cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

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A266816 Numbers whose arithmetic derivative is equal to the product of their digits.

Original entry on oeis.org

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
Offset: 1

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Author

Paolo P. Lava, Feb 10 2016

Keywords

Comments

4 appears to be the only even number in the sequence.

Examples

			4’ = 4;
11’ = 1 = 1*1;
25’ = 10 = 2*5;
329’ = 54 = 3*2*9; etc.
		

Crossrefs

Programs

  • Maple
    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);
  • Mathematica
    Select[Range[3, 10^5], Times @@ IntegerDigits@ # == # Total[#2/#1 & @@@
    FactorInteger@ Abs@ #] &] (* Michael De Vlieger, Feb 10 2016, after Michael Somos at A003415 *)

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

Original entry on oeis.org

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

Views

Author

Ilya Gutkovskiy, Dec 12 2016

Keywords

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

Examples

			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.
		

Crossrefs

Programs

  • Mathematica
    Select[Range[1000000], DivisorSigma[1, #1] - #1 == Total[IntegerDigits[#1]]^2  &]
  • PARI
    is(n) = sigma(n)-n==sumdigits(n)^2 \\ Felix Fröhlich, Dec 13 2016
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