How Many Pairs of Products of Consecutive Integers Have the Same Prime Factors?
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چکیده
e _ 2'/z 2 4 4 6 6 8 1 / 8 2 () (33) 1/4 ( 5577) which is proved as follows . For v > 2, the Pth factor is [2' '. . 2P/(2p'+ 1) . (2 1)]'áz [(2"-'-1)!!z2"!!z/2 .2v-'!!z(2v-1)!!z]'/z where n!!=n(n-2) . . .4.2 if n is even, n(n-2) . . .3 .1 if n is odd. Since 2°!!=2 z " '2° '1 and (2°-1)!!=2°! /2°!!=2°!/2z" '2°'!, this expression becomes [2z"2°-'16/2 .2 zí42'!z]I/z By induction on v, the product of the first v factors is [2 2vz"2 '1z/2' ~211/2" Applying Stirling's formula and letting v--->oo completes the proof .
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