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A178407 a(n+1) = a(n)*1000 + 101^n with a(0)= 0. 1
0, 1, 1101, 1111201, 1112231301, 1112335361401, 1112345871501501, 1112346933021651601, 1112347040235186811701, 1112347051063753867981801, 1112347052157439140666161901, 1112347052267901353207282352001, 1112347052279058036673935517552101 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,3

COMMENTS

Numerators of fractions in partial sums of sum_{n=0...infinity} 101^n/1000^(n+1)converging to fraction 1/899 or sum_{n=0...infinity} 100^n/999^(n+1) whose alternating sum converges to fraction 1/1099 [= alternating sum_99^n/1000^(n+1)].

LINKS

Colin Barker, Table of n, a(n) for n = 0..334

Index entries for linear recurrences with constant coefficients, signature (1101,-101000).

FORMULA

a(n) = (1/899)*(1000^n-101^n), with n>=1. - Paolo P. Lava, Jun 10 2010

From Colin Barker, Oct 02 2015: (Start)

a(n) = 1101*a(n-1) - 101000*a(n-2) for n>2.

G.f.: x / ((101*x-1)*(1000*x-1)).

(End)

EXAMPLE

As interlocking Pascal triangles starting a new triangle to the left of each row:

................1

............1.1.0.1

........1.1.1.1.2.0.1

....1.1.1.2.2.3.1.3.0.1

1.1.1.2.3.3.5.3.6.1.4.0.1

MATHEMATICA

RecurrenceTable[{a[0]==0, a[n]==1000*a[n-1]+101^(n-1)}, a, {n, 15}] (* Harvey P. Dale, Nov 18 2013 *)

PROG

(PARI) concat(0, Vec(x/((101*x-1)*(1000*x-1)) + O(x^30))) \\ Colin Barker, Oct 02 2015

CROSSREFS

Cf. A000930, A007318.

Sequence in context: A283356 A283172 A283253 * A178348 A250796 A271431

Adjacent sequences:  A178404 A178405 A178406 * A178408 A178409 A178410

KEYWORD

nonn,easy

AUTHOR

Mark Dols, May 27 2010

EXTENSIONS

Prepended zero (from definition), changed offset accordingly, and more terms from Harvey P. Dale, Nov 18 2013

STATUS

approved

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Last modified November 16 22:20 EST 2019. Contains 329208 sequences. (Running on oeis4.)