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A104144 a(n) = a(n-1)+a(n-2)+a(n-3)+a(n-4)+a(n-5)+a(n-6)+a(n-7)+a(n-8)+a(n-9). 16
0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 2, 4, 8, 16, 32, 64, 128, 256, 511, 1021, 2040, 4076, 8144, 16272, 32512, 64960, 129792, 259328, 518145, 1035269, 2068498, 4132920, 8257696, 16499120, 32965728, 65866496, 131603200, 262947072, 525375999, 1049716729, 2097364960 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,11

COMMENTS

Might be called the Fibonacci 9-step numbers.

For n >= 8, this gives the number of integers written without 0 in base ten, the sum of digits of which is equal to n-7. E.g. a(11)=8 because we have the 8 numbers : 4, 13, 22, 31, 112, 121, 211, 1111.

The offset for this sequence is fairly arbitrary. - N. J. A. Sloane.

LINKS

T. D. Noe, Table of n, a(n) for n = 0..208

Martin Burtscher, Igor Szczyrba, RafaƂ Szczyrba, Analytic Representations of the n-anacci Constants and Generalizations Thereof, Journal of Integer Sequences, Vol. 18 (2015), Article 15.4.5.

Eric Weisstein's World of Mathematics, Fibonacci n-Step Number

Index entries for linear recurrences with constant coefficients, signature (1,1,1,1,1,1,1,1,1).

FORMULA

a(n) = sum_{k=1..9} a(n-k) for n>8, a(8)=1, a(n)=0 for n=0..7.

G.f.: x^8/(1-x-x^2-x^3-x^4-x^5-x^6-x^7-x^8-x^9). - N. J. A. Sloane, Dec 04 2011

Another form of the g.f. f: f(z)=(z^8-z^9)/(1-2*z+z^(10)), then a(n)=sum((-1)^i*binomial(n-8-9*i,i)*2^(n-8-10*i),i=0..floor((n-8)/10))-sum((-1)^i*binomial(n-9-9*i,i)*2^(n-9-10*i),i=0..floor((n-9)/10)) with sum(alpha(i),i=m..n)=0 for m>n. - Richard Choulet, Feb 22 2010

From N. J. A. Sloane, Dec 04 2011: (Start)

Let b be the smallest root (in magnitude) of g(x) := 1-x-x^2-x^3-x^4-x^5-x^6-x^7-x^8-x^9).

Then b = 0.50049311828655225605926845999420216157202861343888...

Let c = -b^8/g'(b) = 0.00099310812055463178382193226558248643030626601288701...

Then a(n) is the nearest integer to c/b^n. (End)

MAPLE

for n from 0 to 50 do k(n):=sum((-1)^i*binomial(n-8-9*i, i)*2^(n-8-10*i), i=0..floor((n-8)/10))-sum((-1)^i*binomial(n-9-9*i, i)*2^(n-9-10*i), i=0..floor((n-9)/10)):od:seq(k(n), n=0..50); a:=taylor((z^8-z^9)/(1-2*z+z^(10)), z=0, 51); for p from 0 to 50 do j(p):=coeff(a, z, p):od :seq(j(p), p=0..50); # Richard Choulet, Feb 22 2010

MATHEMATICA

a={1, 0, 0, 0, 0, 0, 0, 0, 0}; Table[s=Plus@@a; a=RotateLeft[a]; a[[ -1]]=s, {n, 50}]

LinearRecurrence[{1, 1, 1, 1, 1, 1, 1, 1, 1}, {0, 0, 0, 0, 0, 0, 0, 0, 1}, 50] (* Vladimir Joseph Stephan Orlovsky, May 25 2011 *)

With[{nn=9}, LinearRecurrence[Table[1, {nn}], Join[Table[0, {nn-1}], {1}], 50]] (* Harvey P. Dale, Aug 17 2013 *)

PROG

(PARI) a(n)=([0, 1, 0, 0, 0, 0, 0, 0, 0; 0, 0, 1, 0, 0, 0, 0, 0, 0; 0, 0, 0, 1, 0, 0, 0, 0, 0; 0, 0, 0, 0, 1, 0, 0, 0, 0; 0, 0, 0, 0, 0, 1, 0, 0, 0; 0, 0, 0, 0, 0, 0, 1, 0, 0; 0, 0, 0, 0, 0, 0, 0, 1, 0; 0, 0, 0, 0, 0, 0, 0, 0, 1; 1, 1, 1, 1, 1, 1, 1, 1, 1]^n*[0; 0; 0; 0; 0; 0; 0; 0; 1])[1, 1] \\ Charles R Greathouse IV, Jun 16 2015

CROSSREFS

Cf. A000045, A000073, A000078, A001591, A001592, A066178, A079262 (Fibonacci n-step numbers).

Cf. A255529 (Indices of primes in this sequence).

Sequence in context: A145115 A172318 A234590 * A258800 A194632 A251759

Adjacent sequences:  A104141 A104142 A104143 * A104145 A104146 A104147

KEYWORD

nonn,easy

AUTHOR

Jean Lefort (jlefort.apmep(AT)wanadoo.fr), Mar 07 2005

EXTENSIONS

Edited by N. J. A. Sloane, Aug 15 2006 and again Nov 11 2006

I deleted an incorrect formula and replaced it with a correct one. - N. J. A. Sloane, Dec 04 2011

STATUS

approved

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Last modified December 9 12:25 EST 2016. Contains 278971 sequences.