 999 + 1000 = 1001 * 500 = 500500
...
(End)
		

References

• C. Pickover, Wonders of Numbers, Oxford University Press, NY, 2001, p. 328.

Links

• Brian Hayes, Gauss's Day of Reckoning, American Scientist
• Bill Johnson, The great Gauss summation trick
• C. A. Pickover, "Wonders of Numbers, Adventures in Mathematics, Mind and Meaning," Zentralblatt review
• Index entries for linear recurrences with constant coefficients, signature (110, -1000).

Crossrefs

A subsequence of the triangular numbers A000217.

Cf. A038544.

Programs

Mathematica

LinearRecurrence[{110,-1000},{1,55},20] (* Harvey P. Dale, Oct 11 2023 *)

Formula

a(n) = A000533(n) * A093143(n). - Omar E. Pol, May 03 2015

From Chai Wah Wu, May 28 2016: (Start)

a(n) = 110*a(n-1) - 1000*a(n-2).

G.f.: (1 - 55*x)/((10*x - 1)*(100*x - 1)).

(End)

a(n) = sqrt(A038544(n)). - Bernard Schott, Jan 20 2022

Extensions

Corrected by T. D. Noe, Nov 07 2006