A094925
A hexagonal spiral Fibonacci sequence.
Original entry on oeis.org
1, 1, 2, 4, 7, 12, 20, 34, 55, 90, 148, 240, 394, 638, 1043, 1688, 2750, 4450, 7232, 11736, 19002, 30827, 49884, 80856, 130978, 211982, 343348, 555964, 899706, 1456702, 2358089, 3815834, 6176654, 9996926, 16176330, 26180456, 42368468, 68567892
Offset: 1
a(2) = a(1) = 1,
a(3) = a(1) + a(2) = 2,
a(4) = a(1) + a(2) + a(3) = 4,
a(5) = a(1) + a(3) + a(4) = 7,
a(6) = a(1) + a(4) + a(5) = 12,
a(7) = a(1) + a(5) + a(6) = 20, etc.
Thus:
12----20----34
/ \
/ \
7 1-----1 55
\ / /
\ / /
638 4-----2 90
\ /
\ /
394---240---148
A092360
Spiro-tribonacci numbers: a(n) = sum of three previous terms that are nearest when terms arranged in a spiral.
Original entry on oeis.org
0, 1, 1, 3, 5, 8, 13, 14, 28, 43, 45, 89, 135, 138, 143, 284, 430, 438, 451, 897, 1356, 1404, 1446, 2878, 4352, 4423, 4511, 4645, 9245, 13979, 14203, 14476, 14757, 15184, 30225, 45693, 46407, 47275, 48164, 49512, 98573, 148982, 151235, 153968, 156749
Offset: 0
Terms are written in square boxes radiating spirally (cf. Ulam prime spiral). a(0) = 0, a(1) = 1 and a(2) = 1, so write 0, then 1 to its right, and another 1 below the first 1. The next unfilled box forms a square with the three filled boxes, so a(3) = a(0) + a(1) + a(2) = 0 + 1 + 1 = 2.
.
8--13--14--28
|
5 0---1
| |
3---2---1
.
a(4) = 2 because a(0) + a(1) + a(2) = 0 + 1 + 1 = 2.
A092369
Spiro-tetranacci numbers: a(n) = sum of four previous terms that are nearest when terms arranged in a spiral.
Original entry on oeis.org
0, 1, 1, 1, 3, 5, 9, 15, 25, 41, 68, 111, 181, 294, 299, 597, 900, 1505, 1522, 3041, 4577, 7642, 7691, 7772, 15529, 23367, 39005, 39225, 39585, 79102, 118979, 198556, 199330, 200520, 202316, 404333, 608146, 1013976, 1017903, 1023971, 1033111
Offset: 0
Terms are written in square boxes radiating spirally (cf. Ulam prime spiral). a(0) = 0, a(1) = 1, a(2) = 1 and a(3) = 1, so write 0, then 1 to its right, another 1 below the first 1 and another to the left of that. The next unfilled box forms an incomplete rectangle with the four filled boxes, so a(4) = a(0) + a(1) + a(2) + a(3) = 0 + 1 + 1 + 1 = 3. The next unfilled box forms the complete rectangle with the filled boxes. Since a(2) is nearer than a(3), a(5) = a(0) + a(1) + a(3) + a(4) = 0 + 1 + 3 + 5 = 9. In the case of a tie in nearness, the chronologically nearer value is used.
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