OFFSET
1,3
COMMENTS
Let the decimal expansion of n = d(0)d(1)...d(p). Numbers such that Sum_{k=0..p} d(k)! is square.
Same as A130687 except for the additional 0. - Georg Fischer, Oct 12 2018
LINKS
Jinyuan Wang, Table of n, a(n) for n = 1..9700
EXAMPLE
17 is in the sequence because 1! + 7! = 1 + 5040 = 71^2.
MAPLE
with(numtheory):for n from 0 to 1000 do:l:=length(n):n0:=n:s:=0:for m from
1 to l do:q:=n0:u:=irem(q, 10):v:=iquo(q, 10):n0:=v :s:=s+u!:od: q:=sqrt(s):if
floor(q)= q then printf(`%d, `, n):else fi:od:~
MATHEMATICA
Select[Range[0, 1000], IntegerQ[Sqrt[Total[IntegerDigits[#]!]]]&] (* Jinyuan Wang, Feb 26 2020 *)
PROG
(Magma) [ n: n in [0..1000] | n eq 0 or IsSquare(&+[ Factorial(d): d in Intseq(n) ]) ];
(Python)
from itertools import count, islice, combinations_with_replacement
from math import factorial
from sympy.ntheory.primetest import is_square
from sympy.utilities.iterables import multiset_permutations
def A173687_gen(): # generator of terms
yield 0
for l in count(0):
for i in range(1, 10):
fi = factorial(i)
yield from sorted(int(str(i)+''.join(map(str, k))) for j in combinations_with_replacement(range(10), l) for k in multiset_permutations(j) if is_square(fi+sum(map(factorial, j))))
CROSSREFS
KEYWORD
nonn,base
AUTHOR
Michel Lagneau, Nov 25 2010
STATUS
approved