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A108564
a(0) = 0, a(1) = 1, a(2) = 1, a(3) = 2, a(4) = 4, for n>3: a(n+1) = SORT[a(n) + a(n-1) + a(n-2) + a(n-3)], where SORT places digits in ascending order and deletes 0's.
4
0, 1, 1, 2, 4, 8, 15, 29, 56, 18, 118, 122, 134, 239, 136, 136, 456, 679, 147, 1148, 234, 228, 1577, 1378, 1347, 345, 4467, 3577, 3679, 1268, 11299, 12389, 23568, 24458, 11477, 12789, 22279, 137, 24668, 35789, 23788, 23488, 13377, 24469, 12258, 23579
OFFSET
0,4
COMMENTS
Sorted tetranacci numbers, a.k.a. sorted Fibonacci 4-step sequence.
As found by T. D. Noe: Max=4556699. Cycle period=41652. Cycle starts with the 23944th term.
LINKS
Richard I. Hess, Problem 920: sorted Fibonacci sequence, Pi Mu Epsilon Journal, Vol. 10 (Fall 1998) No. 9, pp. 754-755.
EXAMPLE
a(8) = SORT[a(4) + a(5) + a(6) + a(7)] = SORT[108] = 18.
a(10) = SORT[a(6) + a(7) + a(8) + a(9)] = SORT[221] = 122.
MATHEMATICA
nxt[{a_, b_, c_, d_, e_}]:={b, c, d, e, FromDigits[Sort[DeleteCases[ IntegerDigits[ b+c+d+e], _?0]]]}; NestList[nxt, {0, 1, 1, 2, 4}, 50][[All, 1]] (* Harvey P. Dale, Jan 17 2022 *)
CROSSREFS
KEYWORD
base,easy,nonn
AUTHOR
Jonathan Vos Post, Jun 10 2005
STATUS
approved