1,2

(a_1 + a_2 + .. + a_k)*(a_1^2 + a_2^2 + .. + a_k^2) = n, the k-digit number a_k...a_2a_1, in base 10.

The sequence is finite and all terms are listed. Proof. Each natural number n has not more than log_10(n)+1 digits. Furthermore each digit is, of course, not bigger than 9. This gives the inequality (a1+a2+..ap)(a1^2+a2^2+..+ap^2) <= 9*(log_10(n)+1)*81*(log_10(n)+1) = 729*(log_10(n)+1)^2. On the other hand, for all n > 20582 we have 729*(log_10(n)+1)^2 < n. Therefore all terms of the sequence have to be smaller than this upper bound. A simple computer search shows that indeed all terms are listed. - Stefan Steinerberger

(2+1+9+6)(2^2+1^2+9^2+6^2)=2196

p := proc (K) options operator, arrow; floor((log[10])(K))+1 end proc; A := proc (K) options operator, arrow; convert(K, base, 10) end proc; g := proc (K) options operator, arrow; sum(A(K)[i], i = 1 .. p(K)) end proc p := proc (K) options operator, arrow; floor((log[10])(K))+1 end proc; A := proc (K) options operator, arrow; convert(K, base, 10) end proc; g := proc (K) options operator, arrow; sum(A(K)[i], i = 1 .. p(K)) end proc

fQ[n_] := Block[{id = IntegerDigits@n}, Plus @@ id * Plus @@ (id^2) == n]; lst = {}; Do[ If[fQ@n, AppendTo[lst, n]], {n, 10^8}]; lst (* Robert G. Wilson v *)

For[n = 1, n < 30000, n++, b = IntegerDigits[n]; If[Sum[b[[i]], {i, 1, Length[b]}]*Sum[b[[i]]^2, {i, 1, Length[b]}] == n, Print[n]]] - Stefan Steinerberger

Sequence in context: A050882 A146181 A105984 this_sequence A064903 A070158 A055940

Adjacent sequences: A115515 A115516 A115517 this_sequence A115519 A115520 A115521

base,fini,full,nonn

Aktar Yalcin (aktaryalcin(AT)msn.com), Jun 24 2007

Edited by Robert G. Wilson v (rgwv(AT)rgwv.com) and Stefan Steinerberger (stefan.steinerberger(AT)gmail.com), Jun 28 2007