

A162457


Numbers whose prime factors when sorted and stacked fill an equilateral triangle.


1



2, 3, 5, 7, 22, 26, 33, 34, 38, 39, 46, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, 95, 106, 111, 115, 118, 119, 122, 123, 129, 133, 134, 141, 142, 145, 146, 155, 158, 159, 161, 166, 177, 178, 183, 185, 194, 201, 203, 205, 213, 215, 217, 219, 235
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OFFSET

1,1


COMMENTS

Write the sorted list of all prime factors of each integer k (with multiplicity) underneath, with one prime per row. If there is one prime factor with one digit, one prime factor with 2 digits, one with three, one with four, etc., the tight hull/circumference of the digits looks like an equilateral triangle, and the integer k is added to the sequence. This implies that multiplicities of the prime factors of these k need to be 1 (as in A005117).
We can replace the digits in the stack by dots; if the requirement on the smooth variation in the digit counts is met, the total number of dots is a triangular number A000217.
Conjecture: The number of terms in this sequence is infinite. While this may be true, there are large gaps with no occurrences of factor triangles. Between 700 and 2000 there are no numbers that form factor triangles.


REFERENCES

Mohammad K. Azarian, A Trigonometric Characterization of Equilateral Triangle, Problem 336, Mathematics and Computer Education, Vol. 31, No. 1, Winter 1997, p. 96. Solution published in Vol. 32, No. 1, Winter 1998, pp. 8485.
Mohammad K. Azarian, Equating Distances and Altitude in an Equilateral Triangle, Problem 316, Mathematics and Computer Education, Vol. 28, No. 3, Fall 1994, p. 337. Solution published in Vol. 29, No. 3, Fall 1995, pp. 324325.


LINKS

Table of n, a(n) for n=1..60.


EXAMPLE

218 = 2*109. Stacking these, we have 2 (with 1 digit) and 109 (with 3 digits), but no prime factor with 2 digits, so 218 is not in the sequence.
7777 = 7*11*101. Stacking these, smallest to largest on top of each other, the digits form an equilateral triangle. So 7777 belongs to the sequence.


MAPLE

A055642 := proc(n) max(1, ilog10(n)+1) ; end: omega := proc(n) nops(numtheory[factorset](n)) ; end:
isA162457 := proc(n) local plen, p, e, dlen, i ;
if omega(n) = numtheory[bigomega](n) then plen := [seq(0, i=1..100)] ; for p in ifactors(n)[2] do e := op(2, p) ; if e > 1 then RETURN(false) ; fi; dlen := A055642( op(1, p)) ; if op(dlen, plen) > 0 then RETURN(false) ; fi; plen := subsop(dlen=1, plen) ; od: for i from 1 to nops(plen1) do if op(i, plen) = 0 and op(i+1, plen) = 1 then RETURN(false); fi; od: true;
else false ; fi; end:
for n from 2 to 300 do if isA162457(n) then printf("%d, ", n) ; fi; od: # R. J. Mathar, Sep 16 2009


PROG

(PARI) factortriangle(m, n) =
{
local(x, a, v, j, f, ln, lna, c);
for(x=m, n,
f=0;
a = ifactor(x);
lna=length(a);
for(j=1, lna, if(length(Str(a[j]))!=j, f=1; break); );
if(!f, print1(x", "));
);
}
ifactor(n) = \\ The vector of the prime factors of n with multiplicity.
{
local(f, j, k, flist);
flist=[];
f=Vec(factor(n));
for(j=1, length(f[1]),
for(k = 1, f[2][j], flist = concat(flist, f[1][j]));
);
return(flist)
}


CROSSREFS

Sequence in context: A125665 A046034 A062087 * A084983 A062239 A066483
Adjacent sequences: A162454 A162455 A162456 * A162458 A162459 A162460


KEYWORD

base,nonn


AUTHOR

Cino Hilliard, Jul 04 2009


EXTENSIONS

Comment extended by R. J. Mathar, Sep 16 2009


STATUS

approved



