A282772 Starting from F(n), minimum number, greater than 1, of consecutive Fibonacci numbers whose average is an integer.
4, 2, 3, 12, 2, 13, 3, 2, 6, 5, 2, 12, 4, 2, 3, 12, 2, 24, 3, 2, 6, 24, 2, 12, 4, 2, 3, 12, 2, 5, 3, 2, 6, 13, 2, 12, 4, 2, 3, 5, 2, 24, 3, 2, 5, 24, 2, 12, 4, 2, 3, 12, 2, 24, 3, 2, 6, 24, 2, 5, 4, 2, 3, 12, 2, 24, 3, 2, 6, 5, 2, 12, 4, 2, 3, 12, 2, 24, 3, 2, 6
Offset: 0
Examples
a(0) = 4 because F(0) + F(1) + F(2) + F(3) = 0 + 1 + 1 + 2 = 4 and 4/4 = 1; a(1) = 2 because F(1) + F(2) = 1 + 1 = 2 and 2/2 = 1; a(2) = 3 because F(2) + F(3) + F(4) = 1 + 2 + 3 = 6 and 6/3 = 2; a(3) = 12 because F(3) + F(4) + ... + F(13) + F(14) = 2 + 3 + ... + 233 + 377 = 984 and 984/12 = 82.
Links
- Paolo P. Lava, Table of n, a(n) for n = 0..1000
Programs
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Maple
with(combinat): P:=proc(q) local a,k,n; for k from 0 to q do a:=fibonacci(k); for n from 1 to q do a:=a+fibonacci(k+n); if type(a/(n+1),integer) then print(n+1); break; fi; od; od; end: P(10^3);
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Mathematica
Table[k = 1; While[! IntegerQ@ Mean@ Take[#, n ;; n + k], k++]; k + 1, {n, Length@ # - 24}] &@ Fibonacci@ Range[0, 419] (* Michael De Vlieger, Mar 06 2017 *)
Formula
a(3*k + 1) = 2;
a(12*k + 2) = a(12*k + 6) = 3;
a(12*k) = 4;
a(30*k + 9) = a(30*k + 29) = a(60*k + 44) = 5;
a(60*k + 8) = a(60*k + 20) = a(60*k + 32) = a(60*k + 56) = 6;
a(60*k + 3) = a(60*k + 11) = a(60*k + 15) = a(60*k + 23) = a(60*k + 27) = a(60*k + 35) = a(60*k + 47) = a(60*k + 51) = 12;
a(420*k + 5) = a(420*k + 33) = a(420*k + 117) = a(420*k + 173) = a(420*k + 201) = a(420*k + 257) = a(420*k + 285) = a(420*k + 341) = 13;
a(420*k + x) = 24, with x = 17, 21, 41, 45, 53, 57, 65, 77, 81, 93, 101, 105, 113, 125, 137, 141, 153, 161, 165, 177, 185, 197, 213, 221, 225, 233, 237, 245, 261, 273, 281, 293, 297, 305, 317, 321, 333, 345, 353, 365, 377, 381, 393, 401, 405, 413, 417.
Comments