There is an easy inversion formula for $$a(k)=\sum_{j=0}^k(-1)^j\Big({k\atop $a(k)=\sum_{j=0}^k\Big({k\atop j}\Big)b(j)\ ,$$ namely, $$b(k)=\sum_{j=0}^k (-1)^j\Big({k\atop -1)^{k-j}\Big({k\atop j}\Big)a(j)\ ,$$ and in your case ($a(k):=(-1)^kk^n$) $a(k):=k^n$) this produce the formula shown by Barry. You can see and prove it in several ways: The inverse matrix of the triangular matrix whose entries are the binomial coefficientswith suitable signs is involutory. Or an identity between suitable generating functions. Also, as an instance of the inclusion-exclusion formula. I think I'm quite sure you may find nice hints in Concrete Mathematics by Graham, Knuth, and Patashnik.
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There is an easy inversion formula for $$a(k)=\sum_{j=0}^k\Big({k\atop $a(k)=\sum_{j=0}^k(-1)^j\Big({k\atop j}\Big)b(j)\ ,$$ namely, $$b(k)=\sum_{j=0}^k (-1)^{k-j}\Big({k\atop -1)^j\Big({k\atop j}\Big)a(j)\ ,$$ and in your case ($a(k):=k^n$) $a(k):=(-1)^kk^n$) this produce the formula shown by Barry. You can see and prove it in several ways: The inverse matrix of the triangular matrix whose entries are the binomial coefficients with suitable signs is involutory. Or an identity between suitable generating functions. Also, as an instance of the inclusion-exclusion formula. I'm quite sure I think you may find nice hints in Concrete Mathematics by Graham, Knuth, and Patashnik. |
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There is an easy inversion formula for $$a(k)=\sum_{j=0}^k\Big({k\atop j}\Big)b(j)\ ,$$ namely, $$b(k)=\sum_{j=0}^k (-1)^{k-j}\Big({k\atop j}\Big)a(j)\ ,$$ and in your case ($a(k):=k^n$) this produce the formula shown by Barry. You can see and prove it in several ways: The inverse matrix of the triangular matrix whose entries are the binomial coefficients. Or an identity between suitable generating functions. Also, as an instance of the inclusion-exclusion formula. I'm quite sure you may find nice hints in Concrete Mathematics by Graham, Knuth, and Patashnik. |
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