OFFSET
1,1
COMMENTS
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
LINKS
Fred Schneider and R. Gerbicz, Smooth Power Trios.
EXAMPLE
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.
PROG
(PARI) isok1(k, n) = {r = floor(sqrtn(k, n)); f = factor(k); for (j=1, #f~, if (f[j, 1] > r, return (0)); ); return (1); }
a(n) = {i = 3; found = 0; while (! found, found = (isok1(i, n) && isok1(i-1, n)); if (! found, i++); ); return (i); } \\ Michel Marcus, Jul 12 2013
CROSSREFS
KEYWORD
hard,more,nonn
AUTHOR
Fred Schneider, Sep 09 2006
EXTENSIONS
Edited by R. J. Mathar, Sep 02 2009
STATUS
approved