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A069563
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a(1) = 1, a(2) = 4; for n > 2, a(n) = k*a(n-1) + 1 where k is smallest number > 1 such that k*a(n-1) + 1 is a multiple of n.
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2
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1, 4, 9, 28, 85, 426, 3409, 23864, 167049, 1837540, 18375401, 128627808, 1157650273, 10418852458, 83350819665, 1250262294976, 22504721309569, 382580262262674, 4973543409414763, 64656064322391920
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OFFSET
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1,2
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COMMENTS
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There is no solution to k * 64656064322391920 + 1 = 0 (mod 21). - Sean A. Irvine, May 04 2024
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LINKS
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EXAMPLE
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a(6) = 426, a(7) = 3409 since 3409 = 8*426 + 1 is a multiple of 7.
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MATHEMATICA
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a[1] = 1; a[n_] := a[n] = Module[{k}, If[ Intersection[ Transpose[ FactorInteger[a[n - 1]]] [[1]], Transpose[ FactorInteger[n]] [[1]]] == {}, k = 2; While[ !IntegerQ[(k*a[n - 1] + 1)/n], k++ ]; Return[ k*a[n - 1] + 1], k = 1; While[ !IntegerQ[(k*a[n - 2] + 1)/n], k++ ]; Return[ k*a[n - 2] + 1]]]; Table[ a[n], {n, 1, 23}]
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CROSSREFS
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KEYWORD
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nonn,fini,full
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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