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A106229 Least j > 1 for n > 0 such that j^2 = (n^2 + 1)*(k^2) + (n^2 + 1)*k + 1 where k sequence = A106230. 1
5, 19, 11, 35, 79, 149, 251, 391, 575, 809, 1099, 1451, 1871, 2365, 2939, 3599, 4351, 5201, 6155, 7219, 8399, 9701, 11131, 12695, 14399, 16249, 18251, 20411, 22735, 25229, 27899, 30751, 33791, 37025, 40459, 44099, 47951, 52021, 56315, 60839, 65599, 70601, 75851 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
For j^2 = (n^2 + 1)*(k^2) + (n^2 + 1)*k + 1, there is a sequence j(i,n) with a recurrence.
For n=1, j(1,1) = 1, j(2,1) = 5, j(i,1) = 6*j(i-1,1) - j(i-2,1).
For n=2, j(1,2) = 1, j(2,2) = 19, j(i,2) = 18*j(i-1,2) - j(i-2,2).
For n>2, j(1,n) = 1, j(2,n) = n^3 - 2*n^2 + n - 1, j(3,n) = n^3 + 2*n^2 + n + 1, j(4,n) = (4*n^2 + 2)*j(2,n) + 1 then j(i,n) = (4*n^2+2)*j(i-2,n) - j(i-4,n).
LINKS
FORMULA
For n > 2, a(n) = n^3 - 2*n^2 + n - 1.
MATHEMATICA
LinearRecurrence[{4, -6, 4, -1}, {5, 19, 11, 35, 79, 149}, 43] (* Georg Fischer, Oct 25 2020 *)
PROG
(PARI) a(n) = if(n<3, 14*n-9, n^3-2*n^2+n-1); \\ Jinyuan Wang, Apr 07 2020
CROSSREFS
Sequence in context: A370491 A089082 A217011 * A129734 A109325 A370824
KEYWORD
nonn,easy
AUTHOR
Pierre CAMI, Apr 26 2005
EXTENSIONS
More terms from Jinyuan Wang, Apr 07 2020
STATUS
approved

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Last modified May 13 06:21 EDT 2024. Contains 372498 sequences. (Running on oeis4.)