A126783 Smallest k > 1 such that (sum of digits of k^n)*(sum of digits of k^(n+1)) = k, or 0 if no such k exists.

80, 80, 70, 3905, 4004, 700, 19278, 32761, 5600, 8100, 24940, 10600, 56330, 68040, 81760, 149705, 116180, 126360, 123580, 0, 65500, 311003, 205030, 114400, 454951, 317350, 312170, 296270, 359380, 332750, 699785, 723338, 498150, 499130, 901368

Offset: 1

Author

Paolo P. Lava and Giorgio Balzarotti, Apr 17 2007

Keywords

• hard
• nonn
• base

Comments

For each n there is an upper bound (see A130179) for values of k such that (sum of digits of k^n)*(sum of digits of k^(n+1)) = k, hence the number of such k is finite, possibly zero, (see A130180) and if the number is not zero there is a largest one (see A130181).

Examples

			For n = 2 the smallest such k is 80: 80^2 = 6400 and 6+4+0+0 = 10; 80^3 = 512000 and 5+1+2+0+0+0 = 8; 10*8 = 80. Hence a(2) = 80.
For n = 3 the smallest such k is 70: 70^3 = 343000 and 3+4+3+0+0+0 = 10; 70^4 = 24010000 and 2+4+0+1+0+0+0+0 = 7; 10*7 = 70. Hence a(3) = 70.
		

Links

• Klaus Brockhaus, Table of n, a(n) for n=1..54

Crossrefs

Cf. A130179, A130180, A130181.

Programs

Maple

P:=proc(n) local a,i,j,k,w,x; for a from 1 by 1 to n do for i from 1 by 1 to n*n do w:=0;k:=i^a;j:=0;x:=i^(a+1); while k>0 do w:=w+k-(trunc(k/10)*10); k:=trunc(k/10); od; while x>0 do j:=j+x-(trunc(x/10)*10); x:=trunc(x/10); od; if (i=w*j and i>1) then print(i); break; fi; od; od; end: P(1000);

Extensions

Edited and a(17) to a(35) added by Klaus Brockhaus, May 14 2007