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A037453 Positive numbers whose base-5 representation contains no 3 or 4. 8

%I

%S 1,2,5,6,7,10,11,12,25,26,27,30,31,32,35,36,37,50,51,52,55,56,57,60,

%T 61,62,125,126,127,130,131,132,135,136,137,150,151,152,155,156,157,

%U 160,161,162,175,176,177,180,181,182,185,186,187

%N Positive numbers whose base-5 representation contains no 3 or 4.

%C 5 divides neither C(2s-1,s) = A001700[ s ] (nor C(2s,s) = A000984[ s ], central column of Pascal's Triangle) if and only if s is one of the terms in this sequence.

%C n such that the last decimal digit of C(2n,n) is not zero. - _Benoit Cloitre_, Aug 18 2002

%C Let us recall the plan of Apery's irrationality proof. Consider the recurrence (n + 1)^3 * u_(n+1) = (34n^3 + 51n^2 + 27n + 5)u_n - n^3 * u_(n-1). The solution with starting values u_0 = 1; u_1 = 5 has the peculiar property that it has integral terms, despite the fact that at every recursion step we divide by (n + 1)^3. The n-th term is given by f(n) = sum(i=0,n,binomial(n+i,i)^2*binomial(n,i)^2) (see Beukers link) and m such that if Mod(f(m),5) <>0 is same as 2*a(n). - Mohammed Bouayoun (bouyao(AT)wanadoo.fr), Mar 08 2004

%H Clark Kimberling, <a href="/A037453/b037453.txt">Table of n, a(n) for n = 1..1000</a>

%H Frits Beukers <a href="http://www.math.uu.nl/people/beukers/caen.pdf">Consequences of Apery's work on zeta(3)</a>

%H W. Shur, <a href="https://doi.org/10.37236/1331">The last digit of C(2*n,n) and sigma C(n,i)*C(2*n-2*i,n-i)</a>, The Electronic Journal of Combinatorics, R16, Volume 4, Issue 2 (1997).

%F a(3n)=5a(n), a(3n+1)=5a(n)+1, a(3n+2)=5a(n)+2, where by definition a(0)=0. - _Emeric Deutsch_, Mar 23 2004

%F G.f. satisfies g(x) = 5*(1+x+x^2)*g(x^3) + (x + 2*x^2)/(1-x^3). - _Robert Israel_, Sep 02 2014

%p a:= proc(t) option remember; 5*procname(floor(t/3))+ (t mod 3) end proc:

%p a(0):= 0:

%p seq(a(n),n=1..100); # _Robert Israel_, Sep 02 2014

%t Table[FromDigits[IntegerDigits[k,3],5], {k,60}] - _T. D. Noe_, Apr 18 2007

%t Rest[FromDigits[#,5]&/@Tuples[{0,1,2},4]] (* _Harvey P. Dale_, Aug 31 2016 *)

%o (PARI) f(n)=sum(i=0,n,binomial(n+i,i)^2*binomial(n,i)^2); for (i=1,1000,if(Mod(f(i),5)<>0,print1(i/2,",")))

%o (Julia)

%o function a(n)

%o m, r, b = n, 0, 1

%o while m > 0

%o m, q = divrem(m, 3)

%o r += b * q

%o b *= 5

%o end

%o r end; [a(n) for n in 1:53] |> println # _Peter Luschny_, Jan 03 2021

%Y Cf. A050607, A005836.

%K nonn,base

%O 1,2

%A _Clark Kimberling_

%E Better definition from _T. D. Noe_, Apr 18 2007

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Last modified July 24 06:56 EDT 2021. Contains 346273 sequences. (Running on oeis4.)