1,1

Smooth Power Duos: Search for consecutive numbers a(n)-1 and a(n) such that

the largest prime factor of a(n)-1 raised to the power n remains <= a(n)-1 and such that

the largest prime factor of a(n) raised to the power n remains <= a(n):

( A006530(a(n)))^n <= a(n) and (A006530(a(n)-1))^n <= a(n)-1.

The prime factorization for the a(n) and a(n)-1 are:

n=1: 3=3, 2=2. n=2: 9=3^2, 8= 2^3. n=3 or 4: 2401=7^4, 2400=2^5*3*5^2.

n=5: 5909761 = 11^2*13^2*17^2, 5909760 = 2^8*3^5*5*19.

n=6: 1611308700 = 2^2*3^6*5^2*23*31^2, 1611308699 = 7^4*11*13^2*19^2 .

n=7: 421138799640 = 2^3*3^5*5*13^4*37*41, 421138799639 = 17*19*23^2*31*43^3 .

n=8: 2286831727304145 = 3^15*5*7^3*19*67*73, 2286831727304144 = 2^4*17*23^2*37*41^2*59*61*71 .

n=9: 3948741978036988496 = 2^4*7^5*13*23*43*59^3*67*83, 3948741978036988495 = 5*11*17*31*97^2*101*103*109*113^2 .

Note: All numbers through 2^62 have been searched

Fred Schneider and R. Gerbicz, Smooth Power Trios.

For n=7, a(7)= 421138799640 = 2^3*3^5*5*13^4*37*41 and a(7)-1 =421138799639 = 17*19*23^2*31*43^3

are solutions because 41 <= floor(421138799640^(1/7)) = 45 and 43 <= floor(421138799639^(1/7)) = 45.

Cf. A122464, A122465, A116486.

Sequence in context: A112726 A112725 A060712 this_sequence A072005 A060377 A027896

Adjacent sequences: A122460 A122461 A122462 this_sequence A122464 A122465 A122466

hard,more,nonn

Fred Schneider (frederick.william.schneider(AT)gmail.com), Sep 09 2006

Edited by R. J. Mathar (mathar(AT)strw.leidenuniv.nl), Sep 02 2009