I discovered a new sequence operation by trying to create a formula for one of the sequences in the EIS. %I A007690 M0167 %S A007690 1,0,1,1,2,1,4,2,6,5,9,7,16,11,22,20,33,28,51,42,71,66,100, %T A007690 92,147,131,199,193,275,263,385,364,516,511,694,686,946,925, %U A007690 1246,1260,1650,1663,2194,2202,2857,2928,3721,3813,4866,4967 %N A007690 Partitions of n in which no part occurs just once. %D A007690 R. Honsberger, Mathematical Gems III, M.A.A., 1985, p. 242. %K A007690 nonn,part %O A007690 0,5 %A A007690 njas, Robert G. Wilson v (rgwv@kspaint.com) So what's the formula? Well, if you want to cut to the chase, the generating function is: PRODUCT {k>0 is a multiple of 2 or 3} (1/(1-x^k)). But how in general do we solve problems like this? When dealing with partition problems, the first places I look are the Euler and Weigh transforms described on the transforms page of the EIS. The Euler transform is for partitions in which any part can occur any number of times. The Weigh transform is for partitions in which each part occurs 0 or 1 time. For any sequence a(n) the Euler transform gives g.f. PRODUCT {k>0} (1/(1-x^k)^a(k)). and the Weigh transform gives PRODUCT {k>0} (1+x^k)^a(k). For simplicity, lets make a(n) a 0,1 sequence. The 1/(1-x) term of the Euler transform expands to 1+x+x^2+x^3+... The "1" means I'm allowed to have a part occur 0 times The "x" means I'm allowed to have a part occur 1 time The "x^2" means I'm allowed to have a part occur 2 times The "x^3" means I'm allowed to have a part occur 3 times ... Thus a part can occur any number of times. The corresponding term of the Weigh transform is 1+x meaning a part can occur 0 or 1 time. So if I want to allow a part to occur 0 or 2 or more times (any number other than 1 as in A007690) I can use (1+x^2+x^3+x^4+...)*(1+x^4+x^6+x^8+...)*(1+x^6+x^9+x^12+...)*... or PRODUCT {k>0} ((1-x+x^2)/(1-x^k)). Or we could insist each part occur an odd number of times (or 0). (1+x+x^3+x^5+x^7+...)*(1+x^2+x^6+x^10+x^14+...)*(1+x^3+x^9+x^15+x^21+...)*... Starting at n=0 1 1 1 3 2 5 6 9 9 16 20 25 32 40 54 69... (Not in EIS) or each part occur a prime number of times: 1 0 1 1 1 1 3 2 3 4 4 6 8 8 10 13 13 20 20 24 26 38... (Also not in EIS) I played around with this and came up with a simple algorithm. I haven't actually proven the algorithm works, but it passes the sanity checks at least. (If you prove it, I'd like to hear about it.) Let a be any sequence with a(0)=0 and let b be a sequence of 1's and 0's with b(0)=1. We will produce an output sequence c. a represents a sequence of unlabeled structures. b represents the multiplicities we are allowing. b(n)=1 if I'm allowed to have n copies of the same a-structure. b(n)=0 if I'm not allowed. Let s be the inverse Euler transform of b. Let t have g.f. s(1)*A(x)+s(2)*A(x^2)+s(3)*A(x^3)+... where A(x) is the g.f. of a. Let c be the Euler transform of t and there you have it. When the b sequence is 1 0 1 1 1 (i.e. b(1)=0, and 1 for all other n) t is 0 0 1 1 0 0 -1 0 0 ... i.e. x^2+x^3-x^6. Thus for the special case of multiplicity 1 not allowed we have: c = EULER(A(x^2)+A(x^3)-A(x^6)) hence the formula I gave for A007690. Note also that while I restricted b to being a sequence of 1's and 0's, but the algorithm will give you a result without that restriction. One interpretation (that seems to pass the sanity check) is that b(n) is the number of colors of a cluster that has multiplicity n. Best regards Christian P.S. I need a name for this operation. Right now I'm calling it EOP. E for exponential (for fixed b it's exponential on the g.f. of a). OP for operation.