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 A279459 Numbers n such that sum of the proper divisors of n is the square of the sum of the digits of n. 0
 24, 153, 176, 794, 3071, 3431, 4607, 9671, 15599, 17711, 18167, 19511, 45671, 50927, 56471, 62807, 74639, 119279, 127559, 154199, 165791, 174719, 175871, 695399, 699359 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Subsequence of A073040. Numbers n such that A001065(n) = A118881(n) or A000203(n) - n = (A007953(n))^2. Every term in the sequence is composite (since the only proper divisor of a prime is 1). The sum of the proper divisors of a k-digit composite number n must exceed sqrt(n) >= sqrt(10^(k-1)), but the square of the sum of the digits of a k-digit number cannot exceed (9k)^2 = 81k^2. Since sqrt(10^(k-1)) > 81k^2 for all integers k > 8, every term in the sequence must be less than the smallest 9-digit number, 10^8. An exhaustive search through 10^8 shows that a(25)=699359 is the last term. - Jon E. Schoenfield, Dec 13 2016 LINKS Eric Weisstein's World of Mathematics, Digit Sum EXAMPLE 24 is in the sequence because 24 has 7 proper divisors {1,2,3,4,6,8,12}, 1 + 2 + 3 + 4 + 6 + 8 + 12 = 36 and (2 + 4)^2 = 36; 153 is in the sequence because 153 has 5 proper divisors {1,3,9,17,51}, 1 + 3 + 9 + 17 + 51 = 81 and (1 + 5 + 3)^2 = 81; 176 is in the sequence because 176 has 9 proper divisors {1,2,4,8,11,16,22,44,88}, 1 + 2 + 4 + 8 + 11 + 16 + 22 + 44 + 88 = 196 and (1 + 7 + 6)^2 = 196, etc. MATHEMATICA Select[Range[1000000], DivisorSigma[1, #1] - #1 == Total[IntegerDigits[#1]]^2  &] PROG (PARI) is(n) = sigma(n)-n==sumdigits(n)^2 \\ Felix FrÃ¶hlich, Dec 13 2016 CROSSREFS Cf. A073040, A001065, A118881, A145746. Sequence in context: A039494 A159650 A305160 * A092181 A001702 A004308 Adjacent sequences:  A279456 A279457 A279458 * A279460 A279461 A279462 KEYWORD nonn,base,fini,full AUTHOR Ilya Gutkovskiy, Dec 12 2016 STATUS approved

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Last modified April 3 19:43 EDT 2020. Contains 333198 sequences. (Running on oeis4.)