A108859
Numbers k such that k divides the sum of the digits of k^(2k).
Original entry on oeis.org
1, 3, 5, 9, 18, 63, 72, 74, 104, 111, 116, 117, 565, 621, 734, 1242, 1620, 4596, 4728, 5823, 5956, 21135, 28251, 46530, 46908, 78257, 129619, 277407, 463689, 464706, 599119
Offset: 1
734 is a term because the sum of the digits of 734^(2*734), 19084, is divisible by 734.
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Do[If[Mod[Plus @@ IntegerDigits[n^(2*n)], n] == 0, Print[n]], {n, 1, 10000}]
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from gmpy2 import digits, mpz
def ok(n): return n and sum(map(mpz, digits(n**(2*n))))%n == 0
print([k for k in range(2000) if ok(k)]) # Michael S. Branicky, May 08 2025
A167436
3rd Fibonacci polynomial evaluated at n^n.
Original entry on oeis.org
2, 17, 730, 65537, 9765626, 2176782337, 678223072850, 281474976710657, 150094635296999122, 100000000000000000001, 81402749386839761113322, 79496847203390844133441537, 91733330193268616658399616010
Offset: 1
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Table[Fibonacci[3,n^n],{n,20}] (*and/or*) Table[(n^2)^n+1,{n,17}]
A356568
a(n) = (4^n - 1)*n^(2*n).
Original entry on oeis.org
0, 3, 240, 45927, 16711680, 9990234375, 8913923665920, 11111328602485167, 18446462598732840960, 39346257980661240576303, 104857500000000000000000000, 341427795961470170556885610263, 1333735697353436921058237339402240, 6156119488473827117528057630000587767
Offset: 0
For n=1, the functions are f1: (1,1),(2,1); f2: (1,2),(2,2); f3: (1,2),(2,1).
A085525
a(n) = n^(2*n + 2).
Original entry on oeis.org
0, 1, 64, 6561, 1048576, 244140625, 78364164096, 33232930569601, 18014398509481984, 12157665459056928801, 10000000000000000000000, 9849732675807611094711841, 11447545997288281555215581184, 15502932802662396215269535105521, 24201432355484595421941037243826176
Offset: 0
A259926
a(n) = n^(2*n) - n^(2*n - 1).
Original entry on oeis.org
0, 8, 486, 49152, 7812500, 1813985280, 581334062442, 246290604621824, 133417453597332552, 90000000000000000000, 74002499442581601012110, 72872109936441607122321408, 84676920178401799992368876316, 114656931713301654695784797437952, 178967655284025147557258605957031250
Offset: 1
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[n^(2*n) - n^(2*n - 1): n in [1..20]]; // Vincenzo Librandi, Jul 10 2015
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Table[n^(2 n) - n^(2 n - 1), {n, 15}]
Array[#^(2 #) - #^(2 # - 1)&, 15] (* Vincenzo Librandi, Jul 10 2015 *)
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vector(20, n, n^(2*n) - n^(2*n-1)) \\ Michel Marcus, Jul 09 2015
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[n**(2*n) - n**(2*n - 1) for n in range(1, 20)] # Anders Hellström, Jul 10 2015
A356691
a(n) = n! * Sum_{k=0..n} k^(2*k)/k!.
Original entry on oeis.org
1, 2, 20, 789, 68692, 10109085, 2237436846, 693885130771, 287026057756824, 152677869816810537, 101526778698168105370, 82519543952519610272391, 80487081730821079456710228, 92779662255769290691336848973, 124775610962828705895908497741878
Offset: 0
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a(n) = n!*sum(k=0, n, k^(2*k)/k!);
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a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=i*v[i]+i^(2*i)); v;
A382359
Number of labeled deterministic finite automata with n states and two letters.
Original entry on oeis.org
2, 128, 17496, 4194304, 1562500000, 835884417024, 607687873272704, 576460752303423488, 691636079448571949568, 1024000000000000000000000, 1833841138186726138360895488, 3907429033741066770846918377472, 9769232732262334599652925506494464
Offset: 1
For n = 1, we have two choices (a(1)=2), either the node is an accept state or not. We have no choice but to send both letters of the alphabet to itself, and only one choice for the start state. Therefore 1*2*1 = 2.
For n = 2, we have 2 choices for starting, 4 choices for which states are accepting, and 2^4 choices for transition functions. So a(2) = 2*4*16 = 128.
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