Sequences  math puzzle
What are the next numbers in the following sequences?
a: 77, 143, 221, 323, 437
b: 1, 5, 32, 288, 3413
c: 827, 23, 11, 6, 3
d: 2, 3, 5, 11, 31
e: 100, 121, 144, 202, 244
a: 77, 143, 221, 323, 437
b: 1, 5, 32, 288, 3413
c: 827, 23, 11, 6, 3
d: 2, 3, 5, 11, 31
e: 100, 121, 144, 202, 244
Explanation
Sequence a:
77, 143, 221, 323, 437, 667
because
77 = 7 * 11
143 = 11 * 13
221 = 13 * 17
323 = 17 * 19
437 = 19 * 23
667 = 23 * 29
The numbers in the sequence are the product of two sequentially following prime numbers.
Sequence b:
1, 5, 32, 288, 3413, 50069
because
1 = 1^1
5 = 1^1 + 2^2
32 = 1^1 + 2^2 + 3^3
288 = 1^1 + 2^2 + 3^3 + 4^4
3413 = 1^1 + 2^2 + 3^3 + 4^4 + 5^5
50069 = 1^1 + 2^2 + 3^3 + 4^4 + 5^5 + 6^6
The nth number in the sequence is the sum from k = 1 to n of k^k.
Sequence c:
827, 21, 12, 6, 3, 4
because
827 = eight hunderd twenty seven  23 characters
23 = twenty three  11 characters
11 = eleven  6 characters
6 = six  3 characters
3 = three  5 characters
5
The (n+1)th number is the number of characters of the nth number
Sequence d:
2, 3, 5, 11, 31, 127
because
3 is the second prime number
5 is the third prime number
11 is the fifth prime number
31 is the eleventh prime number
127 is the thirthy first prime number
The prime number to 127 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 83, 89, 91, 97, 101, 103, 107, 109, 113, 127
The (n+1)th number is the xth prime number, where x is the nth number in the sequence.
Sequence e:
100, 121, 144, 202, 244, 400
because
100 in base 10 = 100 in base 10
121 in base 9 = 81 + 18 + 1 = 100 in base 10
144 in base 8 = 64 + 32 + 4 = 100 in base 10
202 in base 7 = 98 + 0 + 2 = 100 in base 10
244 in base 6 = 72 + 24 + 4 = 100 in base 10
400 in base 5 = 100 in base 10
The nth number is is the number which in base (11n) corresponds to 100 in base 10.
77, 143, 221, 323, 437, 667
because
77 = 7 * 11
143 = 11 * 13
221 = 13 * 17
323 = 17 * 19
437 = 19 * 23
667 = 23 * 29
The numbers in the sequence are the product of two sequentially following prime numbers.
Sequence b:
1, 5, 32, 288, 3413, 50069
because
1 = 1^1
5 = 1^1 + 2^2
32 = 1^1 + 2^2 + 3^3
288 = 1^1 + 2^2 + 3^3 + 4^4
3413 = 1^1 + 2^2 + 3^3 + 4^4 + 5^5
50069 = 1^1 + 2^2 + 3^3 + 4^4 + 5^5 + 6^6
The nth number in the sequence is the sum from k = 1 to n of k^k.
Sequence c:
827, 21, 12, 6, 3, 4
because
827 = eight hunderd twenty seven  23 characters
23 = twenty three  11 characters
11 = eleven  6 characters
6 = six  3 characters
3 = three  5 characters
5
The (n+1)th number is the number of characters of the nth number
Sequence d:
2, 3, 5, 11, 31, 127
because
3 is the second prime number
5 is the third prime number
11 is the fifth prime number
31 is the eleventh prime number
127 is the thirthy first prime number
The prime number to 127 are:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 83, 89, 91, 97, 101, 103, 107, 109, 113, 127
The (n+1)th number is the xth prime number, where x is the nth number in the sequence.
Sequence e:
100, 121, 144, 202, 244, 400
because
100 in base 10 = 100 in base 10
121 in base 9 = 81 + 18 + 1 = 100 in base 10
144 in base 8 = 64 + 32 + 4 = 100 in base 10
202 in base 7 = 98 + 0 + 2 = 100 in base 10
244 in base 6 = 72 + 24 + 4 = 100 in base 10
400 in base 5 = 100 in base 10
The nth number is is the number which in base (11n) corresponds to 100 in base 10.
