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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.

User: Dan Dart

Dan Dart's wiki page.

Dan Dart has authored 6 sequences.

A384195 a(n) = tau(n+1) - tau(n-1), where tau(n) = A000005(n), the number of divisors of n.

Original entry on oeis.org

1, 1, 0, 1, 0, 0, 1, 0, -1, 2, 0, -2, 2, 1, -2, 1, 0, 0, 2, -2, -2, 4, 1, -4, 1, 2, -2, 2, 0, -2, 2, -2, 0, 5, -2, -5, 2, 4, -2, 0, 0, -2, 4, -2, -4, 6, 1, -4, 1, 0, -2, 2, 2, 0, 0, -4, -2, 8, 0, -8, 4, 3, -2, 1, -2, -2, 2, 2, -2, 4, 0, -8, 4, 2, -2, 2, -2, 2, 3, -6, -3, 8, 2, -8
Offset: 2

Views

Author

Dan Dart, May 21 2025

Keywords

Comments

Conjecture: a(n)=0 for an infinite number of values.

Examples

			a(10) = tau(11) - tau(9) = 2 - 3 = -1.

Crossrefs

Cf. A000005, A051950, A051951, A067888.

Programs

  • Haskell
    divides ∷ (Integral a) ⇒ a → a → Bool
divides a b = b `mod` a == 0
properFactors ∷ Integral a ⇒ a → [a]
properFactors n = 1 : filter (`divides` n) [2..(n `div` 2)]
factors ∷ Integral a ⇒ a → [a]
factors 1 = [1]
factors n = properFactors n <> [n]
tauSuccMinusTauPred :: Integer -> Integer
tauSuccMinusTauPred n = fromIntegral (length (factors (n + 1))) - fromIntegral (length (factors (n - 1)))
  • PARI
    a(n) = numdiv(n+1) - numdiv(n-1); \\ Michel Marcus, May 21 2025

Formula

a(n) = tau(n+1) - tau(n-1).

A355467 a(n) is the smallest number which is greater than n and has more prime factors (with multiplicity) than n.

Original entry on oeis.org

2, 4, 4, 8, 6, 8, 8, 16, 12, 12, 12, 16, 14, 16, 16, 32, 18, 24, 20, 24, 24, 24, 24, 32, 27, 27, 32, 32, 30, 32, 32, 64, 36, 36, 36, 48, 38, 40, 40, 48, 42, 48, 44, 48, 48, 48, 48, 64, 50, 54, 52, 54, 54, 64, 56, 64, 60, 60, 60, 64, 62, 63, 64, 128, 66, 72, 68, 72, 70, 72, 72, 96, 74, 75, 80, 80, 78, 80, 80, 96, 96
Offset: 1

Views

Author

Dan Dart, Jul 03 2022

Keywords

Comments

Distinct from 2^A073093 because of the proviso that a(n) > n and bigomega(a(n)) > bigomega(n).

Examples

			For n = 1, a(1) = 2, since 2 is the first number satisfying 2 > 1 and bigomega(2) = 1 > bigomega(1) = 0.
For n = 5, a(5) = 8, since 8 is the first number satisfying 8 > 5 and bigomega(8) = 3 > bigomega(5) = 1.
For n = 12, a(12) = 16, since 16 is the first number satisfying 16 > 12 and bigomega(16) = 4 > bigomega(12) = 3.

Crossrefs

Cf. A073093, A001222.

Programs

  • Haskell
    import Data.Numbers.Primes
result :: [Integer]
result = fmap (
  \n -> head (
      dropWhile (
          \m -> length (primeFactors m :: [Integer]) <= length (primeFactors n :: [Integer])
      )
      [n..]
  )
  ) [1..]
  • Maple
    A355467 := proc(n)
    local a,nOmega ;
    nOmega := A001222(n) ;
    for a from n+1 do
        if A001222(a) > nOmega then
            return a;
        end if;
    end do;
end proc:
seq(A355467(n),n=1..80) ; # R. J. Mathar, May 05 2023
  • PARI
    a(n) = my(k=n+1, nb=bigomega(n)); while (bigomega(k) <= nb, k++); k; \\ Michel Marcus, Jul 05 2022

Formula

a(2^n) = 2^(n+1) because the smallest extra factor is 2.
a(3*2^n) = 2^(n+2) because 4 (i.e., 2^2) is the next biggest pair of factors.

A309979 Hash Parker numbers: Integers whose real 32nd root's first six nonzero digits (after the decimal point) rearranged in ascending order are equal to 234477.

Original entry on oeis.org

4, 1191, 2340, 4915, 8101, 8703, 13937, 13952, 14029, 14041, 25111, 25127, 26062, 26203, 26324, 26479, 26490, 27934, 28077, 28195, 50506, 50536, 52216, 52359, 52892, 55703, 55957, 56030, 56059, 56075, 56178, 56244, 56566, 56577, 74747, 75877, 75952, 75996, 80752, 80764, 80765
Offset: 1

Views

Author

Dan Dart, Aug 25 2019

Keywords

Examples

			4^(1/32) = 1.0442737824274138...
Rearranging 442737 in ascending order gives 234477.
1191^(1/32) = 1.2477346... -> 247734 -> 234477;
2340^(1/32) = 1.2743478... -> 274347 -> 234477.

Links

Programs

  • Haskell
    import Data.List
hash :: Double -> Inthash = read . sort . take 6 . filter (/='0') . drop 1 . dropWhile (/='.') . show . (** 0.03125)
main :: IO ()main = print $ map (floor . fst) . filter ((==234477) . snd) $ map (\x -> (x, hash x)) [2..1000000]
  • Mathematica
    Select[Range[81000],With[{c=RealDigits[Surd[#,32],10,20]/.(0->Nothing)},Sort[Take[Drop[c[[1]],c[[2]]],6]]=={2,3,4,4,7,7}]&]//Quiet (* Harvey P. Dale, Jun 22 2025 *)

A308267 Numbers k such that k divides A048678(k).

Original entry on oeis.org

1, 2, 4, 7, 8, 14, 16, 28, 31, 32, 56, 62, 64, 83, 112, 124, 127, 128, 166, 224, 248, 254, 256, 332, 397, 448, 496, 508, 511, 512, 664, 794, 891, 896, 992, 1016, 1022, 1024, 1163, 1328, 1588, 1782, 1792, 1984, 2032, 2044, 2047, 2048, 2326, 2656, 3176, 3441, 3564, 3584, 3968
Offset: 1

Views

Author

Dan Dart, May 17 2019

Keywords

Comments

Apparently includes all powers of 2.
All numbers 2^(2k+1)-1 are in this sequence, and if n is in this sequence then so is 2n. - Charlie Neder, May 17 2019

Examples

			The first few terms of A048678 are 1,2,5,4,9,10,21,8. 2 is a multiple of 2, 5 isn't a multiple of 3, 4 is a multiple of 4, 9 isn't a multiple of 5, 10 isn't a multiple of 6, 21 is a multiple of 7, etc.

Crossrefs

Cf. A048678, A083420 (subsequence).

Programs

  • Haskell
    bintodec :: [Integer] -> Integerbintodec = sum . zipWith (*) (iterate (*2) 1) . reverse
decomp :: (Integer, [Integer]) -> (Integer, [Integer])decomp (x, ys) = if even x then (x `div` 2, 0:ys) else (x - 1, 1:ys)
zeck :: Integer -> Integerzeck n = bintodec (1 : snd (last . takeWhile (\(x, _) -> x > 0) $ iterate decomp (n, [])))
output :: [Integer]output = filter (\x -> 0 == zeck x `mod` x) [1..100]
main :: IO ()main = print output
  • Mathematica
    Select[Range[4000], Divisible[FromDigits[Flatten[IntegerDigits[#, 2] /. {1 -> {0, 1}}], 2], #] &] (* Amiram Eldar, Jul 08 2019 after Robert G. Wilson v at A048678 *)

A275124 Multiples of 5 where Pisano periods of Fibonacci numbers A001175 and Lucas numbers A106291 agree.

Original entry on oeis.org

55, 110, 155, 165, 205, 220, 305, 310, 330, 355, 385, 410, 440, 465, 495, 505, 605, 610, 615, 620, 655, 660, 710, 715, 755, 770, 820, 880, 905, 915, 930, 935, 955, 990, 1010, 1045, 1065, 1085, 1155, 1205, 1210, 1220, 1230, 1240, 1255, 1265, 1310, 1320, 1355, 1395, 1420, 1430, 1435, 1485, 1510, 1515, 1540, 1555, 1595, 1640, 1655, 1705, 1760, 1810, 1815, 1830
Offset: 1

Views

Author

Dan Dart, Jul 18 2016

Keywords

Comments

Multiples of 5 where A001175 and A106291 agree. See 1st comment of A106291.

Examples

			55 is the first multiple of 5 where the Pisano period (Fibonacci) of n = 55 and the Pisano period (Lucas) of n = 55 agree (this is in this case 20).

Links

Crossrefs

Cf. A001175, A106291.

Programs

  • JavaScript
    let bases = [],
    basesd = [],
    baselimit = 2000;
for (let base = 2; base <= baselimit; base++) {
    let fibs = [1 % base,1 % base],
        lucas = [2 % base,1 % base],
        repeatingf = false,
        repeatingl = false;
    while (!repeatingf) {
        fibs.push((fibs[fibs.length - 2] + fibs[fibs.length - 1]) % base);
        if (1 == fibs[fibs.length - 2] &&
            0 == fibs[fibs.length - 1])
            repeatingf = true;
    }
    while (!repeatingl) {
        lucas.push((lucas[lucas.length - 2] + lucas[lucas.length - 1]) % base);
        if ((lucas[0] == (lucas[lucas.length - 2] + lucas[lucas.length - 1]) % base) &&
            (lucas[1] == (lucas[lucas.length - 2] + 2 *lucas[lucas.length - 1]) % base))
            repeatingl = true;
    }
    if (fibs.length != lucas.length)
        bases.push(base);
}
for (let i = 1; i <= baselimit/5; i++) {
    if (!bases.includes(i * 5)) basesd.push(i * 5);
}
console.log(basesd.join(','));

A275167 Pisano periods of A275124.

Original entry on oeis.org

20, 60, 60, 40, 40, 60, 60, 60, 120, 140, 80, 120, 60, 120, 120, 100, 220, 60, 40, 60, 260, 120, 420, 140, 100, 240, 120, 120, 180, 120, 120, 180, 380, 120, 300, 180, 280, 240, 80, 240, 660, 60, 120, 60, 500, 240, 780, 120, 540, 120, 420, 420, 80, 360, 300, 200, 240, 620, 140, 120, 220, 60, 240, 180, 440, 120, 120, 120, 180, 1140, 520, 120
Offset: 1

Views

Author

Dan Dart, Jul 18 2016

Keywords

Examples

			A275124(1) = 55, and PisanoPeriod(55) = 20, etc.

Crossrefs

Cf. A001175, A275124.

Programs

  • JavaScript
    let bases = [],
    Fs = [],
    Ls = [],
    agrees = [],
    baselimit = 2000;
for (let base = 2; base <= baselimit; base++) {
    let fibs = [1 % base,1 % base],
        lucas = [2 % base,1 % base],
        repeatingf = false,
        repeatingl = false;
    while (!repeatingf) {
        fibs.push((fibs[fibs.length - 2] + fibs[fibs.length - 1]) % base);
        if (1 == fibs[fibs.length - 2] &&
            0 == fibs[fibs.length - 1])
            repeatingf = true;
    }
    while (!repeatingl) {
        lucas.push((lucas[lucas.length - 2] + lucas[lucas.length - 1]) % base);
        if ((lucas[0] == (lucas[lucas.length - 2] + lucas[lucas.length - 1]) % base) &&
            (lucas[1] == (lucas[lucas.length - 2] + 2 *lucas[lucas.length - 1]) % base))
            repeatingl = true;
    }
    Fs[base] = fibs.length;
    Ls[base] = lucas.length;
    if (fibs.length != lucas.length)
        bases.push(base);
    //console.log('F', fibs.join(','), 'L:', lucas.join(','));
}
for (let i = 1; i <= baselimit/5; i++) {
    if (!bases.includes(i * 5)) {
        agrees.push(Fs[i * 5]);
    }
}
console.log(agrees.join(','));

Formula

a(n) = A001175(A275124(n)).

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