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A104144
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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).
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16
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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
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OFFSET
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0,11
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COMMENTS
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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.
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LINKS
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T. D. Noe, Table of n, a(n) for n = 0..208
Martin Burtscher, Igor Szczyrba, Rafał 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).
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FORMULA
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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)
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MAPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(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
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CROSSREFS
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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
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KEYWORD
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nonn,easy
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AUTHOR
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Jean Lefort (jlefort.apmep(AT)wanadoo.fr), Mar 07 2005
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EXTENSIONS
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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
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STATUS
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approved
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