A162457 Numbers whose prime factors when sorted and stacked fill an equilateral triangle.

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

Offset: 1

Author

Cino Hilliard, Jul 04 2009

Keywords

• base
• nonn

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.

Examples

			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.
		

Programs

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(plen-1) 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

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)
}

Extensions

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