Robert B Fowler - cp's OEIS frontend

User: Robert B Fowler

Robert B Fowler has authored 20 sequences. Here are the ten most recent ones:

A379607 Denominators corresponding to A379606.

1, 60, 1400, 25200, 17248000, 7207200000, 12713500800000, 38118080000000, 252957982717440000000, 177399104762880000000, 39217856135377920000000000, 314173814535060480000000000, 566078379029271972864000000000000, 8188688361066909696000000000000, 3391339339748110516021960704000000000000

Offset: 0

Robert B Fowler, Dec 27 2024

Cf. A379606 (numerators).

Denominator[CoefficientList[InverseSeries[Series[Surd[(6*(x - Sin[x])), 3], {x, 0, 40}]], x][[2 ;; -2 ;; 2]]] (* Amiram Eldar, Dec 27 2024 *)

Edited by N. J. A. Sloane, Jan 14 2025

A379606 S = (A-sin(A))/2 gives the area of a segment of the unit circle in terms of the arc length A (<= Pi). Expressing A in terms of S we get A = Sum_{n>=0} b^(2n+1)c(n) where b = (12S)^(1/3). Sequence gives numerators of c(n).

Original entry on oeis.org

1, 1, 1, 1, 43, 1213, 151439, 33227, 16542537833, 887278009, 15233801224559, 9597171184603, 1374085664813273149, 1593410154419351, 53299328587804322691259, 1065024810026227256263721, 11374760871959174491194191, 70563256104582737796094772987, 657272463951301325116190773432261

Offset: 0

Robert B Fowler, Dec 27 2024

			 A = b + b^3/60 + b^5/1400 + b^7/25200 + ..., where b = (12*S)^(1/3); the c(n) are 1, 1/60, 1/1400, 1/25200, 43/17248000, 1213/7207200000, ...

Cf. A379607 (denominators).

Numerator[CoefficientList[InverseSeries[Series[Surd[(6*(x - Sin[x])), 3], {x, 0, 40}]], x][[2 ;; -2 ;; 2]]] (* Amiram Eldar, Dec 27 2024 *)

Edited by N. J. A. Sloane, Jan 14 2025

A375862 Number of caesium clock "ticks" in time units: tick, second, minute, hour, day, week, month, year, decade, century, millennium, eon (in Julian years).

Original entry on oeis.org

1, 9192631770, 551557906200, 33093474372000, 794243384928000, 5559703694496000, 24174783028746000, 290097396344952000, 2900973963449520000, 29009739634495200000, 290097396344952000000, 290097396344952000000000000

Offset: 1

Robert B Fowler, Aug 31 2024

The standard SI second has been defined since 1968 as 9192631770 (A230458) transitions ("ticks") of the caesium-133 atom.

The values of a(7) to a(12) are based on the average Julian calendar year of 365.25 days, which appears frequently in astronomical publications, where it is usually called simply "Julian years".

Wikipedia, Atomic clock.

Cf. A053401, A375666 (similar sequences based on units of seconds).
Cf. A213612, A213613, A213614 (use seconds in year).
Cf. A230458 (value of a(2)).

a(n) = A375666(n-1) * 9192631770, n>1.

A375666 Number of seconds in time units: second, minute, hour, day, week, month, year, decade, century, millennium, eon (in Julian years).

Original entry on oeis.org

1, 60, 3600, 86400, 604800, 2629800, 31557600, 315576000, 3155760000, 31557600000, 31557600000000000

Offset: 1

Robert B Fowler, Aug 23 2024

The last six numbers are based on the average Julian calendar year of 365.25 days, which appears frequently in astronomical publications, where it is usually called simply "Julian years".

UnitJuggler, Convert julian years (average) to milliseconds

Cf. A053401 (with Gregorian years of 365.2425 days).

Cf. A213612, A213613, A213614 (uses a(7)).

A375027 Number of occurrences of Easter Sunday on March 22, March 23, ..., April 25 during a 532-year Julian Easter cycle.

Original entry on oeis.org

4, 8, 8, 12, 16, 16, 20, 16, 16, 20, 16, 16, 20, 16, 20, 20, 16, 20, 16, 16, 20, 16, 16, 20, 16, 20, 16, 16, 20, 16, 12, 12, 8, 8, 4

Offset: 1

Robert B Fowler, Jul 28 2024

During any 532-year range of the Julian Calendar, each of the 35 possible dates for Easter Sunday occurs either 4, 8, 12, 16, or 20 times. This cycle is much simpler than the Gregorian Easter cycle (A224110).

Cf. A224110 (frequencies for Gregorian Easter Sunday dates).
Cf. A348924 (algorithms for Paschal Full Moon and Easter Sunday in both Julian and Gregorian Calendars).
Cf. A349710 (algorithms for Paschal Full Moon and Easter Sunday in Julian Calendar).

Use the Julian Easter algorithm in A348924 for years 1 to 532 (or any range of 532 years), and tally the occurrence of each Easter date between March 22 and April 25.

A351862 Denominators of the coefficients in a series for the angles in the Spiral of Theodorus.

Original entry on oeis.org

1, 6, 120, 840, 8064, 4224, 2196480, 199680, 5013504, 74088448, 1568931840, 1899233280, 2411724800, 2831155200, 8757706752, 6968215339008, 76890652016640, 1488206168064, 289223097712640, 74371653697536, 2197648866017280, 10176804748787712, 29785769996451840

Offset: 0

Robert B Fowler, Feb 22 2022

S(i) is the sum of the angles in the first i-1 triangles of the Spiral of Theodorus (in radians). [corrected by Robert B Fowler, Oct 23 2022]
S(i) = K + sqrt(i) * (2 + 1/(6*i) - 1/(120*i^2) - 1/(840*i^3) + ...) where K is Hlawka's Schneckenkonstante = A105459 * (-1) = -2.1577829966... .
The coefficients in the polynomial series are A351861(n)/a(n). The series is asymptotic, but is accurate for even very low values of i.
See A351861 for the numerators, as well as references, links, and crossrefs.

			2/1 + 1/(6*i) - 1/(120*i^2) - 1/(840*i^3) + ...

Cf. A351861 (numerators).

c[0] = 2; c[n_] := ((2*n - 2)!/(n - 1)!) * Sum[(-1)^(n + 1) * BernoulliB[n - k] * k!/(4^(n - k - 1) * (2*k + 1)! * (n - k)!), {k, 0, n}]; Denominator @ Array[c, 30, 0] (* Amiram Eldar, Feb 22 2022 *)

A351861 Numerators of the coefficients in a series for the angles in the Spiral of Theodorus.

Original entry on oeis.org

2, 1, -1, -1, 5, 1, -521, -29, 1067, 13221, -538019, -692393, 2088537, 3155999, -27611845, -33200670659, 1202005038007, 40366435189, -29289910899229, -14754517273097, 1825124640773023, 18449097055233961, -250479143430425927, -1976767636081931863, 1419438523008706978221

Offset: 0

Robert B Fowler, Feb 22 2022

S(i) is the sum of the angles of the first i-1 triangles in the Spiral of Theodorus (in radians). [Corrected by Robert B Fowler, Oct 23 2022]
S(i) = K + sqrt(i) * (2 + 1/(6*i) - 1/(120*i^2) - 1/(840*i^3) + ...) where K is Hlawka's Schneckenkonstante, K = A105459 * (-1) = -2.1577829966... .
The coefficients in the polynomial series are a(n)/A351862(n). The series is asymptotic, but is very accurate even for low values of i.

			2/1 + 1/(6*i) - 1/(120*i^2) - 1/(840*i^3) + ...

P. J. Davis, Spirals from Theodorus to Chaos, A K Peters, Wellesley, MA, 1993.

David Brink, The Spiral of Theodorus and sums of zeta values at half-integers, The American Mathematical Monthly, Vol. 119, No. 9 (November 2012), pages 779-786.
David Brink, The Spiral of Theodorus and sums of zeta values at half-integers, July 2012. There are errors in equation (9): the denominator factors n, n^2, n^3, n^4, n^5 should actually be n, n^3, n^5, n^7, n^9, respectively.
Detlef Gronau, The Spiral of Theodorus, The American Mathematical Monthly, Vol. 111, No. 3 (March 2004), pages 230-237.
Edmund Hlawka, Gleichverteilung und Quadratwurzelschnecke, Monatsh. Math. 89 (1980) pages 19-44. [For a summary in English, see the Davis reference, pages 157-167.]
Herbert Kociemba, The Spiral of Theodorus, 2018.
Wikipedia, Spiral of Theodorus

Cf. A351862 (denominators).
Cf. A105459, A185051 (Hlawka's constant).
Cf. A027641, A027642 (Bernoulli numbers).
Cf. A072895, A224269 (spiral revolutions).

c[0] = 2; c[n_] := ((2*n - 2)!/(n - 1)!) * Sum[(-1)^(n + 1) * BernoulliB[n - k] * k!/(4^(n - k - 1) * (2*k + 1)! * (n - k)!), {k, 0, n}]; Numerator @ Array[c, 30, 0] (* Amiram Eldar, Feb 22 2022 *)

a(n) = {numerator(if(n==0, 2, ((2*n-2)!/(n-1)!) * sum(k=0, n, (-1)^(n+1) * bernfrac(n-k) * k! / (4^(n-k-1) * (2*k+1)! * (n-k)!))))} \\ Andrew Howroyd, Feb 22 2022

Let r(n) = ((2*n-2)! / (n-1)!) * Sum_{k=0..n} ((-1)^(n+1)*B(n-k)*k!) / ((4^(n-k-1) * (2*k+1)! * (n-k)!) ) for n > 0, where B(n-k) are Bernoulli numbers. Then:

a(n) = numerator(r(n)) for n >= 1 and additionally a(0) = 2.

A351582 Decimal expansion of the root of cot(Pi/(s+1)) - csc(Pi/s).

Original entry on oeis.org

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

Offset: 1

Robert B Fowler, Feb 14 2022

For regular unit-sided polygons with number of sides s >= 3, the s-gon fits inside the (s+1)-gon, and hence inside any t-gon where t > s. For s = 3 and s = 4, this is verified by diagram. For s >= 5, it is verified by observing that the s-gon's circumcircle is smaller than the (s+1)-gon's incircle. The difference of the two circles' radii is negative for s <= 4 and positive for s >= 5, and changes sign at non-integer value s = 4.49547...

Diagrams demonstrating this property of regular s-gons are interesting (see links).

			4.4954747887528...

Robert B Fowler, Diagram of Nested Unit-sided Regular Polygons (s=3 to s=12)
Luxor, Diagram of Concentric Unit-sided Polygons (s=3 to s=20). See diagrams near the start and near the end of the article. The triangle, square and pentagon intersect.

Digits:= 120:
fsolve(cot(Pi/(s+1))-csc(Pi/s),s);  # Alois P. Heinz, Feb 16 2022

RealDigits[s /. FindRoot[Cot[Pi/(s + 1)] == Csc[Pi/s], {s, 4}, WorkingPrecision -> 120]][[1]] (* Amiram Eldar, Feb 14 2022 *)

solve(s=4, 5, cotan(Pi/(s+1)) - 1/sin(Pi/s)) \\ Michel Marcus, Feb 14 2022

For integer values of s >= 3:
  c(s) = circumcircle radius of unit-sided regular s-gon = csc(Pi/s) / 2,
  i(s) = incircle radius of unit-sided regular s-gon = cot(Pi/s) / 2,
  d(s) = i(s+1) - c(s),
  d(s) <= 0 for s <= 4, d(s) > 0 for s >= 5.
For real values of s:
  d(1) = -infinity,
  d'(s) > 0 for s > 1,
  d(s) = 0 for s = 4.4954747887528...

A350539 Chronological Julian day number of the first day (Muharram 1) of Tabular Islamic year n.

Original entry on oeis.org

1948440, 1948794, 1949149, 1949503, 1949857, 1950212, 1950566, 1950921, 1951275, 1951629, 1951984, 1952338, 1952692, 1953047, 1953401, 1953755, 1954110, 1954464, 1954819, 1955173, 1955527, 1955882, 1956236, 1956590, 1956945, 1957299, 1957654, 1958008, 1958362, 1958717, 1959071, 1959425

Offset: 1

Robert B Fowler, Jan 04 2022

The Islamic calendar is purely lunar. It starts on Friday 0001-Mulharram-1 AH (Anno Hegirae) = AD 662-Jul-16 (Julian calendar) = AD 622-Jul-19 (Gregorian proleptic) = JDN 1948440. Every 12 months is a lunar year containing either 354 days (regular) or 355 days (leap year). Odd-numbered months are 30 days, even-numbered months are 29 days, except month 12 is 30 days in leap years. Each 30-year cycle contains 19 regular years and 11 leap years. Thus, 1 cycle = 30 lunar years = 360 lunar months = 10631 days, and a(n+30*k) = a(n) + k*10631, for all k. Since 10631 is not a multiple of 7, the calendar repeats after 7 cycles = 210 lunar years.
In various locations, the Islamic new moon is chosen to be dated by either (a) the first sighting of the lunar crescent, (b) astronomical new moon tables, or (c) tabular methods. Only the tabular methods are described here. At least five methods exist, differing only in the distribution of leap years. a(n) are calculated here using the most common method (Fazari or West Islam), in which the leap years within each 30-year cycle (first year of cycle is 1, not 0) are years {2, 5, 7, 10, 13, 16, 18, 21, 24, 26, 29} = {floor((30*k-1-c) / 11), k = 1..11, c = 3}. Three other tabular methods correspond to other values of c, namely, c = 4 (Kushyar or East Islam), c = 0 (Ismaili), c = -2 (Habash). In a fifth method (Fattuh), the leap years are not spaced evenly enough to fit this algorithm.
In a minority of locations, an epoch date of Thursday AD 662-Jul-15 is used; this subtracts one day from each of the five calculation methods.
The chronological Julian day number (JDN) is the number of days since 4713-Jan-1 BC (Julian proleptic calendar), e.g., 2000-Jan-1 (Gregorian) = JDN 2451545. As used by historians, chronologers and calendarists, it is an integer and does not incorporate time or location. The astronomical JDN incorporates both time and location: it equals the chronological JDN at UT (Greenwich) noon, and includes time as a decimal fraction of a day, e.g., JDN 2451545.50 = 2000-Jan-1 24:00 UT.
As of AD 2000, the astronomical synodic month averages 29.530588865 days; the Islamic month averages 10631/360 = 29.5305555555 days, and falls behind the synodic moon by 0.04120 days per century. The astronomical tropical year averages 365.242192 days; the Islamic lunar year averages 12*10631/360 = 354.366666 days, so there are an average of 103.07120 Islamic years per tropical century.
The astronomical new moon of July 622 occurred on July 14 at 05:30 UT = 08:10 Mecca Local Mean Time (MLMT), but the crescent moon was not visible in Mecca until sunset of the next day July 15 (~18:00 MLMT), the start of 0001-Mulharram-1 AH, which is equated with AD 622-Jul-16 (which began 6 hours later at 24:00 MLMT). - Robert B Fowler, Aug 31 2024

			a(1) = floor((1*10631+3)/30) + 1948086 = 1948440 (JDN).
Year 1 has a(2) - a(1) = 354 days (a regular year).
Year 1 began on weekday (a(1) mod 7) = 4 (Friday).
Year 2 has a(3) - a(2) = 355 days (a leap year).

Jean Meeus, Astronomical Algorithms, Willmann-Bell, Richmond, Virginia. Second edition, 1998, chapter 9, pages 73-76.
Edward M. Reingold and Nachum Dershowitz, Calendrical Calculations, Cambridge University, UK. 1st edition, 1997, chapter 6 and appendix B8. 4th edition, 2018. Chapter 7 and Appendix D7.
Edward Graham Richards, Mapping Time, Oxford University, London, 1998. Chapter 15, pages 231-235, 311, 323-324.
Paul Kenneth Seidelmann and Leroy Elsworth Doggett, Explanatory Supplement to the Astronomical Almanac, Mill Valley, 1992. Pages 589-591.

Robert Harry van Gent, Islamic Calendar Converter, Mathematical Institute, Utrecht University, December 2019. An excellent source. The "Origin" section clearly describes how the various leap year methods are constructed; the "Islamic calendar converter" section converts Gregorian/Julian dates simultaneously into eight Islamic dates (four leap year methods in two epochs); the "Other online converters" section indicates which of the eight possible Islamic dates are used by ten online calendar converters, most of which do not divulge their method.
Dieter Koch & Alois Treindl, Astrodienst Planetary Positions for AD 622 July (page 272).
E. G. Richards, The identification of days, Figure 22.1 in: Mapping Time: The Calendar and Its History.
P. K. Seidelmann, The Islamic Calendar, Explanatory Supplement 1992.
Wikipedia, Islamic calendar.
Wikipedia, Tabular Islamic calendar.
Index entries for sequences related to calendars
Index entries for linear recurrences with constant coefficients, signature (1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,-1).

Cf. A057347, A057348, A097105.

a[n_] := Floor[(n*10631 + 3)/30 + 1948086];
Array[a, 32] (* Amiram Eldar, Jan 04 2022 *)
IslamicNewYear[n_] := Module[{},
    date := DateObject[{n, 1, 1, 12},
            CalendarType -> "Islamic",
            TimeZone -> "Europe/London"];
    jl := CalendarConvert[date, "Julian"];
    jd := JulianDate[jl];
    MixedFractionParts[jd][[1]]
]; Table[IslamicNewYear[n], {n, 1, 32}] (* Peter Luschny, Feb 13 2022 *)

a(n) = floor((n*10631+c)/30) + 1948086.
c = 3 is used here; for other calendar methods, see comments section.
The epoch date of July 16 is assumed; for epoch July 15, subtract one from a(n).
Number of days in Islamic year n = a(n+1) - a(n).
Day of week of first day in year n = (a(n) mod 7) = 0 (Monday) to 6 (Sunday).
Julian day number of general Islamic date y,m,d = floor((y*10631+c)/30) + floor(m*59/2) + d + 1948056. Note that this single equation defines the entire Tabular Islamic calendar (for the four tabular methods mentioned in the comments).

A350149 Triangle read by rows: T(n, k) = n^(n-k)*k!.

Original entry on oeis.org

1, 1, 1, 4, 2, 2, 27, 9, 6, 6, 256, 64, 32, 24, 24, 3125, 625, 250, 150, 120, 120, 46656, 7776, 2592, 1296, 864, 720, 720, 823543, 117649, 33614, 14406, 8232, 5880, 5040, 5040, 16777216, 2097152, 524288, 196608, 98304, 61440, 46080, 40320, 40320

Offset: 0

Robert B Fowler, Dec 27 2021

T(n,k) are the denominators in a double summation power series for the definite integral of x^x. First expand x^x = exp(x*log(x)) = Sum_{n>=0} (x*log(x))^n/n!, then integrate each of the terms to get the double summation for F(x) = Integral_{t=0..x} t^t = Sum_{n>=1} (Sum_{k=0..n-1} (-1)^(n+k+1)*x^n*(log(x))^k/T(n,k)).
This is a definite integral, because lim {x->0} F(x) = 0.
The value of F(1) = 0.78343... = A083648 is known humorously as the Sophomore's Dream (see Borwein et al.).

			Triangle T(n,k) begins:
--------------------------------------------------------------------------
n/k         0        1       2       3      4      5      6      7      8
--------------------------------------------------------------------------
0  |        1,
1  |        1,       1,
2  |        4,       2,      2,
3  |       27,       9,      6,      6,
4  |      256,      64,     32,     24,    24,
5  |     3125,     625,    250,    150,   120,   120,
6  |    46656,    7776,   2592,   1296,   864,   720,   720,
7  |   823543,  117649,  33614,  14406,  8232,  5880,  5040,  5040,
8  | 16777216, 2097152, 524288, 196608, 98304, 61440, 46080, 40320, 40320.
...

Borwein, J., Bailey, D. and Girgensohn, R., Experimentation in Mathematics: Computational Paths to Discovery, A. K. Peters 2004.
William Dunham, The Calculus Gallery, Masterpieces from Newton to Lebesgue, Princeton University Press, Princeton NJ 2005.

G. C. Greubel, Rows n = 0..50 of the triangle, flattened
Eric Weisstein's World of Mathematics, Sophomore's dream
Wikipedia, Sophomore's dream

Cf. A000312 (first column), A000169 (2nd column), A003308 (3rd column excluding first term), A000142 (main diagonal), A000142 (2nd diagonal excluding first term), A112541 (row sums).
Values of the integral: A083648, A073009.
Cf. A061711, A350297.

A350149:= func< n,k | n^(n-k)*Factorial(k) >;
[A350149(n,k): k in [0..n], n in [0..12]]; // G. C. Greubel, Jul 31 2022

T := (n, k) -> n^(n - k)*k!:
seq(seq(T(n, k), k = 0..n), n = 0..9); # Peter Luschny, Jan 07 2022

T[n_, k_]:= n^(n-k)*k!; Table[T[n, k], {n, 0,12}, {k,0,n}]//Flatten (* Amiram Eldar, Dec 27 2021 *)

def A350149(n,k): return n^(n-k)*factorial(k)
flatten([[A350149(n,k) for k in (0..n)] for n in (0..12)]) # G. C. Greubel, Jul 31 2022

T(n, 0) = A000312(n).
T(n, 1) = A000169(n).
T(n, 2) = A003308(n), n >= 2.
Sum_{k=0..n} T(n, k) = A112541(n).
T(n, n) = A000142(n).
T(n, n-1) = A000142(n), n >= 1.
T(n,k) = A061711(n) * (n+1) / A350297(n+1,k). - Robert B Fowler, Jan 11 2022

cp's OEIS Frontend

User: Robert B Fowler

A379607 Denominators corresponding to A379606.

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A379606 S = (A-sin(A))/2 gives the area of a segment of the unit circle in terms of the arc length A (<= Pi). Expressing A in terms of S we get A = Sum_{n>=0} b^(2n+1)*c(n) where b = (12*S)^(1/3). Sequence gives numerators of c(n).

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A375862 Number of caesium clock "ticks" in time units: tick, second, minute, hour, day, week, month, year, decade, century, millennium, eon (in Julian years).

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A375666 Number of seconds in time units: second, minute, hour, day, week, month, year, decade, century, millennium, eon (in Julian years).

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A375027 Number of occurrences of Easter Sunday on March 22, March 23, ..., April 25 during a 532-year Julian Easter cycle.

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A351862 Denominators of the coefficients in a series for the angles in the Spiral of Theodorus.

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Mathematica

A351861 Numerators of the coefficients in a series for the angles in the Spiral of Theodorus.

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A351582 Decimal expansion of the root of cot(Pi/(s+1)) - csc(Pi/s).

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A350539 Chronological Julian day number of the first day (Muharram 1) of Tabular Islamic year n.

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A379606 S = (A-sin(A))/2 gives the area of a segment of the unit circle in terms of the arc length A (<= Pi). Expressing A in terms of S we get A = Sum_{n>=0} b^(2n+1)c(n) where b = (12S)^(1/3). Sequence gives numerators of c(n).