A289544 - OEIS

%I #48 Dec 05 2023 16:12:24

%S 2467891035,2497861035,2647891053,2697841053,2847691053,2867491035,

%T 2897461035,2897641053,3247651089,3257641089,3427561098,3467521098,

%U 3478591206,3498571206,3527461098,3567421098,3578491206,3598471206,3647251089,3657241089,4236751098,4256731098

%N Pandigital numbers (each digit 0-9 used exactly once) where the first 3 digits plus the next 3 digits equals the last 4 digits.

%C Leading zeros in the last four digits are not included, else 1246590783, with 124 + 659 = 783 would be the first term.

%H Jonathan Schwartz, <a href="/A289544/b289544.txt">Table of n, a(n) for n = 1..96</a>

%H David A. Corneth, <a href="/A289544/a289544.gp.txt">PARI program</a>

%e 2467891035 is in the sequence as 246|789|1035: 246 + 789 = 1035 and each digit (0-9) is used exactly once.

%t FromDigits /@ Select[Permutations[Range[0, 9]], And[#1 + #2 == #3, #3 >= 1000] & @@ Map[FromDigits, {Take[#, 3], #[[4 ;; 6]], Take[#, -4]}] &] (* _Michael De Vlieger_, Aug 02 2018 *)

%o (Python)

%o from itertools import permutations

%o def t2i(t): return int("".join(map(str, t)))

%o alst = [t2i(p) for p in permutations(range(10)) if p[6] != 0 and t2i(p[:3]) + t2i(p[3:6]) == t2i(p[6:])]

%o print(alst) # _Michael S. Branicky_, May 30 2022

%Y Cf. A050278, A289552, A286846.

%K nonn,base,fini,full

%O 1,1

%A _Jonathan Schwartz_, Aug 02 2017