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A073117
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a(n+1) = a(n) + a(n) mod n; a(1) = 1.
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7
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1, 1, 2, 4, 4, 8, 10, 13, 18, 18, 26, 30, 36, 46, 50, 55, 62, 73, 74, 91, 102, 120, 130, 145, 146, 167, 178, 194, 220, 237, 264, 280, 304, 311, 316, 317, 346, 359, 376, 401, 402, 435, 450, 470, 500, 505, 550, 583, 590, 592, 634, 656, 688, 740, 778
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OFFSET
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1,3
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COMMENTS
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Conjecture (seems provable): More generally let a and b(1) be integers. If b(n+1) = b(n) + b(n) (mod(n+a)) there is an integer x(a,b(1)) such that b(n+1) = b(n) + x(a,b(1)) for n sufficiently large. We have x(0,1) = x(1,1) = x(2,1) = 97, x(3,1) = 1, x(4,1) = 3, x(5,1) = 3, x(6,1) = 6, ..., x(97,1) = 43, x(0,11) = 2, etc. - Benoit Cloitre, Aug 20 2002
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LINKS
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EXAMPLE
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a(397) = 38606 = 2*97*199 = (2*199)*97 = 398*97 = (397+1)*97; a(397) mod 397 = (397*97 + 97) mod 397 = 97, a(398) = a(397) + a(397) mod 397 = (397+1)*97 + 97 = (398+1)*97, etc.: a(n+1) = a(n) + 97 for n >= 397.
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MATHEMATICA
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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STATUS
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approved
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