A014917 a(1)=1, a(n) = n*5^(n-1) + a(n-1).

1, 11, 86, 586, 3711, 22461, 131836, 756836, 4272461, 23803711, 131225586, 717163086, 3890991211, 20980834961, 112533569336, 600814819336, 3194808959961, 16927719116211, 89406967163086, 470876693725586

Offset

1,2

Comments

From Gary Detlefs, Aug 31 2021 (Start)

This is the x=5 member of the x-family of sequences with member a(x, n) = x^n*Sum_{k=1..n} S(x, k), with S(x, k) = Sum_{j=1..k} 1/x^j.

S(x, k) = (x^k - 1)/((x-1)*x^k) = (1/x^k)*Sum_{j=0..k-1} x^j, and a(x,n) = ((n*(x-1) - 1)*x^n + 1)/(x-1)^2.

The sequence {x^k*S(x, k)} with recurrence signature (x+1, -x) leads to sequence {a(x, n)} with recurrence signature (2*x+1, -x*(x+2), x^2). (End) [Rewritten by Wolfdieter Lang, Nov 30 2021]

Links

Vincenzo Librandi, Table of n, a(n) for n = 1..1000

Index entries for linear recurrences with constant coefficients, signature (11,-35,25).

Formula

From Vincenzo Librandi, Oct 23 2012: (Start)

a(n) = 10*a(n-1) - 25*a(n-2) + 1; a(1)=1, a(2)=11.

G.f.: x/((1-x)*(1-5*x)^2). (End)

a(n) = 11*a(n-1) - 35*a(n-2) + 25*a(n-3); a(1)=1, a(2)=11, a(3)=86. - Harvey P. Dale, May 06 2013

a(n) = 5^n*Sum_{k=1..n} (Sum_{j=1..k} 1/x^j) = ((4*n - 1)*5^n + 1)/4^2. See the general comment above, and the first formula. -Gary Detlefs, Aug 31 2021 [Edited by Wolfdieter Lang, Nov 30 2021]

Mathematica

CoefficientList[Series[1/((1 - x)(1 - 5 x)^2), {x, 0, 40}], x] (* Vincenzo Librandi, Oct 23 2012 *)
LinearRecurrence[{11, -35, 25}, {1, 11, 86}, 20] (* Harvey P. Dale, May 06 2013 *)

Prog

(Magma) I:=[1, 11]; [n le 2 select I[n] else 10*Self(n-1)-25*Self(n-2)+ 1: n in [1..30]]; // Vincenzo Librandi, Oct 23 2012

Crossrefs

Cf. A000351, A003463, A014915, A014916, A014917, A014918, A014920, A014921, A014923.

Sequence in context: A320940 A012794 A302954 * A080962 A243879 A298253

Adjacent sequences: A014914 A014915 A014916 * A014918 A014919 A014920

Keyword

nonn,easy

Author

Olivier Gérard

Status

approved