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a(n) = a(n-1) + a(n - 1 minus the number of terms of a(k) == (mod 4) so far).
0

%I #2 Mar 30 2012 17:30:31

%S 1,2,3,4,7,11,14,25,36,50,64,89,125,175,225,314,378,467,592,817,992,

%T 1306,1684,2062,2529,2996,3988,4580,5572,6878,9407,10399,12928,15457,

%U 19445,21507,25495,31067,35647,40227,49634,60033,66911,77310,92767

%N a(n) = a(n-1) + a(n - 1 minus the number of terms of a(k) == (mod 4) so far).

%e a(5) = a(4) + a(4 - the number of terms congruent to 1 (mod 4) so far) = a(4) + a(4-1) = 4 + 3 = 7.

%t m[ 1 ] = 1; m[ 2 ] = 2; m[ 3 ] = 3; a[ 4 ] = 4; m[ n_Integer ] := m[ n ] = Block[ {a = b = c = d = 0}, Do[ Switch[ Mod[ m[ k ], 4 ], 0, a++, 1, b++, 2, c++, 3, d++ ], {k, 1, n - 1} ]; Switch[ Mod[ n, 4 ], 0, m[ n - 1 ] + m[ n - 1 - a ], 1, m[ n - 1 ] + m[ n - 1 - b ], 2, m[ n - 1 ] + m[ n - 1 - c ], 3, m[ n - 1 ] + m[ n - 1 - d ] ] ]; Table[ m[ n ], {n, 1, 50} ]

%K nonn

%O 1,2

%A _Robert G. Wilson v_, Apr 22 2001