A226224 - OEIS

%I #15 Jun 04 2013 13:04:18

%S 1,25,9,64,100,144,49,64,81,225,121,441,169,441,441,256,289,324,361,

%T 1296,1296,484,529,1089,625,676,729,2401,841,2601,961,1024,3025,1156,

%U 2500,4096,1369,1444,4356,3136,1681,4900,1849,5929,3025,2116,2209,6561,2401

%N The largest value of k in base n for which the sum of digits of k = sqrt(k).

%C There are no values of k in base n with more than 3 digits. Proof: such a value with d digits would need to meet the criterion d*(n-1)>=sqrt(n)^d which establishes an upper limit of 4 digits for 2<=n<=6 and 3 for n>6. Because there are no four digit values of k in bases 2 through 6, k has a maximum of three digits in all bases.

%C Because k must be a square, there are only sqrt(n)^3 possible values in any base.

%C From the above, it can be shown that for three-digit fixed points of the form xyz, x <= 6; also x<=4 for n>846. These theoretical upper limits are statistically unlikely, and in fact of the 86356 solutions in bases 2 to 10000, only 6.5% of them begin with 2, and none begin with 3 through 6.

%C a(n)=1 iff A226087(n)=1. Conjecture: this occurs exactly once -- in base 2.

%H Christian N. K. Anderson, <a href="/A226224/b226224.txt">Table of n, a(n) for n = 2..10000</a>

%H Christian N. K. Anderson, <a href="/A226224/a226224.txt">Table of base, max(k), number of k, and all values of k for base 2 to 10000.</a>

%e For a(16) the solutions are the square numbers {1, 36=6^2, 100=10^2, 225=15^2, 441=21^2} because in base 16 they are written as {1, 24, 64, E1, 1B9} and 1 = 1, 6 = 2+4, 10 = 6+4, 15 = 14+1, and 21 = 1+11+9.

%o (R)

%o for(n in 2:500) cat("Base",n,":",which(sapply((1:(ifelse(n>6,7,1)*n^ifelse(n>6,1,2)))^2, function(x) sum(inbase(x,n))==sqrt(x)))^2, "\n")

%Y Cf. A007953, A003132, A055012, A055013, A055014, A055015, A226087.

%K nonn,base

%O 2,2

%A _Christian N. K. Anderson_ and _Kevin L. Schwartz_, May 31 2013