A334745 Starting with a(1) = a(2) = 1, proceed in a square spiral, computing each term as the sum of diagonally adjacent prior terms.

1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 2, 1, 1, 2, 2, 1, 1, 2, 2, 1, 1, 3, 2, 3, 1, 1, 3, 2, 3, 1, 1, 3, 3, 3, 3, 1, 1, 3, 3, 3, 3, 1, 1, 4, 3, 6, 3, 4, 1, 1, 4, 3, 6, 3, 4, 1, 1, 4, 4, 6, 6, 4, 4, 1, 1, 4, 4, 6, 6, 4, 4, 1, 1, 5, 4, 10, 6, 10, 4, 5, 1, 1, 5, 4, 10, 6

Offset: 1

Author

Alec Jones and Peter Kagey, May 09 2020

Keywords

• nonn

Examples

			Spiral begins:
... 3---3---3---3---1
                    |
1---1---2---2---1   1
|               |   |
2   1---1---1   1   3
|   |       |   |   |
2   1   1---1   2   2
|   |           |   |
1   1---2---1---1   3
|                   |
1---3---2---3---1---1
The last illustrated term above is a(35) = 3 = 2 + 1 because diagonally down-right is 2 and diagonally down-left is 1.
		

Links

• Peter Kagey, Table of n, a(n) for n = 1..10000
• Peter Kagey, Bitmap illustrating the parity of the first 2^22 terms. (Even and odd numbers are represented with black and white pixels respectively.)

Crossrefs

Cf. A141481, A278180, A334741, A334742.

The x- and y-coordinates at n-th step are A174344 and A274923 respectively.

Formula

Conjecture: a(2n-1) = A247976(n).