Bitwise shift and the largest integer in Bash

Bash uses intmax_t variables for arithmetic. On your system these are 64 bits in length, so:

$ echo $((1<<62))
4611686018427387904

which is

100000000000000000000000000000000000000000000000000000000000000

in binary (1 followed by 62 0s). Shift that again:

$ echo $((1<<63))
-9223372036854775808

which is

1000000000000000000000000000000000000000000000000000000000000000

in binary (63 0s), in two's complement arithmetic.

To get the biggest representable integer, you need to subtract 1:

$ echo $(((1<<63)-1))
9223372036854775807

which is

111111111111111111111111111111111111111111111111111111111111111

in binary.

If you shift again from -9223372036854775808, the shift wraps around: the left-most digit (1) "falls off" the end and is put back in from the right:

$ echo $((1<<64))
1

This repeats every 64 shifts:

$ echo $((1<<128))
1
$ echo $((1<<192))
1

etc. You can think of shift arithmetic as modulo 64, which explains the results you're seeing: $((1<<1025)) is $((1<<1)), $((1<<1026)) is $((1<<2))...

You'll find the type definitions and maximum values in stdint.h; on your system:

/* Largest integral types.  */
#if __WORDSIZE == 64
typedef long int                intmax_t;
typedef unsigned long int       uintmax_t;
#else
__extension__
typedef long long int           intmax_t;
__extension__
typedef unsigned long long int  uintmax_t;
#endif

/* Minimum for largest signed integral type.  */
# define INTMAX_MIN             (-__INT64_C(9223372036854775807)-1)
/* Maximum for largest signed integral type.  */
# define INTMAX_MAX             (__INT64_C(9223372036854775807))