cp's OEIS Frontend

A067275 Number of Fibonacci numbers A000045(k), k <= 10^n, which end in 4.

%I A067275 #63 Jul 27 2025 19:20:32
%S A067275 0,1,7,67,667,6667,66667,666667,6666667,66666667,666666667,6666666667,
%T A067275 66666666667,666666666667,6666666666667,66666666666667,
%U A067275 666666666666667,6666666666666667,66666666666666667,666666666666666667,6666666666666666667,66666666666666666667,666666666666666666667
%N A067275 Number of Fibonacci numbers A000045(k), k <= 10^n, which end in 4.
%C A067275 Numbers n such that the digits of A000326(n), the n-th pentagonal number, begin with n. [original definition of this sequence]
%C 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
%C A067275 Except for the first two terms, these are the 3-automorphic numbers ending in 7. - _Eric M. Schmidt_, Aug 28 2012
%C A067275 From _Wolfdieter Lang_, Feb 08 2017: (Start)
%C A067275 This sequence appears in curious identities based on the Armstrong numbers 370 = A005188(11), 371 = A005188(12) and 407 = A005188(13).
%C A067275 For such identities based on 153 = A005188(10) see a comment in A246057 with the van der Poorten et al. reference.
%C A067275 For 370 these identities are A002277(n)^3 + a(n+1)^3  + 0(n)^3 = A002277(n)*10^(2*n) + a(n+1)*10^n + 0(n) = A281858(n), where 0(n) means n 0's.
%C A067275 For 371 these identities are A002277(n)^3 + a(n+1)^3 + (0(n-1)1)^3 = A002277(n)*10^(2*n) + a(n+1)*10^n + 0(n-1)1 = A281860(n), where 0(n-1)1 means n-1 0's followed by a 1.
%C A067275 For 407 these identities are A093137(n)^3 + 0(n)^3 + a(n+1)^3 = concatenation(A093137(n), 0(n), a(n+1)) = A281859(n). (End)
%H A067275 Seiichi Manyama, <a href="/A067275/b067275.txt">Table of n, a(n) for n = 0..1001</a>
%H A067275 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (11,-10).
%F A067275 a(n) = ceiling((2/30)*10^n) - _Benoit Cloitre_, Aug 27 2002
%F A067275 From _Paul Barry_, Mar 24 2004: (Start)
%F A067275 G.f.: x*(1 - 4*x)/((1 - x)*(1 - 10*x)).
%F A067275 a(n) = 10^n/15 + 1/3 for n>0. (End)
%F A067275 a(n) = 10*a(n-1) - 3 for n>1. - _Vincenzo Librandi_, Dec 07 2010 [Immediate consequence of the previous formula, _R. J. Mathar_]
%F A067275 From _Eric M. Schmidt_, Oct 28 2012: (Start)
%F A067275 For n>0, a(n) = A199682(n-1)/3 = (2*10^(n-1) + 1)/3.
%F A067275 For n>=2, a(n+1) = a(n) + 6*10^n. (End)
%F A067275 From _Elmo R. Oliveira_, Jul 22 2025: (Start)
%F A067275 E.g.f.: (-6 + 5*exp(x) + exp(10*x))/15.
%F A067275 a(n) = 11*a(n-1) - 10*a(n-2) for n >= 3.
%F A067275 a(n) = A073553(n)/2 for n >= 1. (End)
%e A067275 a(2) = 7 because 7 of the first 10^2 Fibonacci numbers end in 4.
%e A067275 From _Wolfdieter Lang_, Feb 08 2017: (Start)
%e A067275 Curious cubic identities:
%e A067275 3^3 + 7^3 + 0^3 = 370, 33^3 + 67^3 + (00)^3 = 336700, 333^3 + 667^3 + (000)^3 = 333667000, ...
%e A067275 3^3 + 7^3 + 1^3 = 371, 33^3 + 67^3 + (01)^3 = 336701, 333^3 + 667^3 + (001)^3 = 333667001, ...
%e A067275 4^3 + 0^3 + 7^3 = 407, 34^3 + (00)^3 + 67^3 = 340067 , 334^3 + (000)^3 + 677^3 = 334000677, ... (End)
%t A067275 s = Fibonacci@ Range[10^5]; Table[Count[Take[s, 10^n], m_ /; Mod[m, 10] == 4], {n, 0, Floor@ Log10@ Length@ s}] (* or *) Table[Boole[n > 0] Ceiling[10^n/15], {n, 0, 20}] (* or *) CoefficientList[Series[x (1 - 4 x)/((1 - x) (1 - 10 x)), {x, 0, 20}], x] (* _Michael De Vlieger_, Feb 08 2017 *)
%o A067275 (PARI) a(n)=(10^n+13)\15 \\ _Charles R Greathouse IV_, Jun 05 2011
%Y A067275 Cf. A072702, A093137, A109344, A246057, A281858, A281859, A281860.
%Y A067275 Cf. A000045, A000326, A001045, A002277, A005188, A073553, A199682.
%K A067275 nonn,base,easy
%O A067275 0,3
%A A067275 _Joseph L. Pe_, Feb 21 2002
%E A067275 A073552 merged into this sequence by _Eric M. Schmidt_, Oct 28 2012