

A249916


a(n) = 4*(n  1)  a(n3), n >= 3, a(0) = a(1) = 1, a(2) = 5.


0



1, 1, 5, 7, 11, 11, 13, 13, 17, 19, 23, 23, 25, 25, 29, 31, 35, 35, 37, 37, 41, 43, 47, 47, 49, 49, 53, 55, 59, 59, 61, 61, 65, 67, 71, 71, 73, 73, 77, 79, 83, 83, 85, 85, 89, 91, 95, 95, 97, 97, 101, 103, 107, 107, 109, 109, 113, 115, 119, 119, 121, 121, 125, 127
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OFFSET

0,3


COMMENTS

Conjecture: These are the natural numbers of the form 6*j + 1 in which those of the form 12*k + 1 are repeated.
The second conjecture in Formula section is correct.
We can see this from the recurrence: a(n) = 4*(n  1)  a(n3) (1)
Replace n with n+1: a(n+1) = 4*n  a(n2) (2)
Subtract (1) from (2): a(n+1) = 4 + a(n)  a(n2) + a(n3) (3)
Replace n with n+1: a(n+2) = 4 + a(n+1)  a(n1) + a(n2) (4)
Subtract (3) from (4): a(n+2) = 2*a(n+1)a(n)a(n1)+2*a(n2)a(n3) (5)
This also confirms the conjecture in Comment section and the conjecture on the g.f. in Formula section. (End)


LINKS



FORMULA

G.f.: (1  x + 4*x^2  x^3 + x^4)/((1  x)^2*(1 + x^3)). [Confirmed, see Jianing Song in Comment section.]
Recurrence: a(n) = 2*a(n1)  a(n2)  a(n3) + 2*a(n4)  a(n5) for n > 4, a(0)=a(1)=1, a(2)=5, a(3)=7, a(4)=11. [Confirmed, see Jianing Song in Comment section.]


MATHEMATICA

a[0] = a[1] = 1; a[2] = 5; a[n_] := 4*(n  1)  a[n  3]; Table[a[n], {n, 0, 63}]
RecurrenceTable[{a[0]==a[1]==1, a[2]==5, a[n]==4(n1)a[n3]}, a, {n, 70}] (* Harvey P. Dale, Jan 26 2019 *)


CROSSREFS



KEYWORD

nonn,easy


AUTHOR



STATUS

approved



