A073848 - OEIS

%I #7 Dec 05 2013 19:55:31

%S 1,1,6,58,615,7761,117628,4,2151,999999955,25937424546,181398462,

%T 23298085122403,793714773254053,29192926025390520,3976,

%U 48661191875666868345,5642219814759,104127350297911241532670

%N Smallest of n consecutive numbers in A.P. with a common difference of n with a sum that is an n-th power.

%F For n!=4: a(n)=(p_1*p_2*...p_r)^n/n-n*(n-1)/2 with prime factorization n=p_1^b_1*...*p_r^b_r.

%F Proof of formula: If a is the first member in A.P. then there must exist an integer m with n*a+n*n*(n-1)/2=m^n. Therefore all prime divisors of n divide m. For n>4 this gives a positive solution for a.

%e a(4) = 58 and 58 +62 +66 +70 = 256 = 4^4.

%p for n from 1 to 50 do a := ifactors(n); b := 1:for j from 1 to nops(a[2]) do b := b*a[2][j][1]; od; c[n] := b^n/n-n*(n-1)/2; od:c[4] := 58:seq(c[j],j=1..50);

%K nonn

%O 1,3

%A _Amarnath Murthy_, Aug 14 2002

%E Formula and more terms from _Sascha Kurz_, Aug 14 2002