Timothy L. Tiffin - cp's OEIS frontend

A360148 Decimal expansion of the nontrivial number x for which x^sqrt(2) = sqrt(2)^x.

8, 9, 3, 7, 4, 3, 7, 0, 6, 6, 0, 5, 9, 0, 6, 2, 3, 1, 6, 8, 2, 0, 2, 0, 8, 0, 6, 4, 6, 2, 4, 6, 9, 1, 0, 4, 8, 7, 1, 7, 0, 6, 8, 5, 8, 1, 2, 6, 8, 3, 7, 1, 6, 5, 6, 8, 5, 4, 4, 2, 4, 1, 3, 6, 2, 8, 1, 7, 6, 3, 1, 1, 6, 2, 3, 8, 8, 7, 4, 5, 1, 4, 1, 4, 7, 2, 7, 9, 1, 2, 6, 8, 5, 4, 4, 8, 1, 1, 6

Offset: 1

Timothy L. Tiffin, Jan 27 2023

Not surprisingly, x appears to be irrational. If x is also algebraic, then x^sqrt(2) would be transcendental by the Gelfond-Schneider theorem.
x is irrational by the Lindemann-Weierstrass theorem. - Charles R Greathouse IV, Jan 27 2023
x = W(-1,-log(2)/(2*sqrt(2)))*-2*sqrt(2)/log(2) = e^-W(-1,-log(2)/(2*sqrt(2))), where W(-1,z) is branch -1 of the Lambert W function. (Branch 0 returns sqrt(2).) Together with sqrt(2), x is unique over the complex numbers as well as the reals. - Natalia L. Skirrow, Jun 22 2023

			8.937437066059062316820208064624691048717068...

{a, b} = NSolve[x^Sqrt[2] == Sqrt[2]^x, x,
  WorkingPrecision -> 300]; a; RealDigits[N[x /. b, 300]][[1]]
N[LambertW[-1,-Log[2]/(2*Sqrt[2])]*-2*Sqrt[2]/Log[2],300] (* Natalia L. Skirrow, Jun 22 2023 *)

From Natalia L. Skirrow, Jun 22 2023: (Start)
Newton's method gives x' = x - (x^sqrt(2) - sqrt(2)^x)/(sqrt(2)*x^(sqrt(2)-1) - sqrt(2)^x*log(2)/2).
Taking logs first gives x' = x - (sqrt(2)*log(x) - x*log(2)/2)/(sqrt(2)/x - log(2)/2).
Beginning with x^(2/x)=sqrt(2)^sqrt(2) instead gives x' = x - (2^(1/sqrt(2)) - x^(2/x))/(log(x) - 1).
(End)

A359257 First differences of A002476.

Original entry on oeis.org

6, 6, 12, 6, 6, 18, 6, 6, 6, 18, 6, 6, 18, 12, 12, 6, 6, 18, 12, 6, 12, 12, 6, 12, 30, 6, 6, 24, 6, 18, 6, 12, 18, 6, 6, 18, 12, 12, 12, 6, 18, 6, 24, 12, 24, 18, 6, 24, 6, 24, 6, 6, 6, 12, 12, 18, 12, 18, 18, 18, 6, 6, 12, 6, 12, 18, 24, 12, 6, 24, 6, 18, 6, 24, 12, 18, 30, 24, 6

Offset: 1

Timothy L. Tiffin, Dec 22 2022

Each term of this sequence is a multiple of 6.
Since no term of A002476 ends in 5, the longest run of 6's in this sequence will have length 3 (e.g., 61, 67, 73, 79 in A002476), the longest run of 12's will have length 3 (e.g., 397, 409, 421, 433 in A002476), the longest run of 18's will have length 3 (e.g., 673, 691, 709, 727 in A002476), and the longest run of 24's will have length 3 (e.g., 14149, 14173, 14197, 14221 in A002476). This run limit of length 3 also extends to other multiples of 6 that are not divisible by 5.
For multiples of 6 that are divisible by 5, the length of the longest run does not appear to be bounded.
Of course there cannot be 6 or more consecutive 30s in this sequence because then one of the primes must be divisible by 7, but there could be up to 10 consecutive 210s. The first run of four 30s corresponds to the primes 320149, 320179, 320209, 320239, 320269 and the first run of five 30s corresponds to the primes 28204591, 28204621, 28204651, 28204681, 28204711, 28204741. - Charles R Greathouse IV, Jan 27 2023

			a(9999) = A002476(10000) - A002476(9999) = 225217 - 225163 = 54.

Cf. A002476, A007528, A358548.

A002476 := Select[6 Range[10000] + 1, PrimeQ]; Table[A002476[[n+1]] - A002476[[n]], {n, 1, 3013}]

do(lim)=my(v=List(),p=7); forprimestep(q=13,lim\1,6, listput(v,q-p); p=q); Vec(v) \\ Charles R Greathouse IV, Jan 27 2023

a(n) = A002476(n+1) - A002476(n).

A348359 Decimal expansion of the nontrivial number x for which x^phi = phi^x, where phi is the golden ratio (1+sqrt(5))/2.

Original entry on oeis.org

6, 0, 5, 5, 7, 2, 2, 0, 9, 1, 0, 2, 4, 7, 4, 1, 0, 0, 2, 1, 2, 6, 6, 3, 9, 1, 1, 7, 5, 8, 3, 1, 4, 9, 7, 3, 1, 6, 8, 3, 8, 2, 8, 7, 5, 3, 7, 8, 3, 6, 7, 7, 7, 4, 3, 9, 4, 9, 9, 6, 7, 7, 3, 5, 2, 8, 1, 8, 7, 9, 7, 4, 4, 8, 5, 2, 3, 5, 8, 1, 4, 7, 9, 3, 8, 9, 4, 6, 6, 6, 0, 7, 4, 2, 8, 1, 7, 8, 9, 4, 7, 8, 9, 4, 5, 7

Offset: 1

Timothy L. Tiffin, Oct 14 2021

The x-th root of x equals the phi-th root of phi: x^(1/x) = phi^(1/phi) = A185261 = 1.3463608200348694434247534661858... .

Not surprisingly, x appears to be irrational. If x is also algebraic, then x^phi would be transcendental by the Gelfond-Schneider theorem.

			6.055722091024741002126639117583149731683828...
x^phi = phi^x  = 18.431940924839652158136364051482054378959672... .

Wikipedia, Gelfond-Schneider theorem

Cf. A001622 (phi), A094214 (1/phi), A185261 (phi^(1/phi)), A073226 (e^e, see first comment).

RealDigits[x/.FindRoot[x^GoldenRatio==GoldenRatio^x,{x,6},WorkingPrecision->120],10,120][[1]] (* Harvey P. Dale, Dec 09 2024 *)

Prior Mathematica program replaced by Harvey P. Dale, Dec 09 2024

A347203 Numbers k for which sigma(k)/k = 32/13.

Original entry on oeis.org

546, 45136, 739648, 5752422, 3053080576, 781678084096, 12506920910848, 209831713740735643648, 241919495232854688763577028051799638016

Offset: 1

Timothy L. Tiffin, Aug 22 2021

This sequence will contain terms of the form 91*P, where P is a perfect number (A000396) not divisible by 7 or 13. Proof: sigma(91*P)/(91*P) = sigma(91)*sigma(P)/(91*P) = 112*(2*P)/(91*P) = 32/13. QED.

Terms ending in "6" or "48" have this form. Example: a(n) = 91*A000396(n) for n = 1, 5, 6, 7, 8, 9 and a(n) = 91*A000396(n+1) for n = 2, 3.

			5752422 is a term, since sigma(5752422)/5752422 = 14159808/5752422 = 32/13.

G. P. Michon, Multiperfect Numbers and Hemiperfect Numbers
Walter Nissen, Abundancy: Some Resources (preliminary version 4)
Walter Nissen, Primitive Friendly Pairs with friends < 2^34 with denom < 20000

Cf. A000203, A000396.

Subsequence of A005101.

Select[Range[5*10^8], DivisorSigma[1, #]/# == 32/13 &]
Do[If[DivisorSigma[1, k]/k == 32/13, Print[k]], {k, 5*10^8}]

a(8)-a(9) from Michel Marcus, Aug 23 2021

A348148 Numbers k for which sigma(k)/k = 32/9.

Original entry on oeis.org

3780, 66960, 167400, 406224, 1097280, 6656832, 13035330, 29410290, 4529295360, 27477725184, 88071903612, 1159632322560, 7035102756864, 18554223329280, 22385029489560, 54934276752360, 112562288197632, 125356165141536, 307631949813216

Offset: 1

Timothy L. Tiffin, Oct 02 2021

nonn

This sequence will contain terms of the form 135*P and 819*Q, where P is a perfect number (A000396) not divisible by 3 or 5, and Q is a perfect number not divisible by 3, 7, or 13. Proof: sigma(135*P)/(135*P) = sigma(135)*sigma(P)/(135*P) = 240*(2*P)/(135*P) = 32/9 and sigma(819*Q)/(819*Q) = sigma(819)*sigma(Q)/(819*Q) = 1456*(2*Q)/(819*P) = 32/9. QED
Terms ending in "4", "32", or "80" and some terms ending in "60" will have one of these forms:
  a( 1) =            3780 = 135*          28 = 135*A000396(2)
  a( 2) =           66960 = 135*         496 = 135*A000396(3)
  a( 4) =          406224 = 819*         496 = 819*A000396(3)
  a( 5) =         1097280 = 135*        8128 = 135*A000396(4)
  a( 6) =         6656832 = 819*        8128 = 819*A000396(4)
  a( 9) =      4529295360 = 135*    33550336 = 135*A000396(5)
  a(10) =     27477725184 = 819*    33550336 = 819*A000396(5)
  a(12) =   1159632322560 = 135*  8589869056 = 135*A000396(6)
  a(13) =   7035102756864 = 819*  8589869056 = 819*A000396(6)
  a(14) =  18554223329280 = 135*137438691328 = 135*A000396(7)
  a(17) = 112562288197632 = 819*137438691328 = 819*A000396(7).

			167400 is a term, since sigma(167400)/167400 = 595200/167400 = 32/9.

G. P. Michon, Multiperfect Numbers and Hemiperfect Numbers
Walter Nissen, Abundancy: Some Resources (preliminary version 4)
Walter Nissen, Primitive Friendly Pairs with friends < 2^34 with denom < 20000

Cf. A000203, A000396, A211680, A212610.

Subsequence of A005101 and A218416.

Select[Range[5*10^8], DivisorSigma[1, #]/# == 32/9 &]
Do[If[DivisorSigma[1, k]/k == 32/9, Print[k]], {k, 5*10^8}]

A348261 Decimal expansion of the nontrivial number x for which x^Pi = Pi^x.

Original entry on oeis.org

2, 3, 8, 2, 1, 7, 9, 0, 8, 7, 9, 9, 3, 0, 1, 8, 7, 7, 4, 5, 5, 5, 5, 9, 3, 0, 5, 2, 5, 2, 0, 8, 7, 8, 5, 3, 5, 6, 8, 9, 7, 6, 7, 9, 9, 6, 7, 8, 2, 3, 2, 5, 9, 1, 0, 1, 2, 9, 4, 8, 1, 1, 7, 7, 1, 3, 5, 3, 4, 4, 4, 6, 9, 0, 7, 4, 6, 9, 3, 5, 4, 1, 6, 6, 8, 7, 5, 8, 2, 5, 3, 9, 6, 1, 6, 6, 9, 2, 2, 0, 8, 9, 7, 2, 1, 4

Offset: 1

Timothy L. Tiffin, Oct 08 2021

The x-th root of x equals the Pi-th root of Pi: x^(1/x) = Pi^(1/Pi) = A073238 = 1.43961949584759... .

Like Pi, is x also transcendental?

			2.382179087993018774555593052520878...
x^Pi = Pi^x = 15.28621734783496640312486439999472... .

Cf. A000796 (Pi), A049541 (1/Pi), A073238 (Pi^(1/Pi)), A073226 (e^e, see first comment), A231737.

evalf((t-> -LambertW(-t)/t)(log(Pi)/Pi), 120);  # Alois P. Heinz, Oct 13 2021

{a, b} = NSolve[x^Pi == Pi^x, x, WorkingPrecision -> 300]; a; RealDigits[N[x/.a, 300]][[1]]

Equals -Pi*LambertW(-log(Pi)/Pi)/log(Pi). - Alois P. Heinz, Oct 13 2021

A348021 Numbers k for which sigma(k)/k = 832/225.

Original entry on oeis.org

94500, 195300, 1674000, 27432000, 56692800, 325883250, 735257250, 113232384000, 234013593600, 28990808064000, 59914336665600, 463855583232000, 559625737239000, 958634872012800, 1373356918809000, 7782220152472338432000, 16083254981776166092800, 8972288971548182138209587578844217344000

Offset: 1

Timothy L. Tiffin, Sep 24 2021

nonn

This sequence contains terms of the form 3375*P and 6975*Q, where P is a perfect number (A000396) not divisible by 3 or 5, and Q is a perfect number not divisible by 3, 5, or 31. Proof: sigma(3375*P)/(3375*P) = sigma(3375)*sigma(P)/(3375*P) = 6240*(2*P)/(3375*P) = 832/225 and sigma(6975*Q)/(6975*Q) = sigma(6975)*sigma(Q)/(6975*Q) = 12896*(2*Q)/(6975*P) = 832/225. QED
Many terms ending in "00" will have one of these forms:
  a( 1) =                   94500 = 3375*                 28 = 3375*A000396(2)
  a( 2) =                  195300 = 6975*                 28 = 6975*A000396(2)
  a( 3) =                 1674000 = 3375*                496 = 3375*A000396(3)
  a( 4) =                27432000 = 3375*               8128 = 3375*A000396(4)
  a( 5) =                56692800 = 6975*               8128 = 6975*A000396(4)
  a( 8) =            113232384000 = 3375*           33550336 = 3375*A000396(5)
  a( 9) =            234013593600 = 6975*           33550336 = 6975*A000396(5)
  a(10) =          28990808064000 = 3375*         8589869056 = 3375*A000396(6)
  a(11) =          59914336665600 = 6975*         8589869056 = 6975*A000396(6)
  a(12) =         463855583232000 = 3375*       137438691328 = 3375*A000396(7)
  a(14) =         958634872012800 = 6975*       137438691328 = 6975*A000396(7)
  a(16) =  7782220152472338432000 = 3375*2305843008139952128 = 3375*A000396(8)
  a(17) = 16083254981776166092800 = 6975*2305843008139952128 = 6975*A000396(8).

			325883250 is a term, since sigma(325883250)/325883250 = 1205043840/325883250 = 832/225.

G. P. Michon, Multiperfect Numbers and Hemiperfect Numbers
Walter Nissen, Abundancy: Some Resources (preliminary version 4)
Walter Nissen, Primitive Friendly Pairs with friends < 2^34 with denom < 20000

Cf. A000203, A000396, A211680, A212610.

Subsequence of A005101.

Select[Range[5*10^8], DivisorSigma[1, #]/# == 832/225 &]
Do[If[DivisorSigma[1, k]/k == 832/225, Print[k]], {k, 5*10^8}]

More terms from Michel Marcus, Oct 03 2021

A347222 Numbers k for which sigma(k)/k = 12/5.

Original entry on oeis.org

30, 140, 2480, 6200, 40640, 167751680, 42949345280, 687193456640, 11529215040699760640, 13292279957849158723273463079769210880, 957809713041180536473966890421518190654986607740846080, 65820182292848241686198767302293614551117361591934715588918640640

Offset: 1

Timothy L. Tiffin, Aug 23 2021

nonn

This sequence will contain terms of the form 5*P, where P is a perfect number (A000396) not divisible by 5. Proof: sigma(5*P)/(5*P) = sigma(5)*sigma(P)/(5*P) = 6*(2*P)/(5*P) = 12/5. QED

Terms ending in "30", "40", or "80" have this form. Example: a(n) = 5*A000396(n) for n = 1, 2, 3 and a(n) = 5*A000396(n-1) for n = 5..12.

			6200 is a term, since sigma(6200)/6200 = 14880/6200 = 12/5.

G. P. Michon, Multiperfect Numbers and Hemiperfect Numbers
Walter Nissen, Abundancy: Some Resources (preliminary version 4)
Walter Nissen, Primitive Friendly Pairs with friends < 2^34 with denom < 20000

Cf. A000203, A000396.

Subsequence of A005101 and A218407.

Select[Range[5*10^8], DivisorSigma[1, #]/# == 12/5 &]
Do[If[DivisorSigma[1, k]/k == 12/5, Print[k]], {k, 5*10^8}]

a(9)-a(10) from Michel Marcus, Aug 24 2021

a(11)-a(12) from David A. Corneth, Aug 24 2021

A347261 Numbers k for which sigma(k)/k = 32/11.

Original entry on oeis.org

924, 2970, 16368, 18018, 268224, 1107161088, 283465678848, 4535476813824, 76092819268618420224, 87729047721804447573604856326476791808

Offset: 1

Timothy L. Tiffin, Aug 24 2021

This sequence will contain terms of the form 33*P, where P is a perfect number (A000396) not divisible by 3 or 11. Proof: sigma(33*P)/(33*P) = sigma(33)*sigma(P)/(33*P) = 48*(2*P)/(33*P) = 32/11. QED
Terms ending in "24" or "x8" (where x is an even digit) have this form. Example: a(1) = 33*A000396(2), a(3) = 33*A000396(3), and a(n) = 33*A000396(n-1) for 5, 6, 7, 8.
Conjecture: a(n) = 33*A341623(n) for n >= 1. Motivation: If no term of A341623 is divisible by 11 (which appears to be the case), then sigma(33*A341623(n))/(33*A341623(n)) = sigma(11)*sigma(3*A341623(n))/(33*A341623(n)) = 12*(8*A341623(n))/(33*A341623(n)) = 32/11. Does this sequence, though, contain any additional terms that are not generated by A341623?

			2970 is a term, since sigma(2970)/2970 = 8640/2970 = 32/11.
18018 is a term, since sigma(18018)/18018 = 52416/18018 = 32/11.

G. P. Michon, Multiperfect Numbers and Hemiperfect Numbers
Walter Nissen, Abundancy: Some Resources (preliminary version 4)
Walter Nissen, Primitive Friendly Pairs with friends < 2^34 with denom < 20000

Cf. A000203, A000396, A341623.

Subsequence of A005101.

Select[Range[5*10^8], DivisorSigma[1, #]/# == 32/11 &]
Do[If[DivisorSigma[1, k]/k == 32/11, Print[k]], {k, 5*10^8}]

a(9)-a(10) from Michel Marcus, Aug 26 2021

A347169 Numbers k for which sigma(k)/k = 16/7.

Original entry on oeis.org

42, 3472, 56896, 544635, 234852352, 60129083392, 962070839296, 16140901056979664896, 18609191940988822212582848311676895232

Offset: 1

Timothy L. Tiffin, Aug 20 2021

This sequence will contain terms of the form 7*P, where P is a perfect number (A000396) not divisible by 7. Proof: sigma(7*P)/(7*P) = sigma(7)*sigma(P)/(7*P) = 8*(2*P)/(7*P) = 16/7. QED

Terms ending in "2" or "96" have this form. Example: a(n) = 7*A000396(n) for n = 1, 5, 6, 7, 8, 9 and a(n) = 7*A000396(n+1) for n = 2, 3.

			544635 is a term, since sigma(544635)/544635 = 1244880/544635 = 16/7.

G. P. Michon, Multiperfect Numbers and Hemiperfect Numbers
Walter Nissen, Abundancy: Some Resources (preliminary version 4)
Walter Nissen, Primitive Friendly Pairs with friends < 2^34 with denom < 20000

Cf. A000203, A000396.

Subsequence of A005101 and A218409.

Select[Range[5*10^8], DivisorSigma[1, #]/# == 16/7 &]
Do[If[DivisorSigma[1, k]/k == 16/7, Print[k]], {k, 5*10^8}]

a(8)-a(9) from Michel Marcus, Aug 21 2021

cp's OEIS Frontend

User: Timothy L. Tiffin

A360148 Decimal expansion of the nontrivial number x for which x^sqrt(2) = sqrt(2)^x.

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A359257 First differences of A002476.

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A348359 Decimal expansion of the nontrivial number x for which x^phi = phi^x, where phi is the golden ratio (1+sqrt(5))/2.

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A347203 Numbers k for which sigma(k)/k = 32/13.

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A348148 Numbers k for which sigma(k)/k = 32/9.

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A348261 Decimal expansion of the nontrivial number x for which x^Pi = Pi^x.

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A348021 Numbers k for which sigma(k)/k = 832/225.

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A347222 Numbers k for which sigma(k)/k = 12/5.

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