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 A226087 Number of values k in base n for which the sum of digits of k = sqrt(k). 2
 1, 4, 2, 3, 3, 6, 2, 2, 2, 5, 2, 6, 2, 5, 5, 2, 2, 4, 2, 6, 6, 4, 2, 5, 2, 4, 2, 6, 2, 11, 2, 2, 6, 4, 5, 6, 2, 4, 6, 5, 2, 11, 2, 6, 5, 4, 2, 6, 2, 4, 6, 5, 2, 4, 5, 5, 6, 4, 2, 13, 2, 4, 4, 2, 5, 11, 2, 5, 6, 11, 2, 5, 2, 4, 6, 6, 6, 11, 2, 5, 2, 4, 2, 12, 5 (list; graph; refs; listen; history; text; internal format)
 OFFSET 2,2 COMMENTS Values of k in base n have at most 3 digits. Proof: Because sqrt(k) increases faster than the digit sum of k, only numbers with d digits meeting the condition d*(n-1)>=n^(d/2) are candidate fixed points. Using numeric methods, d<3 for n>6. and since there are no fixed points of four or more digits in bases 2 through 5, there are no fixed points in any base with more than 3 digits. 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. LINKS Christian N. K. Anderson, Table of n, a(n) for n = 2..10000 EXAMPLE For a(16)=5 the solutions are the square numbers {1, 36, 100, 225, 441} because in base 16 they are written as {1, 24, 64, E1, 1B9} and sqrt(1) = 1 sqrt(36) = 6 = 2+4 sqrt(100) = 10 = 6+4 sqrt(225) = 15 = 14+1, and sqrt(441) = 21 = 1+11+9 PROG (R) sapply(2:16, function(n) sum(sapply((1:(n^ifelse(n>6, 1.5, 2)))^2, function(x) sum(inbase(x, n))==sqrt(x)))) CROSSREFS Cf. A226224. Cf. digital sums for digits at various powers: A007953, A003132, A055012,A055013, A055014, A055015. Sequence in context: A079623 A177864 A090112 * A140396 A118945 A016692 Adjacent sequences:  A226084 A226085 A226086 * A226088 A226089 A226090 KEYWORD nonn,base AUTHOR Christian N. K. Anderson and Kevin L. Schwartz, May 25 2013 STATUS approved

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Last modified December 16 06:18 EST 2019. Contains 330016 sequences. (Running on oeis4.)