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.

Showing 1-3 of 3 results.

A274406 Numbers m such that 9 divides m*(m + 1).

Original entry on oeis.org

0, 8, 9, 17, 18, 26, 27, 35, 36, 44, 45, 53, 54, 62, 63, 71, 72, 80, 81, 89, 90, 98, 99, 107, 108, 116, 117, 125, 126, 134, 135, 143, 144, 152, 153, 161, 162, 170, 171, 179, 180, 188, 189, 197, 198, 206, 207, 215, 216, 224, 225, 233, 234, 242, 243, 251, 252, 260, 261, 269
Offset: 1

Views

Author

Bruno Berselli, Jun 20 2016

Keywords

Comments

Equivalently, numbers congruent to 0 or 8 mod 9.
Terms of A007494 with indices in A047264. Also, terms of A060464 with indices in A047335.

Links

Crossrefs

Cf. A008591 (first bisection), A010689 (first differences), A017257 (second bisection).
Cf. similar sequences in which m*(m+1) is divisible by k: A014601 (k=4), A047208 (k=5), A007494 (k=3 and 6), A047335 (k=7), A047521 (k=8), this sequence (k=9).
Cf. A301451: numbers congruent to {1, 7} mod 9; A193910: numbers congruent to {2, 6} mod 9.

Programs

  • Magma
    [n: n in [0..300] | IsDivisibleBy(n*(n+1),9)];
  • Mathematica
    Select[Range[0, 300], Divisible[# (# + 1), 9] &]
  • PARI
    for(n=0, 300, if(n*(n+1)%9==0, print1(n", ")))
  • Sage
    [n for n in range(300) if 9.divides(n*(n+1))]

Formula

G.f.: x^2*(8 + x)/((1 + x)*(1 - x)^2).
a(n) = (18*n + 7*(-1)^n - 11)/4. Therefore: a(2*m) = 9*m-1, a(2*m+1) = 9*m. It follows that a(j)+a(k) and a(j)*a(k) belong to the sequence if j and k are not both even.
a(n) = -A090570(-n+2).
a(n) = a(n-1) + a(n-2) - a(n-3).
a(2*r+1) + a(2*r+s+1) = a(4*r+s+1) and a(2*r) + a(2*r+2*s+1) = a(4*r+2*s). A particular case provided by these identities: a(n) = a(n - 2*floor(n/6)) + a(2*floor(n/6) + 1).
E.g.f.: 1 + ((9*x - 2)*cosh(x) + 9*(x - 1)*sinh(x))/2. - Stefano Spezia, Apr 24 2021

A193910 Leap centuries in the revised Julian calendar.

Original entry on oeis.org

2, 6, 11, 15, 20, 24, 29, 33, 38, 42, 47, 51, 56, 60, 65, 69, 74, 78, 83, 87, 92, 96, 101, 105, 110, 114, 119, 123, 128, 132, 137, 141, 146, 150, 155, 159, 164, 168, 173, 177, 182, 186, 191, 195, 200, 204, 209, 213, 218, 222, 227, 231, 236, 240, 245, 249
Offset: 1

Views

Author

Frank Ellermann, Aug 09 2011

Keywords

Comments

Terms divided by 100, e.g., 29 indicates year 2900, which is a revised Julian leap year, but not a Gregorian leap year. Values below 20 are "proleptic" (only based on the formula).

Examples

			20 mod 9 is 2; 2000 was a leap year in the revised Julian calendar.
24 mod 9 is 6; 2400 and 2000 also happen to be Gregorian leap years.
28 is the first integer greater than 16 only contained in A008586.
29 is the first integer greater than 16 not contained in A008586.

Links

Crossrefs

A008586 enumerates "Gregorian leap centuries" (N mod 4 = 0).
A193879 enumerates all differences from A008586.
Cf. A274406: numbers congruent to {0, 8} mod 9; A301451: numbers congruent to {1, 7} mod 9. This sequence lists the numbers congruent to {2, 6} mod 9.

Programs

  • Mathematica
    Table[1/4 (18 m - (-1)^m - 11), {m, 56}] (* Farideh Firoozbakht, Oct 08 2014 *)
  • PARI
    a(n)=(9*n-5)\2 \\ Charles R Greathouse IV, Aug 23 2011
  • Rexx
    do C = 0 to 250; J = C // 9; if J = 2 | J = 6 then say C; end C

Formula

a(n) = a(n-2) + 9. - Charles R Greathouse IV, Aug 09 2011
a(n) = 2 or 6 (mod 9).
For all positive integers n, a(n) = (1/4)*(18*n-17*(-1)^n-11), which implies a(2*n-1) = 9*n-3 and a(2*n) = 9*n-7. - Farideh Firoozbakht, Oct 08 2014
G.f.: x*(2 + 4*x + 3*x^2)/((1 + x)*(1 - x)^2). - Philippe Deléham, Nov 30 2016

A301383 Expansion of (1 + 3*x - 2*x^2)/(1 - 7*x + 7*x^2 - x^3).

Original entry on oeis.org

1, 10, 61, 358, 2089, 12178, 70981, 413710, 2411281, 14053978, 81912589, 477421558, 2782616761, 16218279010, 94527057301, 550944064798, 3211137331489, 18715879924138, 109084142213341, 635788973355910, 3705649697922121, 21598109214176818, 125883005587138789, 733699924308655918
Offset: 0

Views

Author

Bruno Berselli, Mar 20 2018

Keywords

Comments

y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y+2). The corresponding x values are listed in A075841.
y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y+1) are in A002315, and A075870 gives the x values.
y solutions to A000217(x-1) + A000217(x) = A000290(y-1) + A000290(y) are in A046090, and A001653 gives the x values.
Also, indices y for which 4*A000217(y) + 5 is a square. The next integers k such that k*A000217(y) + 5 is a square for infinitely many y values are 11, 20, 22, 29, 31, ...
First differences are in A106329.

Links

Crossrefs

Subsequence of A301451.
Cf. A000217, A000290, A001652, A002315, A033539, A046090, A053142, A075841, A077444, A106329, A241976.

Programs

  • Julia
    using Nemo
function A301383List(len)
    R, x = PowerSeriesRing(ZZ, len+2, "x")
    f = divexact(1+3*x-2*x^2, 1-7*x+7*x^2-x^3)
    [coeff(f, k) for k in 0:len]
end
A301383List(23) |> println # Peter Luschny, Mar 21 2018
  • Magma
    m:=30; R:=PowerSeriesRing(Integers(), m); Coefficients(R!((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3)));
  • Maple
    f:= gfun:-rectoproc({a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3), a(0)=1,a(1)=10,a(2)=61},a(n),remember):
map(f, [$0..50]); # Robert Israel, Mar 21 2018
  • Mathematica
    CoefficientList[Series[(1 + 3 x - 2 x^2)/(1 - 7 x + 7 x^2 - x^3), {x, 0, 30}], x]
  • Maxima
    makelist(coeff(taylor((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3), x, 0, n), x, n), n, 0, 30);
  • PARI
    Vec((1+3*x-2*x^2)/(1-7*x+7*x^2-x^3)+O(x^30))
  • Sage
    m=30; L. = PowerSeriesRing(ZZ, m); f=(1+3*x-2*x^2)/(1-7*x+7*x^2-x^3); print(f.coefficients())

Formula

O.g.f.: (1 + 3*x - 2*x^2)/((1 - x)*(1 - 6*x + x^2)).
a(n) = 7*a(n-1) - 7*a(n-2) + a(n-3) = 6*a(n-1) - a(n-2) + 2.
a(n) = (3/4)*((1 + sqrt(2))^(2*n + 1) + (1 - sqrt(2))^(2*n + 1)) - 1/2.
a(n) = A033539(2*n+2) = A241976(n+1) + 1 = 3*A001652(n) + 1 = 3*A046090(n) - 2.
a(n) = A053142(n+1) + 3*A053142(n) - 2*A053142(n-1), n>0.
2*a(n) = 3*A002315(n) - 1.
4*a(n) = 3*A077444(n+1) - 2.
E.g.f.: (3*exp(3*x)*(cosh(2*sqrt(2)*x) + sqrt(2)*sinh(2*sqrt(2)*x)) - cosh(x) - sinh(x))/2. - Stefano Spezia, Mar 06 2020
Let T(n) be the n-th triangular number, A000217(n). Then T(a(n)-3) + 2*T(a(n)-2) + 3*T(a(n)-1) + 4*T(a(n)) + 3*T(a(n)+1) + 2*T(a(n)+2) + T(a(n)+3) = (A001653(n) + A001653(n+2))^2. - Charlie Marion, Mar 16 2021
Showing 1-3 of 3 results.

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

Content is available under The OEIS End-User License Agreement.