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

%I A162457 #15 Jul 07 2023 14:35:01
%S A162457 2,3,5,7,22,26,33,34,38,39,46,51,55,57,58,62,65,69,74,77,82,85,86,87,
%T A162457 91,93,94,95,106,111,115,118,119,122,123,129,133,134,141,142,145,146,
%U A162457 155,158,159,161,166,177,178,183,185,194,201,203,205,213,215,217,219,235
%N A162457 Numbers whose prime factors when sorted and stacked fill an equilateral triangle.
%C A162457 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).
%C A162457 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.
%C A162457 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.
%e A162457 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.
%e A162457 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.
%p A162457 A055642 := proc(n) max(1, ilog10(n)+1) ; end: omega := proc(n) nops(numtheory[factorset](n)) ; end:
%p A162457 isA162457 := proc(n) local plen,p,e,dlen,i ;
%p A162457 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;
%p A162457 else false ; fi; end:
%p A162457 for n from 2 to 300 do if isA162457(n) then printf("%d,",n) ; fi; od: # _R. J. Mathar_, Sep 16 2009
%o A162457 (PARI) factortriangle(m,n) =
%o A162457 {
%o A162457 local(x,a,v,j,f,ln,lna,c);
%o A162457 for(x=m,n,
%o A162457 f=0;
%o A162457 a = ifactor(x);
%o A162457 lna=length(a);
%o A162457 for(j=1,lna, if(length(Str(a[j]))!=j,f=1;break););
%o A162457 if(!f,print1(x","));
%o A162457 );
%o A162457 }
%o A162457 ifactor(n) = \\ The vector of the prime factors of n with multiplicity.
%o A162457 {
%o A162457 local(f,j,k,flist);
%o A162457 flist=[];
%o A162457 f=Vec(factor(n));
%o A162457 for(j=1,length(f[1]),
%o A162457 for(k = 1,f[2][j],flist = concat(flist,f[1][j]));
%o A162457 );
%o A162457 return(flist)
%o A162457 }
%K A162457 base,nonn
%O A162457 1,1
%A A162457 _Cino Hilliard_, Jul 04 2009
%E A162457 Comment extended by _R. J. Mathar_, Sep 16 2009