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A133897
Numbers m such that binomial(m+7,m) mod 7 = 0.
0
42, 43, 44, 45, 46, 47, 48, 91, 92, 93, 94, 95, 96, 97, 140, 141, 142, 143, 144, 145, 146, 189, 190, 191, 192, 193, 194, 195, 238, 239, 240, 241, 242, 243, 244, 287, 288, 289, 290, 291, 292, 293, 336, 337, 338, 339, 340, 341, 342, 385, 386, 387, 388, 389, 390
OFFSET
0,1
COMMENTS
Also numbers m such that floor(1+(m/7)) mod 7 = 0.
Partial sums of the sequence 42,1,1,1,1,1,1,43,1,1,1,1,1,1,43,... which has period 7.
FORMULA
a(n)=7n+42-6*(n mod 7).
G.f.: g(x)=(42+x+x^2+x^3+x^4+x^5+x^6+x^7)/((1-x^7)(1-x)).
G.f.: g(x)=(42-41x-x^8) /((1-x^7)(1-x)^2).
KEYWORD
nonn,easy
AUTHOR
Hieronymus Fischer, Oct 20 2007
STATUS
approved