26.8

DLMF	Formula	Constraints	Maple	Mathematica	Symbolic Maple	Symbolic Mathematica	Numeric Maple	Numeric Mathematica
26.8.E1	${\displaystyle{\displaystyle s\left(n,n\right)=1}}$ \Stirlingnumbers@{n}{n} = 1	${\displaystyle{\displaystyle n\geq 0}}$	Stirling1(n, n) = 1	StirlingS1[n, n] == 1	Successful	Failure	-	Successful [Tested: 3]
26.8.E2	${\displaystyle{\displaystyle s\left(1,k\right)=\delta_{1,k}}}$ \Stirlingnumbers@{1}{k} = \Kroneckerdelta{1}{k}		Stirling1(1, k) = KroneckerDelta[1, k]	StirlingS1[1, k] == KroneckerDelta[1, k]	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
26.8.E4	${\displaystyle{\displaystyle S\left(n,n\right)=1}}$ \StirlingnumberS@{n}{n} = 1	${\displaystyle{\displaystyle n\geq 0}}$	Stirling2(n, n) = 1	StirlingS2[n, n] == 1	Successful	Failure	-	Successful [Tested: 3]
26.8.E6	${\displaystyle{\displaystyle S\left(n,k\right)=\frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\genfrac{(}{)}{0.0pt}{}{k}{j}j^{n}}}$ \StirlingnumberS@{n}{k} = \frac{1}{k!}\sum_{j=0}^{k}(-1)^{k-j}\binom{k}{j}j^{n}		Stirling2(n, k) = (1)/(factorial(k))*sum((- 1)^(k - j)*binomial(k,j)*(j)^(n), j = 0..k)	StirlingS2[n, k] == Divide[1,(k)!]*Sum[(- 1)^(k - j)*Binomial[k,j]*(j)^(n), {j, 0, k}, GenerateConditions->None]	Aborted	Failure	Successful [Tested: 9]	Successful [Tested: 9]
26.8.E7	${\displaystyle{\displaystyle\sum_{k=0}^{n}s\left(n,k\right)x^{k}=(x-n+1)_{n}}}$ \sum_{k=0}^{n}\Stirlingnumbers@{n}{k}x^{k} = (x-n+1)_{n}		sum(Stirling1(n, k)*(x)^(k), k = 0..n) = x - n + 1[n]	Sum[StirlingS1[n, k]*(x)^(k), {k, 0, n}, GenerateConditions->None] == Subscript[x - n + 1, n]	Failure	Failure	Error	Failed [9 / 9] Result: Plus[1.5, Times[-1.0, Subscript[1.5, 1]]] Test Values: {Rule[n, 1], Rule[x, 1.5]} Result: Plus[0.75, Times[-1.0, Subscript[0.5, 2]]] Test Values: {Rule[n, 2], Rule[x, 1.5]} ... skip entries to safe data
26.8.E8	${\displaystyle{\displaystyle\sum_{n=0}^{\infty}s\left(n,k\right)\frac{x^{n}}{n!}=\frac{(\ln\left(1+x\right))^{k}}{k!}}}$ \sum_{n=0}^{\infty}\Stirlingnumbers@{n}{k}\frac{x^{n}}{n!} = \frac{(\ln@{1+x})^{k}}{k!}	${\displaystyle{\displaystyle|x|<1}}$	sum(Stirling1(n, k)*((x)^(n))/(factorial(n)), n = 0..infinity) = ((ln(1 + x))^(k))/(factorial(k))	Sum[StirlingS1[n, k]*Divide[(x)^(n),(n)!], {n, 0, Infinity}, GenerateConditions->None] == Divide[(Log[1 + x])^(k),(k)!]	Error	Failure	-	Failed [2 / 3] Result: Plus[-0.08220097694658271, NSum[Times[Power[0.5, n], Power[Factorial[n], -1], StirlingS1[n, 2]] Test Values: {n, 0, DirectedInfinity[1]}, Rule[GenerateConditions, None]]], {Rule[k, 2], Rule[x, 0.5]} Result: Plus[-0.011109876001414293, NSum[Times[Power[0.5, n], Power[Factorial[n], -1], StirlingS1[n, 3]] Test Values: {n, 0, DirectedInfinity[1]}, Rule[GenerateConditions, None]]], {Rule[k, 3], Rule[x, 0.5]}
26.8.E9	${\displaystyle{\displaystyle\sum_{n,k=0}^{\infty}s\left(n,k\right)\frac{x^{n}}{n!}y^{k}=(1+x)^{y}}}$ \sum_{n,k=0}^{\infty}\Stirlingnumbers@{n}{k}\frac{x^{n}}{n!}y^{k} = (1+x)^{y}	${\displaystyle{\displaystyle|x|<1}}$	sum(sum(Stirling1(n, k)*((x)^(n))/(factorial(n))*(y)^(k), k = 0..infinity), n = 0..infinity) = (1 + x)^(y)	Sum[Sum[StirlingS1[n, k]*Divide[(x)^(n),(n)!]*(y)^(k), {k, 0, Infinity}, GenerateConditions->None], {n, 0, Infinity}, GenerateConditions->None] == (1 + x)^(y)	Error	Failure	-	Failed [6 / 6] Result: Plus[-0.5443310539518174, NSum[Sum[Times[Power[-1.5, k], Power[0.5, n], Power[Factorial[n], -1], StirlingS1[n, k]] Test Values: {k, 0, DirectedInfinity[1]}, Rule[GenerateConditions, None]], {n, 0, DirectedInfinity[1]}, Rule[GenerateConditions, None]]], {Rule[x, 0.5], Rule[y, -1.5]} Result: Plus[-1.8371173070873836, NSum[Sum[Times[Power[0.5, n], Power[1.5, k], Power[Factorial[n], -1], StirlingS1[n, k]] Test Values: {k, 0, DirectedInfinity[1]}, Rule[GenerateConditions, None]], {n, 0, DirectedInfinity[1]}, Rule[GenerateConditions, None]]], {Rule[x, 0.5], Rule[y, 1.5]} ... skip entries to safe data
26.8.E10	${\displaystyle{\displaystyle\sum_{k=1}^{n}S\left(n,k\right)(x-k+1)_{k}=x^{n}}}$ \sum_{k=1}^{n}\StirlingnumberS@{n}{k}(x-k+1)_{k} = x^{n}		sum(Stirling2(n, k)*x - k + 1[k], k = 1..n) = (x)^(n)	Sum[StirlingS2[n, k]*Subscript[x - k + 1, k], {k, 1, n}, GenerateConditions->None] == (x)^(n)	Failure	Failure	Error	Failed [9 / 9] Result: Plus[-1.5, Subscript[1.5, 1]] Test Values: {Rule[n, 1], Rule[x, 1.5]} Result: Plus[-2.25, Subscript[0.5, 2], Subscript[1.5, 1]] Test Values: {Rule[n, 2], Rule[x, 1.5]} ... skip entries to safe data
26.8.E12	${\displaystyle{\displaystyle\sum_{n=0}^{\infty}S\left(n,k\right)\frac{x^{n}}{n!}=\frac{({\mathrm{e}^{x}}-1)^{k}}{k!}}}$ \sum_{n=0}^{\infty}\StirlingnumberS@{n}{k}\frac{x^{n}}{n!} = \frac{(\expe^{x}-1)^{k}}{k!}		sum(Stirling2(n, k)*((x)^(n))/(factorial(n)), n = 0..infinity) = ((exp(x)- 1)^(k))/(factorial(k))	Sum[StirlingS2[n, k]*Divide[(x)^(n),(n)!], {n, 0, Infinity}, GenerateConditions->None] == Divide[(Exp[x]- 1)^(k),(k)!]	Failure	Failure	Error	Successful [Tested: 9]
26.8.E13	${\displaystyle{\displaystyle\sum_{n,k=0}^{\infty}S\left(n,k\right)\frac{x^{n}}{n!}y^{k}=\exp\left(y({\mathrm{e}^{x}}-1)\right)}}$ \sum_{n,k=0}^{\infty}\StirlingnumberS@{n}{k}\frac{x^{n}}{n!}y^{k} = \exp\left(y(\expe^{x}-1)\right)		sum(sum(Stirling2(n, k)*((x)^(n))/(factorial(n))*(y)^(k), k = 0..infinity), n = 0..infinity) = exp(y*(exp(x)- 1))	Sum[Sum[StirlingS2[n, k]*Divide[(x)^(n),(n)!]*(y)^(k), {k, 0, Infinity}, GenerateConditions->None], {n, 0, Infinity}, GenerateConditions->None] == Exp[y*(Exp[x]- 1)]	Translation Error	Translation Error	-	-
26.8#Ex1	${\displaystyle{\displaystyle s\left(n,0\right)=0}}$ \Stirlingnumbers@{n}{0} = 0		Stirling1(n, 0) = 0	StirlingS1[n, 0] == 0	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
26.8#Ex2	${\displaystyle{\displaystyle s\left(n,1\right)=(-1)^{n-1}(n-1)!}}$ \Stirlingnumbers@{n}{1} = (-1)^{n-1}(n-1)!		Stirling1(n, 1) = (- 1)^(n - 1)*factorial(n - 1)	StirlingS1[n, 1] == (- 1)^(n - 1)*(n - 1)!	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
26.8.E16	${\displaystyle{\displaystyle-s\left(n,n-1\right)=S\left(n,n-1\right)}}$ -\Stirlingnumbers@{n}{n-1} = \StirlingnumberS@{n}{n-1}		- Stirling1(n, n - 1) = Stirling2(n, n - 1)	- StirlingS1[n, n - 1] == StirlingS2[n, n - 1]	Successful	Failure	-	Successful [Tested: 3]
26.8.E16	${\displaystyle{\displaystyle S\left(n,n-1\right)=\genfrac{(}{)}{0.0pt}{}{n}{2}}}$ \StirlingnumberS@{n}{n-1} = \binom{n}{2}		Stirling2(n, n - 1) = binomial(n,2)	StirlingS2[n, n - 1] == Binomial[n,2]	Successful	Failure	-	Successful [Tested: 3]
26.8#Ex3	${\displaystyle{\displaystyle S\left(n,0\right)=0}}$ \StirlingnumberS@{n}{0} = 0		Stirling2(n, 0) = 0	StirlingS2[n, 0] == 0	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
26.8#Ex4	${\displaystyle{\displaystyle S\left(n,1\right)=1}}$ \StirlingnumberS@{n}{1} = 1		Stirling2(n, 1) = 1	StirlingS2[n, 1] == 1	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
26.8#Ex5	${\displaystyle{\displaystyle S\left(n,2\right)=2^{n-1}-1}}$ \StirlingnumberS@{n}{2} = 2^{n-1}-1		Stirling2(n, 2) = (2)^(n - 1)- 1	StirlingS2[n, 2] == (2)^(n - 1)- 1	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
26.8.E18	${\displaystyle{\displaystyle s\left(n,k\right)=s\left(n-1,k-1\right)-(n-1)s\left(n-1,k\right)}}$ \Stirlingnumbers@{n}{k} = \Stirlingnumbers@{n-1}{k-1}-(n-1)\Stirlingnumbers@{n-1}{k}		Stirling1(n, k) = Stirling1(n - 1, k - 1)-(n - 1)*Stirling1(n - 1, k)	StirlingS1[n, k] == StirlingS1[n - 1, k - 1]-(n - 1)*StirlingS1[n - 1, k]	Failure	Failure	Successful [Tested: 9]	Successful [Tested: 9]
26.8.E19	${\displaystyle{\displaystyle\genfrac{(}{)}{0.0pt}{}{k}{h}s\left(n,k\right)=\sum_{j=k-h}^{n-h}\genfrac{(}{)}{0.0pt}{}{n}{j}s\left(n-j,h\right)s\left(j,k-h\right)}}$ \binom{k}{h}\Stirlingnumbers@{n}{k} = \sum_{j=k-h}^{n-h}\binom{n}{j}\Stirlingnumbers@{n-j}{h}\Stirlingnumbers@{j}{k-h}	${\displaystyle{\displaystyle n\geq k,k\geq h}}$	binomial(k,h)*Stirling1(n, k) = sum(binomial(n,j)*Stirling1(n - j, h)*Stirling1(j, k - h), j = k - h..n - h)	Binomial[k,h]*StirlingS1[n, k] == Sum[Binomial[n,j]*StirlingS1[n - j, h]*StirlingS1[j, k - h], {j, k - h, n - h}, GenerateConditions->None]	Error	Failure	-	Failed [11 / 30] Result: 0.16976527263135505 Test Values: {Rule[h, -1.5], Rule[k, 1], Rule[n, 2]} Result: -0.08488263631567752 Test Values: {Rule[h, -1.5], Rule[k, 1], Rule[n, 3]} ... skip entries to safe data
26.8.E20	${\displaystyle{\displaystyle s\left(n+1,k+1\right)=n!\sum_{j=k}^{n}\frac{(-1)^{n-j}}{j!}\,s\left(j,k\right)}}$ \Stirlingnumbers@{n+1}{k+1} = n!\sum_{j=k}^{n}\frac{(-1)^{n-j}}{j!}\,\Stirlingnumbers@{j}{k}		Stirling1(n + 1, k + 1) = factorial(n)*sum(((- 1)^(n - j))/(factorial(j))*Stirling1(j, k), j = k..n)	StirlingS1[n + 1, k + 1] == (n)!*Sum[Divide[(- 1)^(n - j),(j)!]*StirlingS1[j, k], {j, k, n}, GenerateConditions->None]	Failure	Failure	Successful [Tested: 9]	Successful [Tested: 9]
26.8.E21	${\displaystyle{\displaystyle s\left(n+k+1,k\right)=-\sum_{j=0}^{k}(n+j)s\left(n+j,j\right)}}$ \Stirlingnumbers@{n+k+1}{k} = -\sum_{j=0}^{k}(n+j)\Stirlingnumbers@{n+j}{j}		Stirling1(n + k + 1, k) = - sum((n + j)*Stirling1(n + j, j), j = 0..k)	StirlingS1[n + k + 1, k] == - Sum[(n + j)*StirlingS1[n + j, j], {j, 0, k}, GenerateConditions->None]	Failure	Failure	Successful [Tested: 9]	Successful [Tested: 9]
26.8.E22	${\displaystyle{\displaystyle S\left(n,k\right)=kS\left(n-1,k\right)+S\left(n-1,k-1\right)}}$ \StirlingnumberS@{n}{k} = k\StirlingnumberS@{n-1}{k}+\StirlingnumberS@{n-1}{k-1}		Stirling2(n, k) = k*Stirling2(n - 1, k)+ Stirling2(n - 1, k - 1)	StirlingS2[n, k] == k*StirlingS2[n - 1, k]+ StirlingS2[n - 1, k - 1]	Failure	Failure	Successful [Tested: 9]	Successful [Tested: 9]
26.8.E23	${\displaystyle{\displaystyle\genfrac{(}{)}{0.0pt}{}{k}{h}S\left(n,k\right)=\sum_{j=k-h}^{n-h}\genfrac{(}{)}{0.0pt}{}{n}{j}S\left(n-j,h\right)S\left(j,k-h\right)}}$ \binom{k}{h}\StirlingnumberS@{n}{k} = \sum_{j=k-h}^{n-h}\binom{n}{j}\StirlingnumberS@{n-j}{h}\StirlingnumberS@{j}{k-h}	${\displaystyle{\displaystyle n\geq k,k\geq h}}$	binomial(k,h)*Stirling2(n, k) = sum(binomial(n,j)*Stirling2(n - j, h)*Stirling2(j, k - h), j = k - h..n - h)	Binomial[k,h]*StirlingS2[n, k] == Sum[Binomial[n,j]*StirlingS2[n - j, h]*StirlingS2[j, k - h], {j, k - h, n - h}, GenerateConditions->None]	Error	Failure	-	Failed [22 / 30] Result: Plus[-0.08488263631567752, Times[0.08488263631567751, StirlingS2[-1.5, -1.5], StirlingS2[2.5, 2.5]]] Test Values: {Rule[h, -1.5], Rule[k, 1], Rule[n, 1]} Result: Plus[-0.08488263631567752, Times[-0.33953054526271004, StirlingS2[-0.5, -1.5], StirlingS2[2.5, 2.5]], Times[0.04850436360895858, StirlingS2[-1.5, -1.5], StirlingS2[3.5, 2.5]]] Test Values: {Rule[h, -1.5], Rule[k, 1], Rule[n, 2]} ... skip entries to safe data
26.8.E24	${\displaystyle{\displaystyle S\left(n,k\right)=\sum_{j=k}^{n}S\left(j-1,k-1\right)k^{n-j}}}$ \StirlingnumberS@{n}{k} = \sum_{j=k}^{n}\StirlingnumberS@{j-1}{k-1}k^{n-j}		Stirling2(n, k) = sum(Stirling2(j - 1, k - 1)*(k)^(n - j), j = k..n)	StirlingS2[n, k] == Sum[StirlingS2[j - 1, k - 1]*(k)^(n - j), {j, k, n}, GenerateConditions->None]	Failure	Failure	Successful [Tested: 9]	Successful [Tested: 9]
26.8.E25	${\displaystyle{\displaystyle S\left(n+1,k+1\right)=\sum_{j=k}^{n}\genfrac{(}{)}{0.0pt}{}{n}{j}S\left(j,k\right)}}$ \StirlingnumberS@{n+1}{k+1} = \sum_{j=k}^{n}\binom{n}{j}\StirlingnumberS@{j}{k}		Stirling2(n + 1, k + 1) = sum(binomial(n,j)*Stirling2(j, k), j = k..n)	StirlingS2[n + 1, k + 1] == Sum[Binomial[n,j]*StirlingS2[j, k], {j, k, n}, GenerateConditions->None]	Failure	Failure	Successful [Tested: 9]	Successful [Tested: 9]
26.8.E26	${\displaystyle{\displaystyle S\left(n+k+1,k\right)=\sum_{j=0}^{k}jS\left(n+j,j\right)}}$ \StirlingnumberS@{n+k+1}{k} = \sum_{j=0}^{k}j\StirlingnumberS@{n+j}{j}		Stirling2(n + k + 1, k) = sum(j*Stirling2(n + j, j), j = 0..k)	StirlingS2[n + k + 1, k] == Sum[j*StirlingS2[n + j, j], {j, 0, k}, GenerateConditions->None]	Failure	Successful	Successful [Tested: 9]	Successful [Tested: 9]
26.8.E27	${\displaystyle{\displaystyle s\left(n,n-k\right)=\sum_{j=0}^{k}(-1)^{j}\genfrac{(}{)}{0.0pt}{}{n-1+j}{k+j}\,\genfrac{(}{)}{0.0pt}{}{n+k}{k-j}\*S\left(k+j,j\right)}}$ \Stirlingnumbers@{n}{n-k} = \sum_{j=0}^{k}(-1)^{j}\binom{n-1+j}{k+j}\,\binom{n+k}{k-j}\*\StirlingnumberS@{k+j}{j}		Stirling1(n, n - k) = sum((- 1)^(j)*binomial(n - 1 + j,k + j)*binomial(n + k,k - j)* Stirling2(k + j, j), j = 0..k)	StirlingS1[n, n - k] == Sum[(- 1)^(j)*Binomial[n - 1 + j,k + j]*Binomial[n + k,k - j]* StirlingS2[k + j, j], {j, 0, k}, GenerateConditions->None]	Failure	Failure	Successful [Tested: 9]	Failed [3 / 9] Result: StirlingS1[1.0, -1.0] Test Values: {Rule[k, 2], Rule[n, 1]} Result: StirlingS1[1.0, -2.0] Test Values: {Rule[k, 3], Rule[n, 1]} ... skip entries to safe data
26.8.E28	${\displaystyle{\displaystyle\sum_{k=1}^{n}s\left(n,k\right)=0}}$ \sum_{k=1}^{n}\Stirlingnumbers@{n}{k} = 0	${\displaystyle{\displaystyle n>1}}$	sum(Stirling1(n, k), k = 1..n) = 0	Sum[StirlingS1[n, k], {k, 1, n}, GenerateConditions->None] == 0	Failure	Failure	Successful [Tested: 2]	Successful [Tested: 2]
26.8.E29	${\displaystyle{\displaystyle\sum_{k=1}^{n}(-1)^{n-k}s\left(n,k\right)=n!}}$ \sum_{k=1}^{n}(-1)^{n-k}\Stirlingnumbers@{n}{k} = n!		sum((- 1)^(n - k)* Stirling1(n, k), k = 1..n) = factorial(n)	Sum[(- 1)^(n - k)* StirlingS1[n, k], {k, 1, n}, GenerateConditions->None] == (n)!	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
26.8.E30	${\displaystyle{\displaystyle\sum_{j=k}^{n}s\left(n+1,j+1\right)\,n^{j-k}=s\left(n,k\right)}}$ \sum_{j=k}^{n}\Stirlingnumbers@{n+1}{j+1}\,n^{j-k} = \Stirlingnumbers@{n}{k}		sum(Stirling1(n + 1, j + 1)*(n)^(j - k), j = k..n) = Stirling1(n, k)	Sum[StirlingS1[n + 1, j + 1]*(n)^(j - k), {j, k, n}, GenerateConditions->None] == StirlingS1[n, k]	Failure	Successful	Successful [Tested: 9]	Successful [Tested: 9]
26.8.E33	${\displaystyle{\displaystyle S\left(n,n-k\right)=\sum_{j=0}^{k}(-1)^{j}\genfrac{(}{)}{0.0pt}{}{n-1+j}{k+j}\genfrac{(}{)}{0.0pt}{}{n+k}{k-j}\*s\left(k+j,j\right)}}$ \StirlingnumberS@{n}{n-k} = \sum_{j=0}^{k}(-1)^{j}\binom{n-1+j}{k+j}\binom{n+k}{k-j}\*\Stirlingnumbers@{k+j}{j}		Stirling2(n, n - k) = sum((- 1)^(j)*binomial(n - 1 + j,k + j)*binomial(n + k,k - j)* Stirling1(k + j, j), j = 0..k)	StirlingS2[n, n - k] == Sum[(- 1)^(j)*Binomial[n - 1 + j,k + j]*Binomial[n + k,k - j]* StirlingS1[k + j, j], {j, 0, k}, GenerateConditions->None]	Failure	Failure	Successful [Tested: 9]	Failed [3 / 9] Result: StirlingS2[1.0, -1.0] Test Values: {Rule[k, 2], Rule[n, 1]} Result: StirlingS2[1.0, -2.0] Test Values: {Rule[k, 3], Rule[n, 1]} ... skip entries to safe data
26.8.E34	${\displaystyle{\displaystyle\sum_{j=0}^{n}j^{k}x^{j}=\sum_{j=0}^{k}S\left(k,j\right)x^{j}\frac{{\mathrm{d}}^{j}}{{\mathrm{d}x}^{j}}\left(\frac{1-x^{n+1}}{1-x}\right)}}$ \sum_{j=0}^{n}j^{k}x^{j} = \sum_{j=0}^{k}\StirlingnumberS@{k}{j}x^{j}\deriv[j]{}{x}\left(\frac{1-x^{n+1}}{1-x}\right)		sum((j)^(k)* (x)^(j), j = 0..n) = sum(Stirling2(k, j)*(x)^(j)* diff((1 - (x)^(n + 1))/(1 - x), [x$(j)]), j = 0..k)	Sum[(j)^(k)* (x)^(j), {j, 0, n}, GenerateConditions->None] == Sum[StirlingS2[k, j]*(x)^(j)* D[Divide[1 - (x)^(n + 1),1 - x], {x, j}], {j, 0, k}, GenerateConditions->None]	Aborted	Failure	Skipped - Because timed out	Skipped - Because timed out
26.8.E35	${\displaystyle{\displaystyle\sum_{j=0}^{n}j^{k}=\sum_{j=0}^{k}j!S\left(k,j\right)\genfrac{(}{)}{0.0pt}{}{n+1}{j+1}}}$ \sum_{j=0}^{n}j^{k} = \sum_{j=0}^{k}j!\StirlingnumberS@{k}{j}\binom{n+1}{j+1}		sum((j)^(k), j = 0..n) = sum(factorial(j)*Stirling2(k, j)*binomial(n + 1,j + 1), j = 0..k)	Sum[(j)^(k), {j, 0, n}, GenerateConditions->None] == Sum[(j)!*StirlingS2[k, j]*Binomial[n + 1,j + 1], {j, 0, k}, GenerateConditions->None]	Failure	Failure	Successful [Tested: 9]	Successful [Tested: 9]
26.8.E36	${\displaystyle{\displaystyle\sum_{k=0}^{n}(-1)^{n-k}k!S\left(n,k\right)=1}}$ \sum_{k=0}^{n}(-1)^{n-k}k!\StirlingnumberS@{n}{k} = 1		sum((- 1)^(n - k)* factorial(k)*Stirling2(n, k), k = 0..n) = 1	Sum[(- 1)^(n - k)* (k)!*StirlingS2[n, k], {k, 0, n}, GenerateConditions->None] == 1	Failure	Failure	Successful [Tested: 3]	Successful [Tested: 3]
26.8.E38	${\displaystyle{\displaystyle A^{-1}=B}}$ A^{-1} = B		(A)^(- 1) = B	(A)^(- 1) == B	Skipped - no semantic math	Skipped - no semantic math	-	-
26.8.E39	${\displaystyle{\displaystyle\sum_{j=k}^{n}s\left(j,k\right)S\left(n,j\right)=\sum_{j=k}^{n}s\left(n,j\right)S\left(j,k\right)}}$ \sum_{j=k}^{n}\Stirlingnumbers@{j}{k}\StirlingnumberS@{n}{j} = \sum_{j=k}^{n}\Stirlingnumbers@{n}{j}\StirlingnumberS@{j}{k}		sum(Stirling1(j, k)*Stirling2(n, j), j = k..n) = sum(Stirling1(n, j)*Stirling2(j, k), j = k..n)	Sum[StirlingS1[j, k]*StirlingS2[n, j], {j, k, n}, GenerateConditions->None] == Sum[StirlingS1[n, j]*StirlingS2[j, k], {j, k, n}, GenerateConditions->None]	Failure	Failure	Successful [Tested: 9]	Successful [Tested: 9]
26.8.E39	${\displaystyle{\displaystyle\sum_{j=k}^{n}s\left(n,j\right)S\left(j,k\right)=\delta_{n,k}}}$ \sum_{j=k}^{n}\Stirlingnumbers@{n}{j}\StirlingnumberS@{j}{k} = \Kroneckerdelta{n}{k}		sum(Stirling1(n, j)*Stirling2(j, k), j = k..n) = KroneckerDelta[n, k]	Sum[StirlingS1[n, j]*StirlingS2[j, k], {j, k, n}, GenerateConditions->None] == KroneckerDelta[n, k]	Failure	Failure	Successful [Tested: 9]	Successful [Tested: 9]