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

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

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

Table of n, a(n) for n=1..25.

Eric Weisstein's World of Mathematics, Digit Sum

Index entries for sequences related to sums of divisors

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 Fröhlich, 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