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A171646 a(1) = 1, then partial products of Product_{n>=1} (p(n)/p(n-1)*p(n)/p(n-1)) = 1*1*1*(2)*(2)*(3/2)*(3/2)*(5/3)*(5/3)*(7/5)*(7/5)*...*; p = partition numbers, A000041 starting (1, 2, 3, 5, ...).

%I A171646 #11 Feb 09 2022 09:04:01
%S A171646 1,1,1,2,4,6,9,15,25,35,49,77,121,165,225,330,484,660,900,1260,1764,
%T A171646 2352,3136,4312,5929,7777,10201,13635,18225,23760,30976,40656,53361,
%U A171646 68607,88209,114345,148225,188650,240100,307230
%N A171646 a(1) = 1, then partial products of Product_{n>=1} (p(n)/p(n-1)*p(n)/p(n-1)) = 1*1*1*(2)*(2)*(3/2)*(3/2)*(5/3)*(5/3)*(7/5)*(7/5)*...*; p = partition numbers, A000041 starting (1, 2, 3, 5, ...).
%C A171646 A006498 = analogous sequence using the Fibonacci numbers.
%C A171646 A171645 = .............................Primes, analogous formula.
%C A171646 A010551 = .............................Factorial numbers, analogous formula.
%F A171646 a(1) = 1, then partial products of Product_{n>=1} (p(n)/p(n-1)*p(n)/p(n-1)) = 1*1*1*(2)*(2)*(3/2)*(3/2)*(5/3)*(5/3)*(7/5)*(7/5)*...; p = partition numbers, A000041 starting (1, 2, 3, 5, ...).
%e A171646 a(12) = 77 = 1*1*1*2*2*(3/2)*(3/2)*(5/3)*(5/3)*(7/5)*(7/5)*(11/7).
%p A171646 A171646t := proc(n)
%p A171646     local nh;
%p A171646     nh := floor(n/2) ;
%p A171646     combinat[numbpart](nh)/combinat[numbpart](nh-1) ;
%p A171646 end proc:
%p A171646 A171646 := proc(n)
%p A171646     mul(A171646t(i),i=2..n) ;
%p A171646 end proc:
%p A171646 1,seq(A171646(n),n=2..40) ; # _R. J. Mathar_, Jul 21 2015
%Y A171646 Cf. A000041, A171645, A006498, A010551.
%K A171646 nonn
%O A171646 1,4
%A A171646 _Gary W. Adamson_, Dec 13 2009
%E A171646 Corrected by _R. J. Mathar_, Jul 21 2015