login
a(n) is the smallest number k > 1 for which A001142(k)/A002944(k+1)^n is an integer.
1

%I #19 Jan 24 2019 10:00:11

%S 2,3,9,9,15,15,38,45,45,45,61,61,225,225,225,225,225,225,225,225,225,

%T 225,635,635,1545,1545,1545,1545,2137,2137,2137,2137,2137,2137,2137,

%U 2137,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660,2660

%N a(n) is the smallest number k > 1 for which A001142(k)/A002944(k+1)^n is an integer.

%C a(62) > 12500. - _Robert Israel_, Jan 24 2019

%e n=4, A001142(9) = 1*9*36*...*9*1 = 11759522374656,

%e A002944(10) = lcm(1,2,...,10)/10=252 and A001142(9) = 2916*(252^4) = 11759522374656,

%e so a(4)=9, the smallest relevant number.

%p A001142:= proc(n) option remember; procname(n-1)*n^(n-1)/(n-1)! end proc:

%p A001142(0):= 1:

%p A002944:= proc(n) option remember; ilcm(n,procname(n-1)*(n-1))/n end proc:

%p A002944(1):= 1:

%p f:= proc(n) option remember; local k;

%p for k from procname(n-1) do

%p if type(A001142(k)/A002944(k+1)^n, integer) then return k fi

%p od

%p end proc:

%p f(1):= 2:

%p map(f, [$1..61]); # _Robert Israel_, Jan 23 2019

%t Table[fla=1;Do[s1=Apply[Times, Table[Binomial[n, j], {j, 0, n}]]; s2=Apply[LCM, Table[Binomial[n, j], {j, 0, n}]]; If[IntegerQ[s1/(s2^k)]&&!Equal[n, 1]&&Equal[fla, 1], Print[{n, k}];fla=0], {n, 1, 230}], {k, 1, 25}]

%Y Cf. A001142, A002944, A092593.

%K nonn

%O 1,1

%A _Labos Elemer_, Mar 10 2004

%E Corrected and extended by _Robert Israel_, Jan 23 2019