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A394410
Quadratic recurrence sequence with a(1)=1, a(2)=3, and a(n) = (a(n-1)^2 + a(n-2)^2)/2 for n > 2.
1
1, 3, 5, 17, 157, 12469, 77750305, 3022555041534493, 4567919489552793374727018180037, 10432944231518126190458416657023933731548823182242327927044209
OFFSET
1,2
COMMENTS
All terms are odd.
For n >= 4, a(n) mod 8 follows the period-3 cycle (1, 5, 5).
All terms except a(2) = 3 are sums of two squares.
a(n) is prime for n = 2, 3, 4, 5, and 10. The first composite term is a(6) = 37*337.
Thomas Ordowski conjectured in A056826 that if a(n) is prime, then (a(n)^a(n) + 1)/(a(n) + 1) is also prime.
REFERENCES
G. Everest, A. van der Poorten, I. Shparlinski, and T. Ward, Recurrence Sequences, American Mathematical Society, 2003.
LINKS
FORMULA
a(n) = (a(n-1)^2 + a(n-2)^2)/2 for n > 2, with a(1) = 1, a(2) = 3.
a(n) ~ 2*C^(2^n) where C = 1.1462901..., equivalently log(a(n)) ~ 0.1365302*2^n + log(2).
EXAMPLE
a(3) = (3^2 + 1^2)/2 = 10/2 = 5.
a(4) = (5^2 + 3^2)/2 = 34/2 = 17.
a(5) = (17^2 + 5^2)/2 = 314/2 = 157.
a(6) = (157^2 + 17^2)/2 = 24938/2 = 12469.
MAPLE
a:= proc(n) option remember;
`if`(n<3, 2*n-1, (a(n-1)^2+a(n-2)^2)/2)
end:
seq(a(n), n=1..10); # Alois P. Heinz, Mar 20 2026
MATHEMATICA
Module[{a, n}, RecurrenceTable[{a[n] == (a[n-1]^2 + a[n-2]^2)/2, a[1] == 1, a[2] == 3}, a, {n, 12}]] (* Paolo Xausa, Mar 30 2026 *)
PROG
(Python)
from functools import lru_cache
@lru_cache(maxsize=None)
def a(n):
if n == 1:
return 1
if n == 2:
return 3
return (a(n-1)**2 + a(n-2)**2) // 2
print([a(n) for n in range(1, 10)])
(PARI) lista(nn) = {my(va=vector(nn+2)); va[1]=1; va[2]=3; for(n=1, nn, va[n+2]=(va[n]^2+va[n+1]^2)/2; ); va; } \\ Bruce Nye, Mar 20 2026
CROSSREFS
Cf. A000290 (squares), A001481 (sums of two squares), A002315 (NSW numbers), A001333 (Pell-related).
Cf. A056826 (primes in this sequence satisfy a related primality conjecture, per T. Ordowski, 2013).
Sequence in context: A083213 A171271 A056826 * A370879 A278138 A273870
KEYWORD
nonn
AUTHOR
Rayhan Ahmed, Mar 20 2026
STATUS
approved