A092758 Duplicate of A030516.
64, 729, 15625, 117649, 1771561, 4826809, 24137569, 47045881, 148035889
Offset: 1
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.
a(3) = 125; since the 3rd prime is 5, a(3) = 5^3 = 125.
a030078 = a000578 . a000040 a030078_list = map a000578 a000040_list -- Reinhard Zumkeller, May 26 2012
[p^3: p in PrimesUpTo(300)]; // Vincenzo Librandi, Mar 27 2014
Array[Prime[ # ]^3&, 5! ] (* Vladimir Joseph Stephan Orlovsky, Sep 01 2008 *)
a(n)=prime(n)^3 \\ Charles R Greathouse IV, Mar 20 2013
from sympy import prime, primerange def aupton(terms): return [p**3 for p in primerange(1, prime(terms)+1)] print(aupton(35)) # Michael S. Branicky, Aug 27 2021
[p**3 for p in prime_range(100)] # Zerinvary Lajos, May 15 2007
a030514 = (^ 4) . a000040 a030514_list = map (^ 4) a000040_list -- Reinhard Zumkeller, Jun 03 2015
[NthPrime(n)^4: n in [1..100] ]; // Vincenzo Librandi, Apr 22 2011
map(p -> p^4, select(isprime,[2,seq(2*i+1,i=1..100)])); # Robert Israel, Aug 29 2014
Array[Prime[#]^4 &, 5!] (* Vladimir Joseph Stephan Orlovsky, Sep 01 2008 *)
a(n)=prime(n)^4 \\ Charles R Greathouse IV, Mar 21 2013
[p**4 for p in prime_range(100)] # Zerinvary Lajos, May 15 2007
From _David A. Corneth_, May 12 2017: (Start) 1 has prime signature (), the first distinct prime signature. Therefore, a(1) = 1. 2 has prime signature (1), the second distinct prime signature after (1). Therefore, a(2) = 2. 3 has prime signature (1), as does 2. Therefore, a(3) = a(2) = 2. 4 has prime signature (2), the third distinct prime signature after () and (1). Therefore, a(4) = 3. (End) From _Antti Karttunen_, May 12 2017: (Start) Construction of restricted growth sequences: In this case we start with a(1) = 1 for A046523(1) = 1, and thereafter, for all n > 1, we use the least so far unused natural number k for a(n) if A046523(n) has not been encountered before, otherwise [whenever A046523(n) = A046523(m), for some m < n], we set a(n) = a(m). For n = 2, A046523(2) = 2, which has not been encountered before (first prime), thus we allot for a(2) the least so far unused number, which is 2, thus a(2) = 2. For n = 3, A046523(2) = 2, which was already encountered as A046523(1), thus we set a(3) = a(2) = 2. For n = 4, A046523(4) = 4, not encountered before (first square of prime), thus we allot for a(4) the least so far unused number, which is 3, thus a(4) = 3. For n = 5, A046523(5) = 2, as for the first time encountered at n = 2, thus we set a(5) = a(2) = 2. For n = 6, A046523(6) = 6, not encountered before (first semiprime pq with distinct p and q), thus we allot for a(6) the least so far unused number, which is 4, thus a(6) = 4. For n = 8, A046523(8) = 8, not encountered before (first cube of a prime), thus we allot for a(8) the least so far unused number, which is 5, thus a(8) = 5. For n = 9, A046523(9) = 4, as for the first time encountered at n = 4, thus a(9) = 3. (End) From _David A. Corneth_, May 12 2017: (Start) (Rough) description of an algorithm of computing the sequence: Suppose we want to compute a(n) for n in [1..20]. We set up a vector of 20 elements, values 0, and a number m = 1, the minimum number we haven't checked and c = 0, the number of distinct prime signatures we've found so far. [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0] We check the prime signature of m and see that it's (). We increase c with 1 and set all elements up to 20 with prime signature () to 1. In the process, we adjust m. This gives: [1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]. The least number we haven't checked is m = 2. 2 has prime signature (1). We increase c with 1 and set all elements up to 20 with prime signature (1) to 2. In the process, we adjust m. This gives: [1, 2, 2, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0] We check the prime signature of m = 4 and see that its prime signature is (2). We increase c with 1 and set all numbers up to 20 with prime signature (2) to 3. This gives: [1, 2, 2, 3, 2, 0, 2, 0, 3, 0, 2, 0, 2, 0, 0, 0, 2, 0, 2, 0] Similarily, after m = 6, we get [1, 2, 2, 3, 2, 4, 2, 0, 3, 4, 2, 0, 2, 4, 4, 0, 2, 0, 2, 0], after m = 8 we get: [1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 0, 2, 4, 4, 0, 2, 0, 2, 0], after m = 12 we get: [1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 0, 2, 6, 2, 0], after m = 16 we get: [1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 0], after m = 20 we get: [1, 2, 2, 3, 2, 4, 2, 5, 3, 4, 2, 6, 2, 4, 4, 7, 2, 6, 2, 8]. Now, m > 20 so we stop. (End) The above method is inefficient, because the step "set all elements a(n) up to n = Nmax with prime signature s(n) = S[c] to c" requires factoring all integers up to Nmax (or at least comparing their signature, once computed, with S[c]) again and again. It is much more efficient to run only once over each m = 1..Nmax, compute its prime signature s(m), add it to an ordered list in case it did not occur earlier, together with its "rank" (= new size of the list), and assign that rank to a(m). The list of prime signatures is much shorter than [1..Nmax]. One can also use m'(m) := the smallest n with the prime signature of m (which is faster to compute than to search for the signature) as representative for s(m), and set a(m) := a(m'(m)). Then it is sufficient to have just one counter (number of prime signatures seen so far) as auxiliary variable, in addition to the sequence to be computed. - _M. F. Hasler_, Jul 18 2019
A101296 := proc(n) local a046523, a; a046523 := A046523(n) ; for a from 1 do if A025487(a) = a046523 then return a; elif A025487(a) > a046523 then return -1 ; end if; end do: end proc: # R. J. Mathar, May 26 2017
With[{nn = 120}, Function[s, Table[Position[Keys@s, k_ /; MemberQ[k, n]][[1, 1]], {n, nn}]]@ Map[#1 -> #2 & @@ # &, Transpose@ {Values@ #, Keys@ #}] &@ PositionIndex@ Table[Times @@ MapIndexed[Prime[First@ #2]^#1 &, Sort[FactorInteger[n][[All, -1]], Greater]] - Boole[n == 1], {n, nn}] ] (* Michael De Vlieger, May 12 2017, Version 10 *)
find(ps, vps) = {for (k=1, #vps, if (vps[k] == ps, return(k)););}
lisps(nn) = {vps = []; for (n=1, nn, ps = vecsort(factor(n)[,2]); ips = find(ps, vps); if (! ips, vps = concat(vps, ps); ips = #vps); print1(ips, ", "););} \\ Michel Marcus, Nov 15 2015; edited by M. F. Hasler, Jul 16 2019
rgs_transform(invec) = { my(occurrences = Map(), outvec = vector(length(invec)), u=1); for(i=1, length(invec), if(mapisdefined(occurrences,invec[i]), my(pp = mapget(occurrences, invec[i])); outvec[i] = outvec[pp] , mapput(occurrences,invec[i],i); outvec[i] = u; u++ )); outvec; };
write_to_bfile(start_offset,vec,bfilename) = { for(n=1, length(vec), write(bfilename, (n+start_offset)-1, " ", vec[n])); }
write_to_bfile(1,rgs_transform(vector(100000,n,A046523(n))),"b101296.txt");
\\ Antti Karttunen, May 12 2017
[n: n in [1..200] | DivisorSigma(0, n) eq 4]; // Vincenzo Librandi, Jul 16 2015
Select[Range[200], DivisorSigma[0,#]==4&] (* Harvey P. Dale, Apr 06 2011 *)
is(n)=numdiv(n)==4 \\ Charles R Greathouse IV, May 18 2015
from math import isqrt from sympy import primepi, integer_nthroot, primerange def A030513(n): def f(x): return int(n+x-primepi(integer_nthroot(x,3)[0])+(t:=primepi(s:=isqrt(x)))+(t*(t-1)>>1)-sum(primepi(x//k) for k in primerange(1, s+1))) m, k = n, f(n) while m != k: m, k = k, f(k) return m # Chai Wah Wu, Aug 16 2024
[p^10: p in PrimesUpTo(300)]; // Vincenzo Librandi, Mar 27 2014
(Prime@Range@30)^10 (* Vladimir Joseph Stephan Orlovsky, Apr 11 2011 *)
is(n)=isprimepower(n)==10 \\ Charles R Greathouse IV, Jun 19 2016
from sympy import prime def A030629(n): return prime(n)**10 # Chai Wah Wu, Sep 13 2024
[p**10 for p in prime_range(100)] # Zerinvary Lajos, May 16 2007
0.0170700868506365129541...
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