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Numbers m such that binomial(m+9,m) mod 9 = 0.
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%I #6 Nov 21 2013 12:49:07

%S 72,73,74,75,76,77,78,79,80,153,154,155,156,157,158,159,160,161,234,

%T 235,236,237,238,239,240,241,242,315,316,317,318,319,320,321,322,323,

%U 396,397,398,399,400,401,402,403,404,477,478,479,480,481,482,483,484,485

%N Numbers m such that binomial(m+9,m) mod 9 = 0.

%C Also numbers m such that floor(1+(m/9)) mod 9 = 0.

%C Partial sums of the sequence 72,1,1,1,1,1,1,1,1,73,1,1,1,1,1,1,1,1,73, ... which has period 9.

%F a(n)=9n+72-8*(n mod 9).

%F G.f.: g(x)=(72+x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9)/((1-x^9)(1-x)).

%F G.f.: g(x)=(72-71x-x^10) /((1-x^9)(1-x)^2).

%t Select[Range[500],Divisible[Binomial[#+9,#],9]&] (* _Harvey P. Dale_, Apr 03 2011 *)

%Y Cf. A000040, A133620, A133621, A133623, A133630, A133635.

%Y Cf. A133879, A133889, A133890, A133900, A133910.

%K nonn

%O 0,1

%A _Hieronymus Fischer_, Oct 20 2007