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A135298 a(n) = the total number of permutations (m(1),m(2),m(3)...m(j)) of (1,2,3,...,j) where n = 1*m(1) + 2*m(2) + 3*m(3) + ...+j*m(j), where j is over all positive integers.

%I A135298 #53 Nov 23 2023 15:54:52
%S A135298 1,1,0,0,1,1,0,0,0,0,1,2,0,2,1,0,0,0,0,0,1,3,1,4,2,2,2,4,1,3,1,0,0,0,
%T A135298 0,1,4,3,6,7,6,4,10,6,10,6,10,6,10,4,6,7,6,3,4,1,1,5,6,9,16,12,14,24,
%U A135298 20,21,23,28,24,34,20,32,42,29,29,42,32,20,34,24,28,23,21,20,25,20,22,30,38
%N A135298 a(n) = the total number of permutations (m(1),m(2),m(3)...m(j)) of (1,2,3,...,j) where n = 1*m(1) + 2*m(2) + 3*m(3) + ...+j*m(j), where j is over all positive integers.
%C A135298 Does every integer greater than some positive integer N have at least one such representation?
%C A135298 a(n) > 0 for n > 34, a(n) > 1 for n > 56. - _Alois P. Heinz_, Aug 28 2014
%H A135298 Alois P. Heinz, <a href="/A135298/b135298.txt">Table of n, a(n) for n = 0..1770</a>
%H A135298 FindStat - Combinatorial Statistic Finder, <a href="http://www.findstat.org/StatisticsDatabase/St000055/">The rank of the permutation inside the alternating sign matrix lattice</a>
%H A135298 J. Sack and H. Úlfarsson, <a href="http://arxiv.org/abs/1106.1995">Refined inversion statistics on permutations</a>, arXiv preprint arXiv:1106.1995 [math.CO], 2011-2012.
%e A135298 21 has a(21)=3 such representations: 21 = 1*4 + 2*3 + 3*1 + 4*2 = 1*4 + 2*2 + 3*3 + 4*1 = 1*3 + 2*4 + 3*2 + 4*1.
%e A135298 Not all representations of an integer n need to necessarily have the same j. For example, 91 = 1*1 + 2*2 + 3*3 + 4*4 + 5*5 + 6*6 (j=6). And 91 also equals 1*7 + 2*4 + 3*5 + 4*3 + 5*6 + 6*2 + 7*1 (j=7).
%e A135298 1 = 1*1;
%e A135298 4 = 1*2+2*1;
%e A135298 5 = 1*1+2*2;
%e A135298 10 = 1*3+2*2+3*1;
%e A135298 11 = 1*2+2*3+3*1;
%e A135298 11 = 1*3+2*1+3*2;
%e A135298 13 = 1*1+2*3+3*2;
%e A135298 13 = 1*2+2*1+3*3;
%e A135298 14 = 1*1+2*2+3*3;
%e A135298 20 = 1*4+2*3+3*2+4*1;
%e A135298 21 = 1*3+2*4+3*2+4*1;
%e A135298 21 = 1*4+2*2+3*3+4*1;
%e A135298 21 = 1*4+2*3+3*1+4*2;
%e A135298 22 = 1*3+2*4+3*1+4*2;
%e A135298 23 = 1*2+2*4+3*3+4*1;
%e A135298 23 = 1*3+2*2+3*4+4*1;
%e A135298 23 = 1*4+2*1+3*3+4*2;
%e A135298 23 = 1*4+2*2+3*1+4*3;
%e A135298 24 = 1*2+2*3+3*4+4*1;
%e A135298 24 = 1*4+2*1+3*2+4*3;
%e A135298 25 = 1*2+2*4+3*1+4*3;
%e A135298 25 = 1*3+2*1+3*4+4*2;
%e A135298 26 = 1*1+2*4+3*3+4*2;
%e A135298 26 = 1*3+2*2+3*1+4*4;
%e A135298 27 = 1*1+2*3+3*4+4*2;
%e A135298 27 = 1*1+2*4+3*2+4*3;
%e A135298 27 = 1*2+2*3+3*1+4*4;
%e A135298 27 = 1*3+2*1+3*2+4*4;
%e A135298 28 = 1*2+2*1+3*4+4*3;
%e A135298 29 = 1*1+2*2+3*4+4*3;
%e A135298 29 = 1*1+2*3+3*2+4*4;
%e A135298 29 = 1*2+2*1+3*3+4*4;
%e A135298 30 = 1*1+2*2+3*3+4*4;
%p A135298 A135298rec := proc(j,n,notm) local a,m ; a := 0 ; if n = 0 then if max( seq(e,e=notm) ) >= j then RETURN(0) ; else RETURN(1) ; fi ; end: for m from 1 do if n-j*m < 0 then break ; elif not m in notm then a := a+A135298rec(j+1,n-j*m,[op(notm),m] ) ; fi ; od: RETURN(a) ; end: A135298 := proc(n) A135298rec(1,n,[]) ; end: for n from 1 to 140 do printf("%d, ",A135298(n)) ; od: # _R. J. Mathar_, Jan 30 2008
%p A135298 # second Maple program:
%p A135298 n:= 8 : # gives binomial(n+3, 3) terms
%p A135298 with(combinat):
%p A135298 (p-> seq(coeff(p, x, j), j=0..binomial(n+3, 3)-1))
%p A135298 (add(add(x^add(i*l[i], i=1..h), l=permute(h)), h=0..n));
%p A135298 # _Alois P. Heinz_, Aug 29 2014
%t A135298 n = 8; (* gives binomial(n+3, 3)-1 terms *) Function[p, Table[ Coefficient[p, x, j], {j, 1, Binomial[n+3, 3]-1}]] @ Sum[x^(l.Range[h]), {h, 1, n}, {l, Permutations @ Range[h]}] (* _Jean-François Alcover_, Jul 22 2017, after _Alois P. Heinz_ *)
%Y A135298 Cf. A000292, A126972, A175929.
%K A135298 nonn,look
%O A135298 0,12
%A A135298 _Leroy Quet_, Dec 04 2007
%E A135298 More terms from _R. J. Mathar_, Jan 30 2008
%E A135298 a(0)=1 prepended by _Alois P. Heinz_, Nov 23 2023