A271390 a(n) = (2*n + 1)^(2*floor((n-1)/2) + 1).

1, 3, 5, 343, 729, 161051, 371293, 170859375, 410338673, 322687697779, 794280046581, 952809757913927, 2384185791015625, 4052555153018976267, 10260628712958602189, 23465261991844685929951, 59938945498865420543457, 177482997121587371826171875, 456487940826035155404146917

Offset: 0

Author

Ilya Gutkovskiy, Apr 06 2016

Keywords

• nonn
• easy

Comments

All members are odd, therefore:

........................

| k | a(n) mod k |

|.......|..............|

| n+1 | A001477(n) |

| 2*n+2 | A005408(n) |

| 2 | A000012(n) |

| 3 | A080425(n+2)|

| 4 | A010684(n) |

| 6 | A130793(n) |

........................

Final digit of (2*n + 1)^(2*floor((n-1)/2) + 1) gives periodic sequence -> period 20: repeat [1,3,5,3,9,1,3,5,3,9,1,7,5,7,9,1,7,5,7,9], defined by the recurrence relation b(n) = b(n-2) - b(n-4) + b(n+5) + b(n+6) - b(n-7) - b(n-8) + b(n-9) - b(n-11) + b(n-13).

Examples

			a(0) =  1;
a(1) =  3^1 = 3;
a(2) =  5^1 = 5;
a(3) =  7^3 = 343;
a(4) =  9^3 = 729;
a(5) = 11^5 = 161051;
a(6) = 13^5 = 371293;
a(7) = 15^7 = 170859375;
a(8) = 17^7 = 410338673;
...
a(10000) = 1.644...*10^43006;
...
a(100000) = 8.235...*10^530097, etc.
This sequence can be represented as a binary tree:
                                    1
                 ................../ \..................
                3^1                                   5^1
     7^3......../ \......9^3                11^5....../ \.......13^5
     / \                 / \                 / \                 / \
    /   \               /   \               /   \               /   \
   /     \             /     \             /     \             /     \
15^7    17^7        19^9    21^9        23^11   25^11       27^13   29^13
		

Links

• Ilya Gutkovskiy, Table of n, a(n) for n = 0..75

Crossrefs

Cf. A005408, A092503, A109613.

Programs

Maple

A271390:=n->(2*n + 1)^(n - 1/2 - (-1)^n/2): seq(A271390(n), n=0..30); # Wesley Ivan Hurt, Apr 10 2016

Mathematica

Table[(2 n + 1)^(2 Floor[(n - 1)/2] + 1), {n, 0, 18}]
Table[(2 n + 1)^(n - 1 + (1 + (-1)^(n - 1))/2), {n, 0, 18}]

PARI

a(n) = (2*n + 1)^(2*((n-1)\2) + 1); \\ Altug Alkan, Apr 06 2016

Python

for n in range(0,10**3):print((int)((2*n+1)**(2*floor((n-1)/2)+1)))
# Soumil Mandal, Apr 10 2016

Formula

a(n) = (2*n + 1)^(n - 1 + (1 + (-1)^(n-1))/2).

a(n) = A005408(n)^A109613(n-1).

a(n) = (2*n + 1)^(n - 1/2 - (-1)^n/2). - Wesley Ivan Hurt, Apr 10 2016