Yuriy Sibirmovsky - cp's OEIS frontend

User: Yuriy Sibirmovsky

Yuriy Sibirmovsky has authored 15 sequences. Here are the ten most recent ones:

A282497 'Somos expansion' of e: e=a(0)sqrt(a(1)sqrt(a(2)sqrt(a(3)sqrt(...)))). a(n)=floor(x(n)), x(n)=x(n-1)^2/a(n-1)^2, x(0)=e.

2, 1, 3, 1, 1, 2, 1, 3, 1, 2, 1, 3, 1, 1, 2, 2, 1, 1, 1, 3, 1, 1, 2, 1, 1, 3, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 1, 3, 1, 2, 1, 1, 2, 1, 2, 1, 2, 1, 3, 1, 1, 2, 1, 1, 3, 1, 2, 1, 2, 1, 1, 2, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 3, 1, 2, 1, 3, 1, 2, 1, 2, 1, 3, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 1, 3, 1, 1, 1, 1, 3, 1, 1, 1, 1, 2, 1, 1, 3, 1, 1, 2, 1

Offset: 0

Yuriy Sibirmovsky, Feb 16 2017

nonn

1<=a(n)<=3 for all n. Reasoning: for x>1 it follows that 1

			Integer part of e is 2. Integer part of e^2/4 is 1.

Yuriy Sibirmovsky, Table of n, a(n) for n = 0..1999

Cf. A001113 (digits).

$MaxExtraPrecision = 1000;
x0 = E;
Nm = 130;
j = 1;
Res = Table[1, {j, 1, Nm}];
While[j < Nm, Res[[j]] = Floor[x0]; x0 = N[(x0/ Res[[j]])^2, 20000];
  j++];

Product_{k>=0} a(k)^(1/2^k) = e.

A282496 'Somos expansion' of Pi: Pi=a(0)sqrt(a(1)sqrt(a(2)sqrt(a(3)sqrt(...)))). a(n)=floor(x(n)), x(n)=x(n-1)^2/a(n-1)^2, x(0)=Pi.

Original entry on oeis.org

3, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2, 1, 2, 2, 1, 1, 2, 1, 1, 1, 1, 3, 1, 2, 1, 3, 1, 1, 2, 2, 1, 1, 2, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 1, 1, 3, 1, 2, 1, 1, 3, 1, 2, 1, 1, 1, 1, 3, 1, 2, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 3, 1, 1, 1, 1, 1, 3, 1, 1, 2, 1, 1, 3, 1, 1, 3, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 3, 1, 2, 2, 1, 1, 1, 1, 2, 1, 3, 1, 1, 1, 1

Offset: 0

Yuriy Sibirmovsky, Feb 16 2017

nonn

1<=a(n)<=3 for all n. Reasoning: for x>1 it follows that 1

			Integer part of Pi is 3. Integer part of Pi^2/9 is 1.

Yuriy Sibirmovsky, Table of n, a(n) for n = 0..1999

Cf. A000796 (digits), A100044 (Pi^2/9), A001203 (continued fraction), A276459 (another nested radical expansion).

$MaxExtraPrecision = 1000;
x00 = Pi;
x0 = x00;
Nm = 130;
j = 1;
Res = Table[1, {j, 1, Nm}];
While[j < Nm, Res[[j]] = Floor[x0]; x0 = N[(x0/ Res[[j]])^2, 20000];
  j++];
Res

Product_{k>=0} a(k)^(1/2^k) = Pi.

A281873 a(n+1) is the smallest number greater than a(n) such that Sum_{j=1..n+1} 1/a(j) <= 4, a(1) = 1.

Original entry on oeis.org

1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20, 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 200, 77706, 16532869712, 3230579689970657935732, 36802906522516375115639735990520502954652700

Offset: 1

Yuriy Sibirmovsky, Jan 31 2017

The method for any number A is to find the largest harmonic number H(n) smaller than A, then use the greedy algorithm to expand the difference A - H(n).

A140335 is the same sequence for 3. The sequence for 5 consists of 99 terms, the largest of which has 142548 digits.

A. M. Gleason, R. E. Greenwood, and L. M. Kelly, The William Lowell Putnam Mathematical Competition, Problems and Solutions, 1938-1964, MAA, 1980, pages 398-399.

John Scholes, 14th Putnam Mathematical Competition, 1954, Problem B6, after Gleason, Greenwood & Kelly.
Index entries for sequences related to Egyptian fractions

Cf. A140335, A306349.

x0=4-Sum[1/k,{k,1,30}];
Nm=10;
j=0;
While[x0>0||j==Nm,a0=Ceiling[1/x0];
x0=x0-1/a0;
Print[a0];j++]
f[s_List, n_] := Block[{t = Total[1/s]}, Append[s, Max[ s[[-1]] +1, Ceiling[1/(n - t)]]]]; Nest[f[#, 4] &, {1}, 34] (* Robert G. Wilson v, Feb 05 2017 *)

from sympy import egyptian_fraction
print(egyptian_fraction((4, 1))) # Pontus von Brömssen, Feb 10 2019

Sum_{k=1..35} 1/a(k) = 4.

A278288 a(n) is the number of ways to represent a(n-1) as a sum of two distinct terms from a(0) to a(n-2). a(0) = 0. a(1) = a(2) = 1.

Original entry on oeis.org

0, 1, 1, 1, 2, 3, 3, 4, 6, 2, 4, 8, 3, 8, 4, 12, 6, 10, 10, 11, 12, 15, 9, 12, 19, 7, 15, 17, 10, 18, 22, 17, 12, 22, 21, 22, 25, 25, 26, 22, 26, 26, 27, 32, 25, 30, 27, 35, 21, 23, 31, 31, 32, 37, 37, 38, 37, 39, 37, 40, 40, 41, 45, 28, 37, 42, 38, 50, 33, 43, 58, 34

Offset: 0

Yuriy Sibirmovsky, Nov 16 2016

nonn

a(n) is the number of pairs (j, k) such that a(j) + a(k) = a(n-1), with 0 <= j, k < n-1 and j != k.
For large n, a(n) on average follows linear law a(n) ~ 0.7 n with linear spread (see the plot).
Unlike sequences such as A248034 or A276457, this sequence is base-independent.

			a(2) = a(1) + a(0), so a(3) = 1.
a(3) = a(1) + a(0) = a(2) + a(0), so a(4) = 2.
a(4) = a(3) + a(2) = a(3) + a(1) = a(2) + a(1), so a(5) = 3.

Yuriy Sibirmovsky, Table of n, a(n) for n = 0..2999
Yuriy Sibirmovsky, The plot of a(n) for n=0..2999 (with linear regression)

Cf. A248034, A276457.

Nm=100;
A=Table[1,{n,1,Nm}];
A[[1]]=0;
Do[Nc=0;
Do[If[A[[j]]+A[[k]]==A[[n]] && k!=j,Nc++],{j,1,n-1},{k,1,j}];
A[[n+1]]=Nc,{n,3,Nm-1}];
A

A276819 a(n) = (9*n^2 - n)/2 + 1.

Original entry on oeis.org

1, 5, 18, 40, 71, 111, 160, 218, 285, 361, 446, 540, 643, 755, 876, 1006, 1145, 1293, 1450, 1616, 1791, 1975, 2168, 2370, 2581, 2801, 3030, 3268, 3515, 3771, 4036, 4310, 4593, 4885, 5186, 5496, 5815, 6143, 6480, 6826, 7181, 7545, 7918, 8300, 8691, 9091, 9500, 9918, 10345, 10781, 11226, 11680, 12143, 12615

Offset: 0

Yuriy Sibirmovsky, Sep 18 2016

Diagonal of triangular spiral in A051682. The other 5 diagonals are given by A140064, A117625, A081267, A064225, A006137. See the link as well.
First differences are given by A017209.
72*a(n) - 71 is a perfect square. - Klaus Purath, Jan 14 2022

Colin Barker, Table of n, a(n) for n = 0..1000
Yuriy Sibirmovsky, Six diagonals of the triangular spiral.
Index entries for linear recurrences with constant coefficients, signature (3,-3,1).

Cf. A002378, A003215, A006137, A017209, A051682, A064225, A081267, A117625, A140064.

Mathematica
```
Table[(9*n^2-n)/2+1, {n,0,100}]
```

Vec((1+2*x+6*x^2)/(1-x)^3 + O(x^60)) \\ Colin Barker, Sep 18 2016

a(n) = (9*n^2 - n)/2 + 1; \\ Altug Alkan, Sep 18 2016

a(n) = (9*n^2 - n)/2 + 1.
a(n) = a(n-1) + 9*n - 5 with a(0) = 1.
From Colin Barker, Sep 18 2016: (Start)
a(n) = 3*a(n-1) - 3*a(n-2) + a(n-3) for n > 2.
G.f.: (1 + 2*x + 6*x^2)/(1 - x)^3. (End)
From Klaus Purath, Jan 14 2022: (Start)
a(n) = A006137(n) - n.
A003215(a(n)) - A003215(a(n)-3) = A002378(9*n-1). (End)
E.g.f.: exp(x)*(2 + 8*x + 9*x^2)/2. - Stefano Spezia, Dec 25 2022

A276806 Height of the shortest binary factorization tree of n.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 0, 2, 1, 1, 0, 2, 0, 1, 1, 2, 0, 2, 0, 2, 1, 1, 0, 2, 1, 1, 2, 2, 0, 2, 0, 3, 1, 1, 1, 2, 0, 1, 1, 2, 0, 2, 0, 2, 2, 1, 0, 3, 1, 2, 1, 2, 0, 2, 1, 2, 1, 1, 0, 2, 0, 1, 2, 3, 1, 2, 0, 2, 1, 2, 0, 3, 0, 1, 2, 2, 1, 2, 0, 3, 2, 1, 0, 2, 1, 1, 1, 2, 0, 2, 1, 2, 1, 1, 1, 3, 0, 2, 2, 2

Offset: 1

Yuriy Sibirmovsky, Sep 17 2016

nonn

Among all possible binary factorization trees of n we choose a tree with minimal height. The choice may not be unique. a(n) gives the height of the chosen tree.
To compute the terms A001222 and A001221 could be used.
The positions at which numbers (1,2,3) first appear are respectively (4,8,32). The latter sequence can be described by the formula b(n) = 2^(2^(n-1) + 1).

			a(12) = 2 since 12 cannot be factored in a binary factorization tree of height less than 2, but it can be factored in a tree of height 2, e.g.,
      12
      / \
     4   3
    / \
   2   2
Similarly, a(16) = 2:
       16
       / \
      /   \
     4     4
    / \   / \
   2   2 2   2
and a(40) = 2:
       40
       / \
      /   \
     4    10
    / \   / \
   2   2 2   5
and a(84) = 2:
       84
       / \
      /   \
     4    21
    / \   / \
   2   2 3   7

Antti Karttunen, Table of n, a(n) for n = 1..10000
Index entries for sequences computed from exponents in factorization of n

Cf. A001221, A001222, A005171.

a(n)=if(n>1,my(b=bigomega(n),c=(2^logint(b,2)!=b));logint(b,2)+c,0) \\ David A. Corneth, Oct 01 2016

A276806(n) = { my(m=0,h); if((1==n)||isprime(n),0,fordiv(n,d,if((d>1)&&(dA276806(d),A276806(n/d)); if(!m || (h < m),m=h)))); m; }; \\ Antti Karttunen, Aug 12 2017

a(n^2) = a(n) + 1.

A276771 a(n) is the number of runs of an algorithm. Set b_0 = n, if prime or 1 or 0, stop; else, set c_0 = largest divisor of n (!=n); set b_1 = c_0 - b_0/c_0. Run with b_1.

Original entry on oeis.org

0, 0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 2, 0, 1, 1, 2, 0, 1, 0, 2, 2, 2, 0, 2, 1, 1, 2, 3, 0, 1, 0, 2, 2, 2, 1, 3, 0, 1, 2, 2, 0, 1, 0, 3, 3, 3, 0, 3, 1, 1, 2, 3, 0, 2, 2, 2, 3, 3, 0, 4, 0, 1, 2, 2, 2, 1, 0, 3, 3, 3, 0, 3, 0, 2, 3, 4, 2, 1, 0, 2, 3, 3, 0, 3, 3, 1, 2, 2, 0, 1, 2, 4, 4, 4, 2, 4, 0, 1, 2, 4

Offset: 1

Yuriy Sibirmovsky, Sep 17 2016

nonn

For large n: Sum_{k=1..n} a(k) ~ n*log(n)/2 - n/2 (conjectured).

			For n=14: b_0 = 14, not prime or 1 or 0. c_0 = 7. b_1 = 7 - 2 = 5. 5 is prime.
In short: 14 -> {7,2} -> 5. Number of runs a(14) = 1.

Yuriy Sibirmovsky, Table of n, a(n) for n = 1..1000
Yuriy Sibirmovsky, Plot of sum_{k=1..n} a(k) for n = 1..20000

Cf. A000040, A276770.

Nm=100;
a=Table[0,{n,1,Nm}];
Do[b0=n;
j=0;
While[PrimeQ[b0]==False&&b0!=1&&b0!=0,c=Reverse[Divisors[b0]];
b1=c[[2]]-b0/c[[2]];
b0=b1;j++];
a[[n]]=j,{n,1,Nm}];
a

stop(n) = (n<=1) || isprime(n);
a(n) = {b = n; nb = 0; while (! stop(b), d = divisors(b); c = d[#d-1]; b = c - b/c; nb++;); nb;} \\ Michel Marcus, Sep 19 2016

A276770 a(n) is the number of runs of an algorithm. Set b_0 = n, if prime, stop; else, set c_0 = largest divisor of n (!=n); set b_1 = c_0 + b_0/c_0. Run with b_1.

Original entry on oeis.org

0, 1, 0, 2, 2, 1, 0, 3, 0, 3, 3, 2, 0, 1, 0, 4, 2, 1, 0, 4, 2, 4, 4, 3, 0, 1, 0, 2, 4, 1, 4, 5, 0, 3, 3, 2, 0, 1, 0, 5, 2, 3, 0, 5, 4, 5, 5, 4, 0, 1, 3, 2, 2, 1, 0, 3, 0, 5, 5, 2, 2, 5, 0, 6, 5, 1, 0, 4, 0, 4, 4, 3, 2, 1, 0, 2, 2, 1, 0, 6, 2, 3, 3, 4, 0, 1, 5, 6, 2, 5, 5, 6, 0, 6, 6, 5, 0, 1, 0, 2, 4

Offset: 5

Yuriy Sibirmovsky, Sep 17 2016

nonn

a(4) -> infinity. For any other n >= 2, a(n) is finite, however cases n = 2,3 are trivial, so the offset is 5.

For large n: Sum_{k=5..n} a(k) ~ n*log(n)/2 (conjectured).

			For n=14: b_0 = 14, not prime. c_0 = 7. b_1 = 7 + 2 = 9. 9 is not prime.
In short: 14 -> {7,2} -> 9 -> {3,3} -> 6 -> {3,2} -> 5. Number of runs a(14) = 3.

Yuriy Sibirmovsky, Table of n, a(n) for n = 5..1000
Yuriy Sibirmovsky, Plot of sum_{k=5..n} a(k) for n = 5..20000

Cf. A000040, A276771.

Nm=100;
a=Table[0,{n,1,Nm}];
Do[b0=n;
j=0;
While[PrimeQ[b0]==False,c=Reverse[Divisors[b0]];
b1=c[[2]]+b0/c[[2]];
b0=b1;j++];
a[[n]]=j,{n,5,Nm}];
Table[a[[k]],{k,5,Nm}]

stop(n) = (n<=1) || isprime(n);
a(n) = {b = n; nb = 0; while (! stop(b), d = divisors(b); c = d[#d-1]; b = c + b/c; nb++;); nb;} \\ Michel Marcus, Sep 19 2016

A276648 Number of points of norm <= n in the body-centered cubic lattice with the lattice parameter equal to 2/sqrt(3).

Original entry on oeis.org

1, 9, 59, 169, 339, 701, 1243, 1893, 2741, 3943, 5577, 7343, 9409, 12039, 15065, 18421, 22227, 26717, 31879, 37461, 43655, 50557, 58071, 66227, 75121, 85083, 95801, 107227, 119541, 133019, 147271, 161901, 178127, 195481, 214143

Offset: 0

Yuriy Sibirmovsky, Sep 11 2016

nonn

Experimentally observed dense bcc clusters of gold contain 1, 9, 59, 169, 339, 701 and 1243 nanoparticles (N.G. Khlebtsov, Fig. 32 and text on p. 208), exactly matching the first 7 terms of the sequence.

First 5 terms are the same as A276450.

			The origin has norm 0, thus a(0)=1. The distance to the 8 vertices of the cube from the origin is 1, because the edge of the cube is 2/sqrt(3). Thus a(1)=9.

Yuriy Sibirmovsky, Table of n, a(n) for n = 0..50
N. G. Khlebtsov, T-matrix method in plasmonics: An overview, J. Quantitative Spectroscopy & Radiative Transfer 123 (2013) 184-217.

Cf. A276450.

DecM[A_]:=A[[1]]^2+A[[2]]^2+A[[3]]^2;
Do[N1=0;N2=0;
Do[A={l,k,j};
B={l+1/2,k+1/2,j+1/2};
If[DecM[A]<=3/4r^2,N1+=1];
If[DecM[B]<=3/4r^2,N2+=1],{l,-r-1,r+1},{k,-r-1,r+1},{j,-r-1,r+1}];
Print[r," ",N1+N2],{r,0,20}]

A276472 Modified Pascal's triangle read by rows: T(n,k) = T(n-1,k) + T(n-1,k-1), 12. T(n,n) = T(n,n-1) + T(n-1,n-1), n>1. T(1,1) = 1, T(2,1) = 1. n>=1.

Original entry on oeis.org

1, 1, 2, 4, 3, 5, 11, 7, 8, 13, 29, 18, 15, 21, 34, 76, 47, 33, 36, 55, 89, 199, 123, 80, 69, 91, 144, 233, 521, 322, 203, 149, 160, 235, 377, 610, 1364, 843, 525, 352, 309, 395, 612, 987, 1597, 3571, 2207, 1368, 877, 661, 704, 1007, 1599, 2584, 4181

Offset: 1

Yuriy Sibirmovsky, Sep 12 2016

The recurrence relations for the border terms are the only way in which this differs from Pascal's triangle.
Column T(2n,n+1) appears to be divisible by 4 for n>=2; T(2n-1,n) divisible by 3 for n>=2; T(2n,n-2) divisible by 2 for n>=3.
The symmetry of T(n,k) can be observed in a hexagonal arrangement (see the links).
Consider T(n,k) mod 3 = q. Terms with q = 0 show reflection symmetry with respect to the central column T(2n-1,n), while q = 1 and q = 2 are mirror images of each other (see the link).

			Triangle T(n,k) begins:
n\k 1    2    3    4   5    6    7    8    9
1   1
2   1    2
3   4    3    5
4   11   7    8    13
5   29   18   15   21   34
6   76   47   33   36   55   89
7   199  123  80   69   91   144 233
8   521  322  203  149  160  235 377  610
9   1364 843  525  352  309  395 612  987  1597
...
In another format:
__________________1__________________
_______________1_____2_______________
____________4_____3_____5____________
________11_____7_____8_____13________
____29_____18_____15____21_____34____
_76_____47____33_____36____55_____89_

Yuriy Sibirmovsky, T(n,k), read by rows as a linear sequence a(j) for j = 1..5050
Yuriy Sibirmovsky, Symmetrical hexagonal arrangement for initial terms of T(n,k)
Yuriy Sibirmovsky, T(n,k) compared with Pascal's triangle
Yuriy Sibirmovsky, Illustration for T(n,k) mod 3

Cf. A000045, A000204, A001519, A001906, A002878, A005248, A054441, A088305, A122367, A258109, A026671, A026726, A026732, A004187, A049685, A033891, A206351, A092521, A081018, A049684, A058038, A081016, A003482, A049629, A133273, A049664, A049683, A119032, A023039, A007805, A033889.

Nm=12;
T=Table[0,{n,1,Nm},{k,1,n}];
T[[1,1]]=1;
T[[2,1]]=1;
T[[2,2]]=2;
Do[T[[n,1]]=T[[n-1,1]]+T[[n,2]];
T[[n,n]]=T[[n-1,n-1]]+T[[n,n-1]];
If[k!=1&&k!=n,T[[n,k]]=T[[n-1,k]]+T[[n-1,k-1]]],{n,3,Nm},{k,1,n}];
{Row[#,"\t"]}&/@T//Grid

T(n,k) = if (k==1, if (n==1, 1, if (n==2, 1, T(n-1,1) + T(n,2))), if (kMichel Marcus, Sep 14 2016

Conjectures:
Relations with other sequences:
T(n+1,1) = A002878(n-1), n>=1.
T(n,n) = A001519(n) = A122367(n-1), n>=1.
T(n+1,2) = A005248(n-1), n>=1.
T(n+1,n) = A001906(n) = A088305(n), n>=1.
T(2n-1,n) = 3*A054441(n-1), n>=2. [the central column].
Sum_{k=1..n} T(n,k) = 3*A105693(n-1), n>=2. [row sums].
Sum_{k=1..n} T(n,k)-T(n,1)-T(n,n) = 3*A258109(n), n>=2.
T(2n,n+1) - T(2n,n) = A026671(n), n>=1.
T(2n,n-1) - T(2n,n) = 2*A026726(n-1), n>=2.
T(n,ceiling(n/2)) - T(n-1,floor(n/2)) = 2*A026732(n-3), n>=3.
T(2n+1,2n) = 3*A004187(n), n>=1.
T(2n+1,2) = 3*A049685(n-1), n>=1.
T(2n+1,2n) + T(2n+1,2) = 3*A033891(n-1), n>=1.
T(2n+1,3) = 5*A206351(n), n>=1.
T(2n+1,2n)/3 - T(2n+1,3)/5 = 4*A092521(n-1), n>=2.
T(2n,1) = 1 + 5*A081018(n-1), n>=1.
T(2n,2) = 2 + 5*A049684(n-1), n>=1.
T(2n+1,2) = 3 + 5*A058038(n-1), n>=1.
T(2n,3) = 3 + 5*A081016(n-2), n>=2.
T(2n+1,1) = 4 + 5*A003482(n-1), n>=1.
T(3n,1) = 4*A049629(n-1), n>=1.
T(3n,1) = 4 + 8*A119032(n), n>=1.
T(3n+1,3) = 8*A133273(n), n>=1.
T(3n+2,3n+2) = 2 + 32*A049664(n), n>=1.
T(3n,3n-2) = 4 + 32*A049664(n-1), n>=1.
T(3n+2,2) = 2 + 16*A049683(n), n>=1.
T(3n+2,2) = 2*A023039(n), n>=1.
T(2n-1,2n-1) = A033889(n-1), n>=1.
T(3n-1,3n-1) = 2*A007805(n-1), n>=1.
T(5n-1,1) = 11*A097842(n-1), n>=1.
T(4n+5,3) - T(4n+1,3) = 15*A000045(8n+1), n>=1.
T(5n+4,3) - T(5n-1,3) = 11*A000204(10n-2), n>=1.
Relations between left and right sides:
T(n,1) = T(n,n) - T(n-2,n-2), n>=3.
T(n,2) = T(n,n-1) - T(n-2,n-3), n>=4.
T(n,1) + T(n,n) = 3*T(n,n-1), n>=2.

cp's OEIS Frontend

User: Yuriy Sibirmovsky

A282497 'Somos expansion' of e: e=a(0)*sqrt(a(1)*sqrt(a(2)*sqrt(a(3)*sqrt(...)))). a(n)=floor(x(n)), x(n)=x(n-1)^2/a(n-1)^2, x(0)=e.

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A282496 'Somos expansion' of Pi: Pi=a(0)*sqrt(a(1)*sqrt(a(2)*sqrt(a(3)*sqrt(...)))). a(n)=floor(x(n)), x(n)=x(n-1)^2/a(n-1)^2, x(0)=Pi.

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Mathematica

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A281873 a(n+1) is the smallest number greater than a(n) such that Sum_{j=1..n+1} 1/a(j) <= 4, a(1) = 1.

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Python

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A278288 a(n) is the number of ways to represent a(n-1) as a sum of two distinct terms from a(0) to a(n-2). a(0) = 0. a(1) = a(2) = 1.

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Mathematica

A276819 a(n) = (9*n^2 - n)/2 + 1.

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PARI

PARI

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A276806 Height of the shortest binary factorization tree of n.

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PARI

PARI

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A276771 a(n) is the number of runs of an algorithm. Set b_0 = n, if prime or 1 or 0, stop; else, set c_0 = largest divisor of n (!=n); set b_1 = c_0 - b_0/c_0. Run with b_1.

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Mathematica

A282497 'Somos expansion' of e: e=a(0)sqrt(a(1)sqrt(a(2)sqrt(a(3)sqrt(...)))). a(n)=floor(x(n)), x(n)=x(n-1)^2/a(n-1)^2, x(0)=e.

A282496 'Somos expansion' of Pi: Pi=a(0)sqrt(a(1)sqrt(a(2)sqrt(a(3)sqrt(...)))). a(n)=floor(x(n)), x(n)=x(n-1)^2/a(n-1)^2, x(0)=Pi.