A Generalised Bijection from Tuples of Natural Numbers to Natural Numbers












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$begingroup$


I've used the tag "set-theory" in lieu of "bijections".



When we represent (as is done by the OEIS) an infinite two dimensional sequence by the concatenation of antidiagonals, the index of an entry in that sequence is given by a bijection between ℕ×ℕ & ℕ. (ℕ may be ℕ₀ or ℕ₁ - it doesn't affect the essential argument - it's just a matter of offsets ... but I'll stick to casting all this in terms of ℕ₀, as then the formulæ are the simplest.) Explicitly the bijection is $$q=$$$$frac{(m+n)(m+n+1)}{2}+n$$$$=$$$$frac{(m+n)^{overline{2}}}{2!}+n$$$$=$$$$frac{(m+n)^2+m+3n}{2} ,$$ with the notation in the middle formula being the ascending pochhammer notation ... the reason I have bothered to broach this notation here shall be made clear shortly. I once saw the third formula given in it's own right as the simplest possible bijection from duples of natural numbers to natural numbers; and it is sufficient in its own right to serve as a bijection between tuples of naturals of any given order & the naturals - by applying it recursively. (Please mark, though, this does not mean between all tuples & singletons - it means between all tuples of some one given order & singletons!)



But if we are to actually do such a mapping, there might be a 'tidier' (& more disciplined ) way of doing it ... forall its being so very 'handy' a formula, that third one. I'm fairly sure this works ... and the question is, can anyone tell me from their experience or knowledge that it does work? ... I'm not of course asking anyone to do the derivation for me. As far as I can tell it does ... but I do sometimes miss things, and I have not been able to find an exposition that says either yea or nay to it ... and it's this: if you have a tuple ${n_1, ... , n_r}$ would the most economical mapping - which seems to me to be the extension of the logic of the 2-dimensional case to arbitrary number $r$ of dimensions - be $$q=sum_{k=1}^rfrac{bigg(sum_{j=1}^k n_jbigg)^overline{k}}{k!}$$?



The logic is - and I'll merely expound it in a fairly qualitative fashion here so as not very greatly to prolong this post - that $$frac{bigg(sum_{j=1}^k n_jbigg)^overline{k}}{k!} ,$$ being a sum from to of a sum from to ... ($k$ instances thereof (if the termination be sum from to 1) ... triangular numbers, tetrahedral numbers, etc), creates the minimum offset necessary, at each item of the tuple, to ensure that the mapping does not become a surjection.



Actually, using that 'handy' formula recursively would not give rise to any diseconomy if the number of items in the tuple were a power of two, and the recursion applied to the tuple treewise. If you simply apply it to the general tuple at each element consecutively, you would be handling numbers of order of magnitude $$(sum n)^{2^r} ,$$ such unnecessarily high orders of size corresponding to the implied intermediate tuples having a highly uneven distribution of value amongst their items.










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$endgroup$

















    1












    $begingroup$


    I've used the tag "set-theory" in lieu of "bijections".



    When we represent (as is done by the OEIS) an infinite two dimensional sequence by the concatenation of antidiagonals, the index of an entry in that sequence is given by a bijection between ℕ×ℕ & ℕ. (ℕ may be ℕ₀ or ℕ₁ - it doesn't affect the essential argument - it's just a matter of offsets ... but I'll stick to casting all this in terms of ℕ₀, as then the formulæ are the simplest.) Explicitly the bijection is $$q=$$$$frac{(m+n)(m+n+1)}{2}+n$$$$=$$$$frac{(m+n)^{overline{2}}}{2!}+n$$$$=$$$$frac{(m+n)^2+m+3n}{2} ,$$ with the notation in the middle formula being the ascending pochhammer notation ... the reason I have bothered to broach this notation here shall be made clear shortly. I once saw the third formula given in it's own right as the simplest possible bijection from duples of natural numbers to natural numbers; and it is sufficient in its own right to serve as a bijection between tuples of naturals of any given order & the naturals - by applying it recursively. (Please mark, though, this does not mean between all tuples & singletons - it means between all tuples of some one given order & singletons!)



    But if we are to actually do such a mapping, there might be a 'tidier' (& more disciplined ) way of doing it ... forall its being so very 'handy' a formula, that third one. I'm fairly sure this works ... and the question is, can anyone tell me from their experience or knowledge that it does work? ... I'm not of course asking anyone to do the derivation for me. As far as I can tell it does ... but I do sometimes miss things, and I have not been able to find an exposition that says either yea or nay to it ... and it's this: if you have a tuple ${n_1, ... , n_r}$ would the most economical mapping - which seems to me to be the extension of the logic of the 2-dimensional case to arbitrary number $r$ of dimensions - be $$q=sum_{k=1}^rfrac{bigg(sum_{j=1}^k n_jbigg)^overline{k}}{k!}$$?



    The logic is - and I'll merely expound it in a fairly qualitative fashion here so as not very greatly to prolong this post - that $$frac{bigg(sum_{j=1}^k n_jbigg)^overline{k}}{k!} ,$$ being a sum from to of a sum from to ... ($k$ instances thereof (if the termination be sum from to 1) ... triangular numbers, tetrahedral numbers, etc), creates the minimum offset necessary, at each item of the tuple, to ensure that the mapping does not become a surjection.



    Actually, using that 'handy' formula recursively would not give rise to any diseconomy if the number of items in the tuple were a power of two, and the recursion applied to the tuple treewise. If you simply apply it to the general tuple at each element consecutively, you would be handling numbers of order of magnitude $$(sum n)^{2^r} ,$$ such unnecessarily high orders of size corresponding to the implied intermediate tuples having a highly uneven distribution of value amongst their items.










    share|cite|improve this question











    $endgroup$















      1












      1








      1


      0



      $begingroup$


      I've used the tag "set-theory" in lieu of "bijections".



      When we represent (as is done by the OEIS) an infinite two dimensional sequence by the concatenation of antidiagonals, the index of an entry in that sequence is given by a bijection between ℕ×ℕ & ℕ. (ℕ may be ℕ₀ or ℕ₁ - it doesn't affect the essential argument - it's just a matter of offsets ... but I'll stick to casting all this in terms of ℕ₀, as then the formulæ are the simplest.) Explicitly the bijection is $$q=$$$$frac{(m+n)(m+n+1)}{2}+n$$$$=$$$$frac{(m+n)^{overline{2}}}{2!}+n$$$$=$$$$frac{(m+n)^2+m+3n}{2} ,$$ with the notation in the middle formula being the ascending pochhammer notation ... the reason I have bothered to broach this notation here shall be made clear shortly. I once saw the third formula given in it's own right as the simplest possible bijection from duples of natural numbers to natural numbers; and it is sufficient in its own right to serve as a bijection between tuples of naturals of any given order & the naturals - by applying it recursively. (Please mark, though, this does not mean between all tuples & singletons - it means between all tuples of some one given order & singletons!)



      But if we are to actually do such a mapping, there might be a 'tidier' (& more disciplined ) way of doing it ... forall its being so very 'handy' a formula, that third one. I'm fairly sure this works ... and the question is, can anyone tell me from their experience or knowledge that it does work? ... I'm not of course asking anyone to do the derivation for me. As far as I can tell it does ... but I do sometimes miss things, and I have not been able to find an exposition that says either yea or nay to it ... and it's this: if you have a tuple ${n_1, ... , n_r}$ would the most economical mapping - which seems to me to be the extension of the logic of the 2-dimensional case to arbitrary number $r$ of dimensions - be $$q=sum_{k=1}^rfrac{bigg(sum_{j=1}^k n_jbigg)^overline{k}}{k!}$$?



      The logic is - and I'll merely expound it in a fairly qualitative fashion here so as not very greatly to prolong this post - that $$frac{bigg(sum_{j=1}^k n_jbigg)^overline{k}}{k!} ,$$ being a sum from to of a sum from to ... ($k$ instances thereof (if the termination be sum from to 1) ... triangular numbers, tetrahedral numbers, etc), creates the minimum offset necessary, at each item of the tuple, to ensure that the mapping does not become a surjection.



      Actually, using that 'handy' formula recursively would not give rise to any diseconomy if the number of items in the tuple were a power of two, and the recursion applied to the tuple treewise. If you simply apply it to the general tuple at each element consecutively, you would be handling numbers of order of magnitude $$(sum n)^{2^r} ,$$ such unnecessarily high orders of size corresponding to the implied intermediate tuples having a highly uneven distribution of value amongst their items.










      share|cite|improve this question











      $endgroup$




      I've used the tag "set-theory" in lieu of "bijections".



      When we represent (as is done by the OEIS) an infinite two dimensional sequence by the concatenation of antidiagonals, the index of an entry in that sequence is given by a bijection between ℕ×ℕ & ℕ. (ℕ may be ℕ₀ or ℕ₁ - it doesn't affect the essential argument - it's just a matter of offsets ... but I'll stick to casting all this in terms of ℕ₀, as then the formulæ are the simplest.) Explicitly the bijection is $$q=$$$$frac{(m+n)(m+n+1)}{2}+n$$$$=$$$$frac{(m+n)^{overline{2}}}{2!}+n$$$$=$$$$frac{(m+n)^2+m+3n}{2} ,$$ with the notation in the middle formula being the ascending pochhammer notation ... the reason I have bothered to broach this notation here shall be made clear shortly. I once saw the third formula given in it's own right as the simplest possible bijection from duples of natural numbers to natural numbers; and it is sufficient in its own right to serve as a bijection between tuples of naturals of any given order & the naturals - by applying it recursively. (Please mark, though, this does not mean between all tuples & singletons - it means between all tuples of some one given order & singletons!)



      But if we are to actually do such a mapping, there might be a 'tidier' (& more disciplined ) way of doing it ... forall its being so very 'handy' a formula, that third one. I'm fairly sure this works ... and the question is, can anyone tell me from their experience or knowledge that it does work? ... I'm not of course asking anyone to do the derivation for me. As far as I can tell it does ... but I do sometimes miss things, and I have not been able to find an exposition that says either yea or nay to it ... and it's this: if you have a tuple ${n_1, ... , n_r}$ would the most economical mapping - which seems to me to be the extension of the logic of the 2-dimensional case to arbitrary number $r$ of dimensions - be $$q=sum_{k=1}^rfrac{bigg(sum_{j=1}^k n_jbigg)^overline{k}}{k!}$$?



      The logic is - and I'll merely expound it in a fairly qualitative fashion here so as not very greatly to prolong this post - that $$frac{bigg(sum_{j=1}^k n_jbigg)^overline{k}}{k!} ,$$ being a sum from to of a sum from to ... ($k$ instances thereof (if the termination be sum from to 1) ... triangular numbers, tetrahedral numbers, etc), creates the minimum offset necessary, at each item of the tuple, to ensure that the mapping does not become a surjection.



      Actually, using that 'handy' formula recursively would not give rise to any diseconomy if the number of items in the tuple were a power of two, and the recursion applied to the tuple treewise. If you simply apply it to the general tuple at each element consecutively, you would be handling numbers of order of magnitude $$(sum n)^{2^r} ,$$ such unnecessarily high orders of size corresponding to the implied intermediate tuples having a highly uneven distribution of value amongst their items.







      sequences-and-series elementary-set-theory






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      edited Dec 16 '18 at 11:29







      AmbretteOrrisey

















      asked Dec 16 '18 at 11:08









      AmbretteOrriseyAmbretteOrrisey

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