Andrew R. Reiter - cp's OEIS frontend

User: Andrew R. Reiter

Andrew R. Reiter has authored 4 sequences.

A262961 Crandall numbers: (2/Pi)^4 Integral_{t>=0} ([Pi I_0(t)]^2 - [K_0(t)]^2) I_0(t) [K_0(t)]^5 (2t)^(2n-1) dt.

0, 1, 2, 15, 302, 12559, 900288, 98986140, 15459635718, 3251842717671, 885987204390450, 303482789415233775, 127643176985672421000, 64668997044706349592900, 38844990446097247188562800, 27296481783843922533011100000, 22184577644604207037479874293750

Offset: 1

Andrew R. Reiter, Oct 05 2015

nonn

Anton Mellit and David Broadhurst define the sequence to be the "round" of the integral, with the conjecture that this rounding is exact. No one seems to know how to prove that any of the integrals gives a rational number, let alone an integer.
a(0) is not defined: the integral diverges.
Several papers written by Jon Borwein with various coauthors, motivated by work of David Broadhurst, provide recurrence relations for moments of Bessel functions. - M. F. Hasler, Oct 11 2015
Named after the American physicist, mathematician and computer scientist Richard Eugene Crandall (1947-2012). - Amiram Eldar, Jun 23 2021

M. F. Hasler, Table of n, a(n) for n = 1..60; first 49 terms from D. Broadhurst. See also the extended table of 450 terms in the Broadhurst link below.
David H. Bailey, Jonathan M. Borwein, David Broadhurst, M. L. Glasser, Elliptic integral evaluations of Bessel moments, arXiv:0801.0891 [hep-th], 2008.
David H. Bailey, Jonathan M. Borwein, David Broadhurst, M. L. Glasser, Elliptic integral evaluations of Bessel moments, J. Phys. A: Math. Theor., Vol. 41 (2008) 205203.
Jonathan M. Borwein, Bruno Salvy, A Proof of a Recursion for Bessel Moments, Experiment. Math., Vol. 17, No. 2 (2008), pp. 223-230.
David Broadhurst, Crandall Memorial Puzzle, Oct 04, 2015.
David Broadhurst, Crandall Memorial Puzzle. [Cached copy, with permission]
David Broadhurst, Crandall memorial puzzle: solution and heuristics.
David Broadhurst, Crandall memorial puzzle: solution and heuristics. [Cached copy, with permission]
David Broadhurst, Table of n, a(n) for n = 1..450.
David Broadhurst, The largest prime (or noncomposite) factor of A262961(n) for n = 1..94.
David Broadhurst, Feynman integrals, L-series and Kloosterman moments, arXiv:1604.03057 [physics.gen-ph], 2016. See Eq. 147.
Hans Havermann and David Broadhurst, Crandall Numbers Factored.
Yajun Zhou, Hilbert Transforms and Sum Rules of Bessel Moments, arXiv:1706.01068 [math.CA], 2017.
Yajun Zhou, Some algebraic and arithmetic properties of Feynman diagrams, arXiv:1801.05555 [math.NT], 2018.

Cf. A263413 for the largest prime factor of a(n).

See also A265079.

ogf := x * BesselI(0,sqrt(x)/2)^4 * BesselK(0,sqrt(x)/2)^4;
S := convert(simplify(asympt(ogf, x, 25)),polynom):
seq(coeff(S,x,-i),i=0..24); # Mark van Hoeij, Oct 23 2017

a[n_] := (t1 = NIntegrate[(2*t)^(2*n-1)*BesselI[0, t]^3*BesselK[0, t]^5, {t, 0, Infinity}, WorkingPrecision -> 50]; t2 = NIntegrate[(2*t)^(2*n-1) * BesselI[0, t]*BesselK[0, t]^7, {t, 0, Infinity}, WorkingPrecision -> 50]; Round[(2/Pi)^4*(Pi^2*t1-t2)]); Table[Print["a(", n, ") = ", an = a[n]]; an, {n, 1, 16}] (* Jean-François Alcover, Oct 06 2015, adapted from David Broadhurst's PARI script *)

{ default(realprecision,50); infty=[1]; for(n=1,16, t1=intnum(t=0,[infty,2], besseli(0,t)^3*besselk(0,t)^5*(2*t)^(2*n-1)); t2=intnum(t=0,[infty,6], besseli(0,t)*besselk(0,t)^7*(2*t)^(2*n-1)); print(n," ",round((2/Pi)^4*(t1*Pi^2-t2)))); } /* David Broadhurst, Oct 05 2015 */

A262961(n,p=max(2*n,20),a=1)={default(realprecision,p); my(i,k,r=1); forprime(q=3,(n-1)\2,r*=q^(2*ceil(n/q)-4)); n=n*2-1; p=Pi^-2; round(intnum(t=0,[[1],a],((i=besseli(0,t))^3*(k=besselk(0,t))^5-i*k^7*p)*t^n)*2^(n+4)/r/Pi^2)*r} \\ It appears that (in PARI V.2.6.1) the parameter a=1 gives much better results for the numerical integration than the "correct" a=2 (resp. a=6 for the second term); combining all in one integral allows evaluation of the Bessel functions and t^(2n-1) only once. - M. F. Hasler, Oct 11 2015, improved thanks to a suggestion by David Broadhurst, Oct 16 2015

a(n) = (2/Pi)^4 Integral_{t>=0} ([Pi I_0(t)]^2 - [K_0(t)]^2) I_0(t) [K_0(t)]^5 (2t)^(2n-1) dt, where I_0(t) and K_0(t) are Bessel functions.

Floor(a(n+1)/a(n)) = A002943(n-2) = 2(n-2)(2n-3) for n > 7; with round() the relation holds for n = 3, ..., 9. - M. F. Hasler, Oct 11 2015

Offset corrected by David Broadhurst, Oct 05 2015

A262089 a(0) = 0, a(1) = 1, a(n) = a(n-2)^2 + a(n-1)^3.

Original entry on oeis.org

0, 1, 1, 2, 9, 733, 393832918, 61085205568458236705261921, 227933478957258798550715296531141857176484963515197706487620785853119292327685

Offset: 0

Andrew R. Reiter, Sep 10 2015

nonn

Homogeneous cubic recurrence relation of order 2 with +1 coefficients.

Cf. A262088.

I:=[0,1]; [n le 2 select I[n] else Self(n-2)^2 + Self(n-1)^3: n in [1..10]]; // Vincenzo Librandi, Sep 11 2015

RecurrenceTable[{a[0] == 0, a[1] == 1, a[n]==a[n-1]^3 + a[n-2]^2}, a, {n, 10}] (* Vincenzo Librandi, Sep 11 2015 *)

Offset changed and terms a(0)-a(1) prepended by Vincenzo Librandi, Sep 11 2015

A262088 a(0)=0, a(1)=1, a(n) = -a(n-2)^2 - a(n-1)^3.

Original entry on oeis.org

0, 1, -1, 0, -1, 1, -2, 7, -347, 41781874, -72939661777729919216033, 388053169934428398618745564559557538054223536478283729487028027756061

Offset: 0

Andrew R. Reiter, Sep 10 2015

sign

Homogeneous cubic recurrence relation of order 2 with -1 coefficients.

Next term is too large to be included.

Cf. A262089.

f:=proc(n) local i; option remember; if n=0 then 0
elif n=1 then 1
else -f(n-2)^2-f(n-1)^3; fi; end;
[seq(f(n),n=0..10)]; # N. J. A. Sloane, Jun 16 2021

RecurrenceTable[{a[n] == -a[n - 2]^2 - a[n - 1]^3, a[0] == 0, a[1] == 1}, a, {n, 1, 11}] (* Michael De Vlieger, Sep 11 2015 *)

Data and offset corrected by N. J. A. Sloane, Jun 16 2021

A261812 First differences of A098842.

Original entry on oeis.org

-2, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0, -1, 1, 0, 0, 0, -1, 1, 0, 0

Offset: 1

Andrew R. Reiter, Sep 01 2015

Most@ Differences[ Last /@ Tally[ Max[ IntegerLength@#, 1] & /@ Fibonacci[ Range[0, 900]]]] (* Giovanni Resta, Sep 02 2015 *)

import sys
import math
F = [F0,F1] = [0,1]
c = 2
c0 = 0
ll0 = 1
while True:
        T = F[0] + F[1]
        F[0] = F[1]
        F[1] = T
        dfn = 1 + math.log10(T)
        ll = math.floor(dfn)
        if ll0 != ll:
                if c0 != 0:
                        sys.stdout.write("%d," % (c-c0))
                ll0 = ll
                c0 = c
                c = 1
                continue
        c += 1

a(n) = A098842(n+1) - A098842(n).

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User: Andrew R. Reiter

A262961 Crandall numbers: (2/Pi)^4 Integral_{t>=0} ([Pi I_0(t)]^2 - [K_0(t)]^2) I_0(t) [K_0(t)]^5 (2t)^(2n-1) dt.

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A262089 a(0) = 0, a(1) = 1, a(n) = a(n-2)^2 + a(n-1)^3.

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A262088 a(0)=0, a(1)=1, a(n) = -a(n-2)^2 - a(n-1)^3.

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Maple

Mathematica

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A261812 First differences of A098842.

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