diff --git a/dipdip-dirty/bessels.c b/dipdip-dirty/bessels.c
index d1d08de..3d8b778 100644
--- a/dipdip-dirty/bessels.c
+++ b/dipdip-dirty/bessels.c
@@ -4,7 +4,7 @@
 #include <stdio.h>
 #include <string.h>
 
-static const double ln2 = 0.69314718055994531;
+static const double ln2 = 0.693147180559945309417;
 
 
 // general; gives an array of size xxx with TODODESC
@@ -41,7 +41,7 @@ complex double * hankelcoefftable_init(size_t maxn) {
 void hankelparts_fill(complex double *lrt, complex double *srt, size_t maxn,
 								size_t lrk_cutoff, complex double *hct,
 								unsigned kappa, double c, double x) {
-	memset(lrt, 0, (maxn+1)*sizeof(complex double));
+	if (lrt) memset(lrt, 0, (maxn+1)*sizeof(complex double));
 	memset(srt, 0, (maxn+1)*sizeof(complex double));
 	double regularisator = pow(1. - exp(-c * x), (double) kappa);
 	double antiregularisator = 1. - regularisator;
@@ -51,13 +51,13 @@ void hankelparts_fill(complex double *lrt, complex double *srt, size_t maxn,
 	  for(size_t n = k; n <= maxn; ++n) 
 	  	srt[n] += ((k<lrk_cutoff) ? antiregularisator : 1) 
 		  				* xfrac * hankelcoeffs_get(hct,n)[k];
-	  if (k < lrk_cutoff) for (size_t n = k; n <= maxn; ++n)
+	  if (lrt && k < lrk_cutoff) for (size_t n = k; n <= maxn; ++n)
 	  	lrt[n] += regularisator * xfrac * hankelcoeffs_get(hct,n)[k];
 	}
 
 	complex double expix = cexp(I * x);
-	for(size_t n = 0; n <= maxn; ++n){
+	for(size_t n = 0; n <= maxn; ++n)
+					srt[n] *= expix;
+	if (lrt) for(size_t n = 0; n <= maxn; ++n)
 					srt[n] *= expix;
-					lrt[n] *= expix;
-	}
 }
diff --git a/dipdip-dirty/bessels.h b/dipdip-dirty/bessels.h
index 054dfaa..9fba6aa 100644
--- a/dipdip-dirty/bessels.h
+++ b/dipdip-dirty/bessels.h
@@ -16,9 +16,18 @@ hankelcoeffs_get(complex double *hankelcoefftable, size_t n){
 
 
 // general; target_longrange and target_shortrange are of size (maxn+1)
+// if target_longrange is NULL, only the short-range part is calculated
 void hankelparts_fill(complex double *target_longrange, complex double *target_shortrange,
-		size_t maxn, size_t longrange_k_cutoff, // x**(-(k+1)-1) terms go completely to short-range part
+		size_t maxn, size_t longrange_order_cutoff, // x**(-(order+1)-1) terms go completely to short-range part
 		complex double *hankelcoefftable,
-		unsigned kappa, double c, double x);
+		unsigned kappa, double c, double x); // x = k0 * r
+
+
+// this declaration is general; however, the implementation 
+// is so far only for kappa == ???, maxn == ??? TODO
+void lrhankel_recpart_fill(complex  double *target_longrange_kspace,
+	       size_t maxn, size_t longrange_k_cutoff, 
+	       complex double  *hankelcoefftable,
+	       unsigned kappa, double c, double k0, double k);
 
 #endif //BESSELS_H
diff --git a/dipdip-dirty/lrhankel_recspace_dirty.c b/dipdip-dirty/lrhankel_recspace_dirty.c
index 5648e4b..829151f 100644
--- a/dipdip-dirty/lrhankel_recspace_dirty.c
+++ b/dipdip-dirty/lrhankel_recspace_dirty.c
@@ -38,6 +38,18 @@ complex double fk5q1n2l(double c, double k0, double k,
 		-    (e[5] - t*a[5] + t*d[5]*a[5])
 	       )/k0;
 }
+complex double fk5q1n3l(double c, double k0, double k,
+		const complex double *a, const complex double *b, const complex double *d, const complex double *e, const complex double *ash) {
+	double kk3 = 3*k*k;
+	return (
+			-    d[0]*(kk3+4*a[0]*a[0])
+			+  5*d[1]*(kk3+4*a[1]*a[1])
+			- 10*d[2]*(kk3+4*a[2]*a[2])
+			+ 10*d[3]*(kk3+4*a[3]*a[3])
+			-  5*d[4]*(kk3+4*a[4]*a[4])
+			+    d[5]*(kk3+4*a[5]*a[5])
+	       )/(k0*k*k*k);
+}
 complex double fk5q2n0(double c, double k0, double k,
 		const complex double *a, const complex double *b, const complex double *d, const complex double *e, const complex double *ash) {
 	return (
@@ -184,7 +196,7 @@ int main() {
 // complex double fun(double c, double k0, double k, ccd *a, ccd *b, ccd *d, ccd *e)
 				complex double result = 
 					//transfuns_f[kappa][qm][n](c,k0,k,a,b,d,e,ash);
-				 	fk5q2n0s(c,k0,k,a,b,d,e,ash);
+				 	fk5q1n3l(c,k0,k,a,b,d,e,ash);
 				printf("%.16e %.16e ", creal(result), cimag(result));
 			}
 		printf("\n");
diff --git a/notes/BesselTFq3n0test.nb b/notes/BesselTFq3n0test.nb
new file mode 100644
index 0000000..a61fe67
--- /dev/null
+++ b/notes/BesselTFq3n0test.nb
@@ -0,0 +1,380 @@
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diff --git a/notes/ewald-calculations-apr1.lyx b/notes/ewald-calculations-apr1.lyx
index ca02b36..ce4a8e7 100644
--- a/notes/ewald-calculations-apr1.lyx
+++ b/notes/ewald-calculations-apr1.lyx
@@ -291,6 +291,87 @@ extra
 .
 \end_layout
 
+\begin_layout Standard
+
+\lang english
+According to Mathematica, the right sum with 
+\begin_inset Formula $s$
+\end_inset
+
+ going from 0 
+\begin_inset Formula 
+\begin{equation}
+\frac{2^{1-q}}{k_{0}^{q}}\sqrt{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\left(\sigma c-ik_{0}\right)^{2s}\left(-\frac{\text{Γ}\left(\frac{3-q+n}{2}\right)\left(\frac{3-q+n}{2}\right)_{s}\left(\frac{3-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n-1}{2}\right)\text{Γ}\left(\frac{3}{2}+s\right)s!}\frac{\left(\sigma c-ik_{0}\right)}{k}\right)\label{eq:right sum}
+\end{equation}
+
+\end_inset
+
+can be written as (mathematica output) 
+\begin_inset Note Note
+status collapsed
+
+\begin_layout Plain Layout
+
+\lang english
+(2^(2 - q)*k^(-3 + q)*((-I)*k0 + c*sig)*Gamma[(3 + n - q)/2]*Hypergeometric2F1[3
+/2 - n/2 - q/2, 3/2 + n/2 - q/2, 3/2, (k0 + I*c*sig)^2/k^2])/(k0^q*Gamma[(-1
+ + n + q)/2]) 
+\end_layout
+
+\end_inset
+
+
+\begin_inset Formula 
+\[
+\frac{2^{2-q}k^{-3+q}\left(-ik_{0}+c\sigma\right)\text{Γ}\left(\frac{3+n-q}{2}\right)\hgf\left(\begin{array}{c}
+\frac{3-n-q}{2},\frac{3+n-q}{2}\\
+3/2
+\end{array};-\frac{\left(\sigma c-ik_{0}\right)^{2}}{k^{2}}\right)}{k_{0}^{q;}\Gamma\left(\frac{-1+n+q}{2}\right)}
+\]
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+
+\lang english
+Similarly, the left sum
+\begin_inset Formula 
+\begin{equation}
+\frac{2^{1-q}}{k_{0}^{q}}\sqrt{\pi}\sum_{s=0}^{\infty}\left(-1\right)^{s}k^{-2+q-2s}\left(\sigma c-ik_{0}\right)^{2s}\left(\frac{\text{Γ}\left(\frac{2-q+n}{2}\right)\left(\frac{2-q+n}{2}\right)_{s}\left(\frac{2-q-n}{2}\right)_{s}}{\text{Γ}\left(\frac{q+n}{2}\right)\text{Γ}\left(\frac{1}{2}+s\right)s!}\right)\label{eq:left sum}
+\end{equation}
+
+\end_inset
+
+gives (mathematica output) 
+\begin_inset Note Note
+status collapsed
+
+\begin_layout Plain Layout
+
+\lang english
+(2^(1 - q)*k^(-2 + q)*Gamma[(2 + n - q)/2]*Hypergeometric2F1[1 - n/2 - q/2,
+ 1 + n/2 - q/2, 1/2, (k0 + I*c*sig)^2/k^2])/(k0^q*Gamma[(n + q)/2]) 
+\end_layout
+
+\end_inset
+
+ and is equal to
+\begin_inset Formula 
+\[
+\frac{2^{1-q}k^{-2+q}\Gamma\left(\frac{2+n-q}{2}\right)\hgf\left(\begin{array}{c}
+\frac{2-n-q}{2},\frac{2+n-q}{2}\\
+1/2
+\end{array};-\frac{\left(\sigma c-ik_{0}\right)^{2}}{k^{2}}\right)}{k_{0}^{q;}\Gamma\left(\frac{n+q}{2}\right)}
+\]
+
+\end_inset
+
+.
+\end_layout
+
 \begin_layout Subparagraph
 
 \lang english
@@ -393,12 +474,83 @@ so
 \begin_inset Formula 
 \begin{eqnarray*}
 \pht 1{s_{3,k_{0}}^{\textup{L}\kappa>3,c}}\left(k\right) & = & -\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{\kor{2^{-2}}k}{k_{0}^{3}}\kor{\sqrt{\pi}\left(\frac{\sigma c-ik_{0}}{k}\right)}\kor 2\frac{\left(\frac{\sigma c-ik_{0}}{k}\right)\sqrt{1+\left(\frac{\sigma c-ik_{0}}{k}\right)^{2}}+\sinh^{-1}\left(\frac{\sigma c-ik_{0}}{k}\right)}{\kor{\sqrt{\pi}\left(\frac{\sigma c-ik_{0}}{k}\right)}}\\
- & = & -\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k}{2k_{0}^{3}}\left(\left(\frac{\sigma c-ik_{0}}{k}\right)\sqrt{1+\left(\frac{\sigma c-ik_{0}}{k}\right)^{2}}+\sinh^{-1}\left(\frac{\sigma c-ik_{0}}{k}\right)\right)
+(Hq3n1) & = & -\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k}{2k_{0}^{3}}\left(\left(\frac{\sigma c-ik_{0}}{k}\right)\sqrt{1+\left(\frac{\sigma c-ik_{0}}{k}\right)^{2}}+\sinh^{-1}\left(\frac{\sigma c-ik_{0}}{k}\right)\right)
 \end{eqnarray*}
 
 \end_inset
 
 
+\series bold
+což je prej blbě (zjisti proč – blbě opsáno nebo nesprávná větev logaritmu?);
+\series default
+ správný výsledek je (mathematica kód: 
+\begin_inset Note Note
+status collapsed
+
+\begin_layout Plain Layout
+- Sum[(-1)^sig Binomial[kap, sig] (((-I)*k0 + c*sig)*(k0*Sqrt[1 - (k0 +
+ I*c*sig)^2/k^2] + I*c*sig*Sqrt[1 - (k0 + I*c*sig)^2/k^2] + k*ArcSin[(k0
+ + I*c*sig)/k]))/(2*k0^3*(k0 + I*c*sig)) , {sig, 0, kap}]
+\end_layout
+
+\end_inset
+
+ nebo FullSimplify 
+\begin_inset Note Note
+status collapsed
+
+\begin_layout Plain Layout
+(((-I)*k0 + c*sig)*Sqrt[(k^2 - (k0 + I*c*sig)^2)/k^2] - I*k*ArcSin[(k0 +
+ I*c*sig)/k])/(2*k0^3) 
+\end_layout
+
+\end_inset
+
+; snad jsem to tentokrát neopsal blbě)
+\begin_inset Formula 
+\begin{eqnarray*}
+\pht 1{s_{3,k_{0}}^{\textup{L}\kappa>3,c}}\left(k\right) & = & -\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{\left(-ik_{0}+c\sigma\right)\left(k_{0}\sqrt{1+\left(\frac{\sigma c-ik_{0}}{k}\right)^{2}}+ic\sigma\sqrt{1+\left(\frac{\sigma c-ik_{0}}{k}\right)^{2}}+k\sin^{-1}\left(\frac{k_{0}+ic\sigma}{k}\right)\right)}{2k_{0}^{3}\left(k_{0}+ic\sigma\right)}\\
+\mbox{(f.simpl.)} & = & -\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{\left(-ik_{0}+c\sigma\right)\sqrt{1-\left(\frac{k_{0}+ic\sigma}{k}\right)^{2}}-ik\sin^{-1}\left(\frac{k_{0}+ic\sigma}{k}\right)}{2k_{0}^{3}}
+\end{eqnarray*}
+
+\end_inset
+
+
+\end_layout
+
+\begin_layout Subparagraph
+Special case 
+\begin_inset Formula $q=3,n=0$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+Mathematica řiká po fullsimplify zhruba toto 
+\begin_inset Note Note
+status open
+
+\begin_layout Plain Layout
+Sum[((-1)^(1 + sig)*(k*Sqrt[(k^2 - (k0 + I*c*sig)^2)/k^2] + (k0 + I*c*sig)*ArcSi
+n[(k0 + I*c*sig)/k])*Binomial[kap, sig])/k0^3, {sig, 0, kap}]
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula 
+\begin{eqnarray*}
+\pht 0{s_{3,k_{0}}^{\textup{L}\kappa>3,c}}\left(k\right) & = & -\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k\sqrt{1-\left(\frac{k_{0}+ic\sigma}{k}\right)^{2}}+\left(k_{0}+ic\sigma\right)\sin^{-1}\left(\frac{k_{0}+ic\sigma}{k}\right)}{k_{0}^{3}}
+\end{eqnarray*}
+
+\end_inset
+
+
+\begin_inset Formula $\kappa\ge2$
+\end_inset
+
+
 \end_layout
 
 \begin_layout Paragraph
diff --git a/notes/ewald.lyx b/notes/ewald.lyx
index 7a32112..1a97fc1 100644
--- a/notes/ewald.lyx
+++ b/notes/ewald.lyx
@@ -2700,6 +2700,79 @@ for
 \end_inset
 
  
+\end_layout
+
+\begin_layout Paragraph
+Case 
+\begin_inset Formula $n=1,q=3$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+As shown in separate note (check whether copied correctly)
+\begin_inset Formula 
+\[
+\pht 1{s_{3,k_{0}}^{\textup{L}\kappa>3,c}}\left(k\right)=-\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{\left(-ik_{0}+c\sigma\right)\sqrt{1-\left(\frac{k_{0}+ic\sigma}{k}\right)^{2}}-ik\sin^{-1}\left(\frac{k_{0}+ic\sigma}{k}\right)}{2k_{0}^{3}}
+\]
+
+\end_inset
+
+for 
+\begin_inset Formula $\kappa\ge3$
+\end_inset
+
+, 
+\begin_inset Formula $k>k_{0}?$
+\end_inset
+
+ 
+\end_layout
+
+\begin_layout Paragraph
+Case 
+\begin_inset Formula $n=0,q=3$
+\end_inset
+
+
+\end_layout
+
+\begin_layout Standard
+As shown in separate note (check whether copied correctly)
+\lang finnish
+
+\begin_inset Note Note
+status collapsed
+
+\begin_layout Plain Layout
+
+\lang finnish
+Sum[((-1)^(1 + sig)*(k*Sqrt[(k^2 - (k0 + I*c*sig)^2)/k^2] + (k0 + I*c*sig)*ArcSi
+n[(k0 + I*c*sig)/k])*Binomial[kap, sig])/k0^3, {sig, 0, kap}]
+\end_layout
+
+\end_inset
+
+ 
+\begin_inset Formula 
+\begin{eqnarray*}
+\pht 0{s_{3,k_{0}}^{\textup{L}\kappa>3,c}}\left(k\right) & = & -\sum_{\sigma=0}^{\kappa}\left(-1\right)^{\sigma}\binom{\kappa}{\sigma}\frac{k\sqrt{1-\left(\frac{k_{0}+ic\sigma}{k}\right)^{2}}+\left(k_{0}+ic\sigma\right)\sin^{-1}\left(\frac{k_{0}+ic\sigma}{k}\right)}{k_{0}^{3}}
+\end{eqnarray*}
+
+\end_inset
+
+
+\lang english
+for 
+\begin_inset Formula $\kappa\ge2$
+\end_inset
+
+, 
+\begin_inset Formula $k>k_{0}?$
+\end_inset
+
+ 
 \end_layout
 
 \begin_layout Standard
@@ -2964,7 +3037,7 @@ where the spherical Hankel transform
  2)
 \begin_inset Formula 
 \[
-\bsht lg(k)\equiv\int_{0}^{\infty}\ud r\,r^{2}g(r)j_{l}\left(kr\right).
+\bsht lg(k)\equiv\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right).
 \]
 
 \end_inset
@@ -2974,7 +3047,7 @@ Using this convention, the inverse spherical Hankel transform is given by
  3)
 \begin_inset Formula 
 \[
-g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\,k^{2}\bsht lg(k)j_{l}(k),
+g(r)=\frac{2}{\pi}\int_{0}^{\infty}\ud k\, k^{2}\bsht lg(k)j_{l}(k),
 \]
 
 \end_inset
@@ -2987,7 +3060,7 @@ so it is not unitary.
 An unitary convention would look like this:
 \begin_inset Formula 
 \begin{equation}
-\usht lg(k)\equiv\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\ud r\,r^{2}g(r)j_{l}\left(kr\right).\label{eq:unitary 3d Hankel tf definition}
+\usht lg(k)\equiv\sqrt{\frac{2}{\pi}}\int_{0}^{\infty}\ud r\, r^{2}g(r)j_{l}\left(kr\right).\label{eq:unitary 3d Hankel tf definition}
 \end{equation}
 
 \end_inset
@@ -3040,9 +3113,10 @@ where the Hankel transform of order
 
  is defined as
 \begin_inset Formula 
-\begin{equation}
-\pht mg\left(k\right)=\int_{0}^{\infty}\ud r\,g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition}
-\end{equation}
+\begin{eqnarray}
+\pht mg\left(k\right) & = & \int_{0}^{\infty}\ud r\, g(r)J_{m}(kr)r\label{eq:unitary 2d Hankel tf definition}\\
+ & = & \left(-1\right)^{m}\int_{0}^{\infty}\ud r\, g(r)J_{\left|m\right|}(kr)r
+\end{eqnarray}
 
 \end_inset