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Numbers m such that binomial(m+7,m) mod 7 = 0.
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%I #4 Oct 13 2022 13:47:08

%S 42,43,44,45,46,47,48,91,92,93,94,95,96,97,140,141,142,143,144,145,

%T 146,189,190,191,192,193,194,195,238,239,240,241,242,243,244,287,288,

%U 289,290,291,292,293,336,337,338,339,340,341,342,385,386,387,388,389,390

%N Numbers m such that binomial(m+7,m) mod 7 = 0.

%C Also numbers m such that floor(1+(m/7)) mod 7 = 0.

%C Partial sums of the sequence 42,1,1,1,1,1,1,43,1,1,1,1,1,1,43,... which has period 7.

%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (1,0,0,0,0,0,1,-1).

%F a(n)=7n+42-6*(n mod 7).

%F G.f.: g(x)=(42+x+x^2+x^3+x^4+x^5+x^6+x^7)/((1-x^7)(1-x)).

%F G.f.: g(x)=(42-41x-x^8) /((1-x^7)(1-x)^2).

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

%Y Cf. A133877, A133887, A133890, A133900, A133910.

%K nonn,easy

%O 0,1

%A _Hieronymus Fischer_, Oct 20 2007