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13^3 = 2197 = 137*2^4 + 5;

29^3 = 24389 = 762*2^5 + 5 = 381*2^6 + 5;

93^3 = 804357 = 6284*2^7 + 5 = 3142*2^8 + 5 = 1571*2^9 + 5.

13^3 = 2197 = 137*2^4 + 5;

29^3 = 24389 = 762*2^5 + 5 = 381*2^6 + 5;

93^3 = 804357 = 6284*2^7 + 5 = 3142*2^8 + 5 = 1571*2^9 + 5.

For the digits of 5^(1/3), see A323045.

Approximations of p-adic cubic roots:

A322701 (2-adic, 3^(1/3));

this sequence (2-adic, 5^(1/3));

A322934 (2-adic, 7^(1/3));

A322999 (2-adic, 9^(1/3));

A290567 (5-adic, 2^(1/3));

A290568 (5-adic, 3^(1/3));

A309444 (5-adic, 4^(1/3));

A319097, A319098, A319199 (7-adic, 6^(1/3));

A320914, A320915, A321105 (13-adic, 5^(1/3)).

allocated The successive approximations up to 2^n for Jianing Song2-adic integer 5^(1/3).

0, 1, 1, 5, 13, 29, 29, 93, 93, 93, 605, 1629, 3677, 3677, 3677, 20061, 20061, 20061, 151133, 151133, 151133, 151133, 151133, 4345437, 4345437, 21122653, 54677085, 54677085, 188894813, 457330269, 457330269, 457330269, 2604813917, 6899781213, 6899781213

0,4

a(n) is the unique solution to x^3 == 5 (mod 2^n) in the range [0, 2^n - 1].

Wikipedia, <a href="https://en.wikipedia.org/wiki/P-adic_number">p-adic number</a>

For n > 0, a(n) = a(n-1) if a(n-1)^3 - 5 is divisible by 2^n, otherwise a(n-1) + 2^(n-1).

(PARI) a(n) = lift(sqrtn(5+O(2^n), 3))

13^3 = 2197 = 137*2^4 + 5;

29^3 = 24389 = 762*2^5 + 5 = 381*2^6 + 5;

93^3 = 804357 = 6284*2^7 + 5 = 3142*2^8 + 5 = 1571*2^9 + 5.

allocated

nonn

Jianing Song, Aug 30 2019

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allocated for Jianing Song

recycled

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Expansion of x*(1 + 3*x + 20*x^2)/((1 - x^2)*(1 - 10*x^2)).

0, 1, 3, 31, 33, 331, 333, 3331, 3333, 33331, 33333, 333331, 333333, 3333331, 3333333, 33333331, 33333333, 333333331, 333333333, 3333333331, 3333333333, 33333333331, 33333333333, 333333333331, 333333333333, 3333333333331, 3333333333333, 33333333333331

0,3

<a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0,11,0,-10).

G.f.: x*(1 + 3*x + 20*x^2)/((1 - x^2)*(1 - 10*x^2)).

a(n) = 11*a(n-2) - 10* a(n-4).

a(n) = (1/3)*10^floor((n + 1)/2) + (-1)^n - 4/3.

a(n) = 3*(10^n - 1)/9 for n even; a(n) = (10^(n+1) - 7)/3 otherwise.

seq(coeff(series(x*(1+3*x+20*x^2)/((1-x^2)*(1-10*x^2)), x, n+1), x, n), n = 0 .. 30); # Muniru A Asiru, Mar 17 2019

CoefficientList[Series[x (1 + 3 x + 20 x^2) / (10 x^4 - 11 x^2 + 1), {x, 0, 25}], x]

(MAGMA) I:=[0, 1, 3, 31]; [n le 4 select I[n] else 11*Self(n-2)-10*Self(n-4): n in [1..30]];

Bisections give: A002277 (even part), A033175 (odd part).

nonn,easy,changed

recycled

Vincenzo Librandi, Mar 17 2019

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Sum on digits k (0, 1, 3, 4, 6, 7, ...) are congruent to 0 or 1 (mod 3). A032766(n).

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