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a(n+1) = a(n) + a(n minus the number of terms of the same parity as n so far).
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%I #5 Dec 13 2016 16:57:09

%S 1,2,3,4,6,9,12,16,19,25,31,37,49,55,74,83,108,120,145,161,198,217,

%T 266,291,346,377,451,482,565,602,710,765,885,940,1085,1159,1357,1431,

%U 1697,1771,2117,2191,2642,2725,3207,3290,3855,3963,4673,4781

%N a(n+1) = a(n) + a(n minus the number of terms of the same parity as n so far).

%H Ivan Neretin, <a href="/A060729/b060729.txt">Table of n, a(n) for n = 1..10000</a>

%e a(5) = a(4) + a(4 - the number of even terms so far) = a(4) + a(4-2) = 4 + 2 = 6.

%t a[ 1 ] = 1; a[ 2 ] = 2; a[ n_ ] := a[ n ] = Block[ {e = 0}, Do[ If[ EvenQ[ a[ k ] ], e++ ], {k, 1, n - 1} ]; If[ OddQ[ n ], a[ n - 1 ] + a[ n - 1 - e ], a[ n - 1 ] + a[ e ] ] ]; Table[ a[ n ], {n, 1, 50} ]

%Y Cf. A006336, A006604.

%K nonn

%O 1,2

%A _Robert G. Wilson v_, Apr 22 2001