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

0,2

Comments

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

References

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

Links

Table of n, a(n) for n=0..14.

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).

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

Example

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)

Mathematica

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

Crossrefs

A subsequence of the triangular numbers A000217.

Cf. A038544.

Sequence in context: A119081 A119051 A119083 * A116110 A358044 A060204

Adjacent sequences: A037153 A037154 A037155 * A037157 A037158 A037159

Keyword

easy,nonn

Author

Marvin Ray Burns

Extensions

Corrected by T. D. Noe, Nov 07 2006

Status

approved