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 1-5 of 5 results.

A377213 Expansion of 1/(1 - 4*x^3/(1-x))^(3/2).

Original entry on oeis.org

1, 0, 0, 6, 6, 6, 36, 66, 96, 266, 576, 1026, 2246, 4866, 9516, 19598, 41286, 83526, 170048, 351378, 716850, 1458098, 2984028, 6087270, 12380900, 25224222, 51356400, 104380510, 212164362, 431148222, 875353220, 1776567762, 3604752672, 7310374010, 14819370480, 30033014994
Offset: 0

Views

Author

Seiichi Manyama, Oct 20 2024

Keywords

Links

Crossrefs

Cf. A377197, A377204.
Cf. A360309, A377216.

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 35); Coefficients(R!( 1/(1 - 4*x^3/(1-x))^(3/2))); // Vincenzo Librandi, May 08 2025
  • Mathematica
    Table[Sum[(2*k+1)*Binomial[2*k,k]*Binomial[n-2*k-1,n-3*k],{k,0,Floor[n/3]}],{n,0,35}] (* Vincenzo Librandi, May 08 2025 *)
  • PARI
    a(n) = sum(k=0, n\3, (2*k+1)*binomial(2*k, k)*binomial(n-2*k-1, n-3*k));

Formula

a(n) = (2*(n-1)*a(n-1) - (n-2)*a(n-2) + 2*(2*n+3)*a(n-3) - 2*(2*n-2)*a(n-4))/n for n > 3.
a(n) = Sum_{k=0..floor(n/2)} (2*k+1) * binomial(2*k,k) * binomial(n-2*k-1,n-3*k).
a(n) ~ sqrt(n) * 2^(n-2) / sqrt(Pi). - Vaclav Kotesovec, May 03 2025

A360310 a(n) = Sum_{k=0..floor(n/4)} binomial(n-1-3*k,n-4*k) * binomial(2*k,k).

Original entry on oeis.org

1, 0, 0, 0, 2, 2, 2, 2, 8, 14, 20, 26, 52, 98, 164, 250, 426, 762, 1328, 2194, 3682, 6366, 11072, 18878, 32038, 54906, 94860, 163226, 279634, 479806, 826776, 1425542, 2454020, 4223170, 7279164, 12560466, 21671314, 37381714, 64512676, 111414042
Offset: 0

Views

Author

Seiichi Manyama, Feb 03 2023

Keywords

Crossrefs

Cf. A026585, A360309.
Cf. A360292, A360315.

Programs

  • PARI
    a(n) = sum(k=0, n\4, binomial(n-1-3*k, n-4*k)*binomial(2*k, k));
  • PARI
    my(N=40, x='x+O('x^N)); Vec(1/sqrt(1-4*x^4/(1-x)))

Formula

G.f.: 1 / sqrt(1-4*x^4/(1-x)).
n*a(n) = 2*(n-1)*a(n-1) - (n-2)*a(n-2) + 2*(2*n-4)*a(n-4) - 2*(2*n-7)*a(n-5).

A360314 a(n) = Sum_{k=0..floor(n/3)} (-1)^k * binomial(n-1-2*k,n-3*k) * binomial(2*k,k).

Original entry on oeis.org

1, 0, 0, -2, -2, -2, 4, 10, 16, 2, -32, -86, -90, 26, 332, 646, 534, -690, -3040, -4934, -2270, 9066, 27260, 35198, 532, -101946, -232752, -230730, 158986, 1039078, 1899364, 1265370, -2714160, -9926158, -14625008, -4036358, 34062386, 89744810, 104123084
Offset: 0

Views

Author

Seiichi Manyama, Feb 03 2023

Keywords

Crossrefs

Cf. A360313, A360315.
Cf. A360294, A360309.

Programs

  • PARI
    a(n) = sum(k=0, n\3, (-1)^k*binomial(n-1-2*k, n-3*k)*binomial(2*k, k));
  • PARI
    my(N=40, x='x+O('x^N)); Vec(1/sqrt(1+4*x^3/(1-x)))

Formula

G.f.: 1 / sqrt(1+4*x^3/(1-x)).
n*a(n) = 2*(n-1)*a(n-1) - (n-2)*a(n-2) - 2*(2*n-3)*a(n-3) + 2*(2*n-6)*a(n-4).

A377216 Expansion of 1/(1 - 4*x^3/(1-x))^(5/2).

Original entry on oeis.org

1, 0, 0, 10, 10, 10, 80, 150, 220, 710, 1620, 2950, 7010, 16110, 32560, 70682, 156810, 329290, 698540, 1507110, 3189742, 6725150, 14279520, 30141730, 63335960, 133297362, 279996460, 586364410, 1227337710, 2566307410, 5355970048, 11166535430, 23259949980, 48389451510
Offset: 0

Views

Author

Seiichi Manyama, Oct 20 2024

Keywords

Links

Crossrefs

Cf. A377199, A377215.
Cf. A360309, A377213.

Programs

  • Magma
    R:=PowerSeriesRing(Rationals(), 35); Coefficients(R!( 1/(1 - 4*x^3/(1-x))^(5/2))); // Vincenzo Librandi, May 10 2025
  • Mathematica
    Table[Sum[(-4)^k*Binomial[-5/2,k]*Binomial[n-2*k-1,n-3*k],{k,0,Floor[n/3]}],{n,0,35}] (* Vincenzo Librandi, May 10 2025 *)
  • PARI
    a(n) = sum(k=0, n\3, (-4)^k*binomial(-5/2, k)*binomial(n-2*k-1, n-3*k));

Formula

a(n) = (2*(n-1)*a(n-1) - (n-2)*a(n-2) + 2*(2*n+9)*a(n-3) - 2*(2*n+2)*a(n-4))/n for n > 3.
a(n) = Sum_{k=0..floor(n/3)} (-4)^k * binomial(-5/2,k) * binomial(n-2*k-1,n-3*k).
a(n) ~ n^(3/2) * 2^(n-3) / (3*sqrt(Pi)). - Vaclav Kotesovec, May 03 2025

A376884 Number of binary n-sequences ending in 1 with exactly one more occurrence of 11 than 10.

Original entry on oeis.org

0, 0, 1, 1, 1, 4, 7, 10, 23, 46, 79, 157, 315, 588, 1137, 2249, 4337, 8402, 16495, 32179, 62707, 122916, 240837, 471456, 925061, 1816610, 3566865, 7010347, 13789477, 27130956, 53409503, 105205514, 207309743, 408672454, 805989367, 1590166915, 3138371715
Offset: 0

Views

Author

Bruce Levin, Oct 07 2024

Keywords

Comments

It is easy to prove that this sequence is exactly one-half of sequence A360309. The connection to the number of binary n-sequences ending in 1 with exactly one more occurrence of 11 pairs than 10 pairs appears new and is related to a coin-tossing problem posed by Daniel Litt. In particular, the cumulative sum a(0)+...+a(n-1) gives the number of all binary sequences of length n with strictly more 10 pairs than 11 pairs minus the number of all binary sequences of length n with strictly more 11 pairs than 10 pairs. These results are demonstrated in the Levin link below, where more formulas are given for the number of binary n-sequences ending in 1 (or 0) with exactly s more occurrences of 11 pairs than 10 pairs for any integer s.

Examples

			There are a(5)=4 binary 5-sequences ending in 1 with one more occurrence of 11 than 10. They are 11101, 11011, and 10111 (each with two occurrences of 11 and one occurrence of 10); and 00011 (with one occurrence of 11 and zero occurrences of 10).

Links

Crossrefs

One-half of A360309(n+1).
Cf. A217615.

Programs

  • Maple
    a:= proc(n) option remember; `if`(n<4, iquo(n, 2), (2*n*a(n-1)
      -(n-1)*a(n-2)+(4*n-2)*a(n-3)-4*(n-2)*a(n-4))/(n+1))
    end:
seq(a(n), n=0..36);  # Alois P. Heinz, Oct 08 2024
  • Mathematica
    a[n_] := Sum[Binomial[n - 2 k, n + 1 - 3 k]*Binomial[2 k, k]/2, {k, 0, Floor[(n + 1)/3]}];
Table[a[n], {n, 0, 36}] (* Robert P. P. McKone, Dec 28 2024 *)

Formula

a(n) = Sum_{k=1..floor((n+1)/3)} binomial((2*k)-1,k) * binomial(n-2*k,k-1).
a(n) = A360309(n+1)/2.
Showing 1-5 of 5 results.

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