|
|
A069563
|
|
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.
|
|
2
|
|
|
1, 4, 9, 28, 85, 426, 3409, 23864, 167049, 1837540, 18375401, 128627808, 1157650273, 10418852458, 83350819665, 1250262294976, 22504721309569, 382580262262674, 4973543409414763, 64656064322391920
(list;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
1,2
|
|
COMMENTS
|
There is no solution to k * 64656064322391920 + 1 = 0 (mod 21). - Sean A. Irvine, May 04 2024
|
|
LINKS
|
|
|
EXAMPLE
|
a(6) = 426, a(7) = 3409 since 3409 = 8*426 + 1 is a multiple of 7.
|
|
MATHEMATICA
|
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}]
|
|
CROSSREFS
|
|
|
KEYWORD
|
nonn,fini,full,changed
|
|
AUTHOR
|
|
|
EXTENSIONS
|
|
|
STATUS
|
approved
|
|
|
|