This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
a256512 n = n * (1 + (2 * n) ^ n)
Join[{0},Table[n(1+(2n)^n),{n,20}]] (* Harvey P. Dale, Aug 05 2021 *)
de:= diff(y(x),x) = x*y(x)^3/(1-2*x*y(x)):
S:= rhs(dsolve({de, y(0)=2},y(x), series, order=40)):
-1, seq(coeff(S,x,i)*(i+1)!,i=0..39); # Robert Israel, Apr 13 2020
CoefficientList[Series[-1/(1-LambertW[-2*x]), {x, 0, 20}], x] * Range[0, 20]!
my(x='x+O('x^20)); Vec(serlaplace(-1/(1-lambertw(-2*x)))) \\ Michel Marcus, Apr 13 2020
n a(n) a(n)^2 in base n 2 2 4 100 3 24 576 210100 4 16 256 10000 5 280 78400 10002100 6 216 46656 1000000 7 3430 11764900 202000000 8 4096 16777216 100000000 9 19683 387420489 1000000000 10 100000 10000000000 10000000000 11 4348377 18908382534129 6030000000000 12 2985984 8916100448256 1000000000000
for(b=2,12,for(k=1,oo,my(s=k^2,v=digits(s,b));if(sum(k=1,#v,v[k]==0)==b,print1(k,", ");break)))
from numba import njit
@njit # works with 64 bits through a(14)
def digits0(n, b):
count0 = 0
while n >= b:
n, r = divmod(n, b)
count0 += (r==0)
return count0 + (n==0)
from sympy import integer_nthroot
def a(n):
an = integer_nthroot(n**n, 2)[0]
while digits0(an*an, n) != n: an += 1
return an
print([a(n) for n in range(2, 13)]) # Michael S. Branicky, Apr 07 2021
from itertools import product from functools import reduce from sympy.utilities.iterables import multiset_permutations from sympy import integer_nthroot def A342545(n): for a in range(1,n): p, q = integer_nthroot(a*n**n,2) if q: return p l = 1 while True: cmax = n**(l+n+1) for a in range(1,n): c = cmax for b in product(range(1,n),repeat=l): for d in multiset_permutations((0,)*n+b): p, q = integer_nthroot(reduce(lambda c, y: c*n+y, [a]+d),2) if q: c = min(c,p) if c < cmax: return c l += 1 # Chai Wah Wu, Apr 07 2021
Table[((2n+1)^n-1)/(2n),{n,20}]
def A361291(n): return (((n<<1)+1)**n-1)//(n<<1) # Chai Wah Wu, Mar 14 2023
my(N=20, x='x+O('x^N)); Vec(serlaplace(exp(-x)/(1 + lambertw(-2*x))))
a(1) = 0 because 2*1 = 2 is too small to be the sum of two primes. a(2) = 2 because 4^2 = 16 = 3+13 = 5+11. a(3) = 13 because 6^3 = 216 and A180007(3) = Number of Goldbach partitions of 6^3 = 13. a(4) = 53 because 8^4 = 2^12 and A006307(12) = Number of ways writing 2^12 as unordered sums of 2 primes.
A180041 := proc(n) local a,m,p: if(n=1)then return 0:fi: a:=0: m:=(2*n)^n: p:=prevprime(ceil((m-1)/2)): while p > 2 do if isprime(m-p) then a:=a+1: fi: p := prevprime(p): od: return a: end: seq(A180041(n),n=1..5); # Nathaniel Johnston, May 08 2011
f[n_] := Block[{c = 0, p = 3, m = (2 n)^n}, lmt = Floor[m/2] + 1; While[p < lmt, If[ PrimeQ[m - p], c++ ]; p = NextPrime@p]; c]; Do[ Print[{n, f@n // Timing}], {n, 8}] (* Robert G. Wilson v, Aug 10 2010 *)
Table of fractions begins: 2, 3/2, 4/3, 5/4, ... 9/4, 25/16, 49/36, 81/64, ... 64/27, 343/216, 1000/729, 2197/1728, ... 625/256, 6561/4096, 28561/20736, 83521/65536, ... ... Table of denominators begins: 1, 2, 3, 4, ... 4, 16, 36, 64, ... 27, 216, 729, 1728, ... 256, 4096, 20736, 65536, ... ... Triangle of antidiagonals begins: 1; 2, 4; 3, 16, 27; 4, 36, 216, 256; ...
t[n_, k_] := (1+1/(k*n))^n; Table[t[n-k+1, k], {n, 1, 9}, {k, n, 1, -1}] // Flatten // Denominator
Triangle begins: [0] 1; [1] 0, 2; [2] 0, -4, 16; [3] 0, 6, -96, 216; [4] 0, -8, 384, -2592, 4096; [5] 0, 10, -1280, 19440, -81920, 100000; [6] 0, -12, 3840, -116640, 983040, -3000000, 2985984; [7] 0, 14, -10752, 612360, -9175040, 52500000, -125411328, 105413504;
T := (n, k) -> binomial(n, k) * k^n * 2^k * (-1)^(n - k): seq(seq(T(n, k), k = 0..n), n = 0..7);
A385899[n_, k_] := If[k == 0, Boole[n == 0], Binomial[n, k]*k^n*2^k*(-1)^(n - k)]; Table[A385899[n, k], {n, 0, 10}, {k, 0, n}] (* Paolo Xausa, Aug 03 2025 *)
888 is a term because the sum of the digits of (2*888)^888, 13320, is divisible by 888.
Do[If[Mod[Plus @@ IntegerDigits[(2*n)^n], n] == 0, Print[n]], {n, 1, 10000}]
A108860_list = [n for n in range(1,1000) if not sum(int(d) for d in str((2*n)**n)) % n] # Chai Wah Wu, Mar 15 2018
my(N=20, x='x+O('x^N)); Vec(serlaplace(1/((1-x)*(1+lambertw(-2*x)))))
a(n) = n!*sum(k=0, n, (2*k)^k/k!);
a_vector(n) = my(v=vector(n+1)); v[1]=1; for(i=1, n, v[i+1]=i*v[i]+(2*i)^i); v;
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