Dividing n consecutive integers by (1,2,3,...,n)

[Don Reble]

> Do you also solve some
> kind of Perfect Matching / Assignment Problem in your approach?

    My algorithm is straightforward:

    match(N):
	Mark all locations "unused".
	if matchupto(N),
	    Note success; print everything.

    boolean matchupto(Divisor):
	if Divisor = 1, return true.
	for each possible location of Divisor:
	    if the location is marked "unused",
		Mark the location "used".
		if matchupto(Divisor-1), return true.
		Mark the location "unused".
	return false.


> The first disagreement happens for n=6.
> Your value for the number of starting points is 11 while I got 20.
>
> Do you see any errors in this list or was it miscalculation on your side?

    Your list looks good; I must have erred.

-- 
Don Reble  djr at nk.ca


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I give alternative GFs that are peculiar products of eta functions
for 3 seqs in the OEIS and suggest 2 new seqs (which I find natural
to include):


\\ some aux variable for power series:
N=100  \\ power series to this order
x=z+O(z^(N+1));
L=1+sqrtint(N);   \\ avoid "*** eta: valuation (valp) overflow"
L2=ceil(L/2);  \\ dito


etaplus(z)=eta(z^2)/eta(z)  \\ == prod(1+z^n)


\\ A073576 Number of partitions of n into square-free parts.
\\ essentially identical to A058647   McKay-Thompson series of class 36D for Monster.
v=Vec( 1/prod(n=1,N, (1-moebius(n)^2*x^n)) );
v2=Vec( prod(n=1,L, eta(x^(n^2))^(-moebius(n)) ) ); \\ jj's GF
(v-v2)==0 \\ OK for 5000 terms


\\ A087188   Number of partitions of n into distinct square-free parts
v=Vec( prod(n=1,N, (1+moebius(n)^2*x^n)) );
v2=Vec( prod(n=1,L, etaplus(x^(n^2))^(+moebius(n)) ) ); \\ jj's GF
(v-v2)==0 \\ OK for 3000 terms


\\ not in OEIS:  partitions into odd sqrfree parts
v=Vec( 1/prod(n=1,N, 1-moebius(2*n-1)^2*x^(2*n-1) ) );
v2=Vec( prod(n=1,L2, (eta(x^(2*(2*n-1)^2))/eta(x^((2*n-1)^2)))^(+moebius(2*n-1)) ) );
(v-v2)==0 \\ OK up to 3000 terms


\\ seq is similar to A112193 McKay-Thompson series of class 54b for the Monster group.
\\ for which I suggest a GF (correct with all terms given):
v=[1, 1, 1, 2, 2, 3, 4, 5, 6, 7, 9, 11, 13, 16, 19, 23, 27, 32, 38, 44, 52, 61, 71, 82, 95, 110, 127, 145, 167, 191, 218, 249, 283, 322, 365, 414, 469, 529, 597, 673, 757, 851, 955, 1071, 1199, 1341, 1499, 1673, 1865, 2078, 2313, 2572, 2857, 3171, 3517, 3897];  \\ from OEIS
t1=Vec(etaplus(x)/etaplus(x^9))
\\ t2=Vec(eta(x^2)*eta(x^9)/(eta(x)*eta(x^18)))
\\ (t1-t2)==0  \\ true by definition
(vector(#v,j,v[j]-t1[j])==0)


\\  not in OEIS: partitions into distinct odd sqrfree parts
v=Vec( prod(n=1,N, 1+moebius(2*n-1)^2*x^(2*n-1) ) );
v2=Vec( prod(n=1,L2, (etaplus(x^(2*(2*n-1)^2))/etaplus(x^((2*n-1)^2)))^(-moebius(2*n-1)) ) );
(v-v2)==0 \\ OK up to 3000 terms


Submit GFs and new seqs?

Any comments on the crazy eta-products?

cheers,   jj



addendum:

\\ not in OEIS:  partitions of 2*n into even sqrfree parts
v=Vec( 1/prod(n=1,N, 1-moebius(2*n)^2*x^(n) ) );
v2=Vec( prod(n=1,L2, (etaplus(x^((2*n-1)^2)))^(+moebius(2*n-1)) ) );
(v-v2)==0 \\ OK up to 3000 terms


\\ not in OEIS:  partitions of 2*n into distinct even sqrfree parts
v=Vec( prod(n=1,N, 1+moebius(2*n)^2*x^(n) ) );
v2=Vec( prod(n=1,L2, (etaplus(x^(2*(2*n-1)^2))/etaplus(x^((2*n-1)^2)))^(-moebius(2*n-1))) );
(v-v2)==0 \\ OK up to 3000 terms
\\ ***** NOTE: apparently SAME AS partitions into distinct odd sqrfree parts
v=Vec( prod(n=1,N, 1+moebius(2*n-1)^2*x^(2*n-1) ) );
(v-v2)==0 \\ OK up to 3000 terms


I suggest to submit the first.