A082606 n-th powers arising in A082605.
3, 9, 125, 810000, 12181395632886429300000
Offset: 1
Keywords
Formula
Extensions
More terms from David Wasserman, Sep 21 2004
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.
The irregular triangle begins as: 5; 7, 11, 17; 13, 17, 23, 31, 41, 53, 67, 83, 101; 19, 23, 29, 37, 47, 59, 73, 89, 107, 127, 149, 173, 199, 227, 257;
b:= func< n | n eq 1 select 2 else 2^(n-3)*(9-(-1)^n) >; A107448:= func< n,k | b(n) +k^2 +k +1 >; [A107448(n,k): k in [1..b(n)-1], n in [1..8]]; // G. C. Greubel, Mar 23 2024
(* First program *) a[1] = 3; a[2] = 5; a[3] = 11; a[n_]:= a[n]= Abs[1-4*a[n-2]] -2; euler= Table[a[n], {n,10}]; Table[k^2 + k + euler[[n]], {n,7}, {k,euler[[i]] -2}]//Flatten (* Second program *) b[n_]:= 2^(n-3)*(9-(-1)^n) - Boole[n==1]/2; T[n_, k_]:= b[n] +k^2+k+1; Table[T[n,k], {n,8}, {k,b[n]-1}]//Flatten (* G. C. Greubel, Mar 23 2024 *)
def b(n): return 2^(n-3)*(9-(-1)^n) - int(n==1)/2 def A107448(n,k): return b(n) + k^2+k+1; flatten([[A107448(n,k) for k in range(1,b(n))] for n in range(1,8)]) # G. C. Greubel, Mar 23 2024
The irregular triangle begins as: 5; 3, 9, 3; 7, 3, 7, 9, 9, 7, 3, 7, 9; 1, 7, 1, 3, 3, 1, 7, 1, 3, 3, 1, 7, 1, 3, 3;
b:= func< n | n eq 1 select 2 else 2^(n-3)*(9-(-1)^n) >; A107448:= func< n, k | 10 - ((b(n) +k^2 +k +1) mod 10) >; [5,3,9,3] cat [A107448(n, k): k in [1..b(n)-1], n in [3..8]]; // G. C. Greubel, Mar 24 2024
b[n_]:= 2^(n-3)*(9-(-1)^n) -Boole[n==1]/2; T[n_, k_]:= 10 -Mod[k^2+k+1+b[n], 10]; Table[T[n, k], {n,8}, {k,b[n]-1}]//Flatten (* G. C. Greubel, Mar 24 2024 *)
def b(n): return 2^(n-3)*(9-(-1)^n) - int(n==1)/2 def A107449(n, k): return 10 - ((b(n) + k^2+k+1)%10); flatten([[A107449(n, k) for k in range(1, b(n))] for n in range(1, 8)]) # G. C. Greubel, Mar 24 2024
a(n) = 2^(n-1) + (2 + (-1)^n)^((n-2)/2) \\ Charles R Greathouse IV, Jan 29 2013
select(isprime, [seq(n^2 + 5*n + (-1)^n * 3, n=1..1000)]); # Robert Israel, Aug 25 2025
f[n_] := If[Mod[n, 2] == 1, n^2 + 5*n - 3, n^2 + 5*n + 3] b = Flatten[Table[If[PrimeQ[f[n]] == True, f[n], {}], {n, 1, 100}]]
for(n=1, 250, k=n^2+5*n+3-6*(n%2); if(isprime(k), print1(k,", ")))
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