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.

Showing 1-5 of 5 results.

A159995 Denominator of f(A159992(n)/A159993(n)) with f(x)=x^3+2*x^2+10*x-20, numerator=A159994.

Original entry on oeis.org

1, 27000, 46656000000, 46656000000000, 80621568000000000000, 14348907000000000000000, 12694994583552000000000000000000, 812479653347328000000000000000000
Offset: 0

Views

Author

Reinhard Zumkeller, May 01 2009

Keywords

Links

Crossrefs

A159991, subsequence of A051037, the 5-smooth numbers.

A159994 Numerator of f(A159992(n)/A159993(n)) with f(x)=x^3+2*x^2+10*x-20, denominator=A159995.

Original entry on oeis.org

-7, -1219, -193885217, -2512095067, -10152983807749, -249880575515347, -2950249420928771944181, -5129149052857317896107, -1247467412339070464235923, -16941291362994850503969493637
Offset: 0

Views

Author

Reinhard Zumkeller, May 01 2009

Keywords

Comments

a(n)/A159995(n) = f(A159992(n)/A159993(n));
a(n)/A159995(n) < 0; a(n)/A159995(n) <= a(n+1)/A159995(n+1);
a(n)/A159995(n) --> 0.

Examples

			a(0)/A159995(0)=-7;
a(1)/A159995(1)=-1216/27000;
a(2)/A159995(2)=-193885217/46656000000;
a(3)/A159995(3)=-2512095067/46656000000000;
a(4)/A159995(4)=-10152983807749/80621568000000000000;
a(5)/A159995(5)=-249880575515347/14348907000000000000000;
a(6)/A159995(6)=-2950249420928771944181/12694994583552000000000000000000;
a(7)/A159995(7)=-5129149052857317896107/812479653347328000000000000000000;
a(8)/A159995(8)=-1247467412339070464235923/37907050706572935168000000000000000000;
written as decimal fractions:
a(1)/A159995(1) ~= -0.045037037037037037037037;
a(2)/A159995(2) ~= -0.004155633080418381344307;
a(3)/A159995(3) ~= -0.000053842915530692729766;
a(4)/A159995(4) ~= -0.000000125933842017920068;
a(5)/A159995(5) ~= -0.000000017414606946392990;
a(6)/A159995(6) ~= -0.000000000232394697099847;
a(7)/A159995(7) ~= -0.000000000006312956923568;
a(8)/A159995(8) ~= -0.000000000000032908585318.

Links

Crossrefs

A159990.

A159991 Powers of 60: a(n) = 60^n.

Original entry on oeis.org

1, 60, 3600, 216000, 12960000, 777600000, 46656000000, 2799360000000, 167961600000000, 10077696000000000, 604661760000000000, 36279705600000000000, 2176782336000000000000, 130606940160000000000000, 7836416409600000000000000, 470184984576000000000000000
Offset: 0

Views

Author

Reinhard Zumkeller, May 01 2009

Keywords

Examples

			G.f. = 1 + 60*x + 3600*x^2 + 216000*x^3 + 12960000*x^4 + 77600000*x^5 + ... - _Michael Somos_, Jan 01 2019

Links

Crossrefs

Subsequence of A051037.
Cf. A159990, A159993, A159995, A088157, A091649, A125628, A091720, A091721, A091722, A060707, A070197.

Programs

Formula

a(n) = A000400(n)*A011557(n) = A000351(n)*A001021(n) = A000302(n)*A001024(n) = A000244(n)*A009964(n). (Corrected by Robert B Fowler, Jan 25 2023)
From Muniru A Asiru, Nov 21 2018: (Start)
a(n) = 60^n.
a(n) = 60*a(n-1) for n > 0, a(0) = 1.
G.f.: 1/(1-60*x).
E.g.f: exp(60*x). (End)
a(n) = 1/a(-n) for all n in Z. - Michael Somos, Jan 01 2019

A159990 Coefficients in sexagesimal expansion of the positive root of x^3 + 2*x^2 + 10*x = 20, first studied by Leonardo of Pisa (Fibonacci) in 1225.

Original entry on oeis.org

1, 22, 7, 42, 33, 4, 38, 30, 50, 15, 43, 13, 56, 48, 24, 41, 0, 48, 22, 40, 39, 37, 23, 53, 55, 57, 45, 40, 5, 46, 50, 57, 28, 45, 46, 34, 2, 6, 7, 15, 25, 25, 13, 10, 59, 30, 13, 14, 7, 6, 15, 46, 23, 53, 59, 32, 24, 20, 11, 48, 35, 4, 4, 18, 33, 50, 7, 40, 16, 16, 1, 32, 24, 10, 43, 59, 23, 44, 51, 58, 11, 22, 26, 17
Offset: 0

Views

Author

Reinhard Zumkeller, May 01 2009

Keywords

Comments

Leonardo of Pisa (Fibonacci) found a(0), ..., a(5) but gave a(6) as 40.
A159992(n)/A159993(n) = Sum_{k=0..n} a(k)/60^k = (Sum_{k=0..n} a(k)*60^(n-k))/60^n; let f(x) = x^3 + 2*x^2 + 10*x - 20, then for n > 0:
a(n) = Max(k: f(A159992(n-1)/A159993(n - 1) + k/60^n)) < 0),
a(n) + 1 = Min(k: f(A159992(n - 1)/A159993(n - 1) + k/60^n)) > 0);
A159994(n)/A159995(n) = f(A159992(n)/A159993(n)).

Examples

			The root is 1 + 22/60 + 7/60^2 + 42/60^3 + 33/60^4 + 4/60^5 + 38/60^6 + 30/60^7 + 50/60^8 + ...
Leonardo's approximation 1;22.7.42.33.4.40 is to be read as 1 + 22/60 + 7/60^2 + 42/60^3 + 33/60^4 + 4/60^5 + 40/60^6 = A159992(5)/A159993(5) + 40/60^6 = 1596577777 / 1166400000 ~= 1.3688081078532235 and f(1596577777/1166400000) ~= +6.7193226361369/10^10; compare this to A159992(6)/A159993(6) = A159992(5)/A159993(5) + 38/60^6 = 31931555539 / 23328000000 ~= 1.3688081078103566 and f(31931555539/23328000000) ~= -2.3239469709985/10^10.
Assuming that Leonardo did similar calculations, the question may arise: why he didn't find a(6) = 38 instead of 40? Supposedly he just avoided the effort to calculate f(A159992(5)/A159993(5) + k/60^6) for k = 37, 38, or 39: 37/60^6 = 37/46656000000, 38/60^6 = 19/23328000000, or 39/60^6 = 13/15552000000; finally, he did calculate only f(A159992(5)/A159993(5) + k/60^6) for k = 36 and k = 40, the less complex cases concerning sexagesimal fractional arithmetic with 36/60^6 = 1/1296000000 and 40/60^6 = 1/1166400000: f(A159992(5)/A159993(5) + 36/60^6) ~= -1.9999999988632783, f(A159992(5)/A159993(5) + 40/60^6) ~= +0.0000000006719323.
The latter result looks precise enough and could explain and justify Leonardo's 'rounding'.

References

  • Cox, David A., Galois theory. Pure and Applied Mathematics (New York). Wiley-Interscience [John Wiley & Sons], Hoboken, NJ, 2004. xx+559 pp. ISBN: 0-471-43419-1 MR2119052 (2006a:12001). See page 9.
  • A. F. Horadam, Eight hundred years young, The Australian Mathematics Teacher 31 (1975) 123-134.
  • Alfred S. Posamentier & Ingmar Lehmann, The (Fabulous) Fibonacci Numbers. New York: Prometheus Books (2007): 21.

Links

Crossrefs

Cf. A159991, A202300.

Programs

  • Mathematica
    RealDigits[ Solve[x^3 + 2 x^2 + 10 x - 20 == 0, x][[3, 1, 2]], 60, 111][[1]] (* Robert G. Wilson v, May 11 2012 *)
RealDigits[Root[x^3+2x^2+10x-20,1],60,90][[1]] (* Harvey P. Dale, Jun 16 2025 *)
  • PARI
    polrootsreal(x^3+2*x^2+10*x-20)[1] \\ Charles R Greathouse IV, Apr 14 2014

A159992 Numerator of Sum_{k=0..n} A159990(k)/A159991(k).

Original entry on oeis.org

1, 41, 4927, 49277, 5913251, 33262037, 31931555539, 127726222157, 4598143997653, 306542933176867, 827665919577540943, 49659955174652456593, 744899327619786848909, 1862248319049467122273, 446939596571872109345521
Offset: 0

Views

Author

Reinhard Zumkeller, May 01 2009

Keywords

Comments

a(n)/A159993(n) approximates the positive root of x^3+2*x^2+10*x=20:
A159994(n)/A159995(n) = f(a(n)/A159993(n)) --> 0, where f(x) = x^3 + 2*x^2 + 10*x - 20;
a(n)/A159993(n) = a(n-1)/A159993(n-1) + A159990(n)/A159991(n).
Limit can be found at A202300. - Jason Bard, Jul 26 2025

Examples

			a(0)/A159993(0) = 1;
a(1)/A159993(1) = 41/30;
a(2)/A159993(2) = 4927/3600;
a(3)/A159993(3) = 49277/36000;
a(4)/A159993(4) = 5913251/4320000;
a(5)/A159993(5) = 33262037/24300000;
a(6)/A159993(6) = 31931555539/23328000000;
a(7)/A159993(7) = 127726222157/93312000000;
a(8)/A159993(8) = 4598143997653/3359232000000;
and written as decimal fractions:
a(0)/A159993(0) = 1;
a(1)/A159993(1) ~= 1.3666666666666667;
a(2)/A159993(2) ~= 1.3686111111111111;
a(3)/A159993(3) ~= 1.3688055555555556;
a(4)/A159993(4) ~= 1.3688081018518519;
a(5)/A159993(5) ~= 1.3688081069958847;
a(6)/A159993(6) ~= 1.3688081078103566;
a(7)/A159993(7) ~= 1.3688081078210733;
a(8)/A159993(8) ~= 1.3688081078213710.

Links

Crossrefs

Cf. A159990, A159991, A159993 (denominator), A159994, A159995, A202300.
Showing 1-5 of 5 results.

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