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A303029
From a riddle, see Puzzling.SE link.
1
3, 1, 4, 8, 8, 21, 21, 62, 128, 190, 430, 831, 1451, 3030, 6143, 12286, 24361, 48850, 85497, 134347, 268694, 583208, 1071746, 2192342, 3264088, 7514425, 14042601, 24821114, 46378140, 99867664, 171066918, 270934582, 634625444, 1272514976, 2449009584, 0, 2449009584
OFFSET
0,1
LINKS
FORMULA
a(n) = 0 for all n > 362. - Alois P. Heinz, Aug 18 2018
From Jianing Song, Dec 25 2022: (Start)
Let d_k = A000796(k+1) be the k-th digit of Pi, then a(n) = a(n-1) + a(n-2) + ... + a(n-d_{n-3}) for n >= 3.
If there exists consecutive 9 digits ...d_{k}d_{k+1}...d_{k+8}... of Pi such that d_{k+i} <= i for i = 0..8, then a(n) = 0 for all n >= k+3. The 360th to 368th digits of Pi are ...001133053..., so a(n) = 0 for all n >= 363. (End)
EXAMPLE
a(0,1,2) = 3,1,4
To continue, we use the decimal expansion of Pi = 3.14159...:
a(3) = 3+1+4 (3-bonacci) = 8
a(4) = 8 (1-bonacci) = 8
a(5) = 1+4+8+8 (4-bonacci) = 21
a(6) = 21 (1-bonacci) = 21
a(7) = 21+21+8+8+4 (5-bonacci) = 62
...
CROSSREFS
Cf. A000796.
Sequence in context: A329130 A057049 A245672 * A050059 A025121 A025097
KEYWORD
nonn,base,easy,less
AUTHOR
David F. Marrs, Aug 16 2018
STATUS
approved