cp's OEIS Frontend

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.

Previous Showing 11-19 of 19 results.

A005085 Sum of 4th powers of primes = 3 mod 4 dividing n.

Original entry on oeis.org

0, 0, 81, 0, 0, 81, 2401, 0, 81, 0, 14641, 81, 0, 2401, 81, 0, 0, 81, 130321, 0, 2482, 14641, 279841, 81, 0, 0, 81, 2401, 0, 81, 923521, 0, 14722, 0, 2401, 81, 0, 130321, 81, 0, 0, 2482, 3418801, 14641, 81, 279841, 4879681, 81, 2401, 0, 81, 0, 0, 81, 14641, 2401, 130402, 0, 12117361, 81, 0, 923521, 2482, 0, 0, 14722, 20151121, 0, 279922
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A000583, A005065, A005081, A005082, A005083, A005084, A059841.

Programs

  • Mathematica
    Array[DivisorSum[#, #^4 &, And[PrimeQ@ #, Mod[#, 4] == 3] &] &, 69] (* Michael De Vlieger, Jul 11 2017 *)
f[p_, e_] := If[Mod[p, 4] == 3, p^4, 0]; a[n_] := Plus @@ f @@@ FactorInteger[n]; a[1] = 0; Array[a, 100] (* Amiram Eldar, Jun 21 2022 *)
  • PARI
    a(n) = my(f=factor(n)); sum(k=1, #f~, if (((p=f[k,1])%4) == 3, p^4)); \\ Michel Marcus, Jul 11 2017
  • Scheme
    (define (A005085 n) (if (= 1 n) 0 (+ (if (= 3 (modulo (A020639 n) 4)) (A000583 (A020639 n)) 0) (A005085 (A028234 n))))) ;; Antti Karttunen, Jul 11 2017

Formula

Additive with a(p^e) = p^4 if p = 3 (mod 4), 0 otherwise.
a(n) = A005065(n) - A005081(n) - 16*A059841(n). - Antti Karttunen, Jul 11 2017

Extensions

More terms from Antti Karttunen, Jul 11 2017

A351196 Sum of the 8th powers of the primes dividing n.

Original entry on oeis.org

0, 256, 6561, 256, 390625, 6817, 5764801, 256, 6561, 390881, 214358881, 6817, 815730721, 5765057, 397186, 256, 6975757441, 6817, 16983563041, 390881, 5771362, 214359137, 78310985281, 6817, 390625, 815730977, 6561, 5765057, 500246412961, 397442, 852891037441, 256
Offset: 1

Views

Author

Wesley Ivan Hurt, Feb 04 2022

Keywords

Comments

Inverse Möbius transform of n^8 * c(n), where c(n) is the prime characteristic (A010051). - Wesley Ivan Hurt, Jun 22 2024

Links

Crossrefs

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), A005065 (k=4), A351193 (k=5), A351194 (k=6), A351195 (k=7), this sequence (k=8), A351197 (k=9), A351198 (k=10).
Cf. A010051.

Programs

  • Mathematica
    Array[DivisorSum[#, #^8 &, PrimeQ] &, 50]
f[p_, e_] := p^8; a[n_] := Plus @@ f @@@ FactorInteger[n]; a[1] = 0; Array[a, 100] (* Amiram Eldar, Jun 20 2022 *)
  • Python
    from sympy import primefactors
def A351196(n): return sum(p**8 for p in primefactors(n)) # Chai Wah Wu, Feb 05 2022

Formula

a(n) = Sum_{p|n, p prime} p^8.
G.f.: Sum_{k>=1} prime(k)^8 * x^prime(k) / (1 - x^prime(k)). - Ilya Gutkovskiy, Feb 16 2022
Additive with a(p^e) = p^8. - Amiram Eldar, Jun 20 2022
a(n) = Sum_{d|n} d^8 * c(d), where c = A010051. - Wesley Ivan Hurt, Jun 22 2024

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

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

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

Programs

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

Formula

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

Extensions

More terms from Antti Karttunen, Jul 11 2017

A005077 Sum of 4th powers of primes = 2 mod 3 dividing n.

Original entry on oeis.org

0, 16, 0, 16, 625, 16, 0, 16, 0, 641, 14641, 16, 0, 16, 625, 16, 83521, 16, 0, 641, 0, 14657, 279841, 16, 625, 16, 0, 16, 707281, 641, 0, 16, 14641, 83537, 625, 16, 0, 16, 0, 641, 2825761, 16, 0, 14657, 625, 279857, 4879681, 16, 0, 641, 83521, 16, 7890481, 16, 15266, 16, 0, 707297, 12117361, 641, 0, 16, 0, 16, 625, 14657, 0, 83537
Offset: 1

Views

Author

N. J. A. Sloane

Keywords

Links

Crossrefs

Cf. A000583, A005065, A005073, A005074, A005075, A005076, A008472, A079978.

Programs

  • Mathematica
    Array[DivisorSum[#, #^4 &, And[PrimeQ@ #, Mod[#, 3] == 2] &] &, 68] (* Michael De Vlieger, Jul 11 2017 *)
f[p_, e_] := If[Mod[p, 3] == 2, p^4, 0]; a[n_] := Plus @@ f @@@ FactorInteger[n]; a[1] = 0; Array[a, 100] (* Amiram Eldar, Jun 21 2022 *)
  • PARI
    a(n) = my(f=factor(n)); sum(k=1, #f~, if (((p=f[k,1])%3) == 2, p^4)); \\ Michel Marcus, Jul 11 2017
  • Scheme
    (define (A005077 n) (if (= 1 n) 0 (+ (A000583 (if (= 2 (modulo (A020639 n) 3)) (A020639 n) 0)) (A005077 (A028234 n))))) ;; Antti Karttunen, Jul 10 2017

Formula

Additive with a(p^e) = p^4 if p = 2 (mod 3), 0 otherwise.
a(n) = A005065(n) - A005073(n) - 81*A079978(n). - Antti Karttunen, Jul 10 2017

Extensions

More terms from Antti Karttunen, Jul 10 2017

A347158 Sum of 4th powers of distinct prime divisors of n that are < sqrt(n).

Original entry on oeis.org

0, 0, 0, 0, 0, 16, 0, 16, 0, 16, 0, 97, 0, 16, 81, 16, 0, 97, 0, 16, 81, 16, 0, 97, 0, 16, 81, 16, 0, 722, 0, 16, 81, 16, 625, 97, 0, 16, 81, 641, 0, 97, 0, 16, 706, 16, 0, 97, 0, 641, 81, 16, 0, 97, 625, 2417, 81, 16, 0, 722, 0, 16, 2482, 16, 625, 97, 0, 16, 81, 3042
Offset: 1

Views

Author

Ilya Gutkovskiy, Aug 20 2021

Keywords

Crossrefs

Cf. A001159, A005065, A005068, A333806, A333808, A347142, A347156, A347157.

Programs

  • Mathematica
    Table[DivisorSum[n, #^4 &, # < Sqrt[n] && PrimeQ[#] &], {n, 1, 70}]
nmax = 70; CoefficientList[Series[Sum[Prime[k]^4 x^(Prime[k] (Prime[k] + 1))/(1 - x^Prime[k]), {k, 1, nmax}], {x, 0, nmax}], x] // Rest

Formula

G.f.: Sum_{k>=1} prime(k)^4 * x^(prime(k)*(prime(k) + 1)) / (1 - x^prime(k)).

A322080 Square array A(n,k), n >= 1, k >= 0, read by antidiagonals: A(n,k) = Sum_{p|n, p prime} p^k.

Original entry on oeis.org

0, 0, 1, 0, 2, 1, 0, 4, 3, 1, 0, 8, 9, 2, 1, 0, 16, 27, 4, 5, 2, 0, 32, 81, 8, 25, 5, 1, 0, 64, 243, 16, 125, 13, 7, 1, 0, 128, 729, 32, 625, 35, 49, 2, 1, 0, 256, 2187, 64, 3125, 97, 343, 4, 3, 2, 0, 512, 6561, 128, 15625, 275, 2401, 8, 9, 7, 1, 0, 1024, 19683, 256, 78125, 793, 16807, 16, 27, 29, 11, 2
Offset: 1

Views

Author

Ilya Gutkovskiy, Nov 26 2018

Keywords

Examples

			Square array begins:
  0,  0,   0,    0,    0,     0,  ...
  1,  2,   4,    8,   16,    32,  ...
  1,  3,   9,   27,   81,   243,  ...
  1,  2,   4,    8,   16,    32,  ...
  1,  5,  25,  125,  625,  3125,  ...
  2,  5,  13,   35,   97,   275,  ...

Links

Crossrefs

Columns k=0..4 give A001221, A008472, A005063, A005064, A005065.
Cf. A109974, A200768 (diagonal), A285425, A286880, A321258.

Programs

  • Mathematica
    Table[Function[k, Sum[Boole[PrimeQ[d]] d^k, {d, Divisors[n]}]][i - n], {i, 0, 12}, {n, 1, i}] // Flatten
Table[Function[k, SeriesCoefficient[Sum[Prime[j]^k x^Prime[j]/(1 - x^Prime[j]), {j, 1, n}], {x, 0, n}]][i - n], {i, 0, 12}, {n, 1, i}] // Flatten
  • PARI
    T(n,k)={vecsum([p^k | p<-factor(n)[,1]])}
for(n=1, 10, for(k=0, 8, print1(T(n, k), ", ")); print); \\ Andrew Howroyd, Nov 26 2018

Formula

G.f. of column k: Sum_{j>=1} prime(j)^k*x^prime(j)/(1 - x^prime(j)).

A138296 Table T(k,n) read along antidiagonals: sum of the k-th powers of the distinct prime factors of A024619(n).

Original entry on oeis.org

5, 13, 7, 35, 29, 5, 97, 133, 13, 9, 275, 641, 35, 53, 8, 793, 3157, 97, 351, 34, 5, 2315, 15689, 275, 2417, 152, 13, 7, 6817, 78253, 793, 16839, 706, 35, 29, 10, 20195, 390881, 2315, 117713, 3368, 97, 133, 58, 13, 60073, 1953637, 6817, 823671, 16354, 275, 641
Offset: 1

Views

Author

R. J. Mathar, May 07 2008

Keywords

Comments

Row k=1 is A109353. Rows k=2,3 and 4 are subsequences of A005063-A005065.

Examples

			Upper left corner of the table starting at row k=1, column n=1:
1|......5.......7.......5.......9.......8.......5.......7.
2|.....13......29......13......53......34......13......29.
3|.....35.....133......35.....351.....152......35.....133.
4|.....97.....641......97....2417.....706......97.....641.
5|....275....3157.....275...16839....3368.....275....3157.
6|....793...15689.....793..117713...16354.....793...15689.
7|...2315...78253....2315..823671...80312....2315...78253.
8|...6817..390881....6817.5765057..397186....6817..390881.

Links

Programs

  • Maple
    A024619 := proc(n)
    local a;
    if n = 1 then
        RETURN(6);
    else
        for a from A024619(n-1)+1 do
            if A001221(a) > 1 then
               RETURN(a) ;
            fi ;
        od:
    fi ;
end:
A138296 := proc(n,j)
    local f,beta ;
    beta := 0 ;
    for f in ifactors( A024619(n) )[2] do
        beta := beta+op(1,f)^j ;
    od:
    RETURN(beta) ;
end:
for d from 1 to 10 do for n from 1 to d do printf("%d,",A138296(n,d-n+1)) ; od: od: # R. J. Mathar, May 07 2008

Formula

T(k,n) = sum_{d in A000040, d| A024619(n)} d^k.

A199583 a(n) is the smallest number such that the sum of the n-th powers of its distinct prime divisors is divisible by n.

Original entry on oeis.org

2, 2, 3, 2, 5, 70, 7, 2, 3, 33, 11, 1155, 13, 78, 26, 2, 17, 2156564410, 19, 6006, 26, 114, 23, 2156564410, 5, 33, 3, 1365, 29, 110, 31, 2, 62, 15, 201, 2156564410, 37, 30, 14, 961380175077106319535, 41, 1385670, 43, 2805, 26, 266, 47, 961380175077106319535
Offset: 1

Views

Author

Michel Lagneau, Nov 08 2011

Keywords

Comments

a(n) > 1 and a(n) = n if n prime. All terms are squarefree.

Examples

			a(6) = 70 = 2*5*7; 2^6 + 5^6 + 7^6 = 133338 = 22223*6.
a(18)= 2*5*7*11*13*17*19*23*29 = 2156564410 because:
p^18 == 10, 9 (mod 18) for p = 2,3 respectively, and p^18 == 1 (mod 18) for p prime > 3. The minimum sum divisible by 18 is s = 2^18 + Sum_{k=3..10} prime(k)^18 whose residues sum to 10 + 1 + 1 + 1 + 1 + 1 + 1 + 1 + 1 = 18. Hence a(18) = 2156564410.

Crossrefs

Cf. A005063, A005064, A005065, A199167.

Programs

  • Maple
    with(numtheory): T:=array(1..50):for n from 1 to 50 do:q:=0:for k from 2 to 7000 while(q=0)do:x:=factorset(k):s:=sum(x[j]^n ,j=1..nops(x)) :if irem(s,n)=0 then printf ( "%d %d \n",n,k):q:=1:else fi:od:if q=0 then for i from 1 to n do: T[i]:=irem(ithprime(i)^n,n):od:W:=convert(T,set):n1:=nops(W):n2:=W[n1]:n3:=W[n1-1]:
s:=0:p:=1:for a from 1 to n  while(s<>n) do: if T[a]= 1 or T[a]=n2 or (T[a] = n3 and n2+n3

A347160 Sum of 4th powers of distinct prime divisors of n that are <= sqrt(n).

Original entry on oeis.org

0, 0, 0, 16, 0, 16, 0, 16, 81, 16, 0, 97, 0, 16, 81, 16, 0, 97, 0, 16, 81, 16, 0, 97, 625, 16, 81, 16, 0, 722, 0, 16, 81, 16, 625, 97, 0, 16, 81, 641, 0, 97, 0, 16, 706, 16, 0, 97, 2401, 641, 81, 16, 0, 97, 625, 2417, 81, 16, 0, 722, 0, 16, 2482, 16, 625, 97, 0, 16, 81, 3042
Offset: 1

Views

Author

Ilya Gutkovskiy, Aug 20 2021

Keywords

Crossrefs

Cf. A001159, A005065, A005068, A063962, A097974, A098002, A347143, A347158, A347159.

Programs

  • Mathematica
    Table[DivisorSum[n, #^4 &, # <= Sqrt[n] && PrimeQ[#] &], {n, 1, 70}]
nmax = 70; CoefficientList[Series[Sum[Prime[k]^4 x^(Prime[k]^2)/(1 - x^Prime[k]), {k, 1, nmax}], {x, 0, nmax}], x] // Rest

Formula

G.f.: Sum_{k>=1} prime(k)^4 * x^(prime(k)^2) / (1 - x^prime(k)).
Previous Showing 11-19 of 19 results.

Started in 1964 by Neil J. A. Sloane | Maintained by The OEIS Foundation Inc.

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