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A133894
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Numbers m such that binomial(m+4,m) mod 4 = 0.
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0
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12, 13, 14, 15, 28, 29, 30, 31, 44, 45, 46, 47, 60, 61, 62, 63, 76, 77, 78, 79, 92, 93, 94, 95, 108, 109, 110, 111, 124, 125, 126, 127, 140, 141, 142, 143, 156, 157, 158, 159, 172, 173, 174, 175, 188, 189, 190, 191, 204, 205, 206, 207, 220, 221, 222, 223, 236, 237
(list; graph; refs; listen; history; internal format)
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OFFSET
| 0,1
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COMMENTS
| Also numbers m such that floor(1+(m/4)) mod 4 = 0.
Partial sums of the sequence 12,1,1,1,13,1,1,1,13, ... which has period 4.
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LINKS
| Index to sequences with linear recurrences with constant coefficients, signature (1,0,0,1,-1).
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FORMULA
| a(n)=4n+12-3*(n mod 4).
G.f.: 12/(1-x)+x(1+x+x^2+13x^3)/((1-x^4)(1-x)) = (12+x+x^2+x^3+x^4)/((1-x^4)(1-x)) = (12-11x-x^5)/((1-x^4)(1-x)^2).
a(n) = 4*n+3*((1-i)*i^n+(1+i)*(-i)^n+(-1)^n+5)/2, where i=sqrt(-1) - Bruno Berselli, Apr 08 2011
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CROSSREFS
| Cf. A000040, A133620, A133621, A133623, A133630, A133635.
Cf. A133874, A133884, A133890, A133900, A133910.
Sequence in context: A162792 A071589 A083826 * A045879 A112655 A048026
Adjacent sequences: A133891 A133892 A133893 * A133895 A133896 A133897
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KEYWORD
| nonn
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AUTHOR
| Hieronymus Fischer (Hieronymus.Fischer(AT)gmx.de), Oct 20 2007
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