A257988 - OEIS

%I #19 Dec 10 2016 06:23:58

%S 0,0,1,0,1,1,1,0,2,1,1,0,1,1,2,0,1,2,1,1,2,1,1,0,2,1,3,1,1,2,1,0,2,1,

%T 2,0,1,1,2,0,1,2,1,1,3,1,1,0,2,2,2,1,1,3,2,1,2,1,1,1,1,1,3,0,2,2,1,1,

%U 2,2,1,0,1,1,3,1,2,2,1,0,4,1,1,1,2,1,2,1,1,3,2,1,2,1,2,0,1,2,3,2,1,2,1

%N Number of parts that are larger than the number of 1's in the partition having Heinz number n.

%C We define the Heinz number of a partition p = [p_1, p_2, ..., p_r] to be Product(p_j-th prime, j=1...r) (a concept used by _Alois P. Heinz_ in A215366 as an "encoding" of a partition). For example, for the partition [1, 1, 2, 4, 10] we get 2*2*3*7*29 = 2436.

%C In the Maple program the subprogram b yields the number of 1's in the partition having Heinz number n and the subprogram B yields the partition having Heinz number n.

%C The considered partition statistic is needed in the partition statistic "crank" (A257989).

%H Alois P. Heinz, <a href="/A257988/b257988.txt">Table of n, a(n) for n = 1..20000</a>

%H G. E. Andrews and F. G. Garvan, <a href="http://dx.doi.org/10.1090/S0273-0979-1988-15637-6">Dyson's crank of a partition</a>, Bull. Amer. Math. Soc., 18 (1988), 167-171.

%e a(252) = 1 because the partition having Heinz number 252 = 2^2 * 3^2 * 7 is [1,1,2,2,4] and exactly one part, namely 4, is larger than 2 (the number of 1's).

%e a(945) = 5 because the partition having Heinz number 945 = 3^3 * 5 * 7 is [2,2,2,3,4] and all parts are larger than 0 (the number of 1's).

%p with(numtheory): a := proc (n) local b, B, c, i: b := proc (n) if `mod`(n, 2) = 1 then 0 else 1+b((1/2)*n) end if end proc: B := proc (n) local nn, j, m: nn := op(2, ifactors(n)): for j to nops(nn) do m[j] := op(j, nn) end do: [seq(seq(pi(op(1, m[i])), q = 1 .. op(2, m[i])), i = 1 .. nops(nn))] end proc: c := 0: for i to bigomega(n) do if b(n) < B(n)[i] then c := c+1 else  end if end do: c end proc: seq(a(n), n = 1 .. 150);

%p # second Maple program:

%p a:= n->(l->nops(select(x->x>add(`if`(i=1, 1, 0), i=l), l)))(

%p        [seq(numtheory[pi](i[1])$i[2], i=ifactors(n)[2])]):

%p seq(a(n), n=1..100);  # _Alois P. Heinz_, May 10 2016

%t a[n_] := Function[l, Length[Select[l, # > Sum[If[i == 1, 1, 0], {i, l}]&]]][Flatten @ Table[Array[PrimePi[i[[1]]]&, i[[2]]], {i, FactorInteger[n]}] ]; Table[a[n], {n, 1, 100}] (* _Jean-François Alcover_, Dec 10 2016 after _Alois P. Heinz_ *)

%Y Cf. A257989, A215366.

%K nonn

%O 1,9

%A _Emeric Deutsch_, May 16 2015