A005085 -id:A005085 - cp's OEIS frontend

A005065 Sum of 4th powers of primes dividing n.

0, 16, 81, 16, 625, 97, 2401, 16, 81, 641, 14641, 97, 28561, 2417, 706, 16, 83521, 97, 130321, 641, 2482, 14657, 279841, 97, 625, 28577, 81, 2417, 707281, 722, 923521, 16, 14722, 83537, 3026, 97, 1874161, 130337, 28642, 641, 2825761, 2498, 3418801, 14657, 706, 279857, 4879681, 97, 2401, 641, 83602, 28577, 7890481, 97

Offset: 1

N. J. A. Sloane

nonn

Primes are taken without multiplicity, e.g., 12 = 2*2*3, and a(12) = 2^4+3^4 = 97. - Harvey P. Dale, Jul 16 2014

Inverse Möbius transform of n^4 * c(n), where c(n) is the prime characteristic (A010051). - Wesley Ivan Hurt, Jun 22 2024

Antti Karttunen, Table of n, a(n) for n = 1..10000

Column k=4 of A322080.
Cf. A000583, A005068, A005073, A005077, A005081, A005085, A008472, A059841, A079978.
Sum of the k-th powers of the primes dividing n for k=0..10 : A001221 (k=0), A008472 (k=1), A005063 (k=2), A005064 (k=3), this sequence (k=4), A351193 (k=5), A351194 (k=6), A351195 (k=7), this sequence (k=8), A351197 (k=9), A351198 (k=10).
Cf. A010051.

A005065 := proc(n)
        add(d^4, d= numtheory[factorset](n)) ;
end proc;
seq(A005065(n),n=1..40) ; # R. J. Mathar, Nov 08 2011

Join[{0},Table[Total[Transpose[FactorInteger[n]][[1]]^4],{n,2,40}]] (* Harvey P. Dale, Jul 16 2014 *)
Array[DivisorSum[#, #^4 &, PrimeQ] &, 54] (* Michael De Vlieger, Jul 11 2017 *)

a(n) = my(f=factor(n)); sum(k=1, #f~, f[k,1]^4); \\ Michel Marcus, Jul 11 2017

from sympy import primefactors
def a(n): return sum(p**4 for p in primefactors(n))
print([a(n) for n in range(1, 101)]) # Indranil Ghosh, Jul 11 2017

(define (A005065 n) (if (= 1 n) 0 (+ (A000583 (A020639 n)) (A005065 (A028234 n))))) ;; Antti Karttunen, Jul 10 2017

Additive with a(p^e) = p^4.
From Antti Karttunen, Jul 11 2017: (Start)
a(n) = A005068(n) + 16*A059841(n).
a(n) = A005081(n) + A005085(n) + 16*A059841(n).
a(n) = A005073(n) + A005077(n) + 81*A079978(n).
(End)
G.f.: Sum_{k>=1} prime(k)^4*x^prime(k)/(1 - x^prime(k)). - Ilya Gutkovskiy, Dec 24 2018
a(n) = Sum_{p|n, p prime} p^4. - Wesley Ivan Hurt, Feb 04 2022
a(n) = Sum_{d|n} d^4 * c(d), where c = A010051. - Wesley Ivan Hurt, Jun 22 2024

More terms from Antti Karttunen, Jul 10 2017

A005083 Sum of squares of primes = 3 mod 4 dividing n.

Original entry on oeis.org

0, 0, 9, 0, 0, 9, 49, 0, 9, 0, 121, 9, 0, 49, 9, 0, 0, 9, 361, 0, 58, 121, 529, 9, 0, 0, 9, 49, 0, 9, 961, 0, 130, 0, 49, 9, 0, 361, 9, 0, 0, 58, 1849, 121, 9, 529, 2209, 9, 49, 0, 9, 0, 0, 9, 121, 49, 370, 0, 3481, 9, 0, 961, 58, 0, 0, 130, 4489, 0, 538, 49, 5041, 9, 0, 0, 9, 361, 170, 9, 6241, 0, 9, 0, 6889, 58

Offset: 1

N. J. A. Sloane

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000290, A005063, A005079, A005082, A005084, A005085, A059841.

Array[DivisorSum[#, #^2 &, And[PrimeQ@ #, Mod[#, 4] == 3] &] &, 84] (* Michael De Vlieger, Jul 11 2017 *)
f[p_, e_] := If[Mod[p, 4] == 3, p^2, 0]; a[n_] := Plus @@ f @@@ FactorInteger[n]; a[1] = 0; Array[a, 100] (* Amiram Eldar, Jun 21 2022 *)

a(n) = my(f=factor(n)); sum(k=1, #f~, if (((p=f[k,1])%4) == 3, p^2)); \\ Michel Marcus, Jul 11 2017

(define (A005083 n) (if (= 1 n) 0 (+ (if (= 3 (modulo (A020639 n) 4)) (A000290 (A020639 n)) 0) (A005083 (A028234 n))))) ;; Antti Karttunen, Jul 11 2017

Additive with a(p^e) = p^2 if p = 3 (mod 4), 0 otherwise.

a(n) = A005063(n) - A005079(n) - 4*A059841(n). - Antti Karttunen, Jul 11 2017

More terms from Antti Karttunen, Jul 11 2017

A005081 Sum of 4th powers of primes = 1 mod 4 dividing n.

Original entry on oeis.org

0, 0, 0, 0, 625, 0, 0, 0, 0, 625, 0, 0, 28561, 0, 625, 0, 83521, 0, 0, 625, 0, 0, 0, 0, 625, 28561, 0, 0, 707281, 625, 0, 0, 0, 83521, 625, 0, 1874161, 0, 28561, 625, 2825761, 0, 0, 0, 625, 0, 0, 0, 0, 625, 83521, 28561, 7890481, 0, 625, 0, 0, 707281, 0, 625, 13845841, 0, 0, 0, 29186, 0, 0, 83521, 0, 625, 0, 0, 28398241

Offset: 1

N. J. A. Sloane

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000583, A005065, A005078, A005079, A005080, A005085, A059841.

Array[DivisorSum[#, #^4 &, And[PrimeQ@ #, Mod[#, 4] == 1] &] &, 73] (* Michael De Vlieger, Jul 11 2017 *)
f[p_, e_] := If[Mod[p, 4] == 1, p^4, 0]; a[n_] := Plus @@ f @@@ FactorInteger[n]; a[1] = 0; Array[a, 100] (* Amiram Eldar, Jun 21 2022 *)

a(n) = my(f=factor(n)); sum(k=1, #f~, if (((p=f[k,1])%4) == 1, p^4)); \\ Michel Marcus, Jul 11 2017

(define (A005081 n) (if (= 1 n) 0 (+ (if (= 1 (modulo (A020639 n) 4)) (A000583 (A020639 n)) 0) (A005081 (A028234 n))))) ;; Antti Karttunen, Jul 11 2017

Additive with a(p^e) = p^4 if p = 1 (mod 4), 0 otherwise.

a(n) = A005065(n) - A005085(n) - 16*A059841(n). - Antti Karttunen, Jul 11 2017

More terms from Antti Karttunen, Jul 11 2017

A005082 Sum of primes = 3 mod 4 dividing n.

Original entry on oeis.org

0, 0, 3, 0, 0, 3, 7, 0, 3, 0, 11, 3, 0, 7, 3, 0, 0, 3, 19, 0, 10, 11, 23, 3, 0, 0, 3, 7, 0, 3, 31, 0, 14, 0, 7, 3, 0, 19, 3, 0, 0, 10, 43, 11, 3, 23, 47, 3, 7, 0, 3, 0, 0, 3, 11, 7, 22, 0, 59, 3, 0, 31, 10, 0, 0, 14, 67, 0, 26, 7, 71, 3, 0, 0, 3, 19, 18, 3, 79, 0, 3, 0, 83, 10, 0, 43, 3, 11, 0, 3, 7, 23, 34, 47, 19, 3, 0, 7, 14, 0, 0, 3, 103

Offset: 1

N. J. A. Sloane

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A005069, A005078, A005083, A005084, A005085, A008472, A059841.

Table[Total[Select[Transpose[FactorInteger[n]][[1]],Mod[#,4] == 3&]],{n,80}] (* Harvey P. Dale, Jan 18 2015 *)
Array[DivisorSum[#, # &, And[PrimeQ@ #, Mod[#, 4] == 3] &] &, 103] (* Michael De Vlieger, Jul 11 2017 *)

a(n) = my(f=factor(n)); sum(k=1, #f~, if (((p=f[k,1])%4) == 3, p)); \\ Michel Marcus, Jul 11 2017

(define (A005082 n) (if (= 1 n) 0 (+ (* (A079978 (modulo (A020639 n) 4)) (A020639 n)) (A005082 (A028234 n))))) ;; Antti Karttunen, Jul 11 2017

Additive with a(p^e) = p if p = 3 (mod 4), 0 otherwise.
From Antti Karttunen, Jul 11 2017: (Start)
a(1) = 0; for n > 1, a(n) = (A079978(A020639(n) mod 4)*A020639(n)) + a(A028234(n)).
a(n) = A008472(n) - A005078(n) - 2*A059841(n).
(End)

More terms from Antti Karttunen, Jul 11 2017

A005084 Sum of cubes of primes = 3 mod 4 dividing n.

Original entry on oeis.org

0, 0, 27, 0, 0, 27, 343, 0, 27, 0, 1331, 27, 0, 343, 27, 0, 0, 27, 6859, 0, 370, 1331, 12167, 27, 0, 0, 27, 343, 0, 27, 29791, 0, 1358, 0, 343, 27, 0, 6859, 27, 0, 0, 370, 79507, 1331, 27, 12167, 103823, 27, 343, 0, 27, 0, 0, 27, 1331, 343, 6886, 0, 205379, 27, 0, 29791, 370, 0, 0, 1358, 300763, 0, 12194, 343, 357911

Offset: 1

N. J. A. Sloane

nonn

Antti Karttunen, Table of n, a(n) for n = 1..10000

Cf. A000578, A005064, A005080, A005082, A005083, A005085, A059841.

Array[DivisorSum[#, #^3 &, And[PrimeQ@ #, Mod[#, 4] == 3] &] &, 71] (* Michael De Vlieger, Jul 11 2017 *)
f[p_, e_] := If[Mod[p, 4] == 3, p^3, 0]; a[n_] := Plus @@ f @@@ FactorInteger[n]; a[1] = 0; Array[a, 100] (* Amiram Eldar, Jun 21 2022 *)

a(n) = my(f=factor(n)); sum(k=1, #f~, if (((p=f[k,1])%4) == 3, p^3)); \\ Michel Marcus, Jul 11 2017

(define (A005084 n) (if (= 1 n) 0 (+ (if (= 3 (modulo (A020639 n) 4)) (A000578 (A020639 n)) 0) (A005084 (A028234 n))))) ;;  Antti Karttunen, Jul 11 2017

Additive with a(p^e) = p^3 if p = 3 (mod 4), 0 otherwise.

a(n) = A005064(n) - A005080(n) - 8*A059841(n). - Antti Karttunen, Jul 11 2017

More terms from Antti Karttunen, Jul 11 2017

A318281 a(n+1) = a(n-a(n)) if a(n-1) != a(n), otherwise a(n+1) = a(n) + 3; a(1) = a(2) = a(3) = a(4) = 1.

Original entry on oeis.org

1, 1, 1, 1, 4, 1, 4, 1, 4, 4, 7, 1, 7, 1, 7, 1, 7, 4, 1, 4, 1, 4, 4, 7, 7, 10, 1, 10, 4, 7, 4, 1, 4, 4, 7, 10, 10, 13, 7, 1, 7, 4, 13, 7, 10, 7, 7, 10, 13, 10, 1, 10, 4, 13, 7, 10, 7, 10, 10, 13, 7, 13, 13, 16, 10, 7, 10

Offset: 1

Rok Cestnik, Aug 23 2018

nonn

Sequences with an analogous condition a(n+1) = a(n) + s for s != 3 tend towards repetitive patterns:
  for even values of s this is obvious, e.g.:
    s = 2: 1,1,1,3,1,3,1,3,... (1,3 repeats)
    s = 4: 1,1,1,1,1,5,1,5,1,5,1,5,... (1,5 repeats)
  for odd values of s it has been checked up to s <= 19:
    s = 1: 1,1,2,1,2,2,3,1,3,2,1,2,2,3,1,3,... (2,1,2,2,3,1,3 repeats)
    s = 5: settles to a repetitive pattern of 192 terms
    s = 7: settles to a repetitive pattern of 25 terms
    s = 9: settles to a repetitive pattern of 31 terms
    s = 11: settles to a repetitive pattern of 37 terms
    s = 13: settles to a repetitive pattern of 43 terms
    s = 15: settles to a repetitive pattern of 49 terms
    s = 17: settles to a repetitive pattern of 55 terms
    s = 19: settles to a repetitive pattern of 61 terms
  It appears that for further values of s such sequences settle to a repetitive pattern of 4 + 3*s terms.

			a(5) = a(4) + 3 = 4, because a(3) == a(4).
a(6) = a(5-a(5)) = a(1) = 1, because a(4) != a(5).

Rok Cestnik, Table of n, a(n) for n = 1..9999

See A318282 for (a(n)-1)/3.

Cf. A281130, A291598, A004001, A005085, A005229.

#include
#include
#include
int main(void){
  int N = 100; //number of terms
  int *a = (int*)malloc((N+1)*sizeof(int));
  printf("1 1\n2 1\n3 1\n4 1\n");
  a[1] = 1;
  a[2] = 1;
  a[3] = 1;
  a[4] = 1;
  for(int i = 4; i < N; ++i){
    if(a[i-1] != a[i]) a[i+1] = a[i-a[i]];
    else a[i+1] = a[i]+3;
    printf("%d %d\n", i+1, a[i+1]);
  }
  free(a);
  return 0;
}

cp's OEIS Frontend

A005065 Sum of 4th powers of primes dividing n.

Views

Author

Keywords

Comments

Links

Crossrefs

Programs

Maple

Mathematica

PARI

Python

Scheme

Formula

Extensions

A005083 Sum of squares of primes = 3 mod 4 dividing n.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Scheme

Formula

Extensions

A005081 Sum of 4th powers of primes = 1 mod 4 dividing n.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Scheme

Formula

Extensions

A005082 Sum of primes = 3 mod 4 dividing n.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Scheme

Formula

Extensions

A005084 Sum of cubes of primes = 3 mod 4 dividing n.

Views

Author

Keywords

Links

Crossrefs

Programs

Mathematica

PARI

Scheme

Formula

Extensions

A318281 a(n+1) = a(n-a(n)) if a(n-1) != a(n), otherwise a(n+1) = a(n) + 3; a(1) = a(2) = a(3) = a(4) = 1.

Views

Author

Keywords

Comments

Examples

Links

Crossrefs

Programs

C