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%I #16 Aug 01 2023 08:05:33
%S 1,2,8,3,4,12,4,5,6,30,5,6,7,8,105,6,7,8,9,10,210,7,8,9,10,11,12,1155,
%T 8,9,10,11,12,13,14,30030,9,10,11,12,13,14,15,16,1001,10,11,12,13,14,
%U 15,16,17,18,34034,11,12,13,14,15,16,17,18,19,20,323323,12,13,14,15,16,17
%N Triangle read by rows in which the n-th row contains the smallest set of n distinct numbers beginning with n with a product which is a square.
%C Presumably "smallest set" means we start with n-1 consecutive numbers in each row and add the last element to satisfy the requirement on the square, the last element is obtained from the prime factorization of the product of the first n-1 numbers by reducing all prime exponents modulo 2. If the result is <= the (n-1)-st term, multiply by j^2, j=2,3,4,.. until it is. - _R. J. Mathar_, Apr 05 2007
%e 1
%e 2 8
%e 3 4 12
%e 4 5 6 30
%e 5 6 7 8 105
%e ...
%p A083486 := proc(n,m) local fs,k,resu,extr ; if m < n then n+m-1; else fs := ifactors(mul( A083486(n,k),k=1..n-1))[2] ; resu := mul( op(1,op(k,fs))^(op(2,op(k,fs)) mod 2),k=1..nops(fs)) ; extr := 1 ; while extr^2*resu <= A083486(n,n-1) do extr := extr+1 ; od ; RETURN(resu*extr^2) ; fi ; end: for n from 1 to 15 do for m from 1 to n do printf("%a ",A083486(n,m)) ; od ; od ; # _R. J. Mathar_, Apr 05 2007
%t T[n_, m_] := Module[{fs, resu, extr }, If[m < n, n+m-1, fs = FactorInteger[Product[T[n, k], {k, 1, n-1}]]; resu = Product[fs[[k, 1]]^Mod[fs[[k, 2]], 2], {k, 1, Length[fs]}]; extr = 1; While[extr^2*resu <= T[n, n-1], extr++]; resu*extr^2]];
%t Table[T[n, m], {n, 1, 15}, {m, 1, n}] // Flatten (* _Jean-François Alcover_, Aug 01 2023, after _R. J. Mathar_ *)
%Y Cf. A083484, A083485.
%K nonn,tabl
%O 1,2
%A _Amarnath Murthy_ and Meenakshi Srikanth (menakan_s(AT)yahoo.com), May 03 2003
%E Corrected and extended by _R. J. Mathar_, Apr 05 2007
%E Example corrected by _Harvey P. Dale_, Aug 05 2021
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