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first complex reverse compound index function
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@ -159,6 +159,75 @@ subroutine two_e_integrals_index_reverse(i,j,k,l,i1)
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end
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subroutine two_e_integrals_index_reverse_periodic_1(i,j,k,l,i1)
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use map_module
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implicit none
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BEGIN_DOC
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! Computes the 4 indices $i,j,k,l$ from a unique index $i_1$.
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! For 2 indices $i,j$ and $i \le j$, we have
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! $p = i(i-1)/2 + j$.
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! The key point is that because $j < i$,
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! $i(i-1)/2 < p \le i(i+1)/2$. So $i$ can be found by solving
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! $i^2 - i - 2p=0$. One obtains $i=1 + \sqrt{1+8p}/2$
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! and $j = p - i(i-1)/2$.
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! This rule is applied 3 times. First for the symmetry of the
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! pairs (i,k) and (j,l), and then for the symmetry within each pair.
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! always returns first set such that i<=k, j<=l, ik<=jl
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END_DOC
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integer, intent(out) :: i(4),j(4),k(4),l(4)
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integer(key_kind), intent(in) :: i1
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integer(key_kind) :: i2,i3
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i = 0
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i2 = ceiling(0.5d0*(dsqrt(dble(shiftl(i1,3)+1))-1.d0))
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l(1) = ceiling(0.5d0*(dsqrt(dble(shiftl(i2,3)+1))-1.d0))
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i3 = i1 - shiftr(i2*i2-i2,1)
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k(1) = ceiling(0.5d0*(dsqrt(dble(shiftl(i3,3)+1))-1.d0))
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j(1) = int(i2 - shiftr(l(1)*l(1)-l(1),1),4)
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i(1) = int(i3 - shiftr(k(1)*k(1)-k(1),1),4)
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!ijkl a+ib
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i(2) = j(1) !jilk a+ib
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j(2) = i(1)
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k(2) = l(1)
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l(2) = k(1)
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i(3) = i(1) !ilkj a-ib
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j(3) = l(1)
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k(3) = k(1)
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l(3) = j(1)
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i(4) = l(1) !lijk a-ib
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j(4) = i(1)
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k(4) = j(1)
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l(4) = k(1)
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integer :: ii, jj
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do ii=2,4
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do jj=1,ii-1
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if ( (i(ii) == i(jj)).and. &
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(j(ii) == j(jj)).and. &
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(k(ii) == k(jj)).and. &
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(l(ii) == l(jj)) ) then
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i(ii) = 0
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exit
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endif
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enddo
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enddo
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! This has been tested with up to 1000 AOs, and all the reverse indices are
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! correct ! We can remove the test
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! do ii=1,8
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! if (i(ii) /= 0) then
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! call two_e_integrals_index(i(ii),j(ii),k(ii),l(ii),i2)
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! if (i1 /= i2) then
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! print *, i1, i2
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! print *, i(ii), j(ii), k(ii), l(ii)
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! stop 'two_e_integrals_index_reverse failed'
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! endif
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! endif
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! enddo
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end
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