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.

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A046170 Number of self-avoiding walks on a 2-D lattice of length n which start at the origin, take first step in the {+1,0} direction and whose vertices are always nonnegative in x and y.

Original entry on oeis.org

1, 2, 5, 12, 30, 73, 183, 456, 1151, 2900, 7361, 18684, 47652, 121584, 311259, 797311, 2047384, 5260692, 13542718, 34884239, 89991344, 232282110, 600281932, 1552096361, 4017128206, 10401997092, 26957667445, 69892976538, 181340757857, 470680630478, 1222433229262, 3175981845982
Offset: 1

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Keywords

Crossrefs

Formula

a(n) = A038373(n)/2. - Siqi Wang, Jul 15 2022

Extensions

More terms from Stephen A. Silver
More terms from Siqi Wang, Jul 15 2022

A331030 Divide the terms of the harmonic series into groups sequentially so that the sum of each group is minimally greater than 1. a(n) is the number of terms in the n-th group.

Original entry on oeis.org

2, 5, 13, 36, 98, 266, 723, 1965, 5342, 14521, 39472, 107296, 291661, 792817, 2155100, 5858169, 15924154, 43286339, 117664468, 319845186, 869429357, 2363354022, 6424262292, 17462955450, 47469234471, 129034757473, 350752836478, 953445061679, 2591732385596
Offset: 1

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Author

Keywords

Comments

a(n) = A046171(n+1) through a(5), and grows similarly for n > 5.
Let b(n) = Sum_{j=1..n} a(n); then for n >= 2 it appears that b(n) = round((b(n-1) + 1/2)*e). Verified through n = 10000 (using the approximation Sum_{j=1..k} 1/j = log(k) + gamma + 1/(2*k) - 1/(12*k^2) + 1/(120*k^4) - 1/(252*k^6) + 1/(240*k^8) - ..., where gamma is the Euler-Mascheroni constant, A001620). Cf. A081881. - Jon E. Schoenfield, Jan 10 2020

Examples

			a(1)=2 because 1 + 1/2 = 1.5 > 1,
a(2)=5 because 1/3 + 1/4 + 1/5 + 1/6 + 1/7 = 1.0928... > 1,
etc.
		

Crossrefs

Programs

  • PARI
    lista(lim=oo)={my(s=0, p=0); for(i=1, lim, s+=1/i; if(s>1, print1(i-p, ", "); s=0; p=i))} \\ Andrew Howroyd, Jan 08 2020
  • Python
    x = 0.0
    y = 0.0
    z = 0.0
    for i in range(1,100000000000000000000000):
      y += 1
      x = x + 1/i
      z = z + 1/i
      if x > 1:
        print(y)
        y = 0
        x = 0
    

Formula

a(1)=2, a(n) = (min(p) : Sum_{s=r..p} 1/s > 1)-r+1, r=Sum_{k=1..n-1} a(k).

Extensions

a(25)-a(29) from Jon E. Schoenfield, Jan 10 2020
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