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A226224
The largest value of k in base n for which the sum of digits of k = sqrt(k).
4
1, 25, 9, 64, 100, 144, 49, 64, 81, 225, 121, 441, 169, 441, 441, 256, 289, 324, 361, 1296, 1296, 484, 529, 1089, 625, 676, 729, 2401, 841, 2601, 961, 1024, 3025, 1156, 2500, 4096, 1369, 1444, 4356, 3136, 1681, 4900, 1849, 5929, 3025, 2116, 2209, 6561, 2401
OFFSET
2,2
COMMENTS
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.
Because k must be a square, there are only sqrt(n)^3 possible values in any base.
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.
a(n)=1 iff A226087(n)=1. Conjecture: this occurs exactly once -- in base 2.
EXAMPLE
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.
PROG
(R)
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")
KEYWORD
nonn,base
STATUS
approved