|
|
A174460
|
|
Smith numbers of order 2.
|
|
9
|
|
|
56, 58, 810, 822, 1075, 1519, 1752, 2145, 2227, 2260, 2483, 2618, 2620, 3078, 3576, 3653, 3962, 4336, 4823, 4974, 5216, 5242, 5386, 5636, 5719, 5762, 5935, 5998, 6220, 6424, 6622, 6845, 7015, 7251, 7339, 7705, 7756, 8460, 9254, 9303, 9355, 10481, 10626, 10659
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,1
|
|
COMMENTS
|
Composite numbers a(n) such that the sum of digits^2 equals the sum of digits^2 of its prime factors without the numbers of A176670 that have the same digits as its prime factors (without the zero digit).
It seems as though as the order n approaches infinity, the sequence of n-order Smith numbers approaches A176670. Is there a value of n where the only n-order Smith numbers are members of A176670? - Ely Golden, Dec 07 2016
|
|
LINKS
|
Underwood Dudley, Smith numbers, Mathematics Magazine 67(1) (1994), 62-65.
S. S. Gupta, Smith Numbers, Mathematical Spectrum 37(1) (2004/5), 27-29.
A. Wilansky, Smith Numbers, Two-Year College Math. J. 13(1) (1982), p. 21.
|
|
EXAMPLE
|
a(2) = 58 = 2*29 is a Smith number of order 2 because 5^2 + 8^2 = 2^2 + 2^2 + 9^2 = 89.
|
|
MAPLE
|
for s from 2 to 10000 do g:=nops(ifactors(s)[2]): qsp:=0: for u from 1 to g do z:=ifactors(s)[2, u][1]: h:=0: while (z>0) do z:=iquo(z, 10, 'r'): h:=h+r^2: end do: h:=h*ifactors(s)[2, u][2]: qsp:=qsp+h: end do: z:=s: qs:=0: while (z>0) do z:=iquo(z, 10, 'r'): qs:=qs+r^2: end do: if (qsp=qs) then print(s): end if: end do:
|
|
MATHEMATICA
|
With[{k = 2}, Select[Range[12000], Function[n, And[Total@ Map[#^k &, IntegerDigits@ n] == Total@ Map[#^k &, Flatten@ IntegerDigits[#]], Not[Sort@ DeleteCases[#, 0] &@ IntegerDigits@ n == Sort@ DeleteCases[#, 0] &@ #]] &@ Flatten@ Map[IntegerDigits@ ConstantArray[#1, #2] & @@ # &, FactorInteger@ n]]]] (* Michael De Vlieger, Dec 10 2016 *)
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,base
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|