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A263194
4-digit numbers (with leading zeros supplied where necessary) in which the sum of the number consisting of the first two digits and the number consisting of the last two digits equals the number consisting of the middle two digits.
1
0, 109, 110, 219, 220, 329, 330, 439, 440, 549, 550, 659, 660, 769, 770, 879, 880, 989, 990, 1208, 1318, 1428, 1538, 1648, 1758, 1868, 1978, 2307, 2417, 2527, 2637, 2747, 2857, 2967, 3406, 3516, 3626, 3736, 3846, 3956, 4505, 4615, 4725, 4835, 4945, 5604, 5714, 5824, 5934, 6703, 6813, 6923, 7802, 7912, 8901
OFFSET
1,2
COMMENTS
0 can be a leading digit.
REFERENCES
George Bredehorn, The Giant Book of Puzzles, Main Street, 2013, page 12.
FORMULA
The sequence contains stretches where, for some n, a(n) - a(n-1) = 110.
EXAMPLE
Since 19 + 78 = 97, 1978 is a term.
MATHEMATICA
fQ[n_] := Block[{d = PadLeft[IntegerDigits@ n, 4]}, FromDigits@ d[[1 ;; 2]] + FromDigits@ d[[3 ;; 4]] == FromDigits@ d[[2 ;; 3]]]; Select[Range[0, 10^4 - 1], fQ] (* Michael De Vlieger, Oct 26 2015 *)
PROG
(C)
#include <stdio.h>
int main(){
int e = 10; // what base we are using: experiment with different values (values above 10 do not work well)
for (int a = 0; a < e; a++){ // I know these nested loops are inelegant, but they're the easiest way
for (int b = 0; b < e; b++){
for (int c = 0; c < e; c++){
for (int d = 0; d < e; d++){
if ((10*a)+b+(10*c)+d == (10*b)+c){ // if the number formed by the first two digits plus the number formed by the last two digits equals the number formed by the middle two digits
if (e <= 10){
printf("%d%d%d%d, ", a, b, c, d); // print the number
}
else{
printf("%d %d %d %d, ", a, b, c, d); // print the number with extra spaces
}
}
}
}
}
}
printf("\n");
return 0;
}
(PARI) is(n) = n < 10000 && n%100 + n \ 100 == (n \ 10) % 100 \\ David A. Corneth, Oct 14 2017
(Python)
def ok(n): return (n//100) + (n%100) == (n//10)%100
print([m for m in range(10000) if ok(m)]) # Michael S. Branicky, Jan 25 2021
CROSSREFS
Cf. A293686.
Sequence in context: A130705 A253431 A253438 * A231701 A051046 A196667
KEYWORD
base,easy,fini,full,nonn
AUTHOR
Aresh Pourkavoos, Oct 11 2015
STATUS
approved