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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: Yukun Yao

Yukun Yao's wiki page.

Yukun Yao has authored 3 sequences.

A306381 a(n) is the number of semimagic 4 X 3 matrices whose entries are nonnegative integers, with row sum = 3*n and column sum = 4*n.

Original entry on oeis.org

1, 415, 8623, 64405, 289981, 965071, 2626975, 6195673, 13123945, 25572511, 46610191, 80439085, 132644773, 210471535, 323122591, 482085361, 701481745, 998443423, 1393512175, 1911065221, 2579765581, 3433037455, 4509566623, 5853825865, 7516625401, 9555688351
Offset: 0

Views

Author

Yukun Yao, Feb 11 2019

Keywords

Comments

This sequence was found during Doron Zeilberger's Experimental Mathematics class on Feb. 11, 2019. All class members contributed to the discovery of this sequence.

Examples

			For n=1 the a(1)= 415 because there are 415 4 X 3 matrices with nonnegative integer entries, whose row sum is 3 and column sum is 4.

Links

Programs

  • PARI
    a(n) = {(139*n^6 + 417*n^5 + 535*n^4 + 375*n^3 + 154*n^2 + 36*n + 4)/4} \\ Andrew Howroyd, Mar 01 2020

Formula

a(n) = (139/4)*n^6 + (417/4)*n^5 + (535/4)*n^4 + (375/4)*n^3 + (77/2)*n^2 + 9*n + 1.

Extensions

a(0)=1 prepended and terms a(11) and beyond from Andrew Howroyd, Mar 01 2020

A323830 a(0) = 1; thereafter a(n) is obtained by doubling a(n-1) and repeatedly deleting any string of identical digits.

Original entry on oeis.org

1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 636, 1272, 25, 50, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 636, 1272, 25, 50, 1, 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024, 2048, 4096, 8192, 16384, 32768, 636, 1272, 25, 50
Offset: 0

Views

Author

N. J. A. Sloane, Feb 03 2019, following a suggestion from Yukun Yao

Keywords

Comments

Periodic with period length 20.
Conjecture: If we start with any nonnegative number, and repeatedly double it and apply the "repeatedly delete any run of identical digits" operation described here, we eventually reach one of 0, 1, or 5.
In other words, the conjecture is that eventually we reach 0 or join the trajectory shown here or the trajectory shown in A323831.
The number of steps to reach 0, 1, or 5 is given in A323832.

Examples

			After a(15) = 32768 we get 65536 which becomes 636 after deleting "55". Then doubling 636 we get 1272, then 2544 which becomes 25 after deleting "44", then 50, then 100 which becomes 1 after deleting "00", and now the sequence repeats.

Links

Crossrefs

Cf. A000079, A320487, A321801, A321802, A323831, A323832.
See A035615 for a classic related base-2 sequence.

Programs

  • Mathematica
    dad[n_]:=FromDigits[FixedPoint[Flatten[Select[Split[#],Length[#]==1&]]&,IntegerDigits[2n]]];NestList[dad,1,100] (* Paolo Xausa, Nov 14 2023 *)
  • PARI
    Vec((1 + 2*x)*(1 + 4*x^2 + 16*x^4 + 64*x^6 + 256*x^8 + 1024*x^10 + 4096*x^12 + 16384*x^14 + 636*x^16 + 25*x^18) / (1 - x^20) + O(x^40)) \\ Colin Barker, Feb 03 2019

Formula

From Colin Barker, Feb 03 2019: (Start)
G.f.: (1 + 2*x)*(1 + 4*x^2 + 16*x^4 + 64*x^6 + 256*x^8 + 1024*x^10 + 4096*x^12 + 16384*x^14 + 636*x^16 + 25*x^18) / (1 - x^20).
a(n) = a(n-20) for n>19.
(End)
a(n+1) = A321801(2*a(n)). For general numbers, the "repeatedly delete any run of identical digits" operation corresponds to repeatedly applying A321801. - Chai Wah Wu, Feb 11 2019

A318047 a(n) = sum of values taken by all parking functions of length n.

Original entry on oeis.org

1, 8, 81, 1028, 15780, 284652, 5903464, 138407544, 3619892160, 104485268960, 3299177464704, 113120695539612, 4185473097734656, 166217602768452900, 7051983744002135040, 318324623296131263408, 15232941497754507165696, 770291040239888149405944, 41042353622873800536064000, 2298206207793743728251532020
Offset: 1

Views

Author

Yukun Yao, Aug 13 2018

Keywords

Examples

			Case n = 2: There are 3 parking functions of length 2: [1, 1], [1, 2], [2, 1]. Summing up all values gives 2 + 3 + 3 = 8, so a(2) = 8.
Case n = 3: There are 16 parking functions of length 3: [1, 1, 1], [1, 1, 2], [1, 1, 3], [1, 2, 1], [1, 2, 2], [1, 2, 3], [1, 3, 1], [1, 3, 2], [2, 1, 1], [2, 1, 2], [2, 1, 3], [2, 2, 1], [2, 3, 1], [3, 1, 1], [3, 1, 2], [3, 2, 1]. Summing up all values gives a total of 81, so a(3) = 81.

Links

Crossrefs

Cf. A000272, A298593.

Programs

  • Maple
    #Pnax(n,a,x): the sum of x^(sum of all entries in the parking function) over the set of a-parking functions of length n by recurrence relation.
Pnax:=proc(n,a,x) local k:
option remember:
if n=0 then
  return 1:
fi:
if n>0 and a=0 then
  return 0:
fi:
return expand(x^n*add(binomial(n,k)*Pnax(n-k,a+k-1,x),k=0..n)):
end:
seq(subs(x = 1, diff(Pnax(n, 1, x), x)), n = 1 .. 20)
  • Mathematica
    T[n_, k_] := n Sum[Binomial[n-1, j-1] j^(j-2) (n-j+1)^(n-j-1), {j, k, n}];
a[n_] := Sum[k T[n, k], {k, 1, n}];
Array[a, 20] (* Jean-François Alcover, Aug 29 2018, after Andrew Howroyd *)
  • PARI
    \\ here T(n,k) is A298593.
T(n,k)={n*sum(j=k, n, binomial(n-1, j-1)*j^(j-2)*(n+1-j)^(n-1-j))}
a(n)={sum(k=1, n, k*T(n,k))} \\ Andrew Howroyd, Aug 17 2018

Formula

a(n) is the first derivative of P(n,1,x) evaluated at x = 1 where P(n,m,x) satisfies P(n,m,x) = x^n*Sum_{k=0..n} binomial(n,k)*P(n-k, m+k-1, x) with P(0,m,x) = 1 and P(n,0,x) = 0 for n > 0.
a(n) = Sum_{k=1..n} k*A298593(n, k). - Andrew Howroyd, Aug 17 2018

Extensions

Edited by Andrew Howroyd and N. J. A. Sloane, Aug 19 2018

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