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.

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A357555 a(n) is the numerator of Sum_{d|n} (-1)^(d+1) / d^2.

Original entry on oeis.org

1, 3, 10, 11, 26, 5, 50, 43, 91, 39, 122, 55, 170, 75, 52, 171, 290, 91, 362, 143, 500, 183, 530, 215, 651, 255, 820, 275, 842, 13, 962, 683, 1220, 435, 52, 1001, 1370, 543, 1700, 559, 1682, 125, 1850, 61, 2366, 795, 2210, 95, 2451, 1953, 2900, 935, 2810, 205, 3172
Offset: 1

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Author

Ilya Gutkovskiy, Oct 03 2022

Keywords

Examples

			1, 3/4, 10/9, 11/16, 26/25, 5/6, 50/49, 43/64, 91/81, 39/50, 122/121, ...
		

Crossrefs

Cf. A017667, A064027, A098987, A119682, A321543, A357556 (denominators).

Programs

  • Mathematica
    Table[Sum[(-1)^(d + 1)/d^2, {d, Divisors[n]}], {n, 1, 55}] // Numerator
    nmax = 55; CoefficientList[Series[Sum[(-1)^(k + 1) x^k/(k^2 (1 - x^k)), {k, 1, nmax}], {x, 0, nmax}], x] // Numerator // Rest
  • PARI
    a(n) = numerator(sumdiv(n, d, (-1)^(d+1)/d^2)); \\ Michel Marcus, Oct 03 2022
  • Python
    from sympy import divisors
    from fractions import Fraction
    def a(n): return sum(Fraction((-1)**(d+1), d*d) for d in divisors(n, generator=True)).numerator
    print([a(n) for n in range(1, 56)]) # Michael S. Branicky, Oct 03 2022
    

Formula

Numerators of coefficients in expansion of Sum_{k>=1} (-1)^(k+1) * x^k / (k^2 * (1 - x^k)).

A119784 Numerator of Sum[Sum[(-1)^(i+1)*1/(i*j)^2, {i, 1, n}], {j, 1, n}].

Original entry on oeis.org

1, 15, 1519, 23575, 15907111, 5224037, 13034567237, 206747823917, 51412399560631, 10224262569323, 152012226379719803, 151406414338463843, 4372018721271579430163, 4359099317599237566323, 21975413631947472215359
Offset: 1

Views

Author

Alexander Adamchuk, Jun 25 2006

Keywords

Comments

Prime p divides a((p-1)/2) for p>3. p^2 divides a(p-1) for prime p>3.

Crossrefs

Programs

  • Mathematica
    Numerator[Table[Sum[Sum[(-1)^(i+1)*1/(i*j)^2, {i, 1, n}], {j, 1, n}],{n,1,20}]]

Formula

a(n) = numerator[Sum[Sum[(-1)^(i+1)*1/(i*j)^2, {i, 1, n}], {j, 1, n}]]. a(n) = A119682(n)*A007406(n).

A125504 Smallest number k such that the numerator of alternating generalized harmonic number H'(k,n) = Sum[ (-1)^(i+1) * 1/i^n, {i,1,k} ] is a prime.

Original entry on oeis.org

3, 2, 2, 3, 2, 19, 2, 146, 87, 3, 16, 3, 2, 249, 15, 87, 2, 699, 2
Offset: 1

Views

Author

Alexander Adamchuk, Dec 28 2006, Jan 31 2007

Keywords

Comments

a(n) = 2 for n in A000043.
a(n) = 3 for n in {1,4,10,12,24,27,39,...}.
a(n) = 5 for n in {26,76,132,205,238,...}.
a(n) = 9 for n in {100,200,...}.
a(n) = 15 for n in {15,33,65,...}.
a(21) = 18. a(22) = 13. a(41) = 6. a(72) = 11. a(173) = 8.
a(20) > 2100 - Max Alekseyev, Jul 07 2009

Crossrefs

Extensions

More terms from Max Alekseyev, Jul 07 2009

A128820 Numerator of alternating generalized harmonic number H'(p-1,2p) = Sum_{k=1..p-1} (-1)^(k+1)/k^(2*p) divided by p^2 for prime p > 2.

Original entry on oeis.org

7, 2474315503, 53305712401979540402437, 5597916593064896381208777124641713285719656398067086247546781015747740847, 192635872080422175485338764164035657976855166649911323825254242037669356649787653784405726270977624462974729613783
Offset: 2

Views

Author

Alexander Adamchuk, Apr 10 2007

Keywords

Comments

Alternating generalized harmonic number is H'(n,m) = Sum_{k=1..n} (-1)^(k+1)*1/k^m. Numerator of H'(p-1,2n) is divisible by p for all integers n > 0 and primes p > 2. Numerator of H'(p-1,2p) is divisible by p^2 for prime p > 2.

Examples

			prime(2) = 3; a(2) = numerator(1 - 1/2^6) / 3^2 = 63/9 = 7.
prime(3) = 5; a(3) = numerator(1 - 1/2^10 + 1/3^10 - 1/4^10) / 5^2 = 61857887575/25 = 2474315503.
		

Crossrefs

Cf. A119722 (numerator of generalized harmonic number H(p-1, p) = Sum_{k=1..p-1} 1/k^p divided by p^3 for prime p>3).

Programs

  • Mathematica
    Table[ Numerator[ Sum[(-1)^(k+1)*1/k^(2*Prime[n]), {k,1,Prime[n]-1} ] ] / Prime[n]^2, {n,2,10} ]

Formula

a(n) = numerator(Sum_{k=1..prime(n)-1} (-1)^(k+1)/k^(2*prime(n))) / prime(n)^2 for n > 1.
Previous Showing 21-24 of 24 results.