cp's OEIS Frontend

A279459 Numbers n such that sum of the proper divisors of n is the square of the sum of the digits of n.

%I A279459 #14 Feb 16 2025 08:33:37
%S A279459 24,153,176,794,3071,3431,4607,9671,15599,17711,18167,19511,45671,
%T A279459 50927,56471,62807,74639,119279,127559,154199,165791,174719,175871,
%U A279459 695399,699359
%N A279459 Numbers n such that sum of the proper divisors of n is the square of the sum of the digits of n.
%C A279459 Subsequence of A073040.
%C A279459 Numbers n such that A001065(n) = A118881(n) or A000203(n) - n = (A007953(n))^2.
%C A279459 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
%H A279459 Eric Weisstein's World of Mathematics, <a href="https://mathworld.wolfram.com/DigitSum.html">Digit Sum</a>
%H A279459 <a href="/index/Su#sums_of_divisors">Index entries for sequences related to sums of divisors</a>
%e A279459 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;
%e A279459 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;
%e A279459 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.
%t A279459 Select[Range[1000000], DivisorSigma[1, #1] - #1 == Total[IntegerDigits[#1]]^2  &]
%o A279459 (PARI) is(n) = sigma(n)-n==sumdigits(n)^2 \\ _Felix Fröhlich_, Dec 13 2016
%Y A279459 Cf. A073040, A001065, A118881, A145746.
%K A279459 nonn,base,fini,full
%O A279459 1,1
%A A279459 _Ilya Gutkovskiy_, Dec 12 2016