cp's OEIS Frontend

A218030 Numbers k equal to half of the product of the nonzero (base-10) digits of k^2.

%I A218030 #46 Jun 01 2024 09:55:54
%S A218030 2,5,54,648,2160,337169025526136832000,
%T A218030 685506275314921762068267522458966662115416623590907309075726336000000,
%U A218030 46641846972427276691124922228108091690332947069125333309512419901440000000000
%N A218030 Numbers k equal to half of the product of the nonzero (base-10) digits of k^2.
%C A218030 The first 5 terms of the sequence were found by the author around 1980 using his Commodore PET computer. He found the subsequent terms in 1991 by means of an improved program. The author has always referred to these as the "Faithy numbers" after his mother, Faith, who posed the problem.
%H A218030 Michael S. Branicky, <a href="/A218030/b218030.txt">Table of n, a(n) for n = 1..12</a> (all terms < 10^300)
%H A218030 Michael S. Branicky, <a href="/A218030/a218030.py.txt">Python program</a>.
%H A218030 Giovanni Resta, <a href="/A218030/a218030.c.txt">C program for this and related sequences</a>.
%e A218030 For n=5, n^2 is 25; the product of the digits of 25 is 2*5 = 10, which is equal to 2*n.
%t A218030 mx = 2^255; L = {};
%t A218030 p2 = 1; While[p2 < mx, Print["--> 2^", Log[2, p2]];
%t A218030 p3 = p2; While [p3 < mx,
%t A218030   p5 = p3; While[p5 < mx,
%t A218030    n = p5; While[n < mx,
%t A218030     If[2 n == Times @@ Select[IntegerDigits[n^2], # > 0 &],
%t A218030      AppendTo[L, n]; Print[n]]; n *= 7]; p5 *= 5]; p3 *= 3];
%t A218030 p2 *= 2]; Sort[L] (* _Giovanni Resta_, Oct 19 2012 *)
%o A218030 (PARI) is_A218030(n)={my(d=digits(n^2));n*=2;for(i=1,#d,d[i]||next;n%d[i]&return;n\=d[i]);n==1} \\ _M. F. Hasler_, Oct 19 2012
%Y A218030 Special case of A218013 where the ratio of the digit-product to the original number is 2. Related to A218072.
%K A218030 nonn,base
%O A218030 1,1
%A A218030 _Nels Olson_, Oct 18 2012