|
%I #20 Apr 16 2018 03:22:15
%S 1,1,1,2,4,5,6,10,23,38,70,110,196,346,759,1250,2313,3982,8433,14520,
%T 29437,50466,102830,179587,376439,654374,1343540,2352149,4916286,
%U 8654120,18065200,31783592,66233160,117371504,246610521,436972949,913862320,1626523783
%N Let b(n) be the number of permutations {c_1..c_n} of {1..n} for which c_1 - c_2 + ... + (-1)^(n-1)*c_n are pentagonal numbers (A000326). Then a(n) = b(n)/A010551(n).
%C All terms are positive integers (for a proof, cf. comment in A293984).
%C Note that a(1), a(2), a(3), a(4) remain the same, if in the definition the pentagonal numbers are replaced by k-gonal numbers for k >= 3 other than k=4.
%H Peter J. C. Moses, <a href="/A294812/b294812.txt">Table of n, a(n) for n = 1..200</a>
%t polyQ[order_,n_]:=If[n==0,True,IntegerQ[(#-4+Sqrt[(#-4)^2+8 n (#-2)])/(2 (#-2))]&[order]];(*is a number polygonal?*)
%t Map[Total,Table[
%t possibleSums=Range[1/2-(-1)^n/2-Floor[n/2]^2,Floor[(n+1)/2]^2];
%t filteredSums=Select[possibleSums,polyQ[5,#]&&#>-1&];
%t positions=Map[Flatten[{#,Position[possibleSums,#,1]-1}]&,filteredSums];
%t Map[SeriesCoefficient[QBinomial[n,Floor[(n+1)/2],q],{q,0,#[[2]]/2}]&,positions],{n,25}]] (* _Peter J. C. Moses_, Jan 02 2018 *)
%Y Cf. A010551, A293857, A293984, A294811.
%K nonn
%O 1,4
%A _Vladimir Shevelev_ and _Peter J. C. Moses_, Nov 09 2017
|
