Search: a145746 -id:a145746
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A266816
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Numbers whose arithmetic derivative is equal to the product of their digits.
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+10
1
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4, 11, 25, 329, 3383, 4343, 5561, 6623, 12773, 17267, 21479, 57721, 129383, 136259, 142943, 172793, 256631, 292571, 364823, 413663, 413927, 619337, 653291, 1215659, 1218863, 1268951, 1276931, 1483751, 1655219, 1892327, 2952731, 4158719, 4973531, 5418671, 6377663
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OFFSET
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1,1
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COMMENTS
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4 appears to be the only even number in the sequence.
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LINKS
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EXAMPLE
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4’ = 4;
11’ = 1 = 1*1;
25’ = 10 = 2*5;
329’ = 54 = 3*2*9; etc.
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MAPLE
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with(numtheory): P:=proc(q) local a, b, k, n; for n from 1 to q do a:=n; b:=1;
for k from 1 to ilog10(n)+1 do b:=b*(a mod 10); a:=trunc(a/10); od;
if n*add(op(2, a)/op(1, a), a=ifactors(n)[2])=b then print(n); fi; od; end: P(10^9);
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MATHEMATICA
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Select[Range[3, 10^5], Times @@ IntegerDigits@ # == # Total[#2/#1 & @@@
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CROSSREFS
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KEYWORD
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nonn,base
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AUTHOR
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STATUS
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approved
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A279459
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Numbers n such that sum of the proper divisors of n is the square of the sum of the digits of n.
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+10
0
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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)
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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Eric Weisstein's World of Mathematics, Digit Sum
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EXAMPLE
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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.
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MATHEMATICA
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Select[Range[1000000], DivisorSigma[1, #1] - #1 == Total[IntegerDigits[#1]]^2 &]
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PROG
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(PARI) is(n) = sigma(n)-n==sumdigits(n)^2 \\ Felix Fröhlich, Dec 13 2016
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CROSSREFS
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KEYWORD
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nonn,base,fini,full
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AUTHOR
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STATUS
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approved
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