A289382 a(n) = 2^n mod triangular(n).

0, 1, 2, 6, 2, 1, 16, 4, 17, 34, 2, 40, 2, 4, 8, 120, 104, 1, 78, 46, 134, 70, 140, 16, 132, 121, 134, 30, 2, 94, 64, 400, 206, 429, 158, 334, 2, 4, 8, 616, 494, 1, 690, 346, 692, 142, 848, 64, 912, 1024, 8, 796, 797, 379, 1528, 4, 350, 178, 1418, 916, 2, 4, 512, 1056, 32, 2011

Offset: 1

Author

Alex Ratushnyak, Jul 04 2017

Keywords

• nonn

Examples

			a(6) = 2**6 mod (6*7/2) = 64 mod 21 = 1.
		

Links

• Seiichi Manyama, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000079, A000217, A066606, A272934.

Programs

Maple

seq(2^n mod n*(n+1)/2, n=1..100); # Robert Israel, Jul 04 2017

Mathematica

Table[Mod[2^n, n (n + 1)/2], {n, 66}] (* Michael De Vlieger, Jul 04 2017 *)
PowerMod[2,#,(#(#+1))/2]&/@Range[70] (* Harvey P. Dale, Oct 12 2018 *)

PARI

a(n) = lift(Mod(2, n*(n+1)/2)^n); \\ Michel Marcus, Jul 04 2017

Python

for n in range(1,99): print(str(int(2**n % (n*(n+1)/2))), end=', ')

Formula

a(n) = A000079(n) mod A000217(n) = 2^n mod n*(n+1)/2.

a(n) = 1 for n>1 in A272934. - Michel Marcus, Jul 04 2017