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
Keywords
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
Programs
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Maple
seq(2^n mod n*(n+1)/2, n=1..100); # Robert Israel, Jul 04 2017
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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 *)
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PARI
a(n) = lift(Mod(2, n*(n+1)/2)^n); \\ Michel Marcus, Jul 04 2017
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Python
for n in range(1,99): print(str(int(2**n % (n*(n+1)/2))), end=', ')
Formula
a(n) = 1 for n>1 in A272934. - Michel Marcus, Jul 04 2017
Comments