OFFSET
1,2
COMMENTS
0 can be a leading digit.
REFERENCES
George Bredehorn, The Giant Book of Puzzles, Main Street, 2013, page 12.
LINKS
Aresh Pourkavoos, Website with C code to generate sequence
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
KEYWORD
base,easy,fini,full,nonn
AUTHOR
Aresh Pourkavoos, Oct 11 2015
STATUS
approved