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 no 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 *)

Select[Range[2200], Total[IntegerDigits[#]]*Total[IntegerDigits[#]^2]==#&] (* Harvey P. Dale, Jul 10 2014 *)

Sequence in context: A146233 A146181 A105984 * A064903 A259638 A070158

Adjacent sequences: A115515 A115516 A115517 * A115519 A115520 A115521

base,fini,full,nonn

Yalcin Aktar, Jun 24 2007

Edited by Robert G. Wilson v and Stefan Steinerberger, Jun 28 2007

approved