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Number of Fibonacci numbers F(k), k <= 10^n, which end in 4.

Duplicate of A067275.

The sequence 1,7,67.... has a(n)=6*10^n/9+3/9. It is the second binomial transform of 6*A001045(3n)/3+(-1)^n. In general the second binomial transform of k*Jacobsthal(3n)/3+(-1)^n is k*10^n/9+(1-k/9)=1,1+k,1+11k,1+111k,... - Paul Barry, Mar 24 2004

Except for the first two terms, these are the 3-automorphic numbers ending in 7. - Eric M. Schmidt, Aug 28 2012

<a href="/index/Rea#recLCC">Index entries for sequences related to linear recurrences with constant coefficients</a>, signature (11,-10).

a(n)=ceil((2/30)*10^n) - Benoit Cloitre, Aug 27 2002

G.f.: x(1-4x)/((1-x)(1-10x)); a(n)=10^n/15+1/3, n>0. - Paul Barry, Mar 24 2004

a(n)=10*a(n-1)-3, n>1, immediate consequence of the previous formula. [From Vincenzo Librandi, Dec 07 2010]

a(2)=7 because 7 of the first 10^2 Fibonacci numbers end in 4.

(PARI) a(n)=(10^n+13)\15 \\ Charles R Greathouse IV, Jun 05, 2011

Cf. A072702.

base,nonn,easy

dead

Shyam Sunder Gupta (guptass(AT)rediffmail.com), Aug 15 2002

More terms from Benoit Cloitre, Aug 27 2002

Corrected formula by Bruno Berselli, Jun 10 2010

Example corrected by Jonathan Sondow, Jun 04 2011

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Except for the first two terms, these are the 3-automorphic numbers ending in 7. - Eric M. Schmidt, Aug 28 2012

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Example corrected by _Jonathan Sondow (jsondow(AT)alumni.princeton.edu), _, Jun 04 2011

The sequence 1,7,67.... has a(n)=6*10^n/9+3/9. It is the second binomial transform of 6*A001045(3n)/3+(-1)^n. In general the second binomial transform of k*Jacobsthal(3n)/3+(-1)^n is k*10^n/9+(1-k/9)=1,1+k,1+11k,1+111k,... - _Paul Barry (pbarry(AT)wit.ie), _, Mar 24 2004

G.f.: x(1-4x)/((1-x)(1-10x)); a(n)=10^n/15+1/3, n>0. - _Paul Barry (pbarry(AT)wit.ie), _, Mar 24 2004

a(n)=ceil((2/30)*10^n) - _Benoit Cloitre (benoit7848c(AT)orange.fr), _, Aug 27 2002

More terms from _Benoit Cloitre (benoit7848c(AT)orange.fr), _, Aug 27 2002