cp's OEIS Frontend

A226353 Largest integer k in base n whose squared digits sum to sqrt(k).

%I A226353 #10 Jun 08 2013 15:34:35
%S A226353 1,49,169,36,1,1,2601,1089,1,8836,33489,44100,1,149769,128164,96721,1,
%T A226353 156816,1225,40804,12321,831744,839056,1149184,1737124,3655744,407044,
%U A226353 1890625,2208196,1089,1,1466521,6125625,2235025,2832489,1,3759721,6885376,8844676
%N A226353 Largest integer k in base n whose squared digits sum to sqrt(k).
%C A226353 Any d-digit number in base n meeting the criterion must also meet the condition d*(n-1)^2 < n^(d/2). Numerically, it can be shown this limits the candidate values to squares < 22*n^4. The larger values are statistically unlikely, and in fact the largest value of k in the first 1000 bases is ~9.96*n^4 in base 775.
%C A226353 a(n)=1 iff A226352(n)=1.
%H A226353 Christian N. K. Anderson, <a href="/A226353/b226353.txt">Table of n, a(n) for n = 2..1000</a>
%H A226353 Christian N. K. Anderson, <a href="/A226353/a226353.txt">Table of base, all solutions in base 10, and all solutions in base n</a> for bases 2 to 1000.
%e A226353 In base 8, the four solutions are the values {1,16,256,2601}, which are written as {1,20,400,5051} in base 8 and
%e A226353 sqrt(1)   = 1  = 1^2
%e A226353 sqrt(16)  = 4  = 2^2+0^2
%e A226353 sqrt(256) = 16 = 4^2+0^2+0^2
%e A226353 sqrt(2601)= 51 = 5^2+0^2+5^2+1^2
%o A226353 (R) inbase=function(n,b) { x=c(); while(n>=b) { x=c(n%%b,x); n=floor(n/b) }; c(n,x) }
%o A226353 for(n in 2:50) cat("Base",n,":",which(sapply((1:(4.7*n^2))^2,function(x) sum(inbase(x,n)^2)==sqrt(x)))^2,"\n")
%Y A226353 Cf. A226352, A226224.
%Y A226353 Cf. digital sums for digits at various powers: A007953, A003132, A055012, A055013, A055014, A055015.
%K A226353 nonn,base
%O A226353 2,2
%A A226353 _Christian N. K. Anderson_ and _Kevin L. Schwartz_, Jun 04 2013