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.
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
Examples
Since 19 + 78 = 97, 1978 is a term.
References
- George Bredehorn, The Giant Book of Puzzles, Main Street, 2013, page 12.
Links
- Aresh Pourkavoos, Website with C code to generate sequence
Crossrefs
Cf. A293686.
Programs
-
C
#include
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; } -
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 *) -
PARI
is(n) = n < 10000 && n%100 + n \ 100 == (n \ 10) % 100 \\ David A. Corneth, Oct 14 2017
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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
Formula
The sequence contains stretches where, for some n, a(n) - a(n-1) = 110.
Comments