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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A374074 Odd composite numbers k sorted by k/2^(bigomega(k) - 1).

Original entry on oeis.org

9, 27, 15, 81, 21, 45, 25, 243, 63, 33, 135, 35, 75, 39, 729, 189, 49, 99, 405, 51, 105, 55, 225, 57, 117, 125, 65, 2187, 69, 567, 147, 297, 1215, 153, 77, 315, 165, 675, 85, 171, 87, 175, 351, 91, 93, 375, 95, 195, 6561, 207, 1701, 441, 111, 891, 3645, 459
Offset: 1

Views

Author

Friedjof Tellkamp, Jun 27 2024

Keywords

Comments

Sorting by k/2^bigomega(k) would give the same sequence.
It appears that this sequence can be used to approximate the imaginary parts of the nontrivial zeta zeros, that is, A002410(n) is roughly equal to 2*Pi*a(n)/2^bigomega(a(n)) - n/2 + sqrt(n)/2.
Calculations show that the relative error approaches 1.0+-0.005 for the first 3800 zeros (z=2000 in Mathematica code). For further zeros, a better approximation may be useful, e.g. 2*Pi*a(n)/2^bigomega(a(n)) - n/2 + (1/Pi) * n/log(n+1) +- (...).

Examples

			The odd composite numbers (A071904) are: 9, 15, 21, 25, 27, ... .
Divide by 2^(bigomega()-1): 9/2, 15/2, 21/2, 25/2, 27/4, ... .
Sort: 9/2, 27/4, 15/2, 81/8, ... .
Take numerator: this sequence = 9, 27, 15, 81, ... .

Links

Crossrefs

Cf. A001222, A002410, A065091, A071904, A374022, A374603, A374604.
See also A122943.

Programs

  • Mathematica
    (*This algorithm calculates and sorts all noninteger rationals of the form x = k/2^(bigomega(k)-1) up to z, where the numerators of x are returned as the sequence*)
z = 100; op = Prime[Range[2, PrimePi[z]]]; (*A065091, odd primes*)
x = Select[Union[Flatten[Outer[Times, op, op/2]]], # <= z &];
For[i = 1, i < Max[1, Floor[Log[3/2, z/2]]] - 1, i++, x = Select[Union[x, Flatten[Outer[Times, x, op/2]]], # <= z &]]
a = Numerator[x] (*sequence*)
zzaprx = N@Table[2 Pi a[[i]]/2^PrimeOmega[a[[i]]] - i/2 + Sqrt[i]/2, {i, 1, Length[a]}] (*approximation for zeta zeros*)

Formula

A374022(a(n)/2^(bigomega(a(n))-1)) ~ n.

A215405 Largest prime factor of the n-th n-almost prime.

Original entry on oeis.org

2, 3, 3, 5, 3, 7, 5, 3, 7, 11, 5, 5, 13, 3, 7, 11, 5, 17, 7, 5, 19, 13, 3, 23, 7, 7, 11, 5, 17, 7, 11, 5, 19, 29, 13, 31, 5, 13, 3, 23, 7, 7, 37, 11, 5, 17, 11, 7, 41, 11, 5, 17, 19, 43, 29, 7, 13, 13, 31, 5, 47, 19, 13, 3, 23, 53, 7, 7, 37
Offset: 1

Views

Author

Juri-Stepan Gerasimov, Aug 09 2012

Keywords

Comments

Technically, the prime numbers are "1-almost prime."
Prime(m) (m>=1) occurs first at index n = 1, 2, 4, 6, 10, 13, 18, 21, 24, 34, 36, 43, 49, 54, 61, 66, 75, 79, 91, 97, 101, 107, 113, 124, 138, 144, 148, 157, 162, 167, 187, 194, 202, 207, 224, 229,... in the sequence. - R. J. Mathar, Aug 09 2012
n <= a(n) at 1, 2, 3, 4, 6, 10, 13,...
n < 2*a(n) at n = 1, 2, 3, 4, 6, 7, 9, 10, 13, 16, 18, 21, 22, 24, 29, 33, 34, 36, 40, 43, 49, 54, 55, 59, 61, 66, 69,...
Also largest prime factor of A122943(n) for n>1. - Eric Desbiaux, Mar 20 2016

Examples

			a(2) = 3 because the 2nd 2-almost prime (semiprime, A001358) is 6 = 2 * 3, the largest prime factor there being 3.
a(3) = 3 because the 3rd 3-almost prime (A014612) is 18 = 2 * 3^2, the largest prime factor there being 3.
a(4) = 5 because the 4th 4-almost prime (A014613) is 40 = 2^3 * 5, the largest prime factor there being 5.

Crossrefs

Cf. A078841, A101695.

Programs

Extensions

Corrected by R. J. Mathar, Aug 09 2012

A122941 Rectangular table, read by antidiagonals, where the g.f. of row n is Sum_{i>=0} F_i(x)^n / 2^(i+1), where F_0(x)=x, F_{n+1}(x) = F_n(x+x^2), for n>=1.

Original entry on oeis.org

1, 1, 1, 1, 2, 2, 1, 3, 7, 7, 1, 4, 15, 34, 34, 1, 5, 26, 94, 214, 214, 1, 6, 40, 200, 726, 1652, 1652, 1, 7, 57, 365, 1831, 6645, 15121, 15121, 1, 8, 77, 602, 3865, 19388, 70361, 160110, 160110, 1, 9, 100, 924, 7239, 46481, 233154, 846144, 1925442, 1925442, 1
Offset: 1

Views

Author

Paul D. Hanna, Sep 25 2006

Keywords

Comments

A122940(n)/n = Sum_{m=1..n} (-1)^(m-1)*T(m,n-m+1)/m ; where l.g.f. of A122940, L(x), satisfies: L(x+x^2) = 2*L(x) - log(1+x).

Examples

			Table begins:
1, 1, 2, 7, 34, 214, 1652, 15121, 160110, 1925442, 25924260, ...;
1, 2, 7, 34, 214, 1652, 15121, 160110, 1925442, 25924260, ...;
1, 3, 15, 94, 726, 6645, 70361, 846144, 11392530, 169785124, ...;
1, 4, 26, 200, 1831, 19388, 233154, 3139200, 46784118, ...;
1, 5, 40, 365, 3865, 46481, 625820, 9326720, 152426170, ...;
1, 6, 57, 602, 7239, 97470, 1452610, 23739936, 422171622, ...;
1, 7, 77, 924, 12439, 185388, 3029782, 53879148, 1035760670, ...;
1, 8, 100, 1344, 20026, 327296, 5820360, 111889248, 2312153223, ...;
1, 9, 126, 1875, 30636, 544824, 10473576, 216432783, 4784414985, ...;
1, 10, 155, 2530, 44980, 864712, 17868995, 395007850, 9301284465, ...;
Given that A122940 begins:
[1, 1, 4, 17, 106, 796, 7176, 75057, 894100, 11946906, ...],
demonstrate A122940(n)/n = Sum_{m=1..n} (-1)^(m-1)*T(m,n-m+1)/m
at n=4: A122940(4)/4 = 17/4 = 7/1 - 7/2 + 3/3 - 1/4;
at n=5: A122940(5)/5 = 106/5 = 34/1 - 34/2 + 15/3 - 4/4 + 1/5;
at n=6: A122940(6)/6 = 796/6 = 214/1 - 214/2 + 94/3 - 26/4 + 5/5 - 1/6.

Crossrefs

Cf. A122940; rows: A122942, A122943, A122944, A122945; related tables: A122888, A122946, A122948, A122951.

Programs

  • PARI
    /* Get T(n,k) from H(n,), the n-th self-composition of x+x^2: */
{H(n,p)=local(F=x+x^2, G=x+x*O(x^p));if(n==0,G=x,for(i=1,n,G=subst(F,x,G));G)}
{T(n,k)=round(polcoeff( sum(i=0,6*n+100,H(i,k+n-1)^n/2^(i+1)),k+n-1))}

Formula

T(n,k) = [x^k] Sum_{i>=0} F_i(x)^n / 2^(i+1) where F_0(x)=x, F_{n+1}(x) = F_n(x+x^2); a sum involving n-th powers of self-compositions of x+x^2 (cf. A122888).
Showing 1-3 of 3 results.

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